Cengage MECHANICS 1

7.8 Physics for IIT-JEE: Mechanics I pointing vertically downwards and in this situation weight of the object is 111g dirccted vertically downwards (Fig. 7.1_62., Another ? ' II',..---~ ... u! System'\",,//\"'?Ba ~ system: the Earth \" One syslem: you / ',Ball ',. \\ mg \" I \\\"\" I mg I System ::-:: ~ ~Gravitational \\, :' ~,/m~g -\"\" Fig. 7.19 <,~B+~-fon;eofthe ' Earth,\" \\ .. ' Points to Remember \\ Earth J carlh on you 1. Normal force will be perpendicular to the surface of con- tact. II A bal! is projected from earth, this diagram represents the interaction 2. If normal force perpendicular to the surface of contact force of you force between the earth and the ball cannot be drawn, it will act perpendicular to\"the surface on the earth of the body. Fig. 7.16 3. If neither can be done, normal force has to be drawn as two components - one in the X-direction and one in the Normal Force Y-direction. Remember that there is no relation between the forces acting along the X- and Y~directions. They are Whenever two surfaces are in contact, they press (or push) each independent of each other. other by a j{)rce called contact forcc. The component of the 4. The number of normal forces acting on a body depends contact force perpendicular to the surface is normal reaction .on the number oj~E~~:!,~_C?::_.~_:l_~:!~~::_~_~_\"~~X ~:~!~~._.___._,_._. _..__ . and along the surface in contact is frictional' force. 'The nor- mal reaction to lhe contact surface can be computed by using ma.L }' = Figure 7.17 shows action-reaction pairs of normal forces for two physical situations. Normal forces on a block are drawn normal to the contact surface directed towards the block; and N :::: O. ,,, ..ry Force of you Representing Normal Reactions and Weight in Free Bodv N on the surface Diagrams Force-of the Another system: the surface surface on.~...:..,.1I.L,.~~ (b) Let us consider two blocks of mass M and 111 that are placed as shown in Fig, 7.20 and a forcc of magnitude F is acting on you ,, M. Our aim is to represent the forces in free body diagrams in different situations. One system: you (a) N Force of the fre{,.' body diagram of m surface on Y()Ur~44:\"\"0.=-; -Mg - - - =.,- - - Force ofyou TN N on the surfaee (c) Free body diagram of M Fig. 7.17 Representing Normal Reaction in Different Situations . Fig. 7.20 Tension ~NB ~'11 . .. .. . IV~ B \"••••.,.•...,...•.•\".••....•..•-..•...,.... NB It is quite practical that we can pull the objects by a string, but we cannot push the object by a string. This gives us an idea that AA A a string can pull but cannot push. Fig. 7.18 Consider a light rope AB subjected to two equal and opposite forces at ends A and B. Let us. observe the state of affairs at the • Ifdirection of contact force cannot be determined, it should cross-section (or the point) C. For this, consider the equilibrium be shown as two components (as shown in Fig, 7.19). of the parts AC and CB of the rope. Part CB of t'1C rope applies a force on AC at point C, 1·~lC.CfJ' In turn AC also applies a force on CB at point C, FCB.Ac- We introduce a physical quantity, tension, which is a property of rope or a point of the rope, and which equals the magnitude

Newton's laws of Motion 7.9 of the force with which the rope pulls the body connected to it Two hlocks of mass inA and mn are ar- Tension at point C of the rope in the above example: ranged in the diagram as shown in Fig, 7.24, Draw the free body diagram of Tc = IF AC,CIII = II; CB,ACI 1, Block mA String I A F ~4: -EA======:33c---+ FAc.CB = feB, Ae ....\"-E,Co::::=====83--+II F B 2. Block mB ~lC,C/J = FCH. AC String 2 3, Blockmc C IF;,eCBI 4, Pulley Fig. 7.21 Sol. Whenever a body is connected to a light rope, it is pulled with a force which equals the t.ension in the rope at the point where the body is attached to the rope. Consider another example. Let a particle of mass m be hung in the vertical position by a light rope whose onc end is attached to a rigid support. The forces acting on the particle, shown in the free body diagram in Fig. 7.22, are: mg, the weight of the particle, and a force by the rope, T. ,,, r /1/ 1 Mg Fig, 7,22 The rope pulls the particle with a force which is equal to the Fig, 7,24 tension in the rope at the point where the particle is attached to it. Points to Remember Also. the rope pulls the support by a force represented by T, Tension in the rope at the point where it is attached to the support 1. Tension force always pulls a body. is the magnitude of the force, 2. Tension can never push a body or rope. 3. Tension across massless pulley or frictionless pulley re- Two blocl{s of mass m A and m lJ are ar~ mains constant. ranged in the diagram as shown ill Fig, 7,23, Draw the free 4. Ropes become slack when tension force becomes zero, body diagram of Friction 1. Block mA String 3 2, Block mil When a sll1iace ofa solid body slides (or has a tendenc.v to slide Pulley 2 ifil does not actually slide) on a sut/ace ofanother solid body, its 3. Pnlley I String I (sUlface 's) motion (or tendency (~lmotion) is opposed. The force 4, I'ulley 2 which opposes the relative motion (or tendency of the relative motion) hetween the contact .'1m/aces is calledfi·ictionalforce. Pu~ley ! Th~ origin of frictional force is a complicated matter. Friction - SIring 2 is a consequence of molecular interaction~ originating in the realm of molecules and atoms because of electrical reasons. On Sol, an atomic level, both surfaces of contact are irregular. There are many points of contact where the atoms seem to cling together, Fig, 7.23 and when the surface is pulled along, the atoms snap apart and vibration ensues. If we push a h!ock on a whle. the table start~ pushing the block back along the sUl-face in contact (tangentially) (Fig. 7.25). Hence we can call this contact force tangential contact.

7.10 Physics for IIT-JEE: Mechanics I Alg frame of reference. At any time, position vectors of the particle N I ~ Fig. 7.25 with respect to those two frames are --; and r', respectively. At ~ the same moment, position vector of the origin of S' is R with respect to S as shown in Fig. 7.29. Y' Elastic Spring Forces S' YP Tension in an extended (or compressed) spring is proportional to the magnitude of extension (or compression): T ex x, in magni- ,. 0' x' tude, but opposite in direction. So, T = -kx, where k is a pos- itive const~nt, also known as the spring constant of the spring S (Fig. 7.26). R x 0x mass =m Fig. 7.29 --) --) -)- From the vector triangle 00' P, we get r' = r - R Differentiating this equation twice with respect to time we get, Fig. 7.26 Let us connect a body (a block. say) to one end of a spring -. and pull il' other end. We observe that the block receives a pull. Here, a' = Acceleration of the particle P relative to S' If we the spring, the block will receive the push. In other ;; = Acceleration of the particle relative to S ca:~\"\"-s. \".:: ,-;pring get deformed. That means, a deformed spring A= Acceleration' of S' relative to S. can pull or, push an ubject. Multiplying the above equation by m (mass of the panicle) The force exertc! 1 by a deformed spring on the connected Awe get, In ~~ = m C; -m bodies is known u,. \"spring force\" (Fig. 7.27). => pi Ii= Pereal) - m ::;:} F' = +Ferea!) (-mA) --+\" +-a Suppose a box is placed on a railway platform. The only forces acting on it are its weight W acting downward and the normal Elongated spring pulls the block A compressed spring pushes the object reaction R acting on it upward. Wand R are thus equal and opposite. Thus net external fore acting on the box is zero. The Fig. 7.27 atbox is rest. While drawing force diagram, first of all we will mark the points of the connected bodies at which the spring is connected. Let us consider the forces acting on this box and its state with If the spring is really elongated (or assumed to be elongated), respect to three observers, namely it pulls the points (P and Q) of contaet (like a string). If we assume that the spring is compressed, it pushes both the bodies (points P and Q) in contact, such as a rod (Fig. 7.28). An elongated spring pulls P and Q A compressed spring pushes Fig. 7.30 the points P and Q • A person P standing on the platform. • A person Q who is sitting in a train which is moving with Fig. 7.28 uniform velocity in a straight line parallel to the platform. NON-INERTIAL FRAME OF REFERENCE AND • A person M sitting in a train which is moving parallel to PSEUDO (FICTITIOUS) FORCE the platfonn but with some acceleration. Motion of a particle (P) is studied from two frames of references • This box is at rest W,r.t. observer P. Sand 5', S is an inertial frame of reference and 51' is anon-inertial • This box is in uniform motion in a straight line w.r.t. ob- server Q [moving with -v velocity].

Newton's laws of Motion 7.11 • This box is having (-) acceleration w,r,t. the observer M, N For all the three observers (or frames of reference), the force acting on the box is the same: Wand R. As Wand \".~\"~!M\"J R are equal and opposite, the net external force acting on Mg Mg the box is zero. FED of the man w.r.t. observer A FBD of the man w,r.t observer B Let us reconsider the example of the box lying at rest on a railway Fig. 7.33 platform, We saw that it has (- C;) acceleration W.f,t. a train EQUILIBRIUM OF A PARTICLE which is moving with an acceleration (a) even though no net external material force is acting on it. Neither first nor second Equilibrium of a particle in mechanics refers to the situation law remains valid on the box with respect to the accelerated train. when the net external force acting on the particle is zero. But still, if we want to apply Newton's laws of motion, in a The above condition is correct and complete as far as equi- non-inertial reference frame, we are allowed to do so, provided librium of a particle (i.e., a point mass) is concerned. In case of we indude an additional force, the pseudo force, in the free body rigid bodies (i.e., extended bodies), there are two conditions to diagram. be satisfied for such bodies to have equilibrium. First, net extef- nal force acting on the body should be zero. Second, net external If we apply an imaginary force ma in the direction opposite , torque acting on the body should be zero. to the direction of motion of observer A1, the observer Mean wrile equation of motion of box. F pseudo '''' ma Concurrent Forces If two or more forces act on the same particle. we call them concurrent. forces. Free body diagram Free body diagram lamy's Theorem as seen from observer P and Q as seen from observer M Fig. 7.31 Thus, Newton's law can also be applied with respect to a non- If thr'ec CUIH..:urrcnt forces P, Q and R acting on a particle keep inertial reference frame, provided we include an \"extra\" force the particle in equilibrium, then Lamy's (Lami's) theorem states: of (-ma) on the system. This force has no existence, in reality, but has been included only to suit the calculations involved, p Q, R by Newton's second law, while working out a problem w.r.t. a sin y non-inertial reference frame, This imaginary force is known as sina sin fI \"pseudo force\" or \"fictitious force\" or \"inertial force\", [The term \"pseudo\" means something which is not real.] -, A man of mass M stands on a weigh- Here \" = angle opposite to P ing machine an elevator accelerating upwards with an ~ acceleration a(). Draw the free hody diagram of the man as observed hy the observer A (stationary on the ground) and fI = angle opposite to Q observer B (stationary ou the elevator). Also, calculate the -, reading of the weighing machine. y = angle opposite to R ~i , y It_~ \". ...... Q Fig. 7.32 ,,,, R Sol. Using Newton's second law, Fig. 7.34 N-Mg=Ma N = M(g+a) Note'lfthe. concurre/lt!orces are ciJpuujarbutmorethan three, then.itis generallYMllvenienttiJ.resolve.allo/tluun alongtwo .mutuallyP/Jrpendiculardirectiollsandt1!en.the resuTfantoleach set 01 t1!esertsolveacompollentswi?l be zero.

7.12 Physics for lIT-JEE: Mechanics I EF:i=O; E f)= 0 B If th~force are~iJt coplanar, then theycallbe resolved along any three mutually perpendicular directions and thell the followillg c<mditions shaUapply Fig. 7.35 y P(obLem Solving Strategy:.Apptying 31\" Newtoi1's laws TA The following procedure is recommended wtierfdealing with problCllls involving Newton'slaw. 1. Conceptualize: Draw a simple. neat diagram of the system to help establish the mental representation. Es: tablish convenient: coordinate axes' tor each object:in the system. 2. Categorize: If anacceleration component foranob- Fig. 7.36 ject is zero. it ismodelledasa particlein equilibrium in . this directionandr; F = O. If not. the object is. mode dIed as a particle uilder a tict force ill this direction and r;F=.ma. 3... ~ualyses: Isolate the object whose motion is . being E = =F, 0; or 1'8 COS 3T - TA 0 analysed, Draw a-free body'diagram fOf'this obj~ct For 7'.4 = 18C0837° = 4 (i) 1'8'5 systems coniainingmore than one object, draw.sepa- ratdrce body diagrams for each object. Do notinclude and I: f~. = 0; or Tn sin 37° - Tc = 0 (ii) in the free bOdy diagram forces exerted by the object Ql1 Its surroundi:ngs.\" \" Find the components of the forces along the coordi- TB = -Tc- = 300 = 500N - r;F,=nate axes. Apply Newton'ssecol1dlaw, i\"'{, sin 37° 3/5 \"'in component form.Checkyour·dimensions'to··make . . = -45 1'8 .= -45 = 400 N equatIOn (I). sure aU ternis: have units of force. Fro.m we get TA x 500 4•. Solve the component equations for the unknowns.Re- Method II: Usiug Lami's theorem By Lami's theorem, we have member that to obtain a complete SOlutiOl~~, YOll,'ll)tBt have as -m,!-ny independent cqu'ation -as you, have l,lJ~­ TA 1'8 Tc= -C-~:::.......~ knowns. ~~'--cc-_ = 5~ ,FiriaUze:' Make sure,Your.results a~e consistent with the' sin(90 + 37\") sin 90° 8in(180 - 37\") .free body diagram. Also check the predictions of your But Tc ~ 300 N soluti6n~ for extreme values of the 'variables. BY'doing ,SOj YOil can· often detect errors in your results. and Tn = -Tc- = -300 = 500N sin 37° 3/5 TA =Tc ( Sin(90-37\")') = 300 (C-OS-3r) = 400N sin (I 80 - 37\") sin37\" A block of mass 30 kg is suspended by 7r, as shown in Fig. 7.35. Il'lnd the tension in each (180 37') string. .,:, +----ir';c...yJ (90 37\") Sol. Metbod I: Considering equilibrium of each part of sys- tem (Fig. 7.36): 91Y' The whole system is in equilibrium, therefore for part Tc ·'\" 300N Fig. 7.37 r; F= 0 From free body diagram of block C, Tc ~ 300 N Now consider the equilibrium of point 0;

Two particles of masses 10 kg and 35 kg Newton's Laws of Motion 7.13 are connected with four strings at points Band D as shown in Fig. 7.38. \"* T, = 200v'3 N c From equation (i), T4 = ISO + 100v'3 =50 (3 + 2v'3) N. [, m\"Three blocks A, B, and C of masses m2, and m3 are resting one on top of the other as shown in Fig. 7.40. A horizontal force F is applied on block 'B. Assuming all the surfaces are frictionless. IA I T3 I B J ;1\" C lOON I 1 D I I I I I II I II 1111\\11 II I \\ 350N Fig. 7.40 Fig. 7.38 Determine the tensions in various segment~ of the string. Calculate: Sol. Free body diagram of whole system is shown in Fig. 7.39. 1. Acceleration of block A, block B, and block C. 2. Normal reactions between A and B, Band C, and be- tween C and the ground. Sol: Fig. 7.41 This system cannot be in equilibrium because in the horizontal direction, the system has a net external force F. As far as the Fig. 7.39 vertical direCtion is concerned, all the forces are internal action and reaction forces, hence equal and opposit.e, hence will cancel L Fx = 0, or T, si1l45\" - T2 sill 37\" = 0 (i) each other out. and L F\" = 0, or T, cos 45\" + 12 cos 3Y = 350 Oi) Body A: No external force is acting on it, hence it. will remain stationary in equilibrium, hence acceleration of block A, (JA = From equation (i), we have T2 = T, sin4SO O,andR, =m,g. (i) -'-c_ __ Now from equation (ii), sin 37° Body B: There is no external force acting on it vertically, hence it will not have any acceleration in the vertical direct.ion. T, cos 45\" + -1'1 ~sin-4-5° x <) = 350 R2 = R, + \"'2g (ii) sin 37~' cos37 However, there is one external force F which is acting on it. 4 4Jor ~1'-, + -1', x - = 350 or -1', [ I + - = 350 So it will have some acceleration a in the horizontal direction -> of P such lhat .,fi .,fi:1 .,fi 3 F = JJ12afj (iii) ~2)1'1 sin 45° 150.,fi x ( v L wh·ICh gl.ves aceeIcra'tlOn 0 I' block B, (til = -F \"* 1', = 150.,fi N alld 1, = sin 60\" = '--~G)-~ . m2 Body C and earth: Same comments as in caseofhody A above +R3 = R2 l11)g (iv) = 250N +and +R3 = (Ill, /112 nl3)g (v) Analysing the equilibrium ojpoint B: No external force acts on block C in horizontal direction hence L F, = 0; or 1'2 sill 37\" + 1'3 sin 30\" - 1'4 = 0 ae is equal to O. and L P, = 0; or 1'3 cos 30\" - T2cos37\" - 100 = 0 0) Two blocks of masses In, and m, are From equation (ii), 1') cos 3~'' \",250 x -4 - 100 = 0 5 (ii) placed side by side on a smooth horizontal surface as'shown in the Fig. 7.42. A horizontal force F is applied on block m ,. I. Find the acceleration of each block.

7.14 Physics for IJT-JEE: Mechanics I Fig. 7.42 2. Find the normal reaction between the two blocks. Sol. Method 1: Since the two blocks always remain in contact Fig. 7.45 with each other so they must move with the same acceleration. Sol; First we will resolve all the forces acting on the block into Using Newton's second law, (i) x and y components, Forblockm, F-N=tn,a (ii) For block m2 N = m2a L F, = 1', cos 37\" + 1'3 cos 45\" + 1'2 cos 90° L F.v = F\\ sin 37() + F2 cos 0° - F3 sin 45(l N, I' F F, mIg FBD of block 11l! Fig. 7.43 37'-' .... - - - - - - - - - -.. x On adding the two equations, we get 45\" F = (m, +m2) a F a= t ml +m2 Fig. 7.46 Substituting the value of a in equation (ii), we get I: Fy Fm2 I: F, N = m2a = ---'---'''- and ay = ---;;;- Now, ax = ----;;;-- nll +m2 Method 2: The situation may be considered as follows: Instead (90.0) cos 37\" + 128../2 cos 45,0\" 2 of drawing the free body diagrams ofeach block we can draw the 0, = 20.0 = 10 mls ; free body diagram of both blocks together as shown in Fig. 7.44. The net force acting on the system is F and the total mass of a, = (90.0)(sin3T) + 114 - (128../2 sin 45.0°) = 2 m/s2 +the system IS (m, Tn2), thus a = F , 20.0 In! +m2 According to Newton's third law, the force exerted on the block by the boy must be equal to the force exerted on the boy N2 by the block. . ... ---~- ',-+a r a,Therefore, = F!. = 90,0 = 2 m/s2 In 45 a,\\\"= 0 F ,, FBD of both the blocks m2g considered as onc. 1'13'0 of the block m2 Fig. 7.44 Fig. 7.47 To find out the normal reaction N between the two blocks we ~ A block of mass M is being pulled with the help of a string of mass m and length L The horizontal can imagine that, the block m2 is moving with an acceleration a, therefore, the net force acting on it must be (m2a) which is force applied on the string is F. nothing but the normal reaction applied by the block m ,. m2 F Thus, N = tn2a = tnJ +m2 u'lI!Iiitt!lt!!IiJJlJ Three boys, each of mass 45 kg, pull si- multaneously a block on a smooth surface. Mass of block is Fig. 7.48 20.0 kg. 1. Find the acceleration of the block. Determine: 2. Find the instantaneous acceleration of the boy exerting a. force exerted by the string on the block and acceleration the force Fl. of system

Newton's Laws of Motion 7.15 b. tension at the mid-point of the string. a= - -F - and 1'2 = {m(L-x)/L+M)F c. tension at a distance x from the end at which force is '--'-_:-c''-_ _-'- applied. d. assume that the block is kept on a frictionless horizontal M+m M+m surface and the mass is uniformly distributed in the string. Hence, we can say the string having mass tension v charges Sol. along the length. a. The force applied by the string to the block is T. For part ~n In the diagram shown, the blocks A and (a) consider one system is block and other string. Let the B are connected together by a string and placed on a smooth acceleration of the system (block + string) be a inclined plane. B is connected to C (which is suspended ver- System I System II tically) by another string which passes over a smooth pulley fixed to the plane. The mass of A is mA = 1 kg; mass of B, - - - - -- - -- - mB =2kg. Fig. 7.49 a. If the system is at rest. Find the value of mass of C. b. If mass of C is twice the mass calculated in (a) then find the acceleration of the system. Now we apply Newton's law to each of them. (i) For system II : F - l' = ma (ii) For system I : T = M a F Fig. 7.52 After solving equations (i) and (ii), a = - - - ; Sol. a. From the free body diagram of A: M+m T\\ : tension in string between A and B 1'=-M_F .. NA: normal reaction between A and incline M+m As A is at rest, net force parallel and perpendicular to the b. Now we have to redefine our system. Choose system 1 as incline should be balanced. block and half string and system 2 as other half string. On applying Newton's second law to system 1 and system 2, we have System 1 System II ------ \".. 7'1- ml2 - - ... 9 I +-1 j---JIoj.> \"! meg FBI) orc 4 ~ .- ---L/-2--- T, Fig. 7.50 . 1'1 = 'M2 x a (iii) System II : F - Mass per unit length of string = m!M. FBD orA FBD of 11 Hence, mass of LI2 length of string = Lm x 'L2 =::: \"m2 Fig. 7.53 System I : T1 = ( M + ~) a (iv) NA = mAgeosO = lOcos 300 = 5.j3N After solving equations (iii) and (iv) we get, =1'1 = mAg sin 0= 10 sin 30\" 5 N a = _f_'_. 1'1 = c..(M~+:-m~/2:;-):-F From the free body diagram of B: M+m' (M+m) As B is connected to both strings, two tensions 1'1 and T, will act on it. c. Now we can redefine our system in the block and string of T2 = tension (force) of string between Band C acting upwards length (L - x) (system I) and string of length x (system JI) 1'1 = tension of string between A and B acting downwards System! System !l e, eWB sin WB cos are components of weight WB Fig. 7.51 Balancing forces: NB = mEg cos 30\" = 20 cos 30° = IO.j3N 1'2 = h+ mBg sin 30\" =5 + 20sin 30\"=5 + 20 x 2I' = 15 N Ci':System JI : F - 1', = xx) a (v) Force diagram of C: 1'2 is the pulling force of string on block C (vi) =meg = T,. Hence me 1.5 kg System I: 1'2 = lo/;(L-X)+M)a Note: You should remember. thtitpnnponents of .the weight ofa body on the inclined plane are Weos O. Solving equations (v) and (vi) we get

7.16 Physics for IIT-JEE: Mechanics I b. Let the acceleration of the system be a. We can assume an b. Under what circumstances will the balance read 30 N? c. What will be the reading in the balance if the cahle of the arbitrary direction of motion, let the blocks be moving up the incline, elevator breaks? From FBD of A: T, - 10 sin 30\" = I a (i) Sol. From FBD of B: 12 - T, - 20 sin 30\" = 2a (ii) a. Reading in the spring balance is equal the tension in the spring From FBD of C: 30 - 7:' = 3a (iii) =50N. Solving equations (i), (ii), and (iii). a = 2.5 m/s2 i Blocks A and B rest on a horizontal sur- accc!cralion in contact with each other. Pushing forces Fl and 1'2 Fig. 7.57 are applied as shown. Weight of A is 30 N and of B is 20 N. Find the force Fl and the normal reactions hetween all the contact surfaces. As the elevator is accelerating in upward direction with = =2.45, m/s2 the acceleration of the block a 2.5 m/s2 = gj4. T - rng = rna 50 - mg = m(gj4) Fig. 7.54 mg =40N (i) Sol. Force diagram ofB: When the elevator has an upward acceleration, reading is greater than the actual weight. b. Reading of balance = T = 30 N. T Fig. 7.55 Fig. 7.58 N R = normal react.ion between B and the ground As T < mg (Actual weight), the block and elevator must R = normal reaction between B and A (on B by A) have a downward acceleration a, R = F:' = 15 N, Nl! = 20 N Force diagram of A: R = normal force on A by B mg - T = ma NA = normal reaction between A and the ground a = 2,5 m/s2 is in downward direction, It is possible in two ways: N.! 1. the elevator is going up and slowing down, or 2. the elevator is going down and its speed is increasing, ~~~\"\"~ The reading of balance is less than actual weight if ele- FI ~os 10\" vator has a downward acceleration, c. If the cable breaks, the acceleration of the bloek and the ele- \"J'.'-'-'-'+3\"'0\" N vator = g (downwards) Net force = mass x acceleration Fig. 7.56 mg - T = tug, T = 0 Balancing forces: F, cos 30\" = R the reading of the balance = 0 N F, sin 30\" + 30 = NA T F, = 2R/~ = 30/~ = 10~N; mg NA = 1O~:11 + 30 = (30 + 5~) N Fig. 7.59 A hody hangs from a spring balance sup- ported from the ceiling of au elevator. a. If the elevator has an upward acceleration of 2.5 m/s2 and the balance reads 50 N, what is the true weight ofthe body?

Newton's Laws of Motion 7.17 Two masses Ill, and 1Il2 are attached to a A solid sphere of mass 2 kg is resting Ik<ible inextensible massless rope which passes over a fric- inside a cube as shown in Fig. 7.62. The cube is moving with a tionless and massless pulley. Find the accelerations of the masses and tension in the rope. = J)velocity, IT (51 i + 21 m/s. Here I is the time in seconds. All the surfaces are smooth. The sphere is at rest with respect to the cube. What is the total force exerted by the sphere on the cube'! y A Fig. 7.60 o Sol. Fix an inertial reference frame to the ground to observe the Fig. 7.62 motion of the masses. Sol.\" = Stf + 2t} Let the tension in the rope be T. (lnfaet. tension is the property of a point of the rope). Tension may be same at all the points of the C; = 51 + 2} rope or may vary from point to point. In this case with ideal pulley Nx =2x5=10N (massless and frictionless) and ideal rope (incxtensible, massless and nexible), the tension will remain constant throughout the Ny = 2 x 10 + 2 x 2 = 24 N rope). Let the acceleration of m2 be vertically downwards and acceler- fN?Total force = + N): = j(iO)2 + (24)2 = 26 N ation of m I will be vertically upwards. Nr This is because the rope is inextensible; during motion, the length of the rope must not change, and the rope must not slacken either. From the above statement you must notyonclude that the 2g 2x2 acceleration· of the masses connected bia rope, are always Fig. 7.63 equal. The relationship between the accelerations of the ,----1 Concept Application Exercise}.2 f----, luassesdepends onthc' configuration of tile pulley-rope I. As per the diagram given in Fig, 7.64(a), system, which can be obtained from the. fact that the length of an ideal rope must not change and the rope must not Fig. 7.64(a) slacken; Free body diagram of M is (True or f~tlse), Assume all surfaces arc frictionless. maUsing equation L F = for the force diagrams of m I and ~N' T -mig = mja (i) Mg ( ii) nl2g -- T = m2a Fig.7.64(b) Adding equations (i) and (ii), (iii) 2. Mass m is placed on a body of mass M. There is no 1n2g-fnjg=mja+m!a (iv) friction. Force F is applied on M and it moves with acceleration a. Find the force (along X-axis) on the top a - C::~~~:)!i body, Substituting this value of a in equation (i), f11J) +,T=mjg+mj - (11--12 g= 2mlln2 g + Itll 1112 (tn.j fn2) T Fig. 7.65 Here a is acceleration of lower body. 1\"m, lII,g Free body diagnttlls Fig. 7.61

7.18 Physics for IJT·JEE: Mechanics I 3. In the problem shown below, the mass of man is 1', M. The weight of the man as registered by weigh· 1', ing machine will be__ .~~, assuming that the weighing machine, man, and wedge all are stationary. lOkg Fig. 7.70 e 8. Determine the tension T4 Fig. 7.66 4. A small object is suspended at rest from two strings as 10 kg shown in Fig. 7.67. The magnitude ofthe force exerted by each string on the object is 10../2 N. Find the magnitude of the mass of the object. Fig. 7.71 9. Two trolleys A and B are moving with accelerations a and 2a, respectively, in the same direction as shown in Fig. 7.72. To an observer in trolley A, the magnitude of Fig. 7.67 pseudo force acting on a block of mass In on the trolley 5. Figure 7.68 shows a platform on which a man of mass B is \"_..__.__. M is standing and holding a string passing over a system of ideal pulleys. Another mass m is hanging as shown in fj; nA a B m 2a ~figure. Find the force man has to exert to maintain the -+ --+ equilibrium of system. Also fInd the force exerted by the platform on the man. (oj (0) DE) (oj (0) (oJ ( ) Fig. 7.72 10. a. A 10 kg block is supported by a cord that runs to a spring scale, which is suppOlted by another cord from the ceiling as shown in Fig. 7.73. What is the reading on the scale? T m Fig. 7.73 Fig. 7.68 b. In Fig. 7.74 the block is supported by a cord that runs 6. The object in Fig. 7.69 weighs 40 kg and hangs at rest. around a pulley and to a scale. The opposite end of Find the tensions in the three cords that hold it. the scale is attached by a cord to a wall. What is the reading of the scale? 3/':.. TT Fig. 7.69 Fig. 7.74 7. A block of mass In = 10 kg is suspended with the help of three strings as shown in the Fig. 7.70. Find the tensions T\" T2 , and 13.

c. In Fig. 7.75 the wall has been replaced with a second Newton's Laws of Motion 7.19 10 kg block on thc left. What is the rcading on the scale now? 14. Consider the system shown in Fig. 7.80. The system is released from rest, find the tension in the cord connected TT between I kg and 2 kg blocks. Fig. 7.75 Fig. 7.80 11. What is the reading of the spring balance in the following 15. The system shown in Fig. 7.81 is released from rest. Cal- device? culate the tension in the strings and force exerted by the strings on the pulleys. Assuming pulleys and strings are TT massless. Fig. 7.76 Fig. 7.81 12. A block of mass 25 kg is raised by a 50 kg man in two 16. Find acceleration of blocks and tension in the cord in the different ways as shown in Fig. 7.77. What is the action device shown in the Fig. 7.82. on the floor by the man in the two cases? If thc floor yiclds to a normal force of 700 N, which mode should the man adopt [0 lift the block without the floor yielding. ,.-------~ 25kgt~E3~ 25 kg Fig. 7.82 17. Two monkeys of masses 10 kg and 8 kg are moving (a) (b) along a vertical rope as shown in Fig. 7.83. The former Fig. 7.77 climbing up with an acceleration of 2 mis', while the later coming down with a uniform veloyity of 2 m/s. Find 13. Two blocks of masses I kg and 2 kg are placed in contact on a smooth horizontal surface as shown in Fig. 7.78, A the tension in the rope at the fixed support. horizontal force of 3 N is applied a. on I kg block Support Lx 3N ;~kk~.~!q):.• 7 \\ \\ \\ \\ \\ \\ . ~, \\ \\ -- -1 \\ \\. \\ Fig. 7.83 Fig. 7.78 18. A student standing on the large platform of a spring scale b. on 2 kg block. Find force of interaction betwcen the notes his weight. He then takes a step on this platform and notices that the scale reads less than his weight at the blocks (see Fig. 7.79). beginning of the step and more than his weight at the end of the step. Explain.

7.20 Physics for IIT·JEE: Mechanics I 19. A homogeneous rod of length L is acted upon by two 23. A body hangs from a spring balance supported from the forces F, and F2 applied to its ends and directed opposite roof of an elevator. to each other. With what force F will the rod be stretched a. If the elevator has an upward acceleration of 2.45 m/s2 at the cross-section at a distance I from the end where Fl and the balance reads 50 N, what is the true weight of is applied? the body? 20. In the arrangement shown in Fig. 7.84, a wedge of mass b. Under what circumstances will the balance read 30 N? M = 4 kg moves towards left with an acceleration of a = 2 m/52. All surfaces arc smooth. Find the acceleration c. What will the balance read if the elevator cable breaks? of mass m = 1 kg relative to the wedge. 24. An object of mass 5,00 kg attached to a spring scale, rests 60\" on a frictionless, horizontal surface as shown in Fig. 7.87. The spring seale, attached to the front end of a boxcar, has Fig. 7.84 a constant reading of 18.0 N when the car is in motion. 21. A 20· kg monkey has a firm hold on a light rope that passes a. The spring scale reads zero when the car is at rest. over a rrictionless pulley and is attached to a 20- kg bunch Determine the acceleration of the car. of bananas (as shown in Fig, 7,85), The monkey looks up, b. What constant reading will the spring scale show if the sees the bananas, and starts to climb the rope to get them. a. As the monkey climbs, do the bananas move up, down car moves with constant velocity? c. Describe the forees on the object as ohserved by some- ator remain rest? one in the car and by someone at rest outside the car. b. As the monkey climbs, does the distance between the ---+ monkey and the bananas decrease, increase, or remain constant? 5.00 kg Fig. 7.87 20 kg CONSTRAINT RELATION 20 The equations showing the relation of the motions of a system of bodies, in which onee motion is constrained by the others Fig. 7.85 motion, are called the constrained relations. c. The monkey releases her hold on the rope, What hap- First we start our analysis with simple cases of pulleys. Con- pens to the distance between the monkey and the ba- sider the situation shown in Fig, 7,88, Two bodies are connected nanas while she is falling? with a string which passes over a pulley at the corner of a table. d. Before reaching the ground, the monkey grabs the rope Here if string is inextensible, we can directly state that the dis- to stop her falL What do the bananas do? placement of A in downward direction is equal to the displace- 22. A lift is going up, The total mass of the lift and the ment of B in horizontal direction on table, and if displacements passengers is 1,500 kg. The variation in the speed of the of A and B are equal in equal time, their speeds and accelerati0n lift is given by the graph (Fig, 7,86), magnitude must also be equal. ,..1_ _---,13 Case (a) Cnse (,,;) a- (}\"\" 2 4 (, X to 1C2 Fig. 7.88 Fig. 7.86 In this case (c) if the wedge and block are free to move it a. Whot will be the tension in the rope pulling the lift at is obvious that the acceleration of the block and the wedge are related, time f equal to (i) I s (ii) 6 s (iii) II s? b. What will be the average velocity and the average ac- Applying Newton's law alone is not sufficient in most cases. If you apply Newton's laws based on the methods shown above, celeration during the course of the entire motion? you will get a few equations. You may however find that the

Newton's Laws of Motion 7.21 number of unknowns are much larger than the number of equa- the other end B moves downward when the rod makes an tions. That is, you will have a situation where, say the number of angle 0 with the horizontal. variables are three but the number of equations are only two. It is, therefore obvious that you cannot solve the equations to get B the value of the variables. y Look at the diagram in Fig. 7.88(a) In this case let us say that you have to find the acceleration of v A. e x the masses. The number of unknowns will be Fig. 7.90 1. tension T, 2. acceleration a! of the mass, and Sol. Let us first find the relation between the two displacements 3. acceleration of the other body (/2. There are three unknowns. However we will get only two equa- then differentiate with respect to time. Here if the distance from tions - one for one mass and another for the other mass. Clearly you can see that Newton's Jaws are not sufficient to the corner to the point A is x and that up to B is y. Now the left solve the problem. In such cases we need additional equations. These are provided by what are called as constraint equations. velOC.Ity of.pOInt A can be gI.ven as Vii = ~elx and that of B can Constraints mean that two bodies (in this case the bodies elt which are attached to the puliey) are not free to move the way they want. The accelerations between them are dependent on be gI.ven as VB = - -e...l.:y.... (-SI.gn I.nd'lcates, Y I.S decreas.m)g. each other. We need to find out the relationship to be able to elt solve these equations; therefore, constraints provide additional equat.ions. If we relate x and y: x 2 + y' = [2 The need for constraints Dl.\"lterentJ.a.tmg W.ith respect to t = 2xdx- + 2yd-y = 0 Fig. 7.89 elt elt Constraints are the geometrical restrictions imposed on the =::::::} XVA yVB => x motion of a body, which also governs the trajectory of the body. xu = yVs ::::::}v/J = u- = u cote For example, a block placed on the table cannot move normal y to the surface, it is bound to move parallcl to the: surface. Alternatively: In cases where distance between two points is We have to use the method of constraint equations to relate the accelerations between the bodies. always fixed, we can say the relative velocity of one point of an Tn many cases you can write down the relation of accelera- object with respeet to any other point of the same object in the tion by just looking at the situation. In other cases, for complex relationships, we can think of four types of constraints. direction of the line joining them will always remain zero, as 1. General constraints their separation always remains constant. 2. Wedgc constraint. 3. Puliey constraint Here in the above example, the distance between points A 4. Combination of the wedge and pulieys and B of the rod always remains constant, thus the two points General Constraints must have same velocity components in the direction of the line Fig. 7.90 shows a rod of length I resting on a wall and floor. Its lower end A is pulled towards left joining. i.c., along the length of the rod. with a constant velocity u, as a result of this end B starts moving down along the wall. Find the velocity with which If point B is moving down with velocity v11, its component along the length of the rod is VB sinO. Similarly, the velocity component of point A along the length of rod is v cosO. Thus we have V/J sine = u case or VB = U cote ~ InFig.7.91,ahallofmassm,undablock of masS1n2 are joined together with an inextensible string. The ball can slide on a smooth horizontal surface. If v! and V2 are the respective speeds of the ball and the block, determine the constraint relation between the two. Fig. 7.91 Sol. Method 1: Distances are assumed from the center of the pulley as shown in Fig. 7.92. Constraint:Length of the string remains constant. ;;f + hT' = X2 constant

7.22 Physics for IIT-JEE: Mechanics I Case-I ~ Mass A is connected with a string which passes through a fixed ,r------- pulley. The other end of string is connected with a movable pul- hi: ley N. Block B is connected with another string which passes VI : through the pulley N as shown in Fig. 7.95. +- Fig. 7.92 Differentiating both the sides w.r.t. the time, we get 2Xl dXI dx, i +-=0 2Jx + h; dt dt Since the ball moves $0 as to increase Xl with time and block p moves so as to decrease Xl with time, therefore, dXj dX2 Xl = cose - = +Vj and - = -V2 also, r-: dt dt yXj2 +1112 Fig. 7.95 Fig. 7.96 or V2 = VICOS(} Consider the situation shown in Fig, 7.96. If we consider that mass A is going up by distance x, pulley N which is attached Method 2: to the same string will go down by the Same distance x. Due to this the string which is connected to mass B will now have free lengths ab and cd Cab = cd = x) which will go on the side of mass B due to its weight as the other end is fixed at point P. Thus mass B will go down by 2x hence its speed and acceleration will be twice that of block B. Hence, Fig. 7.93 Above example can be understood by another -approach in which we will consider the total length of the string will always Change in the length of segment (1) be constant. = Xl cose - 0 = Xl cose To relate the acceleration of the bodies. assume that the various bodies move by Change in the length of segment (2) = - X2 M Total change in the string length should be zero. a distance Xl, X2, ... , and so on. Calcu- e - eXl cos late the number of segments in the rope, X2 = 0 =} X2 = Xl cos N Hence V2 = v! cosO The segments in the first rope arc Method 3: The problem can be solved very easily if we look at marked 1 and 2. The distance moved the problem from a ditferent viewpoint and identify a different by the various elements are also marked constraint; i.e., velocity of any two points along the string is (as shown in the figure). Note that the same. Obviously, from Fig. 7.94. VI cose = \"2. pulley, which is connected to the ceil- ing, cannot move, \"I Relate the distance moved. (total change in length of the string must be Fig. 7.94 zero). To do this, calculate the changc in length of each segment of the string. Writing Down Constraints-Pulley then add these changes to get the total change in length of the rope. Pulley constraints are applicable when the bodies concerned arc connected through pulleys and the rope connecting them is il1- Change in length of segment 1 = -XI extensible. Change in length of segment 2 = +X2. Therefore total change in length of +string I -XI X2 = () =} XI = x, Once we have the relation between the distances, the relation between acceler- ations is simple, r'or the first string it is OJ = 02·

Newton's Laws of Motion 7.23 For second string Case-III Let us apply steps 3, 4, and 5 for the (a) (b) second inextensible string. The distanccs moved by the pulley Here three blocks A, B, and Care connectcd with strings and N and block Bare X2 and X3 as shown in figure. The ground is at pullcys as shown in figure. rest. Therefore the end of the string that is connected to the ground will Here we develop constraint relation between the motion of not move. Change in length of segment masses A, Band C. Let us assume that masses A and C would 3 = -X2 (Why negative? Because as the pulley moves down, the rope go up hy distance XA and xc, respectively, these lengths of i comes closer to the ground and the length of the segment decreases). the string will slack as length ab .. cd below the pulley Z, Change in length of segment Thus this will go down hy a distance XB as shown in Fig.(b). 4 = +X3 - X,. Thus we have , To see why this is so, let us consider a string with either points moving + +ab + cd = XA Xc or 2xfJ = XA Xc by a distance X2 and X3. This is shown in the figure. differentiat.ing w,r.t. time, we get 2v B = VA + Vc (i) I' diffcrentiating again w,r.t. time 2aB= QA + ac (ii) « ) Because the other end moves by X3, --+ the length of the string increascs by Equations (i) and (ii) arc the constraint relation.s. for motion ----+ Lof masses A, B, and C. .1:2 .1:3 X3· ._ . .__._ J Rope with either PaliS When the other end moves by X2, the length reduces by X2. moving by a distance The change in length is therefore X2 and Xl. Your Shortcut X3 - X2· In the cases where pulley moves along with the blocks connected on both the sides, we can say that the displacement of the pulley The total change in length of the rope is +X2 X3 X2 - 0 is the average of the displacement on both sides of the pulley. :::::} X3 - 2X2 = 0 :::::} X3 = 2X2 Once we get the relation between the distances moved, the ac- celeration relation will be the same. The acceleration relation' x =XA+-2XB- ~_~_lso_,_a3_=_.?:~1_2_ _._ _ _ _ _ _ _ _ _ .._____ p Case-II Fig, 7,97 If onc end of the string is connected with the fixed end, the displacement of that end can be considered as zero, Analysis of Case-I using shortcut method aIn the given situation M is a fixed pullcy and N is movable . Fig, 7,98 pulley. The blocks A and B are tied to strings and axTanged as shown, If mass A goes up by the distance x, wc can observe that the string lengths ab and cd are slack, due to the wcight of block B, this length (ab + cd = 2x) will go on this sidc and block B will descend by a distance 2x. As in equal time duration B has travelled a distance twice that of A. So, VA. = 2vlJ and aA = 2aB

7.24 Physics for IIT-JEE: Mechanics I • As pulley 1 is fixed, hence the displacement should be In the arrangement of three blocks as 0, If the displacement of block A is XA (down) then the in the Fig. 7.101, the string is inextensible. If the di- displacement ofother end should bexA (up)(sec Fig, 7,98), rections of accelerations are as shown in the given figure, then determine the constraint relation. X 1',1 = 0 = 2=>XA +X{) XD = -XA +\"2- • Displacement of bloek B = Displacement of pulley 2 =>XA +0 XA = 2Xp2 :;;:} Xli = 2XB • Xp2 = --2- Analysis of Case-II using Shortcut Method Fig. 7.101 • As pulley I is fixed IXp,,1 = IXAI Sol. Method 1: Let us assume the respective distance of each block as shown in Fig, 7,101. Since the length of the string is If block A moves up by XA, pulley 2 should moves XA in downward direction as shown in Fig, 7.99. constant, therefore, Xl ++ X2 2X3 = constant • For pulley 2 On differentiating twice w,r,t, time, we get x, ~Ia, +- Fig. 7.102 X8 Fig. 7.99 Analysis of Case-Ill using Shortcut Method Sincex\\ and X2 arc assumed to be decreasing with time, there- fore, • As pulley M is fixed XM = 0 (sec Fig, 7, j 00) +;fA XW d2x! d2x2 XW = -XA =>= ·-·2\"~~ ---- = -al and --- = -a2 dt' dt' • Pulley N is also fixed and x] is assumed to be increasing with time, therefore, Xc + Xz -(P-X\"1 =+a1 XN=O= 2 =} xz=-xc dt' \" XIJ = XIV +xz +Thus -al - a2 = 2a3 = 0 or aj a2 = 203_ 2 Method 2: As total change in the length of string should be zero, The summation of the change in the lengths of segment should 2Xl/ = Xw +xz = XA +xc be zero. Xl X2 --+ CD (1) +-- ,XI~ CV X3 XB X2~ CD Fig. 7.100 In) Fig. 7.103

Newton's laws of Motion 7.25 ++-Xl X2 X3 - X2 = 0 1', 1', 1/11 \"\"'2 kg m2\"\"2kg -Xl +2X3 -X2 =0 la, la2 +2X3 = xl Xl Hence, +2a3 = (11 aZ Shortcut method: X3 = Xl +X2 20N 20N 2 1', 1', l In the pUlley-rope-mass arrangement 111] = 4 kg la, T2 Ii shown in the 7.104 tn, = 2 kg, \"'2 = 2 kg, and tn3 = 4 kg. Find the tension in the ropes and the accelerations of the 40N masses when the masses are sct free to move. Assume that the pulley and the ropes are ideal. Take g = 10 m/s2• Fig. 7.106 I- T50 , ,a, = I3T0 2 ,a3 = I7T0 , , = I5T0 , ,a\" m/s m/s m/s m/s 7,T, = -320 N, = -16N0 11 11 Fig. 7.104 Note: You can also establish equatio1ls(1)alld (25 some, Sol. Step 1: Constraint relations whatilUiirectly in thiscase. Now, 11 :;:: constant gives, 1. For the rope length /lnol jochange and jar this 11 =Xj +.'f,,+JTR ropeno~ toslackenaccelerationofml w.r.t the fiXed pulley =:.. acce'erati~1l of themova~lepulley w.r.t.the -d 2/,=-d'x+, -d-2 x p+-d'\"~R fiXed pulley . a,dt' dt' dt' dt' (I) Oral =~ap =} lall=='lupl 0= +a p =} lad = lapl (2) 2; For the rope 'e1lgth I,n,o(10 change and for this rope not .to .slilcken-acceleration of m2 .w.r.t movable and 12 = constant gives, pulley = - acce'erationof\"'3w,r.tthetnovable pulley + +\"I, = (x, - xI') (X3 - xp) R or (a2 .,..up)=~(a3 __ ap) + +\"=} I, = X2 X3 - 2xp R h-d 2 =-d+2x2--d2~X3 ,2-d2+x p -,d~J-rR lI2't\"t:\"').(1;\"\",0 dt 2 dt' dt' cit' dt' ==} 2al a, + a3 =} 0=a,+a3-2ap =} 2(/'=(/2+ a3 3; fOl' can even use any,of the methods given tn previous discussions'. ::t=X±, =i=== III! In the arrangement shown in the tensions in the rope and the acceleration Fig. 7.105 Step 2: Free body diagrams (Fig, 7,106) of the masses m I and m, and pnlleys PI and P2 when the Step 3 : Equations of motion system is selfree to move. Assnme the pulleys to be massless 20 ~ T, = 2a, and strings to he are light and inextensihle. 20 - 7, = 2a, 40 - 7, = 4a3 (3) 2T2 - T, = 0 (4) Solving these equations for the required quantities, (5) (6) Fig. 7.107

7.26 Physics for IIT-JEE: Mechanics I Sol. Step 1: Free body diagrams ---j---- ----- T T2 TT tl'J aPI~ --- ---- all a2~lIJl 1112 Fig, 7,110 T Wedge Constraint mig We can observe that the wedge M can only move in horizontal direction towards left, and the block m can slide on an inclined Fig. 7.108 smiace of M which is always in contact with the wedge. Step 2: Applying the equations of motion for pulleys and blocks • Let us define our x and y axis parallel to the inclined and perpendicular to inclined, respectively, as in Fig. 7. Ill. For m,: m,g - T = m,a, (I) M For 1112: m2g - T2 = InZQ2 (2) For pulley p,: 27' - T ; Oxal\" (3) For pulley pz: Tz - 2T ; 0 xal'2 (4) From equation (3), T ; 0 From equation, (4) 1,;0 and from equations (l) and (2), Q, = g, az = g. Calculating all) and a p2 : Step 3: Constraiut relations: Consider the reference line and the position vectors of the pulleys and masses as shown in the Fig. 7.109. Write the length of the rope in terms of the position vectors and differentiate it to obtain the relations between the accelerations of the masses and the pulleys. t4-- X--..J x Fig, 7,111 T ,nil • We can observe that the displacement of m and M in x'- direction will be same as the block never loses contact with the wedge. • If the wedge moves in horizontal direction by a distance of x during this time the block will move x in x-direction • We can relatc these displacement x and X as '*e-x X =. x = X sinO (i) S111 Fig. 7.109 Hence velocity relation can be written as 1. For the length of the string connecting Pz and mz not to be V, = VsinO (ii) change and for this rope not to slacken. and acceferation relation can be written as a p2 = a2 = g. ax = A sinO (iii) In,2, Length of the string Connecting P, to not to change and for this rope not to Here Vx and ax arc the velocity and acceleration of the slacken: block in the direction perpendicular to the inclined surface. I = (x, - XI'I) + XI'I + (X1'2 - XI\") + XI'2 A block of mass m is placed on the in- :::::} I=X\\--X p1 +2X P2 dined surface of a wedge as shown in Fig. 7,112, Calculate Differentiating this equation w.r.t, twice, the acceleration of the wedge when the block is released. Assume that all the surfaces are frictionless, +2 2 2 So!' e1 [ = d xJ _ d x PI 2 (J2xP2 :::::}O=al Step 1: Constraint relation (Fig. 7.113) dt' <itZ dt' dt' From the previous discussion we can write ax = A sinB :::::} 0 = g - apt + 2ap2 ::::} ap1 = 3g (i)

Me Newton's Laws of Motion 7.27 Fig. 7.112 M Me Fig.7.113 Step 2: Free body diagrams shown in Fig. 7.114. Nsin e Fig. 7.115 ~-:A Neose increase by (x + x cos8). As overall length of the string is con- 1 !!Ig , stant, then we can write T - X + (x + X cos 0) = 0 x = X(1 - cos e). and a, = A(l-cose) (i) /\",!i<g cos 0 From wedge constant it is clear that ~ = sin () x =} Y = X sin 8 and ay = A sin 8 ~mgSine Find the accelerations of rod A and OJ\"____________ 2:\". B in the arrangement shown in the Fig. 7.116 if the mass of the rod is In and that of the wedge is M. The friction Fig. 7.114 hetween all the contact surfaces is negligible. Step 3: Equation of motions is OJ) =eFor M: N sin MA (iii) For In in x' direction Fig. 7.116 mgcose - N = fila, = meA sine) Sol. The rod (mass m) is constrained to move in the vertical direction (with the help of the guides). and the wcdge will move From equations (ij) and (iii) along the surface in the horizontal direction. emg sin cosO Let the acceleration of m w.r.t. ground be a, vertically down- wards and acceleration of M W.r.t. ground be A. horizontally A = +~(M---:m-os-in:2:0c) towards right. Constraint relation: The motion of the system is constrained by Pulley and Wedge Constraint the fact that the \"bottom face of the rod must always be in contact with the inclined plane\". Ifin time t. X oe the displacement of the Consider a block of mass m is placed on the inclined surface of wedge and x be the displacement of the rod. then the constraint the wedgc. Thc block is connccted with a string and arranged as demands that (see Fig. 7.117). shown in Fig. 7.115. Now the system is released from the rest. We can divide the string in two segments (1) and (2). Let at any interval of time the wedge move a distance X towards right and the block move a distance\" x parallel to inclined direction. From the diagrams it is clear that the length of the segment (1) will decrease by X while the length of the segment (2) will

7.28 Physics for IIT-JEE: Mechanics I Motion in vertical direction (2) mg-N cos a = ma ''''\"-- x: /.:::~--- and the forces acting on the wedge arc (Fig. 7.] 20): /':~~ __ ~~~-L________-L____--\" 1. the weight, Mg, 2. N, reaction of N acting on the rod, and Fig. 7.117 3. N\" norma] force by the surface. x N cos ex - = tana X N x = Xtana (1) -A- Differentiating equation (1) w.r.t. t twice, (d2x) (d2X) .d(i = . dl2 tana [tana=constantl Q' cosa = A sina Fig. 7.120 (3) (4) The fact that the rod (or a particle on the wedge) and the wedge The force equations are: must not lose contact is usually called \"wedge constraints\", For Nj-Mg-Ncosa=O,and this, the component of the acceleration of the rod perpendic- Nsina = MA ular to the wedge plane should he equal to the component of acceleration of the wedge perpendicular to the wedge plane. aSina~ a os a £sa A block of mass /1l is placed on an in- a ,,~A clined plane (Fig. 7.121). With what acceleration A towards Fig. 7.118 right should the system move on a horizontal surface so that A sin (X /1l does not slide on the surface of the inclined plane? Assume acasa = A sina that all the snrfaces alee smooth. a=Atana Fig. 7.121 Solving the above equation for a and A. Sol. If the motion of m is analysed from ground, its accelerat.ion tana is A and the forces acting on it are: its weight mg and normal reaction R. a = m--t-a-n\"a-+--M:-e:o-t-a- If the motion of m is analysed from the view point of an ob- mg server standing on the inclined plane (i.e., relative to the plane), and its acceleration is 0 m/52 and the forces acting on it arc: its 1 Fguide weight, the normal reaction, and a pseudo force of magnitude mA towards left. dividing, we get, A = g tan fJ. Nsin a R Ncos a Fig. 7.119 mA (pseudo force) The forces acting on the rod are: mg 1. the weight mg, vertically downwards, 2. the normal force N, normal to the bottom surface of the rod, Fig. 7.122 and 3. the force (Fg\"ido) exerted by the guide to nullify the horizontal component of N as for the rod aUorizonlal= O.

Analysis offorces on In relative to the inclined plane: Newton's laws of Motion 7.29 As In is at rest, forces must balance each other along both direc- Mass of wedge as shown in Fig. 7.126 is and that of the block is m = 5 kg . Neglecting tions friction at all the places and mass of the pulley, calcnlate the acceleration of the wedge. Thread is inextensible. Rcos8=mg (i) R sin e::::::: mA (ii) R cos 8 mA R sin () Smooth !JIg M Fig. 7.123 Fig. 7.126 Find the acceleration of the blocks in Sol. Let the acceleration of the wedge be a rightward then ac, pulley and the strings are massless. celeration of block relative to wedge will also be a but down the incline. Hence, net acceleration of the block will be equal to the vector sum of these two as shown in Fig. 7.127. Fig. 7.124 a Sol. Let a = acceleration of block M and pulley. a a :::: acceleration of m. a sin 3r Let x, X be the co~ordinates of m and pulley as shown. Fig. 7.127 Fig. 7.125 Now considering FBDs (see Fig. 7.128). ,,--1' N, T Length of string passing over the \"pulley is: L=X+X-x L = 2X-x Differentiating twice with respect to time, ------;. Ma 0= 2A - ({ co} a = 2A (I) Note: We tan generalise this resul~ Ifone end ofa string, Fig. 7.128 passing over a moving pulley, isfixed tizenthe acceleration ofother enei is twice the acceleration ofpulley. For the wedge. From the force diagrams of m and iW, we have N2 = N, cos 37° + T sin 37° (i) T = ma, and (2) T - T cos 37\" - N, sin 37\" = Ma (iO F-2T=MA (3) For the block. Comparing equations 1, 2, and 3, we get (iii) F N, sin 37\"- T cos 37° =rn(a + a cos 37°) A = -,'-- and mg - N, cos 37\" - T sin 37\" = rna sin 37\" (iv) M+4m' 2F Solving the above equations. N, = 41.5 N, T = 25.5 N, and a = 0.5 ms-2• +({ = - - - , . - (M 4111)

7.30 Physics for IIT-JEE: Mechanics I 5. A ring A which can slide on a smooth wire is connected to one end of a string as shown in Fig. 7.133. Other end ~---I COl1ceptApplication Exercise 7.3 1--, of the string is connected to a hanging mass B. Find the speed with the ring when the string makes an angle 1. Figure 7.129 shows a system of four pulleys with two masses A and B. Find, at an instant: ewith the wire and mass B is going down with a velocity v. Fig. 7.129 A a. Speed of the block A when the block B is going up at ~ I mls and the pulley Y is going up at 2 m/s. Fig. 7.133 b. Acceleration of block A ifblock B is going up at3 mis' 6. Figure 7.134 shows a block A constrained to slide along and the pulley Y is goiug down at I m/s'. the incline plane of the wedge B. Block A is attached with a string which passes through three ideal pulleys and 2. Figure 7.130 shows a pulley ovcr which a string passes connected to the wedge B.lfthe wedge is pulled towards and is connected to two masses A and B. Pulley moves right with an acceleration 'a'. up with a velocity vp and mass B is also going up at a velocity VB. Find the velocity of mass A if: ~BeA~...... a Fig. 7.134 VI' a. find the acceleration of the block with respect to wedge. b. find the acceleration of the block with respect to ground. A 7. Find the acceleration of the block B as shown in B iVs Fig. 7.135 (a) and (b) relative to the block A and relative Fig. 7.130 to the ground if the block A is moving towards left with an acceleration Q, a. Vp = 5 mls and V8 = 10 mls b. Vp = 5 mls and V8 = -20 mls LZ1 3. Find the relation in the acceleration of the three masses (a) (b) shown in Fig. 7.13I(a) and Fig. 7.13I(b). l<'ig.7.135 Fig. 7.131 8. If all the surfaces are smooth, to keep the block sta- 4. Figure 7.132 shows a small mass In hanging over a pulley. tionary with respect to the wedge, the wcdge should The other end of the thread is being pulled in horizontally be given a horizontal acceleration towards _____. with a uniform speed u. Find the speed with which the The magnitude of the acceleration is given by the hori- zontal force to be applicd on the wedge would be _ __ emass ascend at the instant the string makes an angle ~ with the horizontal. Fig. 7.136' 9. If the string is inextensible, determine the velocity u of each block in terms of v and e. c ,, vv v ~ __e __ u (a) BA Unifonll velocity Fig. 7.137 Fig. 7.132 (i) Fig. 7. 137(a) (A) u = _ .___ (ii) Fig. 7.137(b) u = ____

Newton's Laws of Motion 7.31 10. Calculate the aecelerations of the block A and B in cases 14. The velocities of A and B are shown in the Fig. 7.142. of Figs 7.138 (i), (ii), and (iii). Find the speed (in m/s) of block C. (Assume that the pulleys and the string are ideal.) A~ 3 m/s B (i) -<-- II B Fig. 7.142 (iii) 15. A 20 kg bloek B is suspended from an ideal string Fig. 7.138 attached to a40 kg block A. The ratio of the acceleration 11. The rod of mass m is constrained to move along the guide and always in contact with the wedge of mass M of the block B in cases (I) and (II) shown in Figs. 7.143 as shown in Fig. 7.139. Assume that there is no friction anywhere. Calculate the acceleration of thc wedge and and 7.144 immediately after the system is released [rom rod. rest is 3n Find value of n. (Neglect friction.) 10' 2,,2 / M A /I e Fig. 7.143 Fig. 7.139 12. A man (mass = m) is standing on a platform (mass. = M). / By puJIing the string he causes the motion of platform. ~/ What is the ratio of m/M such that the man and the platform move together? ~~ A Fig. 7.140 ~ 13. In Fig. 7.141 no relative motion takes place between '// / the wedge and the block placed on it. The rod slides Fig. 7.144 downwards oyer the wedge and pushes the wedge to SPRING FORCE AND COMBINATIONS OF move in the horizontal direction. The mass of the wedge SPRINGS is the same as that of thc block and is equal to M. Springs can be of many types such as helical [Fig. 7.145(a)] or spiral [Fig. 7.145(b)]. These springs are either (b) and are '7:3'e I stretchable or compressible. If tan = find the mass of the rod. (Negleet the Regarding springs it is worth nothing that rotation of the rod), (a) (b) -go5 Fig. 7.145 Rod IZl Springs are assumed to be the same everywhere. o e Smooth Fig. 7.141

7.32 Physics for IIT-JEE: Mechanics I Springs can be stretched or compressed and the stretch or y, y, y, compression is always taken to be positive. rA string cannot be ~ ~ ~ compressed. I k, F, F k, Y k2 F, +- F2 +-F! +F- Y2-1> F~FF~F ~ (a) (b) mg mg Yi F ~ Stretching a spring Compressing a spring ~ Y, (a) (b) k, R ~, Fig. 7.146 -'\"\"0\"\" k2 To produce extension Of compression in a spring two equal FJ +- and opposite forces are applied and equilibrium restoring force \"'00',,\" is developed due to the elasticity of spring which is equal to ~ either force. +-F2 i.e.• F = F' and always opposite to applied force. (e) mg Fig. 7.148 For small stretch or compression, springs obey Hook's law, Spring in Parallel Le., for a spring i k~ \\ This situation is shown in Fig. 7.148(a). If the force F pulls the I mass m by y, the stretch in each spring will be y, Hyperbola )\" = y, = y (i) pi Now as for a spring F = ky and as ks are not equal so F\\ # F2 Fo<:y but for equilibrium e K=tanfJ F = F, + F\" i.e., ky = k, y, + k2 + Y2 [as F = ky] y (Stretch or compression) Length of spring I which in the light of equation (i) reduces to (b) (a) k = k, + k2 + ... Fig. 7.147 This is like capacitors in parallel or resistance in series. Force ()( stretch (or compression), i.e., F = ky (i) Spring in Series This means that the restoring force is linear. This force in a This situation is shown in Fig. 7.148(b), as springs are massless, spring is not constant and depends on stretch (or compression) so force in these must be the same, i.e., y, Greater the stretch (or compression) greater will be the force F=~=F (i) and vice-versa. Now as F = ky and ks are not equal so stretches will not be k is called the force constant of the spring and is equal to equal, i.e., y # Y2 the slope of the force versus the stretch curve. It has dimensions But, Y = y, + Y2 F F, F, or,-.-:.=-+- [F / L] which is [MT-'I and units N/m. Greater the force, con- k k, k, stant of a spring lesser will be the stretch (or compression) for =[as for F = ky, y (Flk)] a given force and more stiffer is said to be the spring. The force which in the light of equation (i) reduces to constant k of a spring depends on wire (its length, radius r, and -1 =-1 +-1+ ... material) used to make the spring (radius of spring R and length k k, k2 I of spring). it is well established that f(,r a given spring k ()( Oil) (ii) Spring in Series with a Mass between them This means that smaller the length of the spring, greater will be As shown in Fig. 7.148 (c). if force F stretches a spring by y, the force constant and vice-versa, the other will be compressed by the same amount, so YI = Y2 = Y (i) Now as F = ky and ks are not equal so, F\\ # Fi Force Constant of Composite Springs But, F = F, + f2' i.e., ky = k\\Yl + k2 y, [as F = ky] If a number of springs are connected to a body and we want to which in the light of equation (i) reduces to produce it to a single spring, following three cases of common k = k\\ = k2 interest are possible

Newton's laws of Motion 7.33 . t l a l a Two blocks of masses ml and mz are in Forces of Friction When an object moves either on a surface or equilibrium. The block m2 hangs from a fixed smooth pulley through a viscous medium such as air or water, there is resistance by an inextensible string that is fitted with a light spring of stiffness k as shown in the Fig. 7.149 neglecting the friction to the motion because the object interacts with its surroundings. and mass of the string. Find the acceleration of the bodies just after the string S is cut. We call such resistance as force of friction. Fig. 7.149 Imagine you are trying to slide a heavy box on horizontal =Sol. FBD: Let the spring forces be F kx just after cutting the concrete surface by applying a force F. You try to drag the box spring. Hence, at that instant the forces acting on m1 arc T = kx across the surface of your concrete /loor. The concrete surface -+ mig t and Nt; on tn2 the forces are m2g t and T t· is real, not an idealized, frictionless surface in a simplification s kx T model. If we apply an external horizontal force F to the box acting to the right, the b_ox remains stationary if F is smal1. The force that counteracts F and keeps the box from moving acts to the left and is called the force of static friction j,. As long as the box is not moving, it is modeled as a particle in equilib- .f,rium and j~ ;:;:; F. Therefore, if f' Qccrease, also decreases. Experiments show that the friction force arises from the nature of the two surfaces; because of their roughness, contact is made only at a few points, as shown in the magnified surface view in Fig.7.151(a). (b) Fig. 7.150 Force equation: Initially. all the pal1icles are stationary; hence al = a2 = O. Applying Newton's second law /\",max 1::For Inl: F = T' - kx = 0 (i) ,, For 1n2: 1:: F = T - m2g = 0 (ii) .....- Static region ~+- Kinetic region ___ (e) For spring 1:: F = T - kx = 0 (iii) Fig. 7.151 Solving equations (i), (ii), and (iii), we have T' = T = kx = m2g as shown in the Fig. 7.150 just after the F,If we increase the magnitude of as in Fig. 7.ISl(b) the box string S is cut, tension T' vanishes immediately, but the spring eventually slips. When the box is on the verge of slipping, f, is a cannot regain its shape and size instantaneously. Therefore, the spring force remains as it is then, just after cutting the string, maximum as shown in fig. If F excess f\"max the box moves and the net force acting on m! is equal to kx whereas the net force accelerates to the right. While the box is in motion, the friction acting on m2 is equal to T - m2g. Hence, the acceleration of m 1 and m2 just after cutting the string is given by force is less than f,.max- The friction force for an object in motion is called the force of kinetic friction. The net force F - .I, in 1::F =m,al =kx=m2g,1::F=m2a2 = T -m2g=0 the x direction produces an acceleration to the right, accor9ing Tlll.S .gives, a! = (mm]2\" ) gand a2 =0. to Newton's second law. If we reduce the magnitude of F so ANALYSIS Of FRICTION fORCE that F = ib the acceleration is zero and the box moves to the In the previous section, we had introduced Newton's laws of right with a constant speed. If the applicd force is removed, the motion and applied them to situations in which we ignored fric- friction force acting to the left provides an acceleration of the tion. In this section, we shall expand OUf investigation to objects box in the - x direction and eventually brings it to rest. moving in the presence of friction, which will allow us to model situations more realistically. Experimentally, one finds that, to a good approximation, both j~,max and fk for an object on a surface are proportional to the normal force exerted by the surface on the object; thus, we adopt a simplification model in which this approximation is assumed

7.34 Physics for IIT-JEE: Mechanics I to the exact. The assumption in this simplification model can be 2. How could we find the coefficient of kinetic friction? summarized as follows: Sol. • The magnitude of the force of static friction between any 1. The forces on the block, as shown in Fig. 7.152, arc the grav- two surfaces in contact can have the values. g,itational force m the normal force';, and the force of static (i) friction f~, As long as the block is not moving, these forces where the dimensionless constant fLs is cal~ed the coef- ficient of static friction and N is the magnitude of 'the ~re balanced and the block is in equilibrium. We choose a co- normal force. The equality in equation (i) holds when the surfaces are on the verge of slipping, that is, when ordinate system with the positive x-axis parallel to the incline f = f~',max == jJ.,sN. This situation is called impending ma-' and downhill and the positive y-axis upward pell'endieular tion. The inequality holds when the component of the ap- to the incline. Applying Newton's second law in component plied force parallel to the surfaces is less than this value. form to the block gives (I) • The magnitude of the foree of kinetic friction acting be- twecn two surfaces is a. L F, = mg sin e - f, = 0 J, = \"k l1 Lh. Fl\" = II - mgcosO = () (2) where Ilk is the coefficient of kinetic friction, In our sim- These equations are valid for any angle of inclination e. plification model, this coefficient is independent of the rel- ative speed of the surfaces. e,.At the critical angle at which the block is on the verge \"k.• The values of \", and \"., depend on the nature of the sur- of slipping, the friction force has its maximum magnitude faces, but \", is generally less than • The direction of the friction force on an object is opposite /J-sn, so we rewrite (a) and (b) for this condition as to the actual motion (kinetic friction) or the impending c. mg sin Bc = JLsn (3) motion (static friction) of the object relative to the surface with which it is in contact. d. mg cosOc = n (4) e,Dividing equations (e) and (d), we have tan = fL.,' Therefore, the coefficient of static friction is equal to the tangent of the angle of the incline at which the block begins to slide. laws of Limiting Friction 2. Once the block begins to move, the magnitude of the friction The force of friction always acts along a direction so as to oppose force is the kinetic value J1.kfl, which is smaller than that of the the motion of the body relative to the other. force of static friction. As a result, if the angle is maintained The force of iriction depends upon the nature and the rough- at the critical angle, the block accelerates down the incline. ness of the two surfaces in contact. The rougher the surface the To restore the equilibrium situation in equation (1), with f~ more will be the fj'ictional forces. replaced by fl.:, the angle must be reduced to a value e~ such that the block slides down the incline at constant speed. In Frictional forces arc independent of the area of contact be- this situation, equations (3) and (4), with 0, replaced by 0', tween the two bodies, as long as the nonnal force remains the same. e:,and /J-s by J1.b give us tan = ILk. Angle of Friction Normal reaction N and friction force f are two components of the total reaction force of the surface on the block. Angle The following illustration shows a simple between total reaction R and normal reaction N is called angle method of measuring coefficient of friction. Suppose a block of triction. is placed on a rough surface inclined relative to the horizon- ¢ angle of friction RN tal, as shown in Fig. 7.152. The incline angle 0 is increased until the block starts to move. (y p F -n, x -, mg I Fig. 7.153 ·IS .111.uupend1' 11g IL,N ('\" I.S -iv-When A, mgsin e block state, tan = = /J,S VIs VIS ; mgcos-6 ; maximum angle of friction), ; When block is static, </> :50 </>s. ... ... Im-g} \"SWhen block is sliding, tan </>, = I'~N = \"k. Since > \"\" Fig. 7.152 it follows that </>s > </>,. 1. How is the coefficient of static friction related to the crit- Figure 7.153 shows a block kept on an incline whose angle of ical angle 0, at which the block begins to move? einclination can be varied. At a certain value of sliding motion of the block is impending. The equilibrium equations are

Newton's Laws of Motion 7.35 EllI!MmtiEm A block of weight 100 N lying on a hor- izontal surface just begins to move when a horizontal force of 25 N acts on it. Determine the coefficient of static friction. Sol. As the 25 N force brings the block to the point of sliding, the frictional force = Its N from the force diagram (see Fig. 7.156). N = 100 N ItsN = 25 Its= 0.25 N Fig. 7.154 tOO N I>', = -mg sine + 1t.,N = 0 Fig. 7.156 L F,. = - mg cos e + N = 0 ~ A block lying on an inclined plane has a weight of 50 N. It just beings to slide down when the inclina- On eliminating N between the two equations, we get, tion of plane with the horizontal is 300. Find /1-,. tan 8max = fJ.,; 8max = tan- 1 j.l. Sol. The block reaches the point of sliding when the plane makes an angle 01'30' with the horizontal (see Fig. 7.157). Friction force opposes relative motion between two surfaces. In order to decide the direction of static triction, try to imagine N thc likely direction in which the body will tend to move; friction force is opposite to it. In Fig. 7.154, force P pulls block B towards left and A is pulled towards right. Friction force on B is towards right and A is towards left. Important point to notice is that for two contact surfaces friction force is in opposite direction. It is internal force for two contact surfaces, so it must be an action-reaction pair. Fig. 7.155 Fig. 7.157 ~_---jr Points to Remember. '. 11---, Hence in this situation, frictional force = ItsN Baiancing the forces: 1. It has no unit. 2. If frictional force be limiting frictional force then It = It., N = W cos 30° ItsN = W sin 30° (static frictional coefficient). Its W cos 30\" = W sin 30° 3. If frictional force be kinetic then It = Itk (kinetic fric- liS = 1/.)3 tional coefficient). 4. Static friction::; limiting friction, i.e., the frictional force Note: Minimum. angle for .which a lilock starts sliding down an inclined pianeis known as angle of repose. acting on a body at rest can be anything from zero to limiting friction. A block of weight 100 N lying on a hori- zontal surface is pushed hy a force F acting at an angle 30Q 7 IV .S. It is incorrect to write = J.L Since\"] is nor parallel with horizontal. For what value of F will the block begin to move if /1-., = 0.25? to N. Sol. Consider the force diagram (Fig. 7.158) of the block at the 6. If the body is at rest with respect to the surface then moment when it has just started to move. f <1t.,N. 7. If the body is just in motion with respect to the surface .f = 1t.,N. 8. If the body is in motion with respect to the surface .f = ItkN.

7.36 Physics for IIT-JEE: Mechanics I lOON A 5 kg block is projected upwards with an initial speed of 10 mls from the bottom of a plane inclined Fig. 7.158 at 300 to the horizontal. The coefficient of kinetic friction between the block and the plane is 0.2. Balancing the forces: N = mgF sin 30\" 1. How far does the block move up the plane? F cos 30\" = It., N 2. How 100ig does it move np the plane? 3. After what time from its projedion does the block again F cos 30° = It.Jmg + F sin 30°) come back to the bottom'! With what speed does it arrive? F = J-Lsmg = 0.25(1 (0) • While the block is moving up, the frictional force acts cos 30° - It., sin 30\" vS - 0.25 down the incline. F = 33.74 N • As the block is slowing down, the velocity and the accel- eration must be in opposite directions. 1I11i2llllflml A 5 kg block slides down a plane inclined • Velocity in this case is upwards, so the acceleration is in at 300 to the .horizontal. I>'ind 1. the acceleration of the block if the plane is frictionless. downward direction and hence negative. The magnitude 2. the acceleration if the coefficient ofkineticfriction is \\\" of acceleration = sin 30\" + cos 30\" 2\",3 Sol. III 1. N = mg cos 30\" = g(sin 30' + Itmg cos 30\") = a = -g(sin 30' mg sin 30° = rna + Itmg cos 30\") = -6.6 m/s2 N For the motion of block from the bottom to up the plane: mg sin30\" N Fig. 7.159 a = g sin 30\", down the plane if plane is smooth. mg sin 30\" a =g/2 = 5 mis' ,s N Fig. 7.161 mg sin30\" U = + 10 mIs, using v2 = u2 + 2as, we get. Fig, 7,160 02 = 102 + 2(-6.6)(5) 1, N = mg cos 30\" =} S = 7.58 m mg s\"in 30° - f.IkN = rna a = g sin 30° - It,g cos 30° v = u + at a = 5/2 mis' O=IO-6.6xl =} 1=1.5s Hence the block moves up the plane for 1.5 s covering 7.58 m. For the motion '!f block down the plane: .. . mg sin JO(' + IImg 'cos 30() the magmtude of acceleratIOn = ---\"---'\"---,,--.~--~--\" 1/1 = g(sin 30' - Itg cos 30G ) =} a = 3.2 mis' N mgcos 30\" mg sin 30\" Fig. 7.162

Newton's Laws of Motion 7.37 As acceleration is in downward direction, a = -3.2 m/s2 Sol. Let us first calculate the force F required to bring III into s = -7.58 m (down the plane) and u = Om/s emotion in terms of angle s = ut + 112 at2 Equation using laws of motion are as follows: - 7 .58 = (0) + I /2( - 3.2)t2 =? t = 2.18 s N = mg - F sin I! So the total time taken to come back: eF cos = fJ.,N fup + !doWIl ::;: 1.5 +2.18 = 3,68 s eF cos = J.l (mg - F sine) v = u + at F = f-lsmg v = -6.8 m/s. So the block arrives at the bottom with a speed of 6.8 m/s. cos e + /-is sin e eWe have to find the angle for which this force F is minimum. Find the least pu!ling force which is act- ing at an angle of' 45° with the horizontal, will slide a hody Substituting jJ..,s = tan f-l(for simplification), we get weighing 5 kg along a rough horizontal surface. The coef- tiden! of friction 1', '\" I'k = 1/3. If a force of double this is mg tan A mg sin A applied along the same direction, find the resulting acceler- F = cos e + tan Asin e = ....:.:.::-:=-':--:- alion of the block. cos(8 - A) Sol. When the block is about to start sliding, frictional force is at its limiting value == /J..,s R F is minimum if cos ((-) - ).) is maximum. Hence Fis mini- Balancing forces in vertical direction: emum for = tan-I f-Ls and FmilJ == mg sinA. ,'.ITo bring m into motion with least effort. force should be ap- plied at an angle tan-' and should have a magnitude equal to: . f-l.I,mg Fmin = mg SlIlA = )1C\"+==Jc.l2i=S P sin 45° p A hlock of mass III = 10 kg is to he pulled on a horizontal rough surface with the minimum force. R e1. The block should be pulled at an angle which is equal p cos 45° to _ _~_ Fig. 7.163 2. The magnitude of the force F is equal to _.___ R + P sin 45\" = mg r-l4'F R = tll-g - P sin 4SO ~--. Balancingforces ill horizontal direction: P cos 45\" = {LsR 11- 0.75 Fig. 7.165 P cos 45\" = J.ls(mg - P sin 45\") Sol. 1. The block should be pulled at the angle of friction, i.e.• =? P = _ _.\".c:.._..~.___._. e = tan '(0.75) = 37\" cos 45\" + I\"., sin 45' G)2. The magnitude of external force is If applied force is 2 P: fm;,,=mg sinc/>=(IO)OO)sin37\"=(lOO) = 60N then R = mg - 2P sin 45(l and 2P cos 45° - P\"k R = rna t~ Determine the magnitude of frictional 2P cos 45\" -- fJ.k(lIlg - 21' sin 45\") = lila force and acceleration of the block in each of the following cases: =? a= 2P + I\")g = 5.72 m/s2 ----;;;,0 1'1'1\"12 A block of mass III lying on a horizontal Skg 5 kg 12 kg tOON_ 500N_ surface (coefficient of static friction = It.,) is to be hrought into motion by a pulling force F. At what angle 0 with the horizontal should the force F he applied so that its magni- tude is minimum'? Also find this minimum magnitude. I~--- /---- f~--- Case-I Casc-U . Case-III Fig. 7.166 Fig. 7.164 Sol. Case-I: Limiting value of friction force fm\" = ILN = (0.4)(100) = 40N

7.38 Physics for IlT-JEE: Mechanics I face. If the coefficient of friction between the blocks is It, then determine the minimum value of the horizontal force Force responsible force sliding of two blocks (parallel to wall F required to hold the blocks together. surface) = rng = 5 x 10 = 50 N \\.\\.!':~..J~\\\\lCb\".\"J , Smooth The force parallel to the surface of the wall will act as driv- Fig. 7.168 ing force. (r:lriving = mg = 50 N). Here the driving force is the weight which is greater than the maximum resisting for a F fmax = Fresisting = fllln = fJ.-sN = 0.4 x 100 = 40 N. Sol. Acceleration of system a = --_. ff -S kg M+rn N Free body diagrams as shown in Fig. 7.169 SOON lOON +---- --+ -<Of.-- ----.. N SOON N mg= 120N Fig. 7.167 3 :5FII = 500 cos 37° = 400 N FII = 500 x = 300 N Here Fdriving > Frcsisting Hence the block will slide and friction will be (M Fm)From the free body diagram of rn , N = rn+·' (i) (ii) .ik;OOl;' = I = I\"kN = 0.3 x 100 = 30 N (upward) Hence the equation of the motion of the block m is not sliding, therefore, I = rng For no sliding case I :'0 I\"N Mg - I = rna =} 50 - 30 = 5 x a That gives acceleration of the block mF 50 - 30 mg :'0 I\" (M + m) a = = 4 m/s2 (downward) F> -(, M--'..+_rn-'.)cg:. 5 Case II: In this case the resisting force is - I\" Fresisting = f max = frail = Its N = 0.4 X 500 = 200 N Using the method of pseudo force: Observing m in frame ofM Driving force (force parallel to the wall surface) Fdriving = mg = 50 N Here, Fdriving < f~esisting Therefore, the friction will be ofstatic nature. As static friction force is self adjusting, it takes the value between 0 and I\"sN (0:'0 Is :'Ol\"sN). Hence friction force in this case mg f = rng = 50 N (upward) Fig. 7.170 Hence the acceleration of the block will be zero. In the frame of M the block m is not sliding. Case III: In this case the resisting force is N = rna (i) (ii) Fresisting :::: fmax = frail = I-isN = 0.4 x 400 = 160 N I =rng If f is of static nature.f :'0 I\"N Driving force parallel to the wall; component of 500 N is :smg j..ima FII = 300 N (upward) and the weight of the block, mg = 120 N = =Since I /LN /LIna, therefore, (downward) f mg Net driving force Fdd,;ng = 300 - 120 = 180 N (upward) Here Fdriving > Fresisting Hence friction is of kinetic nature. F = /k;n = I\"kN = 0.3 x 400 = 120 N (downward) The friction will act downward as driving force on the block acts upward. Hence the acceleration of the block is ma . 180 - 120 , a = 12 = 5 m/s- (upward) U!!l;!I11fllIiIiJ The Fig. 7.168 shows two blocks m and .Fig.7.171 M which are pushed together on a smooth horizontal sur- /-una=mga =!£ I\"

Newton's Laws of Motion 7.39 Forbloek M: F - N =Ma or F =(M +m)a imum and maximum values of mass of block m to keep the g heavy block M stationary. Then F = (M +m)- I' m,block of mass A block of mass mj is placed on another as shown in Fig. 7.172. The blocks have initial velocities Vo and vOz' respectively. If VOl> vo2, find !Is - 0.60 11K\"\" 0.50 1. the acceleration of the blocks. 37\" Fig. 7.172 Fig. 7.174 2. the variation of velocities of the blocks versus time. Sol. Since angle of inclination of the plane is more than the angle ofrepose, i.e., 30' > tatC '0.5 ortan 30° > 0.5, therefore, Sol. the block M has a tendency to slide down. In order to keep it stationary the necessary force is applied by the tension in the 1. Since. the body I moves with a velocity v021 (vo, vo,) to- string. wards right relative to the body 2, the kinetic friction on the body 1 is directed towards left the same amount of friction If the block M also has a tendency to slide down, the friction acts on the body 2, towards right. Apart from this, we have will act in up the plane. shown the weighs mIg and m28 and the normal contact forces N andN'. Ny# ~:;,<;:,:,':,:;; mg MgSine~ /<, - - 8 --M-g-c-o-s8 -'---- FBD Fig. 7.175 For equilibrium of M: Mg sin 0 = T + f J=MgsinO-T (1) fi N = MgcosO Oi) f---+V2 For equilibrium of m: T = mg (iii) From equations (i) and (ii), f = Mg sin 0 - mg (iv) For friction to be static, f < f mar.. Fig. 7.173 MgsinO -mg < I's(MIlCOSO) Equation of motion: M(sinO - I'scosO) < m Forml:I: F, = -f, =mlal (i) Hence m > M (sin e - I's cos 0) (v) For tn,: I: F,. = fk = m,a, (ii) Or m > 50 (~ - ~ x ~) =? In > 6 Law of kinetic friction kk = 1', N, where N = mg 555 Hence minimum value of mass, mmin = 6 kg. This gives fk = IL,,,, 1g (iii) If the block M has a tendency to slide down, the friction Substituting fkfrom eqs. (iii), (i), and (ii) we have will act in downward direction. By this relation, we will get the = a2 = In] maximum value ofrn. This value of m can be obtained by simply I'k-g m,a, and -I'kg 2. Kinematics: Applying kinematics equation for m 1, we have substituting (-I's) in place of ILS in equation (v), iinally we will get the relation v = u + at m < M(sinO - (-I's)cosO) Then by substituting, v = VI, U = vo, and a = -I'kg, we have VI = VOl - f-Lkgt Similarly, applying V = u + at for m, by . tn! =? In < M (sin 0 + I's cos 0) m < 50 (5~+5~ x5~) substituting v = V2U = vOz and a = {Lk ~ g, =? m < 54 kg + m,rn, '\"' Hence maximum value of mass, mmax :;;;;; 54 kg. we have, V2 = VOz f.Lk- gt. ['!mOOtMiID A bar of mass m is placed on a trian· llm!I!!M!ilID Two blocks M and m are arranged gular block of mass M as shown in Fig. 7.176. The friction shown in Fig. 7.174. If M = 50 kg, then determine the min· coefficient between the two surfaces is /l and the ground is

7.40 Physics for lIT-JEE: Mechanics I NF smooth. Find the minimum and maximum horizontal force F required to be applied on block so tbat tbe bar will not slip on the inclined surface of block. m 37° lOON F Fig. 7.178 M 1. the weight, 100 N, acting vertically downwards, e 2. the normal force, acting perpendicular to the surface as Fig. 7.176 shown, 3. the applied force F, and Sol. If both the masses are moving together then the accel- 4. a frictional force, static in nature because the object remains F at rest. eration of the system will be ~~-. If we observe the mass The law for the static fIiction is fl' < fJ,sN. The static friction M+m m relative to M, it experiences a pseudo force rna towards left. can take all the values in the range 0 - I'.,N, either along the Along the incline it experiences two forces, mg sin (j downward plane upwards or along the plane downwards, depending upon and rna cos e upward. If mg sin e is more than ma cos e, it has which way the object tends to move under the combined action a tendency of slipping downward so friction on it will act in of the applied force F and the component of weight down the the upward direction. Here if block m is in equilibrium on the inclined surface, we must have = =e,incline mg sin (mg sin 0 10 x 10 x 0.6 60 N). If mg sin o > F, the object tenqs to move downwards. the frictional force e -mg sin rna cosO::: 11 (mg cose + ma sin 0) on it must be upwards along the incline; and when mg sine < F, sine - fLcosO the ohject will tend to move upwards, the frictional force on it or a> g will be downwards along the incline; and j, < fL.,N. You can - cos 0 + fL sin 0 conclude from this that many distinct possibilities develop. F?:c sinO-fLeosO 0) N or . (M+m)g cos e + fL sm 0 If force is more than the value obtained in equation (0. erna cos will increase on In and the static friction on it will decrease. Ata = g tan 0 Iwhen.F = (M +m)g tan 0], we know ethat the force mg sin 0 will be balanced by ma cos at this ac- celeration no friction will act on it. If applied force increases e ebeyond this value, ma cos will exceed mg sin and friction starts acting in upward direction, An object of mass 10 kg is kept at rest on lOON lOOsin37°'='80N 37\" an inclined plane making an angle of 37\" with the horizontal by applying a force F along the plane upwards as shown in Fig. 7.179 Fig. 7.177. The coefficient of static fiction between the object and the plane is 0.2. Find the magnitude of force F. [Take When mg sin 0 > F and the object is about to move down the incline, the static friction on it will be at its peak value (limiting =g 10 m/s'.] friction). Under this condition for the equilibrium of the object, x N - 80 = O. and (I) F F - 60+ 16 = 0 (2) .;A~-J1s \"'\" 0.2 Solving for F, 37\" F=44N sin 37\" = 0.6 Whenmg sinO> F and(mg sine - F) '\" fL,N (Fig. 7.180). cos 37\" = O.R the frictional force will be equal to (mg sin 0 - F), and Fig. 7.177 o < j, 16 N. Sol. For the ohject to remain at rcst on the incline the resul- 60N tant force on the object must be equal to zero L F = 0 is. and equivalent to L F, = O. and L F, = 0 . Choose the coordinate Fig. 7.180 system with x-axis along the plane upwards and y-axis perpen- dicular to it. as shown in the Fig. 7.178. The forces acting on the object are:

or F = mg sinO - 1,·(0 < I, < 16 N) Newton's Laws of Motion 7.41 F = 60 - i\" 44 N < I\" < 60 N (0 < i, < 16 N) 1. Block C is removed from its position and placed on block When mg sine = F, i, = 0, F = 60 N. A, shown in Fig. 7.183(b). What is now the acceleration of block C? When mg sin 0 < F and (I\" - mg sin 0) < fL.,N, the fric- 2. The positions of the blocks A and B is subsequently in- tional force will be i, = F - mg sin 0 (Fig. 7.181). terchanged. Find the new acceleration of C. The coeffi- cient of friction is the same for all the contact surfaces F (Fig. 7.184). 60N liN j; 37\" Fig, 7,181 or F = i, + mg sinO F = I, + 60 (0 < i, < 60 N) 60 N < F < 76 N (0 < I, < 16 N) and when mg sinO < F, and the object is about to move up JAg willthe incline, the frictional force on it be at its peak value Fig. 7.184 (= fL.,N) along the plane downwards, and for the equilibrium, N - 80 = 0, and (3) Sol. It is given that A moves with an acceleration of 2.45 m/s2 Acceleration of Band C each will be 2.45 mis' downwards. N The force equation for A, B, and C put together arc N-4xlO=O (I) T - fLN = 4 x 2.5 (2) 5xlO-T=3x2.5 (3) 100 N 100 sin 37° \"'\" 80 N Solving these equations for fL = 0.25 Fig, 7,182 I. Fig. 7.185 shows the system after block c is removed from its position and placed on A. Assume the accelerations of A, F - 60 - 16 = 0 (4) B, and C and tension in the string. Calculate the numerical values of the accelerations and the tension. A close scrutiny F=76N will show that the calculated values contradict the very basics on which the accelerations wcrc'assumcd. Therefore, 44 N < F < 76 N Now see how the frictional force on the object varies between Fig. 7.185 o and 16 N along the plane upwards or downwards depending The value of T which can overcome the frictional force on A and make it move is on the requirement, (-16 N ::0 i ::0 + 16 N). And that is why T > 0.25 x (3.6 + 1.2) x 9.8 N the popular phrase \"static friction is self adjusting in nature\" has and, as B will never move upward in the given condition. been coined. T < 1.2 x 9.8 N Therefore, A and B will not move, so will C, ac = 0 illrtijijlltdlD The masses of the blocks A, B, and C 2. For the motion of blocks in this arrangement, there are two distinct possibilities, shown in the Fig. 7.183 weigh 4 kg, 1.5 kg, and 1.5 kg a. C slips on B. [Fig. 7.183(a)] respectively. Block A moves with an accel- eration of 2.5 mis2• Fig. 7.183

7.42 Physics for IIT-JEE: Mechanics I Fig. 7.186 Mg Fig. 7.190 b. C does not slip on B. Let C slip on B 3.6 x 9.8 - T = 3.6 a • Static friction force is a self adjusting force 0:'0 1 T - IL x 1.2 x 9.8 - IL x 2.4 x 9.8 = 1.2a :'0 IL.,N, 1m,,, = IL.,N (Fig. 7.190). and 0.25 x 1.2 x 9.8 = 1.2ac From equations (4) and (5) (4) After friction force reaches limiting value, the motion starts and (5) the friction force becomes fk = ILk N N (6) • Assume SYStCI1?- moves together. Le., no sliding between M and til (Fig. 7.191). T mg T III N, N F (i) 1.2 g (M +m) (ii) Fig. 7.191 Fig. 7.187 a = x 9.8 = 5.5125 m/s2 Acceleration of system in this case a and from equation (6) Ge = 0.25 x 1.2 x 9.8 = 2.45 mls2 NowFBDofm (see Fig. 7.192). 1.2 Mg Now 5.5125 m/s2 > 2.45 m/s2 =} a >ac will slip on A N Therefore. ae = 2.45 mls2 Fig. 7.192 FED of m 11Jlltj!IZ1!1l11l1l Two blocks of mass M and m are ar- Equation of motion of m. ranged as shown in Fig. 7.188. There is no friction between ground and M. 1 = rna =m F -c:-c:---,- Fig. 7.188 (M+m) Coefficient of friction between M and m is /L, and ILk. If there is no sliding between M and mf :'0 f,· 1. Calculate the maximum possible value of F so that both + +mF the bodies move together. 2. Maximum possible friction force between the surfaces (M m) :'0 IL.,(mg) =} F :'0 IlAM m)g Sol. Free body diagrams of til and M (Fig. 7.189) If F > IL., (M +m)g N • there will be relative sliding between M and m. Fig. 7.189 Calculate the acceleration of M and m in this case. When relative sliding between M and In starts Free body diagram of M (Fig. 7.193). Equation of motion of m. ILk(mg) = mal =} al = l\"kl! Equation of motion of M. F - ILkmg = Maz M

Newton's Laws of Motion 7.43 .-'---1 C()~cepfApplicatioI1ElCercise7,4 f------. Mg F 1. A block weighing 20 N rests on a horizontal surface. The coefficient of static friction between block and surface is Fig.7.193 FBD of M 0.4 and the coefficient of kinetic friction is 0.20. a. How much is the friction force exerted on the block? Case II b. How much will the friction force be if a horizontal force of 5 N is exerted on the block? M.There is no friction between the ground and The coefficient e. What is the minimum force that will start the block in motion? of static and kinetic friction are fl., and fl.k, respectively. d. What is the minimum force that will keep block in mo- tion once if has been started? • What is the maximum possible value of F so that system e. If the horizontalforceis 10 N. what is the friction force? moves together? 2. A block of mass m rests on a rough fioor. The coefficient • If there is relative sliding between M and m then calculate of friction between the block and the floor is fl.. acceleration of M and m? ea. Two boys apply force P at an angle to the horizon- Let the systems moves together tal. One of them pushes the bloek; the other one pulls. a= --=F -- Which one would require less efforts to cause impend- (M +m) ing motion of the block b. What is the minimum force required to move the block FBDof m andM by pulling it? e. Show that if the block is pushed at a certain angle 80 it mg N mg cannot be moved whatever the value of P be. .----+--l==;--+f 3. What is the value of friction f for the following value of F applied force F . f (i) Fig. 7.196 a.IN b.2N c.3N d.4N e.20N N Assume the coefficient of friction to be fl.., = 0.3; fl., = 0.25. Mass of the body is In = I kg. (Assume Fig. 7.194 g = 10 m/s2) From FBD of M.f = rna 4. A block of mass 5 kg rests on a rough horizontal surface. It is found that a force of 10 N is required to make the f=M(M:m) block just move. However, once the motion begins, a force of only 8 N is enough to maintain the motion. Find As there is no sliding between M and m. c. the coefficients of kinetic and static friction between the block and the horizontal snrface. mg N mg F Fig. 7.197 M 5. A body of mass m is kept on a rough horizontal surface of N' friction coefficient fl.. A force is applied horizontally, but Fig. 7.195 the body is not moving. Find the net force'F exerted by the surface on the body. (M :m):::M fl.,mg =} F::: fl., :(M +m)g 6. Determine the magnitude of frictional force in each of the following cases: If the relative sliding between the block starts. 5 kg 5 kg =} al = F - fl.kmg =} a2 = I' ~ 0.2 'm fl.kmg Fig. 7.198 f = - - - J =~--f=-- M

7.46 Physics for lIT-JEE: Mechanics I I: Fy = T cosO - mg = 0 (i) (ii) 4. When a real force pushes a particle radially outward, we T cosO = mg cannot call it centrifugal force of the tirst kind. L F.y = T sin e = mac = mv2 5. The centrifugal force is directed \"radially\" outward from I' the axis of rotation of the reference frame (or observer) along the line drawn from the particle perpendicular to Divide equation (li) by equation (i) and use the axis of rotation. sinB/cosH = tanO, tanO = 6. The centrifugal force is a pseudo-force which is equal to rg -m a~:p and centripetal force is areal force. The centrifu- gal force is adopted to solve the problems in the rotating Solve for v: v = .JrRtan8 frame. e,Incorporate. r = L sin from the geometry in the Fig. 7.213. A small ball of mass m is suspended from v = v'Lg sin etan e a string oflength L. The hall revolves with a constant speed v in a horizontal circle of radins r as shown in Fig. 7.211. A particle of lIlass In is moving with a (Because the string sweeps out the surface of a cone, the system is known as a conical pendulum.) _Find an expression constant v in a circular path in a smooth horizontal for v. plane (plane of the paper) by a spring force as shown in Fig. 7,213. Ifthenaturalleugth ofthe spring is 10 and stiffness of the spring is k, find the elongation of the spring. \\ Fig. 7.213 \\ \\ L \\ Sol. If the particle executes uniform circular motion then its \\ \\ speed must be uniform because there is no tangential force to I\\ speed it up. Here, the spring force kx acting on the particle is the .. __ c._J b._---1----_..\\ centripetal force caused by the elongation x of the spring. III -- ______ --/ Equation of motion: L r~. = Fsl' = mar (i) v ~.... . Fig. 7.211 -- m Sol. Imagine the motion of the ball in Fig. 7.212 and convince kx v yourself that the string sweeps out a cone and that the ball moves in a circle. Fig. 7.214 The ball in Fig. 7.212 does not accelerate vertically. Therefore, But centripetal acceleration ar = v2 we I}10del it as a particle in equilibrium in the vertical direction. It - experiences a centripetal acceleration in the horizontal direction, so it is modeled as a particle in uniform circular motion in this R direction. (where R = radius of circular path) (ii) (iii) Let 0 represents the angle between the string and the vertical. Radius of rotation: R = 10 + x 7:In the FBD shown in Fig. 7.212, the force exerted by the string Using the above equations, we have the quadratic equations, ais resolved il1to a vertical component T cos and a horizontal ecomponent T sin acting towards the ccnter of the circular path. k:(2 + klox - III v2 = () 1,\"v'0\\ Tcos() . . Jx216 + ~~~k - klo I\\ This gIVes x = ------- 2k ).,T,, A coin is pushed down tangentially from an angular position () on a cylindrical surface, with a velocity v as.shown in the Fig. 7.215. If the coefficient of friction between the coin and the surface is JL. find the tangential acceleration of the coin. Tsin e mg Fig. 7.215 Fig. 7.212 Apply the particle in equilibrium model in the vertical direction:

Newton's Laws of Motion 7.47 Sol. As the coin slides down, the friction is kinetic and acts up T, cos 60\" = 7, cos 60\" + mg (ii) along the plane. mg J T,-T2=---=2mg . cos 60° Dividing equation (i) by (ii) -TA + 1 ' (/),'[ - -\" '* --- - - '*T, + T2 \"l[, 4 + 1 w2[ T2 =T-, ---:T~2 - 2g 4 - I 2g T, _ 1 - 2g T2 mg '*w2 = 109 = 10 x 9,8 x 3 = 196 w = 14 rads-' Fig. 7.216 3[ , 3 x 50 X 10-2 Equation of motion: Two wires AC and BC are tied at C of a small sphere of mass 5 kg, which revolves at a constant speed I: F.. = N - mgcosO = ma,· (i) v in the horizontal speed v in the horizontal circle of radius 1.6 m. Find the minimum value of v. I: F, = mg sin 0 - fk = rna, (ii) A But f, = /kN (iii) mv2 (iv) and centripetal acceleration is G,. = - - R Substituting f, from equation (iii), a, from equation Ov), in equation Oil, We get at' =gsm. 0 -/k-N (v) III c Now, substituting N from equation (i) in equation (v), L6m We get a, = g(sine - f-Imv 2 Fig. 7.219 Sol. From force diagram shown in Fig, 7,220, /kcosO),- - - R T, cos 30' + T2 cos 45' = mg A small block is connected to one end TI cos30\" (i) 2 +T2 cos45° 3of two identical massless strings of length 16 cm each with their other ends fixed to a vertical rod. If the ratio of tensions T,ITz'be 4: 1, then what will be the angnlar velocity IV of the block. T) sin30° +T2 sin45° mg Fig. 7.220 T, sm. 30\"' + 72,s.m45' = -mv-2 (ii) r Fig. 7.217 m g -m-v-2 Sol. For horizontal equilibrium of the block, After solving equations (i) and (ii), T, = r T1 sin 60° + 12 sin 60° = mw2r = mui (~- I) T, , m g -m-v-2 2 But T, > 0, h r >0 mv2 := :IT,,,Wlr,TJ sin60\" 60°1 mg > --, v < Jrif y3-1 12 sin6Qo 60\": mg r !T, ' 2 T2 cos60° Fig. 7.218 VmO> = ..;rg = v'L6 x 9,8 = 3,96m/s T, + T2 = mw2 [ Bending of a Cyclist For vertical equilibrium of the block, When a cyclist takes a tum, he also requires some centripetal force, If he keeps himself vertical while turning, his weight is (i) balanced by the normal reaction of the ground, In that event, he has to depcnd on the force of friction between the tyres and the

7.48 Physics for IlT-JEE: Mechanics I road for obtaining the necessary centripetal force. As the force of fi\"iction is small and uncertain, dependence on it ,is not safe. mg Horizontal Fig. 7.222 o·'-'------x Banking to Avoid FrictionaL Wear and Tear What we really wish is that even if there is no friction between the tyres and the road, we should be able to take a round-turn, eLet us attend to Fig. 7.223. Vertical N cos component of the normaLreaction N will be equal to mg and the horizontal N sin ~'ig. 7.221 () component will provide for the necessary centripetal force. To avoid dependence on the force of friction for obtaining [Please note that as we are assuming JL to be zero here, the centripetal force. the cyclist has to bend a little inward from his total reaction of the road will be the normal reaction.] Frictional vertical position, while turning. By doing so, a component of forces will not act in such a case. normal reaction in the horizontal direction provides the neces- N cos 0 = mg (i) sary centripetal force. To calculate the angle of bending with e. mv2 (ii) N S1l1 = - - vertical, suppose, r v2 m = mass of the cyclist, v = velocity of the cyclist while Dividing equation (ii) by (i), we get tanil = - ; where 0 is rg eturning, r = radius of the circular path, = angle of bending the angle of banking. with vertical In Fig. 7.222, we have shown weight of the cyclist (mg) acting vertically downwards at the center of gravity C. R is force of ,N ...Vertical ereaction of the ground on the cyclist. It acts at an angle with 8', the vertical. -------.;+----» R can be resolved into two rectangular components: Horizol1tal R cos e, along the vertical upward direction, R sin e, along e the horizontal, towards the center of the circular track. f\"\"' )l\"N mg eIn equilibrium, R cos balances the weight of the cyclist, i.e.• Rease = mg (i) Fig, 7.223 eand R sin provides the necessary centripetal forcc(m v2/ r) mv2 (ii) R sin 0 = - - r • This particular formula is very important because it comes Dividing equation (i) by equation (ii), we get back again and again, It had given the angle of tilt of the cyclist and the minimum value of It required also \\vhile frW2 negotiating a turn, Rsine r v2 • The vahle of v here will be the -maximum value of the --- = --; tane = - (iii) vclocity of the vchicle permissible on this banked road, Rcose mg rg i.e., if we assume It = O. Banking of Roads Motion ALong a CircuLar Track Perhaps you have noticed that when a road is straight, it is hori- When a vehiclc negotiates a curved (circular) track., centripetal zontal too. However, when a sharp turn comes, the surface of the force must be active acting t.owards the centre of the circular road does not remain horizontal. Figure 7.222 depicts it. This is path. called banking of the roads. Consider the situation shown in Fig. 7,224, where a vehicle Purpose of Banking negotiates a horizontal circular curve of radius of curvature r with a uniform speed v. The forces acting on the vehicle are (i) Banking is done: its weight mg, (ii) normal reactioll N offered by a rod on to the 1. to reduce frictional wear and tear of tyres. 2. to avoid skidding. and .rvehicle. and (iii) friction between the road and the tyres of the 3. to avoid overturning of vehicles. vehicle.

Newton's Laws of Motion 7.49 --1---rlo- -m,v-2. T,,O Overtuming of car '''''''f--- r -/--+1 Fig. 7.226 Fig. 7.224 A 1,500-kg car moving on a flat, hori- For vertical equilibrium of the vehicle road negotiates a curve as shown in Fig. 7,227, If the radius of the curve is 20,0 m and the coefficient of static N=mg (i) friction between the tires and dry pavement is 0.50, find the maximum speed the car can have and still make the turn For horizontal equilibrium of the vehicle (as, shown in successfully. Fig. 7.225). (Ii) / =m-v2 r v () r m-g, (b) .. mv2 (i) The force of static friction directed r = -r- toward the center of '.he curve keeps f=mg the car moving in a circular path. (ii) Free body diagram for the car. Fig. 7.225 Fig. 7.227 Evidcntly, the speed of a vehicle is variable and so also the Sol. Imagine that the curved roadway is part of a large circle so . frictional force which is a \"self adjusting force\". For speeds that the car is moving in a circular path. within a maximum value, the frictional force automatically sets Based on the conceptualize step of the problem, we model its value, so as to just prevent the vehicle from skidding radially the car as a particle in uniform circular motion in the horizontal outwards. However for speeds too large, the centrifugal force, direction. The car is not accelerating vertically, so it is modeled exceeds to such an extent, that even the limiting friction (max- as a particle in equilibrium in the vertical direction. imum static friction) is also unable to prevent the vehicle from skidding. The force that enables the car to remain in its circular path is the force of static tfiction. (It is static because no slipping Thus, for horizontal equilibrium (with respect to a rotating occurs at the point of contact between road and tires. If this frame), the speed v should be such that force of static friction were zero. for example, if the car were on an icy road, the car would continue in a straight pne and slide mv2 .:::: Jiimiting :.::.::> rnv 2 ::; f(,mg rwhere fl-s, is the static CO~ off the road.) The maximum speed Vmax the car can have around the curve is the speed at which it is on the verge of skidding -r- -r- outward. At this point, the friction force has its maximum value efficient of friction hetween lyres and ground], j~,max = ItS N, '* '*:sv 2 fJ..,.gr :sv ,jfJ.\"gr (iii) Thus, the maximum speed limit for a vehicle with coefficient (i) of frictioml'ls, for successful negotiation, of a curve of radius of curvature r is given by, Vmax . = .J/-L.\\.gr. Overturning Apply the particle in equilibrium model to the car in the ver- tical direction: L F,. = 0 -\" N - mg = 0 -\" N = mg With the increase in the speed of the car N2 increases while NJ Solve equation (i) for the maximum speed and substitute for decreases. For a particular value of speed v = Vmax • NJ become zero, and the car is about to overturn. Thus to prevent overturn- N: ,/'~\"v\"'''' = ing, N, 2: O. Nr = jl'smgr (Ii) - - = -//l.,.gr til m ·ga v <- -h- f for -m-g ( 1--V-2-h) 2:0 or = J(0.50)(lOm/s2)(20m) = 10.0 m/s 2 rga

7.50 Physics for IIT-JEE: Mechanics I Dividing equation (i) by equation (ii) A civil. engineer wishes to redesign a v2 (iii) tane = - curved roadway in such a way that the car will not have to rely on friction to round the curve without skidding. In rg other words, a car moving at a designated speed can nego~ !iate the curve even when the road is covered with ice. Such Solve for the angle e (I)2 a ramp is usually banked, which means that the roadway is tilted toward the inside of the curve. Suppose the designated e , ( (IO,om/s ) ) , 2 speed for the ramp is to he 10.0 m/s and the radius of the curve is 20.0 m. At what angle should the curve be banked'! = tan- (20.0m)(1O.Om/s') = tan- ,---1 Concept Application Exercise 7.5 1-----, 1. Three masses are attached to strings rotating in the hori- zontal plane. The strings pass over two nails as shown in the Fig, 7.229, Will this system be in equilibrium? ___ _~, _..J ~ / F:r: / I eA car rounding a curve on a road banked at an angle to I I the horizontal. In the absence of friction the force that causes /---- the centripetal acceleration and keeps the car moving in its circular path is the horizontal component of the norma! force. m m '\"--- 2m Fig. 7.229 Fig. 7.228 2. Can force determine the direction of motion and direction Sol. The difference between this example and the previous il- of acceleration? lustration is that the car is no longer moving on a flat roadway. Figure 7.228 shows the banked roadway, with the center of the 3. You are riding on a Ferris wheel that is rotating with a circular path of the car far to be the left of the Fig. 7.228. Notice constant speed. The car in which you are riding always that the horizontal component of the normal force participates maintains its correct upward orientation; it does not invert. in causing the car's centripetal acceleration. 3. What is the direction of the normal force on you .from the scat when you are at the top of the wheel? As in previous illustration, the car is modeled as a particle in (i) upward (ii) downward (iii) impossible to determine equilibrium in the vertical direction and a particle in uniform b. From the same choices, what is the direction of the net circular motion in the horizontal direction. force on you when you are at the top of the wheel? On a level (unbanked) road, the force that causes the cen- 4. A bead slides freely along a curved wire lying on a hori- tripetal acceleration is the force of static friction between the zontal surface at a constant speed as shown in Fig. 7.230. 3. Draw the vectors representing the force exerted by the car and the road as we saw in the preceding example. If the road wire on the bead at points A, Band C, Suppose the bead in Fig, 7.230 speeds up with a constant tangential eis banked at an angle as in Fig. 7.229, however, the normal acceleration as it moves toward the right. Draw the vectors representing the force on the bead at point A, force -;; has a horizontal component towards the center of the Band C. curve. Because the ramp is to be designed so that the force of ® estatic friction is zero, only the component Nx = N sin curves ©A bead slides along a curved curve the centripetal acceleration. Fig. 7.230 5. A smooth block loosely fits in a circular tube placed on a Write Newton's second law for the car in the radial direction, horizontal surface. The block moves in a uniform circular which is the x direction motion along the tube (Fig. 7.231). Which wall (inner or outer) will exert a nonzero normal contact force on the I: r~ = N sine = mv2 (i) block? - r Apply the particle in equilibrium model to the car in the ver- tical direction LF,. = Neose -mg = 0 N cose = mg (ii)

Newton's Laws of Motion 7.51 Fig. 7.231 Fig. 7.233 6. An amusement park ride consists of a large vertical a. the tension in the string; cylinder that spins about its axis fast enough that any b. the radial force acting on the puck, and person inside is held up against the wall when the floor drops away. The coefficient of the static friction between c. the speed of the puck? the person and the wall is /).,.\" and the radius of the d. Qualitatively describe what will happen in the motion cylinder is R (see Fig. 7.232). of the puck if the value of l1l2 is somewhat increased. by placing an additional load on it. e. Qualitatively describe what will happen in the motion of the puck if the value of 1112 is instead decreased by removing a part force the hanging load. 10. A sleeve A can slide freely along a smooth rod bent in the shape of a half circle of radius R. The system is set in rotation with a constant angular velocity (;) about a vertical axis 00'. Find the angle 0 corresponding to the steady position of the sleeve (sec Fig. 7.234 for reference). Fig. 7.232 0' a. Show that the maximum period of revolution necessary Fig. 7.234 to keep the person from falling is T = (47T' Ii,l lg)'/2. 11. A ball suspended by a thread swing in a vertical plane so h. Obtain a numerical value for T, taking R = 4,00 III and that its acceleration values in the extreme and the lowest 1.1.,-, = 00400. How many revolutions per minute does the position are equal. Find the thread deflection angle in the cylinder make? extreme position. c. If tbe rate of revolution of the cylinder is made to be 12. A simple pendulum is oscillating with an angular displace- somewhat larger, what happens to the magnitude of ment of 90°. For what angle with the vertical the acceler- each one of the forces acting on the person? What hap- ation of bob directed horizontally? pens to the motion of the person? 13. A ceiling fan has a diameter (of the circle through the outer d. If instead the cylinder's rate of revolution is made to edges of the three blades) of 120 em and rpm 1,500 at full be somewhat smaller, what happens to the magnitude speed. Consider a particle of mass I .R sticking at the outer of each one of the forces acting on the person. What end of a blade. happens to the motion of the person? How much force docs it experience when the fan runs at full speed? 7. Tarzan (m = 85.0 kg) tries to cross a river by swinging on a Who exerts this force on the particle? vine. The vine is 10.0 m long, and his speed at the bottom How much force docs the particle exert on the blade along of the swing (as he just clears the water) will be 8.00 mIs, its surface? 1~u'zan doesn't know that the vine has a breaking strength of 1000 N. Does he make it across the river safely? 14. A blockofmassm is kept on '\\ horizontal ruler. The friction coefficient between the ruler and the block is 111. The ruler 8. A roller-coaster car has a mass 0[500 kg when fully loaded is fixed at one end and the block is at a distance L from with passengers. the fixed end. The ruler is rotated about the fixed end in a. If the vehicle has a speed of 20.0 mls at point A, what the horizontal plane through the fixed end. is the force exerted by the track on the car at this point? a. What can the maximum angular speed be for which the b., What is the maximum speed the vehicle can have at block does not slip? point Ii and still remain on the track? h. If the angular speed of the ruler is uniformly increased from zero at an angular acceleration a, at what angular 9. An air puck of mass m! is tied to a string and allowed to speed will the block slip') revolve in a circle of radius R on a rrjctionlcs~ horizontal table (see Fig. 7.233). The other end of the string passes through a small hole in the center of the table,. and a load of mass 1112 is tied to the :-;tring'? The suspended load remains in the equilibrium while the puck on thc tabletop revolves. What is

7.52 Physics for IIT·JEE: Mechanics I F 15. An old record player of 15.0 em radius turns at 33.0 ·c rev/min while mounted on' a 30° incline as shown in the Fig. 7.235. a. If a mass m can be placed anywhere on the rotating record, which is the most' critical place on the disc where slipping might occur? Fig. 7.235 Fig. 7.238 b. Calculate the least possible coefficient of friction that upward force F is applied on the pulley and maintained at must exist if no slipping occurs. a constant. Calculate the acceleration al and az of the 5 kg and 2 kg masses, respectively, when F is 16. A 60 kg woman is on a large vertical swing of radius 20 ill. The swing rotates with a constant speed. I. 30 N, 2.60 N, 3.110 N (g = 10 ms-'). a. At what speed would she feel weightless at the top? Sol. Apart from the constraint that the string is unstretchable, the additional constraint is that neither of the masses can go b. At this speed. what is her apparent weight at the bot- downward. So the block will be lifted only when the tension of tom? the string exceeds the gravitational pull on them. I. Considering the FED of the pulley (Fig. 7.239) 17. A rod OA rotates about a horizontal axis through 0 with a constant anti clockwise velocity (0;:::: 3 rad/sec. As it passes 30 - 2T = 0 (pulley is massless) or T = 15 N ethe position = 0 a small block of mass m is placed on F it at a radial distance r = 450 mm (sec Fig. 7.236). If the eblock is observed to slip at = 50°, find the coefficient of static friction between the block and the rod. (Given that sin 50\" = 0.766, cos 50\" = 0.64). A o. 8 TT Fig. 7.239 Fig. 7.236 So tension is less than gravitational pull on both the blocks. 18. A particle rests on the top of a smooth hemisphere of radius r. It is imparted a horizontal velocity of ~rygr (see So no acceleration is produced in them. Fig. 7.237). Find the angle made by the radius vector joining the particle with the vertical, at the instant, the Cll = Q2 = 0 particle losses contact with the sphere. 2. Now 60 - 2T = 0, or T = 30 N .,f.if.i.i...i' So the 5 kg weight will not be lifted but the 2 kg weight will I be lifted. 30 - 20 = 2 x \"2 or (/2 = 5 ms-2 I/ Thus \", = 0 and (/2 = 5 ms- 2 rI // e 3. Now 140 - 27' = 0, or T = 70 N I So both the weights arc lifted. [)/ 70 - 50 = Sa\" ora, = 4 ms-' and 70 - 20 = 202, o or Q2 == 25 ms,-2 Fig. 7.237 m1 =1 In tbe arrangement shown in Fig, 7.240 =m, 2 kg. Pulleys are massless and strings are light. For wbat value of M the mass m 1 moves witb a constant Two blocks of masses 5 kg and 2 kg (see velocity. Fig. 7.238) are initially at rest on tbe floor. They are connected Sol. Mass m! moves with a constant velocity if tension in the by a light string, passing over a ligbt frictionless pulley. An lower string is

Newton's Laws of Motion 7.53 M 5N ~ 30N El.\"-I 5N . Fig. 7.243 a,. AcceleratIOn of b5lock, = 1 = 5m/s2 Fig. 7.240 .' -3-02--5=225111182 . Acceleration of plank, a2 = T, =ml!?=(I)(lO)= JON Tension in the upper string is (i) Relative acceleration of plank, a = ([2 ~ al or, (ii) T2 = 2T, =20N (iii) a= 25 - 5= , Acceleration of block M is, therefore, - 7.5m1s 2 7; 20 )2 \"2fiit = V-;; = a=--=- = 0.73 s 7.5 MM Consider a system ofa small body ofmass T~ ION m kept on a body of mass M placed over an inclined plane of angle of inclination 0 to the horizontal. Find the ac- celeration of'm when the system is set in motion. Assume in~ ,, clined plane to'be fixed, All the contact surfaces are smooth. 0t' d\",\"-~ l;l~ m21J = 20 N '><2' Fig. 7.241 \",'J:\"Yi This is also the acceleration of pulley 2. ',';~t~ Absolute acceleration of mass m 1 is zero. Thus, acceleration of In I relative to pulley 2 is a (upwards) or acceleration of m2 Fig. 7.244 wilh rcspcci to pulley 2 is a (downwards). Draw free body dia- Sol. (i) Analysis in an inertial reference frame attachec\\ to gram of m, with respect to pulley 2 (Fig. 7.241). the ground: Equation of motion gives Let a be the acceleration of In W.Lt. M directed horizontally 40 40 towards right. Let A be the acceleration of M w.r.t. the ground directed along 20 - -M - 10 = 2a =M- the incline m downward direction, =Solving this, we get, M 8 kg. ~ 7r:' A cos fJ a A block of mass 1 kg is placed over a plank of mass 2 kg. The length of the plank is 2 m. Coefficient f:;,l'\\! of friction between the block and the plank is 0.5 and the ground over which the plank is placed is smooth. A constant \"\"\",n'g force F = 30 N is applied on the plank in horizontal direction. Fig. 7.245 Find the time after which the block will separate from the plank. Acceleration of m w.r.t ground will be the vector sum of the ~/77A77///////:::/b;: 30 N acceleration of m W.f.t M and acceleration of M W.f.t ground. 1<1 1-1 Force equations [or marc, 2m O=m(a-Acose) (I) Fig. 7.242 mg - N = meA sin 0) (2) Sol. Maximum frictional force between the block and the plank Force equations for Marc, (3) is N' sine + Mgsine = MA 1m\" = I\"mg = (O.5)(l)(I 0) = 5 N A The FED of the block and the plank are shown in Fig. 7.243. Mg Fig. 7.246

7.54 Physics for IIT-JEE: Mechanics I N'cos&+MgcosH-N=O (4) The forces acting on Mare From equation (2) a. the weight Mg, N' = mg - mA sin 0 b. N', normal force exerted by m, Substituting N' in equation (3). c. N, normal force exerted by the incline, and d. (MA), the inertial force along the plane upwards. (mg - rnA sine)sine + Mg sinO = MA acceleration of M w.r.t. M (w.r.t. itsel!) = 0, (M +rn)gsine Mg Sin e + N' sinO - MA = 0 (3) e=} A = '-:-c-'-~'-;--.,-- (along the incline) M +msin2 Mf( cos 0 + N' cosO - N = 0 (4) eAcceleration of In w.r.t ground is A sin (since, eA cos -a =0) thus the acceleration of m w.r.t ground (perpendicular to the incline) is From equations (1), (3), and (4), (M +m)gsine (M + m)g sine M+msin2 & A= - e- (ii) Analysis of motion of m in the non interracial reference + sin 2 frame attached to M: M m Let the acc.eleration of M w.r.t. ground be A along the inclined plane downwards. Consider that ground is an inertial reference __ a( --~ frame, the reference frame attached to M will be non-inertial. For applying Newton's second law of motion to any object w.r.t. A ~ .. M, you have to apply an inertial force (also called, pseudo force) on the object which equals mass of the object times acceleration Fig. 7.249 of M, directed opposite to A. Acceleration of 111 relative to ground, Force equations for m: Let the acceleration of m w,r.t. M be a, in the horizontal +umG = GmM aMo direction towards right: e,Rut a = A cos (from equation 2), therefore N' iliA e(M + m)g sin2 e timG = A sine, vertically downwards M +msitlO mg Pulleys shown in the system of the Fig. 7.247 The forces acting on mare: Fig. are massless and frictionless. Threads are in- a. weight of m, mg, acting vertically downwards, extensible. Mass of blocks A. B, and Care m1 = 2 kg, h. normal force on m by -M, N', vertically upwards, and m2 = 4 kg and rn3 = 2.75 kg, respectively. Calculate accel- c. (mA), the inertial force acting along the incline upwards, as eJ'ation of each block. the acceleration of M is A along the incline downwards. Fig. 7.250 From L;F=ma Sol. Let acceleration of blocks A and B be a and b verti- N' + (mA)sine - mg = 0 (I) (2) cally upwards, respectively. Then according to geometry of the [L;Fy = may] given figure downward acceleration of block C will be equal to e(mA) cos = ma 2a + 4b. Now considering the FRDs (see Fig. 7.251). [L; F, = max] 1', 1'1 1', T2 1', 1', Force equations for M: 4- A 4- C f113(2a+4b) ,,IJIl(/! ,m2b i 11I3g , Fig. 7.251 (i) \\ For block A, T! - mig = mj{i (ii) N' For block E, 2T] - nI2/{ = 1n 2b (iii) '. For pulley F, 1', = 272 Mg Fig. 7.248

7,For block C, m3g - = m3 (2a + 4b) (iv) Newton's Laws of Motion 7.55 Solving above equations, T, = 22 N, T, = II N, (N)'N' =mgcosa - Imi gR ~ a = I ms-', h = I ms-2 I mgR mg Hence, acceleration of block A, a = I ms\"' (t) N = mgcosa - -- = mgcosa - - acccleration of block B, b = I Ins\"' (t) 2R 2 acceleration of block C, = (2a + 4b) = 6 InS\"\"' (t) If he does not loose contact with road, N' >0 A motorcycle has to move with a constant 7r mgcosa - m2g> 0 =} cosa > 2: =} on an overbridge which is in the form of a circular arc Hence arc length, l' = \"R:3rr from top. a= - of radius R and has a totalleng!h L. Suppose the motorcycle starts from the highest point. 3 1. What can its maximum velocity be for which the contact 3. The equation of motion of motor cyclist in function of a mea- with the road is not broken at the highest poin!'! sured from vertical 2. If the motorcycle goes at speed 112 times the maximum mvl!2 \"fouud in part (a), where will it lose the contact with the road? mgcosa - Nfl = - - R 3. What maximum uniform speed can it maintain on the bridge ifit does not lose contact anywhere on the bridge? The normal reaction will be minimum at the end of the arc, Hence critical position is at the end of the arc. where, oL a=-=-· 2 2R v \"'-P-mg cos (2~) N\" = . ( L) R=} N =mgcos 2R - mv\" If he does not loose contact even at the end of the are, Nil> 0 L)mg cos - - -mv-..2 > 0 Fig. 7.252 (2R R Sol. ( L) (;~):::::}gcos - >v'-\" Arc length L 2R R =} u\" < gRcos 1.0= radius = R Free body diagram (Fig. 7.253) at top position. A track consists of two circular partsABC mv2 and CDE of equal radius 100 m and joined smoothly as REquation of motion, mg - N = shown in Fig. 7.254. Each part subtends a right angle at its centre. A cycle weighing 100 kg together with the rider N' travels at a constant speed of 18 kglh on the track. ,,,,,,'/' // / / / // E D Fig. 7.253 mv2 Fig. 7.254 N=mg-- 1. Find the normal contact force by the road on the cycle R when it is at B and at D. If contact at highest point does not loose N > 0, 2. Find the force of friction exerted by the track on the tyres when the cycle is at B, C, and D. mv2 v' < gR =} v < ,fiR mg> -- 3. Find the normal force between the road and the cycle jnst before and just after the cycle crosses C. R 4. What should be the minimum friction coefficient between 2. .. v'I2 v = yv'g'2R the road and the tyre, which will ensure that the cyclist can move with constant speed? If velocIty of motor cyclIst v' = At position A, mv'2 mgcosa - N' = -- R

7.56 Physics for IIT-JEE: Mechanics I Sol. v = 18kmjh = 1800 = 5 mls ii. normal force Nd of thc track along, 60 x 60 Hence, from the dynamics of' circular motion, mv2 1. a. At point B, various forces act.ing on the cyclist are: i the weight mg, vertically downward. and Nd - mg cos ex = - - r ii. normal force NB by the road upward. Since, under the action of these two forces, the cyclist moves (gCOS\" ~2)=} in a circular path of radius r, hence from the dynamics of Nd = m = = 732.11 N circular motion, £)= 100 (10 cos 45\" + 1Il~2 =mg-NB =} NB=m(g_~2) 100 4. The tendency of the cyclist to skid is maximum just before £)= 100 (10 - = 975 N the point C, the point where the radius of curvature changes 100 its direction. This is so because the curve is steepest at this point. h. At point 1), various forces act.ing on the cyclist are: Force along the track at this point = mg sin a, frictional forcc = mNb i. the weight mg, vertically downward, and where N\" normal force of the track just before the point C. The cyclist can move with constant speed only 'if there ii. normal force ND by the road upward. two forces are equal and opposite, fLNb = mg sin\" =} fL x 632.11 mv2 N[) -mg =-- = 100 x 10 x sin 45\" =} fL = 1.037 ND V') £)r =} =m(g+ r 100(10- 100 =975N A plank of mass m rests symmetrically on two wedges Band C of mass M. What is the acceleration 2. At the points Band [), the tracks are almost horizontal. there- of the plank? Neglect the friction between all the contact fore, there is no components of g along the track at those surfaces. points that will change the speed of the cyclist. Since the cy- clist moves with a constant speed, the frictional force at these two points is necessarily zero. At point C, hqwever, the component of rng along the Am track that tends to accelerate the motion of the cyclist = mg sma, BC MM But since his speed remains constant, a force, equal and opposite to mg sin a, must be acting on it and this force is the force of friction mNe, where m is the friction coefficient Fig. 7.256 between the road and the tyre and Nc is the normal reaction at the point C (sec Fig. 7.255). Sol. When the plank is released it falls through a distance y and both the wedgcs moves through a distance x. From Fig. 7.257. eY = x tan (i) mgsin a -tY ... N N ... IIIg N I I X I I I Fig. 7.255 A+ - lYsin 0 Hence frictional force at Cis, /LNc = mg sina = 100 x 10 x sin45\" = 707.12N mg 3. Just before the cyclist crosses the point C, various forces IIIg a acting on him are: Fig. 7.257 i. Wcight mg, vertically downward, and ii. Normal force Nb of the track along. On differentiating this expression twice, we obtain a=AtanO Hence, from the dynamics of circular motion, we have mv2 N mgcosa - Nb = - - Ns-in»e r I I I ~2)Nb = (gCOS\" _ = 682.IIN A t Just after the cyclist crosses the point C; various forces mg acting on him are: Fig. 7.258 i. weight mg, vertically downward, and

---A Newton's Laws of Motion 7.57 Equations of wedge: (ii) LJ;~ = Nsine = MA LFy=N'-Ncos8-Mg=O (iii) Equations of block: (iv) T (v) L F, = N sin e - N sin e = 0 e o eT cos +--\"'CL;~-~ mg cos L Fy = mg - 2N cos e = nia Fig. 7.262 MA From equation (ii), N = -.- sme J .--N, Y~X A cos e Nsin e ! 0, N 1:\\I!- Xt \\ ~ N a A¥ X \\ A sin e \\ \\ a Fig. 7.259 a mg '<\"f>',p\". F mg On substituting expression for N and a in equation (v), we '\"\"\" obtain e2MAcose mA sin Fig. 7.263 sine = cose (am)x = a - Acose ettlg sin 0 cos e({lm)y = A sin e + eA = m-~sin~2 -~-M ;C0:52­ Equations of wedge: In Fig. 7.260, mass In is being pulled on (iii) of mass M. All the surfaces are smooth. L F, = N sin e = M A (iv) L Fy = N' - Nease - Mg = 0 Find the acceleration of the wedge. m Equations of block: (v) (vi) L F, = F +mg sin G = mea - A cos 0) L F,. = mgeose - N = mAsinG M eFrom .equatI.On(... ) N = -M.A-\" o F Ill, sm Fig. 7.260 On substituting expression for N in equation (vi). we obtain Sol. Figure 7.261(a) shows force diagram of the wedge and the MA mgcos8 - -.- = mAsine, abJock. Let acceleration of block relative to wedge be Q'mM = SIn () aMand acceleration of wedge on ground is = li. mgcose = . + -M.A- mA SIne sme ,,,,,,,r-.,, ' oy=Asin8 rng sin ecos e F In A= In sin'!! + M m In the arrangement shown in the o Fig. 7.264, mass of the rod M exceeds the mass In of the ball. The ball has an opening permitting it to slide along tbe (a) (b) thread with some friction. The mass of the pnlley and the friction in its axle are negligible. At the initial moment, the Fig. 7.261 ball was located opposite the lower end of the rod, When set free, both bodies began moving with constant accelerations. Free body diagrams as in Figs. 7.262 and 7.263 Find the friction force between the ball and the thread if t seconds after the beginning of motion, the ball got opposite Equation of motion of M 0) the upper end of the rod. The rod length equals I. N sin e = F - F sin e = M A Equation of motion of m In y-direetion (ii) Sol. The friction force between the ball and the thread is j,.. e - emg cos N = ma), = In A sin Substituting the value of N from equation (ii) in equation (i) Acceleration of both the bodies is downward. we can get the value of A; mg - j, = ma, (i) am = amM +aM

7.58 Physics for IIT-JEE: Mechanics I During this time acceleration of both blocks M al = a, = F + - -a-t - (ml +m2) (ml +m2) \"' +If t \":. -ft-m(2Inl m2)g Fig. 7.264 ami J,. M Friclion will be of kinetic nature. Mg Afg R \",g e+r\"\"gFree body diagrams (Fig. 7.268):- F \"\"aN ~N \"\" N Fig. 7.265 Fo at ' Fig. 7.268 Acceleration of m 1: ftN ftm,g aj=-=-- ml Inl . at - IJJn?g Acceleration of m2: a2 = - 111.2 Draw the acceleration of nt, Vs time a,Graph Mg-T= Maz (ii) but j, = T e time az _ al = ( Mg;; j, ) _ (mg ::: j, ) Fig, 7.269 According lo the problem, to = ftm, + fIl2)g --'(ml I 2 f221 1= Z(a2 - al) = amI allt or (a, - ato /JJn2g =aj = - - m(Mg - f,) - M(mg - f,) 21 +(m! m2) m] Mm t2 Acceleration v.; lime graph for m, (Fig. 7.271) 21mM 2ltnM a, (M -m)I,· = - t-2 f,=---- =} . (M -m)t2 In the Fig. 7.266 shown force F = IX! ap- plied on the block 11Iz. Here IX is a constant and t is the time. Find the acceleration of the block. 1\"'0~£ \"', Fig. 7.270 Fig. 7.266 a,. at - J1.,m·2g :sSol. If F ft tn2 (ml + m2)g then both blocks will move to- After time to; = ---'-':= gether. \"'I nI2 :sHere, t +m2 at ftm,g ft--(ml Inz)g a2=----; ant! nI2 fn2 ~F=at ~I ml I Find the acceleration aI, az, and a3 of the three blocks shown in Fig, 7,271, if a horizontal force of 10 Fig, 7.267 N of is applied on =(1) 2 kg block (2) 3 kg block (3) 7 kg block (take g 10 m/82) Sol. 1. When force of 10 N is applied on 2 kg block.

Newton's Laws of Motion 7.59 The acceleration of the whole system, a = 10 = 2+3+7 5 m / s 2 _ 6 Fig. 7.271 tON The limiting frictional force between 2 kg and 3 kg blocks Fig. 7.274 I, = 0.2 x2g = 0.2 x 2 x 10 = 4 N ;-----~---------' 2 x 5 5' . The pseudo force on 2 kg block = - - = - N WhICh is less ION ~ 4N! 63 -, ,-,'.\\ \" I than the frictional force between 2 kg and 3 kg block, so they -----------------------, move together. Now check whether 2 kg and 3 kg blocks move together over 7 kg block. The . force on (2 + 3) kg is = 5 x 65 = 25 pseudo 6N Fig. 7.272 which is also less than frictional force bctween 3 kg and 7 kg The limiting frictional force between 3 kg and 7 kg block, so all the blocks move together with a common acceler- blocks that can be 52 h = 0.3 x 5g = 0.3 x 5x 10 = 15 N ation of (im/s . As applied force ION is greater that /, but less than h, Therefore a = ell = Q2 = Q3 = 2 10 + 7-- 5 n / s.2. so 2 kg hlock will slide ovcr 3 kg +3 6-r 10 -4 A smooth semicircular wire~track of raw Thus we have: al = --2- = 3m/s2; dius R is fixed in a vertical plane (see Fig. 7.275). One end +4 2 of a massless spriug of natural length (~) R is attached to Cl2 = ell = 3 7 = OAm/s the lowest point 0 of the wire track. A small ring of mass JON m, which can slide on the track. is attached to the other end of the spring. Tbe ring is held stationary of point P such m:.that the spring makes an angle of 600 with the vertical. The spring constant k = Consider the instant when the ring is released, and 1. The FBD of the ring is drawn. Fig. 7.273 , 2. When force of 10 N is applied on 3 kg hlock. c~~:------------ As the applied forcc is less than the friction between 3 kg p~, and 7 kg blocks (that can be 15 N) so that block will move together as one unit with an acceleration Fig. 7.275 a = _ .1_0 - = -5m/s2 2+3+7 6 2. Determine the tangential acceleration of tbe ring and the Now find pseudo force on 2 kg block because of accel- eration of 3 kg block normal reaction. (IIT-JEE. 1996) .Fp,cudo=2x 65 =35N Sol. As the F~seud{l is smaller than the friction between 2 kg 1. The FED of the ring is shown in the Fig. 7.276. The forces and 3 kg (that can be 4 N). So 2 kg will move together will acting on the ring arc: other blocks. a. the weight, mg acting vertically downwards, and 3. When force of 10 N is applied on 7 kg block. Suppose 3 kg and 2 kg blocks move togethcr with a 7 kg bock.