1. Engineering Mechanics Full Material

Chapter 1 Introduction – Forces, Equilibrium CHAPTER HIGHLIGHTS ☞☞ Introduction ☞☞ Resultant of Multiple Forces Acting at a Point ☞☞ Newtonian Mechanics ☞☞ Collinear Forces ☞☞ Branches of Newtonian Mechanics ☞☞ Coplanar Concurrent Force System ☞☞ Deformation of Body ☞☞ Coplanar Parallel Force System ☞☞ Action and Reaction ☞☞ Coplanar Non-concurrent, Non-parallel Force ☞☞ Resolution of a Force into a Force and a Couple System ☞☞ Resultant of a System of Coplanar Forces ☞☞ Distributed Force System Introduction Some of the definitions of the idealizations used in engi- neering mechanics are as follows: In Physics, the branch which deals with the study of the state of rest or motion caused by the action of forces on the 1. Continuum: It is defined as continuous non special bodies is called as mechanics. whole which has no empty spaces and no part is dis- tinct from the adjacent parts. Considering objects in Engineering mechanics applies the principles and laws this way ignores that the matter present in the object is of mechanics to solve problems of common engineering made up of atoms and molecules. elements. 2. Particle: A particle is a body which has finite mass but Newtonian mechanics: Newtonian mechanics or classical the dimensions can be neglected. mechanics deals with the study of the motion of macroscopic objects under the action of a force or a system of forces. 3. System of particles: When a group of particles which are inter-related are dealt together for studying the Branches of Newtonian behavior, it is called as a system of particles. Mechanics 4. Rigid body: A solid body which does not undergo any 1. Statics: It is the study of forces and conditions of deformations under the application of forces is called a equilibrium of material bodies, at rest, subjected to rigid body. In reality, solid bodies are not rigid but are the action of forces. assumed as rigid bodies. 2. Dynamics: It is the branch of mechanics which deals with 5. Matter: It is anything which occupies space. the study of motion of rigid bodies and the co-relation 6. Mass: It is a measure of inertia. The mass of body is with the forces causing and affecting their motion. Dynamics is divided into Kinematics and Kinetics. the quantity of matter contained in it and is the sum of the masses of its constituent’s mass points. 3. Kinematics: Kinematics deals with space time relationship of a given motion of body and not at all Deformation of Body with the forces that cause the motion. A body which changes its shape or size under action of 4. Kinetics: The study of the laws of motion of material external forces is called deformable body. bodies under the action of forces or kinetics is the study of the relationship between the forces and the Action and Reaction resulting motion. Action and reaction occurs when one body exerts a force on another body, the later also exerts a force on the former. These forces are equal in magnitude and opposite in direction.

3.6  |  Part III  •  Unit 1  •  Engineering Mechanics lie in the same plane, then the corresponding forces constitute a non-coplanar force system (Figure (b)). Tension yy It is the pulling force which is acting through a string when it is tight. It acts in the outward direction. Tension F 1F2F3 F4 F1 F2 F3 F4 Thrust x x It is acting in the inward direction and it is the pushing force z  z transferred through a light rod. (a) (b) Thrust 3. Collinear force system: In a collinear force system Force (figure), the lines of action, of the entire constituting forces, will be along the same line. Force may be defined as any action that tends to change the state of rest or uniform motion on of a body on which it is y applied. The specifications or characteristics of a force are: x 1. Magnitude 2. Point of application F 3. Direction, (force is a vector quantity). z 4. Line of action 4. Concurrent and non-concurrent force systems: If Force is a vector quantity since it has amagnitude and a direc- the lines of action of all the forces in a force system tion (scalar quantities have only magnitudes and no directions). pass through a single point, then the force system is called as a concurrent force system (Figure (a)) else it The direction of a force is the direction, along a straight is called as a non-concurrent force system (Figure (b)). line passing through its point of application, in which the force tends to move the body on which it is applied. The straight y F1 y line is called the line of action of the force. For the force of F4 F2 gravity, the direction of the force is vertically downward. F3 F2 x F3 F1 1. System of forces: A system of forces or a force F4 x system is the set of forces acting on the body or a group of bodes of interest. Force system can be classified, according to the orientation of the lines of action of the constituting forces, as follows: System of forces z  z Coplanar Non-coplanar (a) (b) (Space forces) 5. Parallel and non-parallel (general) force systems: In Collinear Concurrenet Concurrent a parallel force system (Figure (a)), the lines of action, of the entire constituting forces, are parallel to each other. Non-concurrenet Non-concurrenet If the line of action of at least one constituting force is not parallel to the line of action of another constituting Parallel Parallel General force in a force system, then the force system is called ‘non-parallel force system’ (Figure (b)). Like Non-parallel Unlike y y F1F2F3 parallel general parallel F1F2F3F4 F4 Unlike parallel Like parallel xx 2. Coplanar and non-coplanar (spatial) force z (a) z (b) Systems: In a coplanar force system (Figure (a)), the constituting forces have their lines of action lying in the same plane. If all the lines of action do not

Chapter 1  •  Introduction – Forces, Equilibrium  |  3.7 6. Like parallel and unlike parallel force systems: In indicate the point of application of the force. Note that a like parallel force system (Figure (a)), the lines of all the forces involved must be represented consistently action, of the entire constituting forces, are parallel to as shown in figures below. each other and act in the same direction. In an unlike parallel force system (figure), the lines of action, of 30° 30° the entire constituting forces, are parallel to each 0 0 other where some of them act in different directions. (a) (b) y F1F2F3F4 x Resolution of a Force into a Force and a Couple z A given force ‘P’ applied to a body at any point ‘A’ can System of Forces always be replaced by an equal force applied at another point ‘B’ together with a couple which will be statically Force System Examples equivalent to the original force. To prove this let the given force ‘P’ act at ‘A’ as shown below. Then at B, we introduce 1. Collinear Forces on a rope in a tug of war two oppositely directed collinear forces each of magnitude ‘P’ and parallel to the line of action of the given force P at A. 2. Coplanar parallel System of forces acting on a beam subjected to vertical loads including PP 3. Coplanar like reactions. parallel Weight of a stationary train on the A aPA rail when track straight. P 4. Coplanar Forces of a rod resting against a concurrent wall. B Forces on a ladder resting against P 5. Coplanar, a wall when a person stands on a non–concurrent rung which is not at its centre of (a) (b) (c) forces (ii) (b) gravity. The weight of the benches It follows from the law of superposition that the system 6. Non–coplanar in a classroom in figure (b) is statically equivalent to that in figure (a). parallel Forces on a tripod carrying However, we may now regard the original force P at ‘A’ and the oppositely directed force ‘P’ at B as a couple of moment 7. Non–coplanar Forces acting on a moving bus. M = Pa. Since this couple may now be transformed in any concurrent manner in its plane of action so long as its moment remains forces (II) (b) unchanged, we may finally represent the system as shown in the figure (c), where the couple is simply indicated by a 8. Non–coplanar curved arrow and the magnitude of its moment. It will be non–concurrent noted that the moment of the couple introduced in the above forces (iv) manner will always be equal to the product of the origi- nal force ‘P’ and the arbitrary distance ‘a’ that we decide NOTE to move its line of action. This resolution of a force into a force and couple is very useful in many problems of statics. Force systems can also be classified, according to the magnitude of the constituting forces as: (a) System of Resultant of a System equal forces—all the constituting forces has the same of Coplanar Forces magnitude; (b) System of unequal forces—all the constituting forces do not have the same magnitude. Parallelogram Law of Forces 7. Representation of a force: Graphically, a force may When two concurrent forces P and Q are represented in be represented by the segment of a straight line with magnitude and direction by the two adjacent sides of a par- arrow head at one end of the line segment. The straight allelogram as shown in fig., the diagonal of the parallelo- line represents the line of action of the force 1 kN gram concurrent with the two forces, P and Q represents the and its length represents its magnitude. The direction resultant R of the forces in magnitude and direction. of force is indicated by placing an arrow head on this straight line. The arrow head at one end of the straight line segment indicate the direction of the force along the line segment. Either the head or tail may be used to

3.8  |  Part III  •  Unit 1  •  Engineering Mechanics If P and Q are two forces making in angle θ with each Coplanar Force System other, then It can be classified into collinear, concurrent, parallel, non- R = P2 + Q2 + 2QP cos θ concurrent, and non-parallel type of force system. q−a The resultant of a general coplanar system of forces may be (a) single force, (b) a couple in the system’s plane or in a Q RQ parallel plane or (c) zero. qa Pq Collinear Forces (180 − q) The resultant of a collinear force system (R) can be deter- tan α = ⎡ P Q sin θ θ ⎤ mined by algebraically adding the forces. ⎣⎢ + Q cos ⎦⎥ R = ΣF = F1 + F2 + F3 α = tan -1 ⎡ P Q sin θ ⎤ A F1 B F2 C F3 D ⎢⎣ + Q cos ⎥⎦ θ QR P Coplanar Concurrent Force System sin α = sin θ = sin(θ - α) y Resultant of Multiple Forces P Acting at a Point Q Let ΣH = algebraic sum of resolved part of the forces along the x-axis. q1 x q2 ΣV = algebraic sum of resolved part of the forces along the y-axis. q3 S R = (∑ H )2 + (∑V )2 The analytical method procedure consists of resolving the forces into components that coincide with the two arbitrar- y y ily chosen axes. F1 ∑ Fx = P cos θ1 + Q cos θ2 + S cos θ3 0 x ΣV R ∑ Fy = P sin Q1 + Q sin θ2 - S sin θ3 F 2   0 ΣH x And the resultant R = (�Fx )2 + (�Fy )2 . Its angle with tan θ = ∑V where θ is the angle which the resultant vector respect to the x-axis is given by α= tan - 1⎛⎝⎜ ∑ Fy ⎞ . ∑H R makes with the x-axis. ∑ Fx ⎠⎟ Coplanar Parallel Force System Triangle Law R x AO xp The resultant of two forces can be obtained by the triangle xQ law of forces. The law states that if two forces acting at a point are represented by the two sides of a triangle, taken xs in order, the remaining side taken in an opposite order will give the resultant. B CO g P Q 180 − g 180 − a P gb Q a 180 − b a Rb PS  Q R Resultant of the parallel forces P, Q and S are R = ΣF = P Q = P = R +Q+S sin α sin γ sin β Now, ΣM0 = Rx

Chapter 1  •  Introduction – Forces, Equilibrium  |  3.9 So, Being the angle between the forces. ΣM0 → sum of the moments of the forces P, Q and S, at 1 point 0. 2R = [(6P)2 + (2P)2 + 2 × 6P × 2P × cos α]2 Rx = Pxp + QxQ + Sxs 1 Coplanar non-concurrent, non-parallel force system: As in the case of an unlike parallel force system, the resultant = [40 + 24 cos α]2 (2) may be a single force, a couple in the plane of the system or From equations (1) and (2) we have zero. The resultant is given by 11 2P[13 +12 cos α]2 = P[40 + 24 cos α]2 ( )R = (∑ fx)2 + ∑ f y 2 and its angle α with the x-axis or 2[13 +12 cos α] = [40 + 24 cos α] ∑ Py cos α = - 1 , α = 120° ∑ Fx 2 is given by tan α = y Example 3:  A weight ‘W’ is supported by two cables. At what value of ‘θ’ the tension is cable making be minimum? o b c dx a PQ S q 60° W Distributed force system: Distributed forces (or loads) are Solution: forces which act over a length, area or volume of a body. On T1 the other hand, a concentrated force (point load) is a force T2 which acts of a point. Solved Examples q 60° Example 1:  Three forces P, 2P and 3P are exerted along the direction of the three sides of an equilateral triangle as shown in the following figure. Determine the resultant force. Solution:  ΣFx = P - 2P cos 60° - 3P cos 60° W = -1.5P = - 3 P ΣFj = 0 2 T1 sin θ + T2 sin 60° = W ΣFy = 2P sin 60° − 3P sin 60° T1 cos θ = T2 cos 60° = -0.866P - 3 P  T2 = T1 cos θ = 2T1 cos θ 2 cos 60° R = ∑ Fx2 + ∑ Fy2 = T1 sin θ + 2T1 cos θ ⋅ sin 60° = W = T1 sin θ + 2T1 cos θ ⋅ sin 60° = W R= 9 P2 + 3 P2 C T1 sin θ + 3T1 cos θ = W 4 4 dT1 3P 2P dθ = 0 = T1 cos θ+ 3T1(- sin θ) = 0 12 P2 = 3P B T1 cos θ = 3T1 sin θ 4  A tan θ = 1 P 3 Example 2:  The resultant of two concurrent forces 3P and θ = 30°. 2P is R. If the first force is doubled, the resultant is also doubled. Determine the angle between the forces. Example 4:  An electric fixture weighing 18 N hangs from a point C by two strings AC and BC as shown in the 1 following figure. The string AC is inclined to the vertical wall at 40° and BC is inclined to the horizontal ceiling at Solution:  R = [(3P)2 + (2P)2 + 2 × 3P × 2P × cos α]2 50°. Determine the forces in the strings. 1 (1)   = P × [13 +12 cos α]2

3.10  |  Part III  •  Unit 1  •  Engineering Mechanics D B Direction for solved examples 6 and 7: 50° P 30° 60° Q T1 T 1 40 T 2 A 40° C 60° 30° T PR 60° 30° T QR R 600 N   600 N E 18 N Solution:  It can be deducted that ∠DCA = 40 and ∠BCD = Example 6:  The tension in the wire QR will be 40° so that ∠ACB = 80°. Now, (A) 519.6 N (B) 625 N ∠ACE = 180° - 40 = 140°; (C) 630 N (D) 735 N ∠BCE = 180° - 40 = 140° Solution: Using sine rule; TQR = TPR 30) = 600 sin(180 - 60) sin(180 - sin 90 T1 = T2 = 18 TQR = TPR = 600 sin 40 sin 40 sin 80 sin 60 sin 30 sin 90 18 × sin 40 The tension in the wire QR, sin 80 T1 = = 11.75 N TQR = 300 3 = 519.6 N T2 = 18 × sin 40 = 11.75 N Example 7:  The tension in the wire ‘PR’ will be sin 80 (A) 575 N (B) 300 N Example 5:  Determine the resultant of the coplanar (C) 275 N (D) 400 N concurrent force system shown in the following figure. Solution: B 100 N The tension in the wire PR, 150 N TPR = 600 sin 30 = 300 N C 30° 20° X Example 8:  A point is located at (−6, 2, 16) with respect to 70° O 45° the origin (0, 0, 0). Specify its position. (i) In terms of the orthogonal components. 250 N (ii) In terms of the direction cosines. (iii) In terms of its unit vector. 200 N Solution:  ΣFx = 100 cos 20 + 250 cos 45 - 200 cos 70 Solution:  - 150 cos 30 = 72.44 N A(−6, 2, 16) ΣFy = 100 sin 20 - 250 sin 45 - 200 sin 70 + 150 sin 30 = -255 N O(0, 0, 0) Now, The components of the vector OA are Resultant R = (∑ Fx )2 + (∑ Fy )2 (−6 − 0) = −6 along the x-axis (2 − 0) = 2 along the y-axis = 72.442 + 255.52 = 265 N (16 − 0) = 16 along the z-axis Its inclination α = tan -1 ∑ Fy z ∑ Fx A = tan -1 ⎛ 255.5⎞ = 74° k ⎜⎝ 72.44 ⎠⎟ 16 i x 2 −6 O Since Σy is negative, the angle falls in the fourth quadrant. j ∴ Angle made with x-axis is 360 - 74 = 286° y (Counter clock wise).

Chapter 1  •  Introduction – Forces, Equilibrium  |  3.11 Position vector, r = −6i + 2j + 16k F (in terms of the orthogonal components.) T Magnitude r = (-6)2 + 22 + 162 = 17.2 C 2r − h Direction cosines are r−h r P h h = 8 cm = cos θx = - 6 = -0.3488 Wb 17.2 m = cos θy = 2 = 0.1163 17.2 Solution:  The free body diagram given above shows the n = cos θz = 16 = 0.9302 horizontal force applied by the man, the weight W acting 17.2 at the centre of the wheel and the reaction R at the point P. (The reaction at O will be zero at the instant the wheel being r = (17.2 ) i + (17.2 m)j + (17.2 n) (in terms of the direction lifted up). cosines) From the geometry Example 9:  Consider a truss ABC with a force P at A as shown in the following figure, b2 = r2 - (r - h)2 = 2rh - h2 = 2 × 0.25 × 0.08 - (0.08)2 AP B 45° 30° C ∴ b = 0.1833 m Taking moments about the point P, -F (2r - h) + wb = 0 The tension in member CB is Or, (A) 0.5P (B) 0.63P F = Wb 2r - h (C) 0.073P (D) 0.87P Solution:  Consider point A. For equilibrium, resolving the Where W = load on one wheel = 1500 forces, 2 P F = Force applied on one wheel. A P 1500 × 0.1833 137.475 TAB 4 6 3 2 0.42 ∴F = = = 327.32 N. 45° 2 × 0.25 - 0.08 B C TBC Moment of a Force TAB cos 45° + TAC cos­60 - P = 0 TAB sin 45 = TAC sin 60 The product of a force and the perpendicular distance of the line of action of the force from a point or axis is defined as the moment of the force about that point or axis. Solving, OP TAB = 2 3P r ( 6 + 2) Balancing of forces at point B give TAB cos 45 = TBC In the figure, the moment of force P about the point O or about the y axis is p × r. Moment may be either clockwise TBC = ⎛ 6 2 ⎞ P = 0.633P. or anti clockwise. In the Figure 1.15 the, moment tends to ⎜⎝ 6+ ⎠⎟ rotate the body in anti clockwise direction. Example 10:  A man sitting on a wheel chair tries to roll The right hand rule is a convenient tool to identify the up a step of 8 cm. The diameter of the wheel is 50 cm. The direction of a moment. The moment M about an axis may wheel chair together with the man weighs 1500 N. What be represented as a vector pointing towards the direction of force he will have to apply on the periphery of the wheel? the thumb of the right hand, while the other fingers show the direction of turn, the force offers about the axis (clockwise or anti-clockwise)

3.12  |  Part III  •  Unit 1  •  Engineering Mechanics y y But OA + AB + BO = 0 M d2 OA - OB = AB F1 F2   F x ∴ (1) Becomes M = AB × F. o d1 d F A B Varignon’s Theorem of Moments X OF QA RC The resultant force is zero, but the displacement ‘d’ of the force couple creates a couple moment. Moment about some O PB arbitrary point is 0. It can be proved that ΔOXA + ΔOXB = ΔOXC. This illus- M = F1d1 + F2d2 = F1d1 − F1d2 = F1 (d1 − d2). If point trates Varignon’s theorem of moments. O is placed in the line of action of force Fz (or F1), then M = Fid (or F2d). Moment of the force Q about X = Twice the area of ΔOXA. Moment of force P about X = Twice the area of ΔOXB. R is Orthogonal components (scalar components) of force F the resultant of P and Q. Moment of the resultant about X is along the rectangular axises, x, y and z axis’s, are Fx, Fy and Fz respectively. = Twice the area of the triangle OXC. The theorem may be stated as follows—the moment of a Fx = |F| cos θx, Fy = |F| cos θy, Fz = |F| cos θ2, where cos θx (zl), cos θy(zm) and cos θ2(zn) are the direction cosines of force about any point is equal to the sum of the moments of the force F and |F| is the magnitude of the force F. components about the same point. Now, F = (Fx )2 + (Fy )2 + (Fz )2 Example: As shown in figure resultant force F and its We have, components F1 and F2; d1 and d2 are the respective normal distances between F, F1 and F2 from any point O. Varignon’s F = Fxi + Fy j + Fz k = | F | (cos θni + cos θy j + cos θz k) theorem states that Fd = F1d1 + F2d2 . Moment of a Couple = |F| (lI + mj + nk). where i, j and k are vectors of unit length along the positive Two parallel forces having the same magnitude and acting in the opposite directions form a couple. x, y and z directions. Unit vector corresponding to the force Moment of the couple is the algebraic sum of moment of vector F, F = | F | ⋅ If nˆ is a unit vector in the direction of the forces involved in it about a point. F F1 the force F, then F = |F| nˆ. A d B d1 Equilibrium of Force Systems d2 F2 A body is said to be acted upon by a system of forces in o equilibrium if the force system cannot change the body’s stationary or constant velocity state. When the resultant is Moment of the force F1 about O = OA × F1 neither a force nor a couple, i.e., Moment of the force F2 about O = OB × (-F2 ) = -OB × F2 Algebraic sum of the moments ΣF = 0  (1) M = OA × F1 - OB × F2 (But F1 = F2 for a couple F1 = F2 ΣM = 0  (2) = F) Then, = (OA - OB) × F (1) ΣF → vector sum of all forces of the system. ΣM → vector sum of the moments (relative to any point) of all the forces of the system.

Chapter 1  •  Introduction – Forces, Equilibrium  |  3.13 Scalar equation equivalent to vector equation (1), in a ΣFx, ΣFy and ΣFz are algebraic sums of the components of rectangular coordinate system, are all the forces in the x, y, and z directions and θx, θy, θz are the angles which the resultant vector R makes with the x, y, ΣFx = 0 and z axes respectively. ΣFy = 0 ΣFz = 0 Parallel Spatial Force System Scalar equation equivalent to the vector equation (2) as ΣMx = 0 The resultant ΣMy = 0 ΣMz = 0 R = ∑ F1Rx = ∑ M x Rz = ∑ M z Now, ΣFx, ΣFy and ΣFz → algebraic sum of forces in the x, y and z directions respectively. ΣMx, ΣMy and ΣMz where x and z are the perpendicular distances of the result- → algebraic sum of moments in the x, y and z directions ant vector from the xy and yz plane respected and ΣMz, ΣMx respectively. are algebraic sums of the moments of forces of the force system about the x and z axes respectively. Equilibrium Equations for Different Coplanar Force Systems If ΣF = 0, the resultant couple can be evaluated 1. Concurrent coplanar force system As C = (∑ Mx )2 + (∑ Mz )2 So, ΣFx = 0, ΣFy = 0 2. Concurrent non co-planar force system tan ϕ = ∑ Mz ∑ Mx ΣFx = 0, ΣFy =0, ΣFz = 0 Where φ is the angle made by the couple. 3. Non-concurrent coplanar force system ΣFx = 0, ΣFy = 0 and ΣM = 0 at any suitable point. Non-concurrent Non Parallel 4. Non-concurrent non-coplanar force system Force System ΣFx = 0, ΣFy = 0; ΣFz = 0 and ΣMx = 0, ΣMy = 0, ΣMz = 0 ( )The resultant, R = (∑ Fx )2 + ∑ Fy 2 + (∑ Fz )2 and cor- Analysis of a System of Forces in Space responding direction cosines are A spatial force system may consist of a set of concurrent cos θx = ∑ Fx , cos θy = ∑ Fy forces, parallel forces or non-concurrent non parallel forces. R R The resultant of a spatial force system is a force R and a couple C, where and, cos θz = ∑ Fz R R = Σ (forces) and C = Σ (moments) Concurrent Spatial Force System Now, Resultant R is given by C = (∑ Mx )2 + (∑ M y )2 + (∑ Mz )2 and the corre- sponding direction cosines are R = (∑ Fx )2 + (∑ Fy )2 + (∑ Fz )2 With the direction cosines given by cos θx = M xCc , cos θy = ∑ My C C1 ∑ Fx cos θx = Ri ∑ Mz C1 ∑ Fy cos θz = Ri cos θy = cos θz = ∑ Fz Where θx, θy, and θz are the angles which the vector rep- Ri resenting the couple C makes with the x, y and z axes respectively.

3.14  |  Part III  •  Unit 1  •  Engineering Mechanics Exercises Practice Problems 1 (C) Force/cos θ is constant (D) Each force is proportional to the sine of angle be- Direction for questions 1 to 10:  Select the correct alterna- tive from the given choices. tween the other two 1. Concurrent forces in a plane will be in equilibrium if 6. For a particle to be in equilibrium under the action of (A) Sum of the forces is zero two forces, the forces must be (B) Algebraic sum is zero (C) Sum of resolved parts zero (A) Concurrent and parallel (D) Sum of the resolved parts in any two (B) Unequal non concurrent perpendicular direction are zero. (C) Equal parallel non collinear (D) Equal, opposite and collinear 2. A free body diagram is a representation of (A) The forces on the body 7. A number of co-planar forces will be in equilibrium when (B) The reactions on the body (A) ΣFx = 0 and ΣFy = 0 (C) Both the active and reactive forces (B) ΣFM0 = 0 (D) Neither the active nor the reactive forces (C) ΣFx = 0 (D) ΣFx = 0, ΣFy = 0, ΣM0 = 0 3. When a driver operates the gear shift lever it is (A) Coplanar force 8. Two cylinders having diameter 50 mm and 100 mm (B) Non-coplanar force (C) Moment respectively are held in a container, bigger one at bot- (D) Couple tom and the smaller on top. Container has a width of 4. Three coplanar forces acting at a point can be in equi- librium only if 110 mm. What will be the reaction at the point of con- (A) All the three forces are parallel tact of bigger cylinder to the container wall (neglect (B) All the three forces are concurrent (C) All the three forces are parallel or concurrent friction). Weight of small cylinder is W and weight of (D) All the three forces are spatial big cylinder is 2 W. 5. Lami’s theorem gives the following when three concur- rent forces acting on a body kept in equilibrium (A) 0.472 W (B) 0.528 W (A) Force divided by tan of angle is zero (C) 0.163 W (D) 0.725 W (B) Force is proportional to tan θ 9. The resultant of coplanar non-concurrent forces constitutes (A) Force (B) Couple (C) Moment or couple (D) Force or couple 10. Turning the cap of a pen is example of (A) Moment (B) Force (C) Couple (D) Impulse Practice Problems 2 3. A weight of 1900 N is supported by two chains of Direction for questions 1 to 10:  Select the correct alterna- lengths of 4m and 3m as shown in figure. Determine tive from the given choices. the tension in each chain 1. Two forces are acting at a point O as shown in figure 5m O 200 N AB ab 4 m T1 T2 a 30° 100 N q1 q2 3m 15° P O Chain No 1 Chain No 2 Determine the resultant in magnitude and direction. 1900 N (A) 84.644 kN, 35° (B) 90.0 kN, 33° (A) 1200 N, 1300 N (B) 1100 N, 100 N (C) 1100 N, 1200 N (D) 1520, 1140 N (C) 100 kN, 28° (D) 120 kN, 25° 2. Two equal forces are acting at a point with an angle of 60° between them. If the resultant force, is equal to Direction for questions 4 and 5: Three forces in magni- tude 80 kN, 30 kN, 40 kN are acting at a point O as shown. 40 × 3N ⋅ Find the magnitude of each force. The angles made by 80 kN, 30 kN and 40 kN forces with x axis are 60°, 120° and 240° respectively. (A) 20 N (B) 40 N (C) 50 N (D) 30 N

Chapter 1  •  Introduction – Forces, Equilibrium  |  3.15 4. Determine the magnitude of the resultant force. 8. A beam simply supported at A and B of span 10 m is carrying a point load of 10 kN at a distance of 4 m from Y A. Determine the reactions at the supports 30 kN 120° 80 kN (A) 7 kN, 8 kN (B) 4 kN, 6 kN 240° 60° (C) 5 kN, 4 kN O (D) 10 kN, 8 kN X 40 kN 9. Four forces of magnitudes 20 N, 40 N, 60 N and 80 N are acting respectively along the four sides of a square ABCD (A) 60.82 kN (B) 50.8 kN as shown in figure. Determine magnitude of resultant. (C) 30.5 kN (D) 62.8 kN 5. Find the direction of the resultant force. 60 N D 40 N C (A) 85.28° (B) 60° (C) 84.38° (D) 15.5° Direction for questions 6 and 7: Four forces of magni- A B 80 N 20 N tude 20 kN, 30 kN, 40 kN and 80 kN are acting at a point (A) 40 2 N (B) 50 2 N O as shown in figure. The angles made by 20 kN, 30 kN, 40 kN and 80 kN with x-axis are 30°, 60°, 90° and 120° respectively. 6. Find the magnitude of the resultant force. (A) +5 kN (B) −6 kN (C) 45 2 N (D) 60 2 N (C) −7.679 kN (D) +8 kN 1 0. Find the direction of the resultant referred to the direc- 7. Find the direction of the resultant force. tion of 20 N forces (A) −86.97° (B) 98° (A) 50° (B) 45° (C) 97.5° (D) −85° (C) 60° (D) 80° Previous Years’ Questions 1. If point A is in equilibrium under the action of the A applied forces, the value of tensions TAB and TAC are 2.5 m respectively. [2006] T AB T AC 3 m 2.5 m 60° A 30° W P 4m B 3. A rigid ball of weight 100 N is suspended with the 600 N help of a string. The ball is pulled by a horizontal force F such that the string makes an angle of 30° (A) 520 N and 300 N (B) 300 N and 520 N with the vertical. The magnitude of force F (in N) is (C) 450 N and 150 N (D) 150 N and 450 N _______. [2016] 2. A ladder AB of length 5 m and weight (W) 600 N is resting against a wall. Assuming frictionless contact at the floor (B) and the wall (A), the magnitude of the force P (in Newton) required to maintain equilibrium 30º of the ladder is ____ [2014] F 100 N

3.16  |  Part III  •  Unit 1  •  Engineering Mechanics Answer Keys Exercises Practice Problems 1 1. D 2. C 3. D 4. C 5. D 6. D 7. D 8. B 9. C 10. C 7. A 8. B 9. A 10. B Practice Problems 2 1. A 2. B 3. D 4. A 5. A 6. C Previous Years’ Questions 3.  57.5 – 58.0 1. A 2.  399 to 401

Chapter 2 Free Body Diagrams – Trusses CHAPTER HIGHLIGHTS ☞☞ Analysis of Roof Trusses ☞☞ Plane Truss ☞☞ Free Body Diagram ☞☞ Nodes ☞☞ Composition and Resolution of Forces ☞☞ Perfect Frame ☞☞ Resolution of a Force ☞☞ Supports ☞☞ Equilibrium Law ☞☞ Assumptions: Analysis of Trusses ☞☞ Internal and External Forces ☞☞ Method of Members: Analysis of Plane Frames ☞☞ Superposition and Transmissibility ☞☞ Equilibrium of Concurrent Forces in a Plane Free Body Diagram Free body diagram (FBD) is a sketch of the isolated body, A W which shows the external forces on the body and the reac- RA tions exerted on it by the removed elements. A general pro- W cedure for constructing a free body diagram is as follows: S 1. A sketch of the body is drawn, by removing the W supporting surfaces. We isolate the body from its supports and show all forces 2. Indicate on the sketch all the applied or active forces, acting on it by vectors, both active (gravity force) and reac- which tend to set the body in motion, such as those tive (support reactions) forces. caused by weight of the body, etc. We then consider the conditions this system of forces 3. Also indicate on this sketch all the reactive forces, must satisfy in order to be in equilibrium, i.e., in order that such as those caused by the constraints or supports they will have no resultant. that tend to prevent motion. W W 4. All relevant dimensions and angles; reference axes H are shown on the sketch. A smooth surface is one A RA whose friction can be neglected. Smooth surface H prevents the displacement of a body normal to both the contacting surfaces at their point of contact. The reaction of a smooth surface or support is directed normal to both contacting surfaces at their point of contact and is applied at that point. Some of the examples are shown in the following figures. W W B RB A RA

3.18  |  Part III  •  Unit 1  •  Engineering Mechanics B F1 A R A DE B E C F2 A B C F4 C F3 D DE Figure 1  Beam with roller support at one end Resolution of a Force A CE D B The replacement of a single force by several components, AM which will be equivalent in action to the given force, is called the problem of resolution of a force. In the general case of Figure 2  Beam with hinged end and fixed end resolution of a force into any number of coplanar components intersecting at one point on the line of action, the problem Composition and Resolution will be indeterminate unless all but two of the components are of Forces completely specified in both their magnitudes and directions. The reduction of a given system of forces to the simplest Equilibrium Law system that will be its equivalent is called the problem of composition of forces. If several forces F1, F2, F3, applied Two forces acting at a point can be in equilibrium only if to a body at one point, all act in the same plane, then they they are equal in magnitude, opposite in direction and col- represent a system of forces that can be reduced to a single linear in action. Let us consider the equilibrium of a body in resultant force. It then becomes possible to find this result- the form of a prismatic bar on the ends of which two forces ant by successive application of the parallelogram law. Let are acting as shown in the figure below: us consider, for example, four forces F1, F2, F3, and F4 act- ing on a body at point A, as shown in the following figure. A m B To find their resultant, we begin by obtaining the resultant S S AC of the two forces F1 and F2. Combining this resultant with force F3 we obtain the resultant AD which must be n equivalent to F1, F2, and F3. Finally, combining the forces AD and F4, we obtain the resultant ‘R’ of the given system Neglecting their own weights, it follows from the principle of forces F1, F2, F3, and F4, This procedure may be carried just stated that the bar can be in equilibrium only when the on for any number of given forces acting at a single point forces are equal in magnitude, opposite in direction and col- in a plane. linear in action which means that they must act along the line joining the points of application. Considering the equilibrium of a portion of the bar ‘AB’ to the left of a section ‘mn’, we conclude that to balance, the external force S at A the portion to the right must exert on the portion to the left an equal, opposite and collinear force S as shown in the above figure. The magnitude of this internal axial force which one part of a bar in tension exerts on another part called the tensile force in the bar or simply the force in the bar, since in gen- eral it may be either a tensile force or a compressive force. Such an internal force is actually distributed over the cross sectional area of the bar and its intensity, i.e., the force per unit of cross section area is called the stress in the bar. F1 A F4 Internal and External Forces F3 Internal forces are the forces which hold together the par- B F2 ticles of a body. For example, if we try to pull a body by applying two equal, opposite and collinear forces an inter- C R nal force comes into play to hold the body together. Internal D E forces always occur in pairs and equal in magnitude, oppo- site in direction and collinear. Therefore, the resultant of all of these internal forces is zero and does not affect the

Chapter 2  •  Free Body Diagrams – Trusses  |  3.19 external motion of the body or its state of equilibrium. forces shown in figure (b) is same as the action on the body External forces or applied forces are the forces that act on the body due to contact with other bodies or attraction forces by the single force P shown in figure (a). P1 from other separated bodies. These forces may be surface forces (contact forces) or body forces (gravity forces). Let AA us consider the equilibrium of a prismatic bar on each end P of which two forces are acting as shown below. PB F2 B       P11 (a) (b) F1 RA This proves that the point of application of a force may be A A transmitted along the line of action without changing the effect of the force on any rigid body to which it may be applied. This statement is called the theorem of transmissi- bility of a force. F3 B RB = −RA F4 RB Equilibrium of Concurrent Forces in a Plane RB If a body known to be in equilibrium is acted upon by sev- Other examples of two force members held in equilibrium eral concurrent coplanar forces then these forces or rather are shown below. their free vectors, when geometrically added must form a closed polygon. This statement represents the condition of AB equilibrium for any system of concurrent forces in a plane. In the figure (a), we consider a ball supported in a vertical plane by a string ‘BC’ and a smooth wall ‘AB’. The free body diagram in which the ball has been isolated from its supports and in which all forces acting upon it both active and reactive, are indicated by vectors as shown in figure (b). A B S a RA aC RA O B RB = RA A W      W A force which is equal, opposite, and collinear to the result- (a) (b) ant of the two given forces is known as equilibrant of the given two forces. Wa S Sa RA O Superposition and Transmissibility RA O     W When two forces are in equilibrium (equal, opposite and collinear) and their resultant is zero and their combined action (c) (d) on a rigid body is equivalent to that—there is no force at all. A generalization of this observation gives us the third princi- The three concurrent forces W, S and RA are a system of forces ple of statics. Sometimes,it is called the law of superposition. in equilibrium and hence their free vectors must build a closed polygon, in this case, a triangle as shown in figure (c). Law of Superposition If numerical data are not given, we can still sketch the The action of a given system of forces on a rigid body will closed triangle of forces and then express: in no way be changed if we add to or subtract from them another system of forces in equilibrium Let us consider now RA = Wtana and S = Wseca a rigid body ‘AB’ under the action of a force ‘P’ applied at ‘A’ and acting along BA as shown in the figure (a). From the Lami’s Theorem principles of superposition we conclude that the application at point ‘B’ of two oppositely directed forces, each equal to If three concurrent forces are acting on a body, kept in equi- and collinear with P will in no way alter the action of the librium, then each force is proportional to the sine of the given force P. That is, the action on the body by the three angle between the other two forces and the constant of pro- portionality is the same. Consider forces P, Q and R acting

3.20  |  Part III  •  Unit 1  •  Engineering Mechanics at a point ‘O’ as shown in Figure (a). Mathematically Lami’s Nodes  The joints of a frame are called as nodes. A frame is theorem is given by the following equation. designed to carry loads at the nodes. P = Q = R = k Perfect frame  A pin joined frame which has got just the sin a sin b sin γ sufficient number of members to resist the loads without undergoing appreciable deformation in shape is called a Since the forces are in equilibrium, the triangle of forces perfect frame. should close. Draw the triangle of forces DABC, as shown in Supports Figure (b), corresponding to the forces P, Q, and R acting at A truss or a framed structure is held on supports which exert a point ‘O’. From the sine rule of the triangle, we get reaction on the truss or framed structure that they carry. Reactions are to be considered for finding the stresses in the P a) = Q b) = R γ) various members of the structures. The types of supports sin(p - sin(p - sin(p - commonly used are QP 1. Simple supports 2. Pin joint and roller supports g b P Bb 3. Smooth surfaces g R 4. Fixed on encaster and fixtures o a C The reactions of the supports are analytically or graphically Q evaluated. 1. In a simply supported truss, the reactions are always R aA vertical at the supports. sin(p - a) = sina 2. At a pin joint support, the reaction passes through the sin(p - b) = sinb sin(p - γ) = sinγ joint. 3. At a roller surface, the support reaction is vertically When feasible, the trigonometric solution, or Lami’s the- orem is preferable to the graphical solution since it is free upwards at the surface. from the unavoidable small errors associated with the 4. The reaction at a support which has a smooth surface graphical constructions and scaling. is always normal to the surface. Assumptions: Analysis of Trusses Analysis of Roof Trusses Each truss is assumed to be composed of rigid members to be all lying in one plane. This means that co-planar force Definitions systems are involved. Forces are transmitted from one member to another through smooth pins fitting perfectly in Truss the members. These are called two force members. Weights of the members are neglected because they are negligible in A ‘truss’ or ‘frame’ or ‘braced structure’ is the one con- comparison to the loads. sisting of a number of straight bars joined together at the extremities. These bars are members of the truss. W1 W2 Plane truss  If the centre line of the members of a truss lies A CD B in a plane, the truss is called a plane truss or frame. If the centre line is are not lying in the same plane, as in the case Pin joint Roller support of a shear leg, the frame is called a space frame. Figure 4  Pin joint and roller support W1 W2 B A Smooth Pin joint surface Figure 3  Plane trusses Figure 5  Pin joint and smooth surface Strut and tie  A member under compression is called a strut W1 W2 W3 and a member under tension is called a tie. Figure 6  Fixed support Loads  A load is generally defined as a weight or a mass supported. Trusses are designed for permanent, intermittent or varying loads.

Chapter 2  •  Free Body Diagrams – Trusses  |  3.21 Free-body Diagram of a Truss and the Joints EF DE EF FE DE BE EF AB D L BE CE = 0 Joint E EF C Figure 7  Truss AF BF BF AF F Joint F E FE AB CD A D R1 L R2 B C Figure 8  Free-body diagram of truss as a whole R1 L R2 compression Figure 10  Free-body diagram of joints Tension Solution by method of joints  To use this technique, draw a R1 free-body diagram of any pin in the truss provided no more Figure 9  Free-body diagram of point A than two unknown forces act on that pin. This limitation is imposed because the system of forces is a concurrent one AF for which of course, only two equations are available for a solution. From one pin to another, until the unknown is found out, the procedure can be followed. Working Rule AB 1. Depending on the nature of support, provide the AF R1 reaction components. R1 AB Joint A (a) For hinged support, provide horizontal and vertical reaction components. (b) For roller support, provide vertical reaction com­ BF BC ponents only. BE BE 2. Considering the external loads affecting the truss only, AB BC L apply the laws of statics at equilibrium to evaluate the L BF Joint B support reactions. CE = 0 AB 3. Give the values of the support reactions at appropriate joints. 4. Take the joint which contains the minimum number of members (minimum number of unknowns) and apply the conditions of equilibrium to evaluate the forces in the members. BC CD For example: DE Joint C Q 2m P C D CD 3m 3m CD AB DE R2 AH R2 Joint D 2m AV BV

3.22  |  Part III  •  Unit 1  •  Engineering Mechanics The reaction components at A are AH and AV (because A For the free body diagram BD, taking moments about B is a hinged joint). YE 2 + X E 2 - (1 - x)F = 0 The reaction component at B is BV only (no horizontal For the free body diagram AC, taking the moments about A reaction since it is a roller). YE 2 - X E 2 + X F = 0. Now evaluate the reaction components considering the ∴ XE = F and YE = (1 –2x) F. external loads only i.e., The resultant is RE = X E2 + YE2 = F 1+ (1 - 2x)2 AV + BV = P + Q ∴ By knowing the numerical data for ℓ, x and the load AH = 0 F the unknowns RD, RC and RE can be found. To find the reactions at A and B i.e., RA and RB take the moments about Taking algebraic sum of the moments of all forces about the point E for both the free body diagrams and solve for ‘A’ and equating to zero, we get two more equations. These RA and RB. three equations are sufficient to evaluate the support reac- tions. Once the support reactions are evaluated, joints can be Solved Examples considered one by one, to evaluate the force in the members. Example 1:  The magnitude and nature of stresses in the Method of Members: Analysis of Plane Frames member ED of the truss, loaded as shown below, is Frames differ from trusses principally in one aspect, i.e., the (A) 30.4 kN (T) (B) 20.2 kN (T) action of forces are not limited to their ends only and so the members are subjected to bending also with tension or (C) 18.69 kN (C) (D) 15.7 kN (T) compression. In this method the members are isolated as a free body and analyzed with the forces acting on them by 20 kN 40 kN vectors. BC Consider the folding stool with the dimensions as shown in the figure resting on a horizontal floor and a force F is A 60° 60° 60° 60° D acting at a distance of xℓ form the end A. 1.5 m E 1.5 m x AB E 3m 3m /2 /2 Solution: 40 kN RD RC C 20 kN The floor is considered as smooth floor therefore the reac- B 40.41 kN tions at C and D are vertical. ∴ By taking moments at C and D. RD = (1 – x) F and Rc = x F. A 60° 60° 60° 60° D 1.5 m E 1.5 m 20.2 kN Now we separate the members AC and BD and analyze the forces acting on each member. 3m R2 = 35 kN YB YA Free-body diagram B A XB XA YE E XE Taking moments about (A) for equilibrium, ΣmA = 0 D XE E -20 × 1.5 - 40 × 4.5 + R2 × 6 = 0 YE 6R2 = 30 + 180 6R2 = 210 C R2 = 35 kN But R1 + R2 = 60 RD = (1 − x) F RC = x F ∴ R1 = 25 kN

Chapter 2  •  Free Body Diagrams – Trusses  |  3.23 Take the joint D, force on the member CD, Example 3:  The force in the member RQ of the truss, as FCD = 40.41 kN because FCD sin60° = 35 kN given in the figure below, is ∴ Force on ED = FCD (A) 27 kN (Tensile) cos60° = 20.2 kN (T ) (B) 15 kN (Compressive) (C) 20 kN (Compressive) Example 2:  All the members of the truss shown below are (D) 7 kN (Tensile) of equal length and the joints are pinned smooth. It carries a load F at S whose line of action passes through V. The T 10 kN reaction at V is 15 kN F 20 kN Qq U S QU P V P R V W R q 7 kN 5 kN T Solution: S (A) Zero 12 m (B) Vertically upwards and equal to F/4 (4 × 3) (C) Vertically upwardsand equal to F/2 (D) Vertically upwards and equal to F Solution: F F sinq U 60° qS 60° Q qV a F cosq RV 60° P 60° T RW Let a = length of one member Consider the junction R. It must be in equilibrium. The force 7 kN can be balanced only by the member QR. From the above figure, sin 60° = SW . ∴ The force in the member QR SR FQR = 7 kN (Tensile). SW = 3 a (∵ SR = a) Example 4:  The figure is a pin jointed plane truss loaded 2 at the point C by hanging a weight of 1200 kN. The member DB of the truss is subjected to a load of Also cos 60° = RW ⇒ RW = a , SR 2 TW = RT - RW ∴ a- a = a AB C 2 2 VW = VT + TW = a+ a = 3 2 2a D ⎡ 3a ⎤ ⎢ ⎥ Em tanθ SW ⎢ 2 ⎥= 1 1200 kN = VW = ⎢ 3a ⎥ 3 (A) Zero ⎣⎢ 2 ⎥⎦ (B) 500 kN in compression (C) 1200 kN in compression θ = 30° (D) 1200 kN in tension Taking moments about P for equilibrium, ΣMP = 0 Solution: Member DB is perpendicular to AC. Resolving the vertical -F sinθ × 3a - F cosθ × 3 a + RV × 30 = 0 component of the forces at B, we observe that no force can 2 2 be present in member DB. F × 1 × 3a + F × 3 3 a = Rv × 3a 2 2 2× 2 Rv = F/2

3.24  |  Part III  •  Unit 1  •  Engineering Mechanics Example 5:  Find the force in the member EC of the truss (A) Perfect (B) Deficient shown in the figure (C) Redundant (D) None of above D 2 kN Solution:  The number of joints, J = 6 The number of member, n = 10 4 kN Then, 2j – 3 = 2 × 6 – 3 = 9 E Since n > (2j – 3), it is a redundant truss. Example 7:  A weight 200 kN is supported by two cables as shown in the figure. AC 2 kN T1 T2 A C 10 m B 10 m q 60° B (A) 8 kN (B) 4 kN (C) 3.5 kN (D) 2 kN 200 kN Solution: The tension in the cable AB will be minimum when the D 2 kN angle θ is: (A) 0° (B) 30° 4 kN (C) 90° (D) 120° E Solution: VD T1 = T2 + θ) = 200 kN 60) 11.547 sin150 sin(90° sin180° - (θ + 4 kN 200 sin 30° sin(120° - θ) HC 2√3 2√3 30° 2 kN T1 = 6.928 C 10 m 10 m A B = 200 sin30° (∵sin30° = sin150°) sin[120° - θ] Equating vertical forces VD = 8 T1 is minimum when 1 is minimum, i.e., Taking moment about D, sin(120° - θ) sin(120° - θ) is maximum. -2 × 20 - 4 × 10 + HC × 11.54 = 0 80 = 11.54 HC sin(120° - θ) = 1 ∴ HC = 6.928 kN Consider the equilibrium of joint A 120° - θ = 90° FAE sin30° = 2 θ = 30°. FAE = 4 kN (T) FAE cos30° = FBA Example 8:  A 200 kN weight is hung on a string as shown in the figure below. The tension T is: (A) 200 kN (B) 300 kN 43 (C) 160 kN (D) 207.1 kN 2 = 2 3FBA T 15° FBE = 0, FCB = 2 3 QO Consider point C 200 Net horizontal force = 6.928 - 2 3 = 3.463 kN Solution: The three forces T, RB and 200 kN are in equilibrium at It is to be balanced by the force on EC point O. FEC cos30 = 3.463 T ∴ FEC = 4 kN. 15° Example 6:  The type of truss, shown in the figure below, is DC QO RQ W A B E F

Chapter 2  •  Free Body Diagrams – Trusses  |  3.25 TW Resolving the vertical components of the forces at P, Ry = 0 sin 90 = sin(90° +15°) Resolving the horizontal component of the forces at T =W sin 90 = 200 ×1 = 207.1 kN RQP, RH = Tsin 63.4 = 80 N sin105 0.965 Example 11: A truck of weight Mg is shown in figure Example 9:  A uniform beam PQ pinned at P and held by below. A force F (pull) is applied as shown. The reaction at a cable at Q. If the tension in the cable is 5 kN then weight the front wheels at location P is of the beam and reaction at P on the beam are respectively (A) 10 kN and 8.86 kN (B) 5 kN and 13 kN F (C) 10 kN and 10 kN (D) 15 kN and 13 kN b RT RP Q Mg P 60° 60° Q aa (A) Mg + Fb (B) Mag + F2b 2a 2 P 30° (C) Mg + Fb (D) M2ag + F2 2 2a W Solution: Solution: Taking moment about Q, Let w be the weight of the beam and RP be the reaction in ΣMQ = 0 = RP × 2a - Mg × a - F × b the beam at P. Since it is the case of 3 forces forming a sys- Mg ×a+ F ×b Mg Fb . tem in equilibrium all the three forces must be either con- RP = 2a = 2 + 2a current or parallel; in this case, they are to be concurrent. W = T = RP Direction for questions 12 to 13:  All the forces acting on a sin 90 sin150 sin120 particle situated at the point of origin of a two dimensional reference frame. One force has magnitude of 20 N acting in W = T = RP the positive x direction. Where as the other has a magnitude 1 0.5 0.886 of 10 N at an angle of 120° with force directed away from the origin with respect to the positive direction to the direc- W = 5×1 = 10 kN tion of 20 N: 0.5 SQ RP = 8.86 kN. Example 10: R 120° R 10 N 60° a P q T 150 mm O 20 N √3002 + 1502 = 335 mm Example 12:  The value of the resultant force, R, will be Ry P Q (A) 18 N (B) 20 N RH (C) 15 N (D) 21 N 300 mm Solution: 40 N R = 202 + 102 + 2 × 20 × 5 × cos120° = 500 -100 = 20 N A mass of 40 N is suspended from a weight less bar PQ which is supported by a cable QR and a pin at P. At P on the bar, the horizontal one vertical component of the reaction, Example 13: The value of a made by the resultant force respectively, are with the horizontal force will be: (A) 80 N and 0 N (B) 75 N and 0 N (A) 25.65 (B) 13 (C) 60 N and 80 N (D) 55 N and 80 N (C) 14.5 (D) 15 Solution: Solution: From triangle OQP θ = cos-1 150 = 63.4° 335 10 R 20 Resolving the vertical components of the forces at Q, sin a = sin 60 = 0.866 = 23.09 40 = T cos 63.4° T = 40 sin a = 10 = 0.433 cos 63.4 23.09 T = 89.336 N a = 25.65.

3.26  |  Part III  •  Unit 1  •  Engineering Mechanics Exercises Practice Problems 1 4. Consider the truss ABC loaded at A with a force F as Direction for questions 1 to 10:  Select the correct alterna- shown in the figure below. tive from the given choices. F 1. If point P is in equilibrium under the action of the A applied forces, then the values of the tensions TPQ and TPR are respectively B 45° 30° C TPQ TPR 60° 30° The tension in the member BC is P (A) 0.5 F (B) 0.63 F (C) 0.73 F (D) 0.87 F 500 N Direction for questions 5 and 6:  Two steel members PQ and QR each having cross sectional area of 200 mm2 are (A) 250 N and 250 3 N (B) 250 3 N and 250 N subjected to a horizontal force F as shown in figure. All the (C) 300 3 N and 300 N joints are hinged. (D) 280 N and 280 3 N Q 2. For the loading of a truss shown in the figure below 45° P F (length of member PU = length of member UT = length 60° of member TS), the reaction at S (RS) is R QR 5. If F= 1 kN the magnitude of the vertical reaction force developed at the point R in kN is P U TS (A) 0.543 kN (B) 2 kN 2 kN 2 kN (C) 0.634 kN (D) 1 kN 6. The maximum value of the force F in kN that can be (A) 2 kN (B) 3 kN applied at P such that the axial stress in any of the truss (C) 2.57 kN (D) 4 kN members does not exceed 100 Pa is (A) 22.3 kN (B) 54 kN 3. A truss consists of horizontal members (PU, UT, TS, (C) 43.6 kN (D) 43.28 kN QR) and vertical members (UQ, TR) all having a length B each. 7. Bars PQ and QR, each of negligible mass support a load F as shown in the figure below. In this arrange- ment, it can be deciphered that Q B R B B B R P B B S U T The members PQ, TQ and SR are inclined at 45° to PQ the horizontal. If an uniformly distributed load ‘F’ per F unit length is present on the member QR of the truss (A) Bar PQ is subjected to bending but bar QR is not shown in the figure above, then the force in the mem- subjected to bending. ber UT is (B) Bars PQ and QR are subjected to bending (C) Neither bar PQ nor bar QR is subjected to bending. (A) FB (B) FB (D) Bar QR is subjected to bending but bar PQ is not 2 subjected to bending. (C) 0 (D) 2FB 3

Chapter 2  •  Free Body Diagrams – Trusses  |  3.27 8. P P 9. A cantilever truss of 4 m span is loaded as shown in the figure. X X CB 200 mm pin 60° 60° 100 mm DA YY 4m 20 kN Determine the farce in member BC (A) 32.1 kN (B) 11.55 kN (C) 46.2 kN (D) 23.1 kN 100 N 1 0. In the figure shown below, the force in the member BD is, The figure shows a pair of pin jointed gripper tongs 10 kN 5 kN holding an object weighing 1000 N. The co-efficient of D C friction (m) at the gripping surface is 0.1. xx is the line of action of the input force P and yy is the line of appli- A B 2.5 m cation of gripping force. If the pin joint is assumed 5m to be frictionless, then the magnitude of the force P (A) 5 kN 5m required to hold the weight is: (C) 15 kN (B) 10 kN (A) 500 N (B) 1000 N (D) 20 kn (C) 2000 N (D) 2500 N Practice Problems 2 5. P and Q are two forces acting at a point and their result- ant force is R. When Q is doubled the resultant force is Direction for questions 1, 2 and 3:  Two collinear forces of doubled. If again the resultant force is doubled when Q is reversed, then P : Q : R is: magnitudes 400 N and 200 N act along with a force 600 N, (A) 2 : 3 : 2 (B) 1: 3 : 2 acting at an angle of 30° with the former forces, as shown (C) 3 : 2 : 6 (D) 3 : 5 : 2 below: 6. A force of 200 N is acting at a point making an angle of D 30° with the horizontal. The components of this force along the x and y directions, respectively, are: 600 (A) 173.2 N and 100 N (B) 200 N and 130 N A BC 30 (C) 250 N and 180.2 N (D) 135.5 N and 160 N 400 200 1. The resultant force is (B) 1424 N Direction for questions 7 and 8:  A small block of weight (A) 1512 N (D) 1108 N 200 N is placed on an inclined plane which makes an angle (C) 1342 N θ = 30° with the horizontal. 2. The angle made by the resultant force with the former 7. The component of the weight perpendicular to the forces is inclined plane is (A) 15° (B) 18° (A) 160.2 N (B) 173.2 N (C) 22° (D) 26° (C) 140.2 N (D) 183.2 N 3. If the forces 200 N and 600 N are collinear and acts 8. The component of the weight perpendicular to the at 30° with the force 400 N, then the resultant force inclined plane is shall be (A) 70 N (B) 78 N (A) 1812 N (B) 1526 N (C) 98 N (D) 100 N (C) 1423 N (D) 1164 N 9. A circular disk of radius 20 mm rolls without slipping at a velocity v. The magnitude of the velocity at the Direction for questions 4 to 6:  Select the correct alterna- point P is tive from the given choices. 4. A force of 500 N is acting on a body. The magnitude of P 60° the force to be acted on the same body at 120° with the v first force, so that the net force on the body is still 500 30° O N acting along the bisector of 120°, is v (A) 300 N (B) 500 N (C) 750 N (D) 100 N

3.28  |  Part III  •  Unit 1  •  Engineering Mechanics (A) v 3 v ((BD)) 23v 2v 0.2 Kg (C) 23 f 1 0. A stone with a mass of 0.2 kg is catapulted as shown in the figure below. The total force Fx (in N) exerted by Stone of mass F the rubber band as a function of the distance x (in m) is given by Fx = 300 x2. If the stone is displaced by 0.2 (A) 0.02 J (B) 0.3 J m from the unstretched position (x = 0) of the rubber (C) 0.8 J (D) 10 J band, the energy stored in the rubber band is Previous Years’ Questions 1. The figure shows a pin-jointed plane truss loaded at (A) 0.63 (B) 0.32 the point M by hanging a mass of 100 kg. The mem- (C) 1.26 (D) 1.46 ber LN of the truss is subjected to a load of:[2004] 4. The maximum force F in kN that can be applied at C K LM such that the axial stress in any of the truss members N DOES NOT exceed 100 MPa is: [2012] Om (A) 8.17 (B) 11.15 (A) 0 N (B) 490 N in compression (C) 14.14 (D) 22.30 (C) 981 N in compression (D) 981 N in tension 5. A two member truss ABC is shown in the figure. 2. Consider a truss PQR loaded at P with a force F as The force (in kN) transmitted in member AB is  [2014] shown in the figure. The tension in the member QR is:  [2008] A 1m B 0.5 m 10 kN F PC Q 45° 30° R 6. For the truss shown in the figure, the forces F1 and F2 are 9 kN and 3 kN, respectively. The force (in kN) in (A) 0.5F (B) 0.63F the member QS is (All dimensions are in m): (C) 0.73F (D) 0.87F  [2014] Direction for questions 3 and 4:  Two steel truss mem- F1 F2 bers, AC and BC, each having cross sectional area of 33 100 mm2, are subjected to a horizontal force F as shown PQ R in figure. All the joints are hinged. 2 3. If F = 1 kN, the magnitude of the vertical reaction S T 1.5 3 force developed at the point B in kN is: [2012] A 45° C F (A) 11.25 tension 60° (B) 11.25 compression (C) 13.5 tension B (D) 13.5 compression

Chapter 2  •  Free Body Diagrams – Trusses  |  3.29 7. Two identical trusses support a load of 100 N as shown (A) 433 N and 250 N (B) 250 N and 433 N in the figure. The length of each truss is 1.0 m; cross- (C) 353.5 N and 250 N sectional area is 200 mm2, Young’s modulus E = (D) 250 N and 353.5 N 200 GPa. The force in the truss AB (in N) is ______ 11. A two-member truss PQR is supporting a load W. The  [2015] axial forces in members PQ and QR are respectively.  [2016] A C 30° (A) 2W tensile and 2W compressive B 30° 100 N (B) 3W tensile and 2W compressive 8. 100 kN L PQ 60° P 30º 60º RW Q 45° R 4m (C) 3W compressive and 2W tensile (D) 2W compressive and 3W tensile For the truss shown in figure, the magnitude of the 12. A force F is acting on a bent bar which is clamped at force in member PR and the support reaction at R are one end as shown in the figure. [2016] respectively [2015] (A) 122.47 kN and 50 kN 60º F (B) 70.71 kN and 100 kN (C) 70.71 kN and 50 kN (D) 81.65 kN and 100 kN 9. For the truss shown in the figure, the magnitude of the force (in kN) in the member SR is: [2015] 30 kN WV SR 1m The CORRECT free body diagram is PU T Q (A) 1m 1m 1m 60º F (A) 10 (B) 14.14 (C) 20 (D) 28.28 10. A weight of 500 N is supported by two metallic ropes Ry Rx as shown in the figure. The values of tensions T1 and T2 are respectively. [2015] 30° M T2 T1 90° 120° 500 N

3.30  |  Part III  •  Unit 1  •  Engineering Mechanics (D) 60º F (B) Ry 60º Rx F Ry Rx M 60º (C) F Ry M Answer Keys Exercises Practice Problems 1 1. B 2. A 3. A 4. B 5. C 6. A 7. A 8. D 9. A 10. C 4. B 9. A 10. C Practice Problems 2 4. B 9. C 10. A 1. C 2. A 3. D 5. A 6. A 7. B 8. D Previous Years’ Questions 1. A 2. B 3. A 5.  18 to 22 6. A 7. 100 8. C 11. B 12. A

Chapter 3 Friction, Centre of Gravity, Moment of Inertia CHAPTER HIGHLIGHTS ☞☞ Friction ☞☞ Area Moment of Inertia ☞☞ Polar Moment of Inertia ☞☞ Laws of Friction ☞☞ Perpendicular Axis Theorem ☞☞ Centroid of Solids ☞☞ Cone of Friction ☞☞ Mass Moment of Inertia ☞☞ Mass Moment of Inertia and Radius of Gyration ☞☞ d etermination of the Center of Gravity of a Thin Irregular Lamina ☞☞ Integration Method for Centroid Determination in a Thin Lamina or Solid Friction 5. Coefficient of static friction Definitions W 1. Static friction: It is the friction between two bodies S P which is a tangential force and which opposes the fR sliding of one body relative to the other. F F 2. Limiting friction: It is the maximum value of the static friction that occurs when motion is impending. f S R 3. Kinetic friction: It is the tangential force between two bodies after motion begins. Its value is less than F the corresponding static friction. It is defined as the ratio of the limiting force of 4. Angle of friction: It is the angle between the action friction (F) to the normal reaction (R) between two line of the total reaction of one body on another and bodies (see above figure, where a solid body rests on the normal to the common tangent between the bodies a horizontal plane). It is denoted by m. when motion is impending.   It is also defined as the angle made by the resultant (S) of the normal reaction (R) and the limiting force of friction (F) with the normal reaction R (see the figure given below). It is denoted by f. From the figure, we have: tan l = F = mR m = Limiting force of friction = F R R Normal reaction R = m = coefficient of friction \\ F = mR

3.32  |  Part III  •  Unit 1  •  Engineering Mechanics 7. Angle of repose Fourth law: When motion takes place as one body slides over the other, the magnitude of the frictional force or W resistance will be slightly less than the offered force at that condition of limiting equilibrium. The magnitude of the F fR frictional force will depend only on the nature of the slid- aS ing surfaces and independent of the shape or extent of the contact surfaces. The above figure shows a block of weight W on a rough and plane inclined at an angle a with the Force Determinations for Different Scenarios horizontal. Let R be the normal reaction and F be the force of friction. From applying the condition Least force required to drag a body on a rough horizon- of equilibrium, algebraic sum of the forces resolved tal plane: along the plane: WP F = μR q   = W sin a = F (1) R Algebraic sum of the forces resolved perpendicular to Force P is applied, at an angle q  to the horizontal, on a block the plane: of weight W such that the motion impends or the block tends   = W cos a = R (2) to move. From equations (1) taannda(2FR) W sin a cos (q - f) F Force, P = R But tan f = Least force required, Pleast = W sin f \\ Angle of plane = Angle of friction Force acting on a block (weight = ) along a rough inclined plane: Suppose the angle of the plane a is increased to a value f, so that the block is at the point of sliding WP or the state of impending motion occurs, then at this angle, Motion F = μR direction R m = tan l = tan a  \\ l = a a Hence, the angle of repose is defined as the angle to which an inclined plane may be raised before an For motion down the plane, P = W sin(a - f ) object resting on it will move under the action of the cos f force of gravity and the reaction of the plane.   Hence, angle of repose = angle of plane Laws of Friction For motion up the plane, P = W sin (a + f) cos f First law: Friction always opposes motion and comes into play only when a body is urged to move. Frictional force Force acting horizontally on a block (weight = ) resting on will always act in a direction opposite to that in which the a rough inclined plane: body tends to move. For motion down the plane, P = W tan (a  - f) Second law: The magnitude of the frictional force is just For motion up the plane, P = W tan (a  + f) sufficient to prevent the body from moving. That is, only as much resistance as required to prevent motion will be W F = μR offered as friction. Motion P Third law: The limiting frictional force or resistance bears direction R a constant ratio with the normal reaction. This ratio depends on the nature of the surfaces in contact. The limiting fric- tional resistance is independent of the area of contact. a

Chapter 3  •  Friction, Centre of Gravity, Moment of Inertia  |  3.33 Force acting, at an angle q to the plane, on a block 3 kN (weight = ) resting on a rough inclined plane: P W q Motion F = mR (A)  0.96 kN W = 2 kN direction R (C)  0.75 kN 30° a Solution: (B)  0.86 kN (D)  0.65 kN For motion down the plane, P = W sin(a - f ) 3 kN cos(q + f) For motion up the plane, P = W sin(a + f) 2 kN cos(q - f) Applying the conditions of equilibrium and summing the force parallel and perpendicular to the plane, we have: Cone of Friction Σ F(parallel to the plane) = 0 SR = -F - 2 sin 30° + 3 cos 30° = 0 Axis F = -2 × 1 + 3× 0.866 2 ff = -1 + 2.598 = 1.598 kN Σ F(perpendicular to the plane) = 0 R - 2 cos 30° - 3 sin 30° = 0 R = 2 × 0.866 + 3 × 0.5 = 1.732 + 1.5   = 3.232 kN Direction of B A Direction This indicates that the value of F necessary to hold the frictional in which force motion block from moving up the plane is 1.598 kN. However, the O impends maximum value obtainable as the frictional force, Point of contact F = mR = 0.3 × 3.232 = 0.9696 kN This means that the block will move up the plane. Let OR represent the normal reaction offered by a surface on Example 2: An effort of 2 kN is required just to move a a body and let the direction of impending motion be along certain body up an inclined plane of angle 15°, the force OA while the direction in which the frictional force acts is acting parallel to the plane. If the angle of inclination of the in the opposite direction, i.e., along OB. Assuming that the plane is made 20°, the effort required, again applied parallel body is in a state of limiting equilibrium, the resultant reac- to the plane, is found to be 2.3 kN. Find the weight of the tion S makes an angle of f with the normal OR. If the body body and the coefficient of friction. slides in any other direction, the resultant reaction S will (A)  3.9 kN, 0.258 (B)  4.5 kN, 0.26 still make the same angle f with the normal. It is, therefore, seen that when limiting equilibrium is maintained, then (C)  3.8 kN, 0.24 (D)  3.8 kN, 0.268 the line of action of the resultant reaction should always Solution: lie on the surface of an inverted right circular cone; whose Let W be the weight of the body, m be the coefficient of fric- tion and P be the effort when the inclination of the plane is a. semi-vertical angle is f. This cone is known as the cone of friction. Applying the conditions of equilibrium and summing the forces parallel and perpendicular to the plane, we have, Σ F(parallel to the plane) = 0 Solved Examples P - mR - W sin a = 0 (1) (2) Example 1:  Determine whether the 2 kN block, shown in Σ F(perpendicular to the plane) = 0 the figure, will be held in equilibrium by a horizontal force of 3 kN? The coefficient of static friction is 0.3. R - W cos a = 0 Eliminating R from equations (1) and (2) we have, P = m W cos a + W sin a or

3.34  |  Part III  •  Unit 1  •  Engineering Mechanics P = W (m cos a + sin a)(3) To illustrate the principle of work, let us consider a body at equilibrium at a point A. A force F acts on the body and When a = 15°, P = 2 kN and when a = 20°, P = 2.3 kN. displaces it to the point A′, where the displacement consists Substituting in equation (3) we have, of the following: 2 = W(m cos a + sin a) 1. A very small rotation through the angle a about the 2 = W(m cos 15° + sin 15°)(4) origin of the rectangular 2-D coordinate system, say 2.3 = W(m cos 20° + sin 20°)(5) origin O in the xy plane. Dividing equation (5) by (4), we have, 2. A very small displacement h along the x-axis, and, 3. A very small displacement k along the y-axis. 2 = mcos15° + sin15° 2.3 mcos 20° + sin 20° If the components of the force F along the x-axis and y-axis are Fx and Fy respectively, then work done by the 2 = m × 0.966 + 0.258 or force F when its point of application is displaced from point 2.3 m × 0.939 + 0.342 A to A′ m[(2.3 × 0.969) - (2 × 0.939)] = hFx + kFy + a(xFy - yFx ) = [(2 × 0.342) - (2.3 × 0.258)] If a system of forces act on the body where h, k and a are or 0.3507 m = 0.0906 the same for every force, then work done by all the forces: m = 0.0906 = 0.258 = h∑ Fx + k ∑ Fy + a∑ (xFy - yFx ), 0.3507 Where, ∑ FX and ∑ FY are the sums of the resolved parts From equation (5), of the forces along the x-axis and y-axis respectively, and ∑ (xFy - yFx ) is the moments of the forces about 2.3 = W [0.258 × 0.939 + 0.342] the origin O. = W (0.242 + 0.342) = 0.584W Since the system is in equilibrium, all the three terms in the above expression, for the work done by all the forces, is W = 2.3 = 3.938 kN zero. Hence, the sum of the virtual works done by the forces   0.584 is zero. Virtual Work Lifting Machine Virtual displacement: Virtual displacement is defined Lifting machines are defined as those appliances or as an infinitesimal (exceedingly small) and displacement, machines which are used for lifting heavy loads. They are also given hypothetically to a particle or to a body or a system called simple machines. Some commonly used machines are: of bodies in equilibrium consistent with the constraints. The displacement is only imagined and it does not have to take 1. Lever place for which it is called virtual displacement. 2. Inclined plane 3. Wedge Virtual work: Virtual work is defined as the work done, 4. Wheel and axle by a force on a body due to a small virtual (i.e., imaginary) 5. Winch crab displacement of the body. 6. A pulley and system of pulleys 7. Screw jack Principle of  Virtual Work Screw jack is the most important among all the above sim- If a system of forces acting on a body or a system of bodies ple machines. be in equilibrium and if the system be assumed to undergo a small displacement consistent with the geometrical condi- Load or resistance: A machine has to either lift a load or tions, then the algebraic sum of the virtual work done by the overcome a resistance. It is usually denoted by W and its forces of the system is zero. unit is N. Examples: A lifting device lifts a load or heavy weight whereas a bicycle overcomes the frictional resist- y A′ ance between the wheels and the road. k Efforts: It is the force which is applied to a machine to lift Bh C a load or to overcome the resistance against a movement. It is usually denoted by P and its unit is N. Examples: Force r F applied on the pedals of a bicycle or on the handle of a r Ax screw jack. a y q x x

Chapter 3  •  Friction, Centre of Gravity, Moment of Inertia  |  3.35 Input of a machine: It is defined as the amount of total Loss in load lifted due to friction = P × VR - W. work done on the machine. This is measured by the product Here P is the actual effort required to overcome resist- of the effort and the distance through which it moves. ance W or lift load W. Input = Effort × Distance moved by the effort = P × y Reversible and Irreversible Machine It has the unit of Nm. A machine is said to be reversible when the load W gets Output of a machine: It is defined as the amount of work lowered on the removal of the effort. In such a case, work is got out of a machine or the actual work done by the machine done by the machine in reverse direction. Output of the machine: A machine is said to be irreversible when the load W = Load × Distance through which load is lifted = W × x does not fall down on the removal of the effort. In such a It has the unit of Nm. case, work is not done in the reverse direction. Velocity Ratio (VR): It is defined as the ratio of the dis- The condition of irreversibility or self locking of a tance moved by the effort and to the distance moved by the machine is that its efficiency should be less than 50%. load during the same interval of time. Compound Efficiency Distance moved by the effort y VR = Distance moved by the load = x It is defined as the overall efficiency of the combination of machines and it is the product of the efficiencies of the indi- NOTE vidual machines. In all machines y > x. The noverall efficiency h of n machines coupled together Mechanical advantage (): It is defined as the ratio of the load or weight lifted to the effort applied. ∏is h = hi , where, η i is the efficiency of the ith machine. i =1 Law of a Machine MA = Weight lifted = W It is defined as the relationship which exists between the Effort applied P effort applied and the load lifted. P = mW + C NOTE P is the effort applied, W is the corresponding load, m and In all machines W > P. C are coefficients which are determined in any machine after Ideal machine: It is defined as the machine which is abso- conducting a series of tests and plotting the W versus P graph. lutely free from frictional resistances. In such a machine, The expression for maximum mechanical advantage is 1 input = output. given by ( MA)max = m . For an ideal machine, VR = MA The expression for maximum efficiency is given by Efficiency of a machine: It is the ratio of output of the hmax = m 1 . machine to the input of the machine. × (VR) h = output of the machine ×100 Screw Jack input of the machine It is a device for lifting heavy loads by applying compara- = useful work done by the machine ×100 = W × x × 100 tively a smaller effort at the end of the handle. The screw energy supplied to the machine P × y jack works on the principle of inclined plane. For an ideal machine, h = 100%. For an actual machine, l Ideal effort = Actual load . W Actual effort Ideal load h= Relation between MA, VR and h Screw head Handle W Nut h = W × x = P = MA P × y y VR x Frictional losses Output = Input - Losses due to friction Effort lost in friction = P - W d VR

3.36  |  Part III  •  Unit 1  •  Engineering Mechanics It mainly consists of a nut which forms the body of the The principle of working of this jack is similar to the one as jack and a screw is fitted into it. The threads are generally described in the above figure. square. The load W is placed on the head of the screw. By LLpsee2tt=pthps1eit=clehpvioetcrfhtlheonefgttthhhreebatehdrsleoaadnnsdSo2thneSe1ffort be applied at the rotating the screw with a handle the load is lifted or lowered. Let W be the load lifted, a be the angle of helix of the screw and f be the angle of friction. end of this lever. tan a Here, efficiency = tan(a + φ) , which shows that effi- When the lever is moved by one revolution, the distance covered by the effort P is 2p l and correspondingly the load ciency is independent of the load lifted or lowered. distance is equal to ps1 - ps2. Then, velocity ratio (VR) = Assuming that the effort is applied at the end of the han- 2pl . dle, let us consider the following two cases. pS1 - pS2   Let the weight W be lifted: Let PE be the effort applied at NOTE the end of the handle. Let l be the length of the handle and let d be the mean diameter of the screw. pms1ecishaanlwicaaylsagdrveaantteargtehaans pwse2.llDause to this difference, the the velocity ratio will Σ m about, the axis is zero. be more. Let p be the pitch and m be the coefficient of friction, then: tan a = p Direction for questions 3 and 4: A screw jack has a pitch pd of 12 mm with a mean radius of thread equal to 25 mm a tan f = m lever 500 mm long is used to raise a load of 1500 kg. The coefficient of friction is 0.10. PE Wd ⋅ p + mpd 2l pd - mp Example 3:  Find the helix angle a  and q (i.e., friction angle)   Let the weight W be lowered: Let Q be the effort applied (A)  6.2°, 4.5° (B)  4.85°, 5.7° at the circumference of the screw and let QE be the actual (C)  4.85°, 5.7° (D)  4.36°, 5.7° effort applied at the end of the handle. Solution:  Given P = 12 mm, d = 2r = 25 × 2 = 50 mm, Q = W tan(f - a  ) l = 500 mm Wd mpd - p W = 1500 kg, m = 0.10, tan f = m = 0.10, 2l pd + mp QE = ⋅ f = 5.71° For an n-threaded screw, tan a = np/p d. tan a = P = 12 = 0.076 pd p × 50 Differential Screw Jack a = 4.36 ° Instead of only one threaded spindle as in the case of a sim- Example 4:  What force is necessary when applied normal ple screw jack it has two threaded spindles S1 and S2. The spindle S1 is screwed to the base which is fixed. to the lever at its free end W (A)  13.319 kg (B)  12.8 kg. (C)  14.5 kg (D)  18.3 kg. Solution:  P = wd tan(a + φ ) = 1500 × 50 × tan (4.36 + 5.71) 2l 2 × 500 P = 13.319 kg. S2 l Direction for questions 5, 6 and 7: A uniform ladder of weight 500 N and the length 8 m rests on a horizontal ground and leans against a smooth vertical wall. The angle made by the ladder with the horizontal is 60°. When a man of weight 500 N, stands on the ladder at a distance of 4 metre from the top of the ladder, the ladder is at the point of sliding. S1 Example 5:  Find the coefficient of friction in terms of RB. B RB This spindle carries both internal as well as external RA A threads. The spindle S2 is engaged to spindle S1 by means 60° of an internal thread. When spindle S1 ascends, the spindle S2 descends. This is also known as ‘Differential Screw’ jack. W+w μRA

Chapter 3  •  Friction, Centre of Gravity, Moment of Inertia  |  3.37 (A) mm==510R000B0RB ((DB)) mm==154R00B00 RB The above figure shows an irregular lamina of total area (C) A whose center of gravity is to be determined. Let the lam- ina be composed of small areas a1, a2 … etc. Such that: Solution:  Resolving all the forces RB = mRA A = a1 + a2 + … = Sai RA = W + w = 500 + 500 = 1000 RB = m × RA = m × 1000 Let the distances of the centroids of the areas a1, a2, … etc. from the x-axis be y1, y2, … etc. respectively and from the   m = RB . y-axis be x1, x2, … etc. The sum of moments of all the small 1000 areas about the y-axis Example 6:  Find the reaction at B (i.e., RB)­ = a1 x1 + a2 x2 + … = S ai xi (A) 289 (B) 300 Let xG and yG be the coordinates of the centre of gravity G (C) 350 (D) 400 from the y-axis and x-axis respectively. From the principle of moments, it can be written that: Solution:  Taking moment at A, MA = 0 RB × 8 × 3 = 500 × 8 × 1 + 500 × 4× 1 AxG = Saixi 2 2 2 2    RB = 500 × 2 +1000 = 289 or          xG = ∑ ai xi = ∑ ai xi 6.92 A ∑ ai Example 7:  Find the value of coefficient of friction Similarly, it can be shown that: (A) 0.370 (B) 0.486 (C) 0.289 (D) 0.355 ∑ ai yi ∑ ai Solution:  From equation m = RB 289 = 0.289 yG = 1000 = 1000 Centre of Gravity NOTES The centre of gravity of a body is the point, through which 1. The axis of reference of a plane figure is generally taken the whole weight of the body acts, irrespective of the posi- as the bottom most line of the figure for determining yG tion in which body is placed. This can also be defined as the and the left most line of the figure for calculating xG. centre of the gravitational forces acting on the body. It is denoted by G or c.g. 2. If the figure is symmetrical about the x-axis or y-axis, then the centre of gravity will lie on the axis of Centroid: It is defined as that point at which the total area symmetry. of a plane figure (like rectangle, square, triangle, quadrilat- eral, circle etc.) is assumed to be concentrated. The centroid 3. For solid bodies, elementary masses m1, m2, etc., are and the centre of gravity are one and the same point. It is considered instead of the areas a1, a2, etc., and the also denoted by G or c.g.. coordination of centre of gravity are given as follows: Centroidal axis: It is defined as that axis which passes through xG = ∑ mi xi , yG = ∑ mi yi the centre of gravity of a body or through the centroid of an area. ∑ mi ∑ mi Lamina: A very thin plate or sheet of any cross-section is Example 8:  Determine the position of the center of gravity known as lamina. Its thickness is so small that it can be con- for the following figure. sidered as a plane figure or area having no mass. Determination of the Centre 10 m 2m of Gravity of a Thin Irregular Lamina 3m 5m 5m y 3m 10 m x1 a1 xG G x2 a2 y1 yG y2 2m ox

3.38  |  Part III  •  Unit 1  •  Engineering Mechanics Solution:  DE yx Plane region y dA B C C′ yG G• y G′ G F xG x OA H x Let dA be a differential (i.e., infinitesimal) area located at The x-axis and y-axis of reference are chosen as shown in the point (x, y) in the plane region area A. the above figure such that the origin O coincides with the Here, A = ∫ dA point A of the figure and the axes coincide with the left most A and bottom most lines of the figure respectively. The posi- First moments of the area about the x-axis and y-axis are respectively: tion of the centre of gravity is determined with respect to M X = ∫ y dA the origin O. A The figure is broken down into the three areas AHGG′, MY = ∫ x dA G′FC′B and CC′ED A For rectangle AHGG′, The coordinates (xG , yG) of the centre of gravity of the plane m2 region is given by: Area A1 = 3 × 2 = 6 c.g. coordinates, x1 = 2 = 1 m 2 ∫ xdA   y1 = 3 = 1.5 m xG = My = 2 A A For rectangle G′FC′B, ∫ dA A Area A2 = (2 + 10) × (5 - 3) = 24 m2 ∫ ydA c.g. coordinates, x2 = (2 + 10) = 6m yG = MX =A 2 A ∫ dA (5 - 3) A y2 = 3+ 2 = 4m For rectangle CC′ED, NOTES Area A3 = 3 × 2 = 6 m2 2 1. If the x-axis passes through the centre of gravity, then c.g. coordinates, x3 = 10 + 2 = 11m Mx = 0. Similarly, My = 0, when the y-axis passes through the centre of gravity. y3 = 5 + 3 = 6.5 m 2 2. If the plane region is symmetric about the y-axis, then My = 0 and xG = 0, i.e., the centre of gravity would lie c.g. of the figure coordinates, somewhere on the y-axis. Similarly, Mx = 0 and yG = 0, if the plane region is symmetric about the x-axis, i.e., xG = A1x1 + A2 x2 + A3 x3 =6 m the centre of gravity would lie somewhere on the x-axis. A1 + A2 + A3 If instead of a plane region, we have a plane curve of yG = A1 y1 + A2 y2 + A3 y3 = 4m length L and on which a differential length dL is considered   A1 + A2 + A3 which is located at the point (x, y) on the curve, then the coordinates of the centre of gravity for the planar curve is Integration Method for Centroid given as follows: Determination in a Thin Lamina or Solid xG = My = ∫ xdL In this method, the given figure is not split into shapes of L figures of known centroid as done in the previous section. L The centroid is directly found out by determining S aiyi or S aixi and S ai by direct integration. ∫ dL L First moment of area: Consider a plane region of area A as shown in the following figure. yG = MX ∫ ydL A =L ∫ dL L

Chapter 3  •  Friction, Centre of Gravity, Moment of Inertia  |  3.39 Example 9:  The centre of gravity of the following shown Example 10:  The centre of gravity of the following hatched area OBC, where the curve OC is given by the equation figure with respect to the point E is: y = 0.625 x2, with respect to the point O (0, 0) is: y y B C O 4B x A C 60 40 40 (A)  (6, 5) (B)  (6, 3) (C)  (3, 5) (D)  (3, 3) E D Fx Solution: 30 y 80 (A)  (20, 30) (B)  (37.84, 27.45) (C)  (20, 27.45) (D)  (37.84, 30) C Solution: 1 For Δ ABC, 2 area A1 =  80 × (60 - 40) = 800 y c.g. coordinates, x1 = 2 × 80 = 160 3 3 O (0, 0) x x B(4, 0) 1 140 y1 = 40 + 3 × (60 - 40) = 3 dx Let us consider an elementary rectangular area of height y For ACFE, area A2 = 40 × 80 = 3200 and width dx as shown in the above figure. c.g. coordinates, x2 = 80 = 40 Area of the elementary rectangle, dA = ydx = 0.625x2 dx 2 44 y2 = 40 = 20 2 Area of OBC, A = ∫ dA = ∫ 0.625x2dx 00       = 0.625 × 43 For Δ CFD, area A3 =  1 30 × 40 = 600   3 2 Moment of area about x-axis, c.g. coordinates, x3 = 50 + 2 × 30 = 70 3 4 dA y 4 02 0.625 x 2 1 40 ∫ ∫M X 0.625x2 dx 2 3 3 = = y3 = × 40 = 0    = 0.6252 × 45 Since Δ CFD is cut out from the figure ABFE to obtain the   2 5 hatched figure, the area of Δ CFD is assigned a negative sign. Moment of area about y-axis, \\ A3 = -600 ∫ ∫Mx= 4 dAx = 4 0.625x2dxx = 0.625 × 44 Let xG and yG be the x and y coordinates of the centre 0 0 4 of gravity of the hatched figure with respect to the point E, then: Let xG and yG be the x and y coordinates of the centre of xG = A1x1 + A2 x2 + A3 x3 = 37.84 gravity of OBC with respect to the point O. A1 + A2 + A3 Then, Mx = A yG and My = AxG yG = A1 y1 + A2 y2 + A3 y3 = 27.45 A1 + A2 + A3 0.6252 45 3 yG 2 × 5 × 0.625 × 43 = 3 xG = 0.625 × 44 × 3 = 3 Theorems of Pappus–Guldinus 4 0.625 × 43 A surface of revolution is a surface which can be generated by rotating a plane curve about a fixed axis.

3.40  |  Part III  •  Unit 1  •  Engineering Mechanics y In the figure, generating area =  1 p r2. 2 y Distance traveled by the centroid of the area while the L 4r A • B body is being generated = 2p × 3p (circumference of a cir- r G x r cle of radius  4r ) r 3p x \\ Volume of the sphere generated = 1 pr2 × 2p × 4r = 4pr3 2 3p 3 L Example 11:  A quartered circular arc AB when rotated For example, in the above figure, the curved surface of a about the y-axis generates a surface of area A y. The same cylinder is obtained by rotating the line AB about the x-axis. y Theorem I r r The area of a surface of revolution is equal to the product A of the length of the generating curve and the distance trave- led by the centroid of the curve while the surface is being r generated. NOTE Bx The generating curve must not cross the axis about which arc when rotated about the x-axis generates a surface of it is rotated. eaqreuaatAiox.nI  fAAtyxhe=rkartino, Ay : Ax is related to the length r by the In the above figure, length of the generating curve = L.  where k, n are constants, then the value Distance traveled by the centroid while the surface is being generated = 2p r (circumference of a circle of radius r) of k and n respectively are \\ Area of the surface of the cylinder generated = L × 2p r = 2p rL (A)  0.27 and 0 (B)  0.27 and 1 A body of revolution is a body which can be generated by rotating a plane area about a fixed axis. (C)  3.75 and 0 (D)  3.75 and 1 y Solution: the arc =  1 p  r; Length of 2 y x coordinate of the centroid of the arc = 2r - 2r . Distance travelled by the centroid when theparc is rotated A about the y-axis = 2p × 2r(p -1) p Using Pappus–Guldinus theorem I, Ay = ⎛ 2r - 2r ⎞ × 2p × 2r(p - 1) = 2r 2 p ( p - 1) ⎝⎜ p ⎟⎠ p B r rx O r x r y coordinate of the centroid of the arc = r - 2r . p (a)            (b) Distance travelled by the centroid when the arc is rotated For example, in the above figure, the volume of a sphere about the x-axis = 2p × r(p - 2) . is obtained by rotating the semi-circle OAB about the x-axis. p Theorem II Using Pappus–Guldinus theorem I, The volume of a body of revolution is equal to the product Ax = ⎛ r - 2r ⎞ × 2p × r(p - 2) = r2p(p - 2) of the generating area and the distance traveled by the cen- ⎜⎝ p ⎠⎟ p troid of the area while the body is being generated. \\ Ay = kr n = 2(p -1) Ax p-2 NOTE ⇒ n = 0 and k = 2(p -1) . The theorem does not apply if the axis of rotation inter- p-2 sects the generating area.

Chapter 3  •  Friction, Centre of Gravity, Moment of Inertia  |  3.41 Example 12:  A solid ring (torus) of circular cross-section ∫J0 = r2dA is obtained by rotating a circle of radius 25 mm about the A x-axis as shown in the following figure. J0 = Ix + Iy y The above equation states that the polar moment of inertia 100 mm of an area about a point O is the sum of the moments of iner- tia of the area about two perpendicular axes that intersect x at O. Radius of Gyration In the above figure, the radii of gyration of an area about the x-axis, y-axis and the origin O are: If the density of the material making up the circular cross- kx = Ix , ky = Iy and ko = Jo A A A section is 7800 kg/m3, the weight of the ring generated is: (A)  82.6 N (B)  94.4 N (C)  123.4 N (D)  90.6 N Parallel Axis Theorem Solution: The moment of inertia of a plane region area about an axis, y coordinate of the centroid of the circle = 100 mm = 0.1 m say AB, in the plane of area through the centre of gravity of the plane region area be represented by IG, then the moment Area of the circle = p × (0.025)2 of inertia of the given plane region area about a parallel axis, say OX, in the plane of the area at a distance d from the Distance traveled by the centroid of the circle while gen- centre of gravity of the area is IX = IG + Ad 2, erating the ring = 2p × (0.1) (circumference of a circle of radius 0.1 m) Plane region, area = A Using Pappus–Guldinus theorem II, Volume of the ring generated AG B  = p × (0.025)2 × 2p × (0.1) = 0.001233 m3 Weight of the generated ring = 7800 × 0.001233 × 9.81 = 94.4 N. Area Moment of Inertia d In a plane region of area A, a differential area dA located at OX the point (x, y) is considered as shown in the below figure. Where, y Plane IX = moment of inertia of the given area about the OX region axis IG = moment of inertia of the given area about AB axis x dA A = area of the plane region d = p erpendicular distance between the parallel axes AB Y and OX r G = centre of gravity of the plane region Ox Perpendicular Axis Theorem The moment of inertia of the area about the x-axis and If IOX and IOY are the moments of inertia of a plane region y-axis respectively are: area about two mutually perpendicular axes OX and OY in the plane of the area, then the moment of inertia of the plane ∫ ∫I x = y2dA and I y = x2dA region area IOZ about the axis OZ, perpendicular to the plane AA and passing through the intersection of the axes OX and OY is: Ix and Iy are also called as the second moments of the area. IOZ = IOX + IOY Polar Moment of Inertia In the above figure, the polar moment of inertia of the area about the point O (actually, about an axis through the point O, perpendicular to the plane of the area) is

3.42  |  Part III  •  Unit 1  •  Engineering Mechanics Z Moment of inertia of the rectangle PQRS about the axis OX d Y Plane region ∫AB,IG = 2 by2dy = bd 3 -d 12 NOTE IOZ is also called as the polar moment of inertia and the IX 2 axis OZ is called as the polar axis. IG \\ = 4. Example 14:  The moment of inertia for the following hatched figure about the axis AB (which passes through the centroid of the figure), where AB = DC = 30 m, PQ = SR = 20 m, BC = AD = 20 m and QR = PS = 10 m, is: AB Example 13:  In the below figure, the axes AB and OX P Q are parallel to each other. If the moments of inertia of the B rectangle PQRS along the axis AB, which passes through R the centroid of the rectangle, and the axis OX are IG and IX A respectively, then the value of IX /IG is QR S A G• B DC OP SX (A) 6.78 × 104 m4 (B) 5.41 × 103 m4 (C) 1.83 × 104 m4 (D) 2.6 × 105 m4 (A) 4   (B) 12   (C) 3   (D) 0.25 Solution: Moment of inertia of the hatched ofifgur eP=QmRSoment of inertia Solution: of  ABCD - Moment of inertia From parallel axis theorem, we have IX = IG + A (perpen- 1 ( DC AD3 SR QR3 ) dicular distance between axes)2, 12 = × × - × Let PQ = d and QR = b, then the perpendicular distance 1 d 12 between the axes = 2 = × (30 × 203 - 20 × 103 ) A d2 d2 = 18333.33 m4 . 4 4 \\ IX = IG + = IG + bd Example 15:  A circular section of diameter d is lying on the xy–plane where the centre of the circular section coincides So, IX bd 3 with the origin O as shown in the following figure. IG 4IG = 1+ z To determine IG, let us consider a rectangular strip of thick- Ox ness dy at a distance y from the axis AB as shown below: b d QR 2 dy d y A yB If the moments of inertia of the circular section along the x, P S −d2 y and z axes are IX, IY and IZ respectively, then which of the following statements is NOT correct? Area of the rectangular strip = bdy Moment of inertia of the strip about the axis (A) IX = pd4 (B) IX = IY 32 AB = (bdy) y2 (C) IZ = pd4 (D) IY = p6d44 32

Chapter 3  •  Friction, Centre of Gravity, Moment of Inertia  |  3.43 Solution: z Area of the elementary ring = 2p rdr . Moment of inertia of the elementary ring about the z-axis dr r = 2p rdr × r2 = 2p r3dr. Moment of inertia of the whole circular section about the O• ∫z-axis =D /2 2 p r 3 dr = pd4 . 32 x From th0e symmetry of the circular section, it can be writ- ten that IX = IY. From the perpendicular axis theorem,we have, y i.e., IZ = 2IX IZ = IX + IY Let us consider an elementary ring of thickness dr and \\IX pd4 = IY located at a distance r from the origin O. 64 Description Shape L xc yc y Horizontal line a a 0 Vertical line 2 Inclined line with q Semicircular arc ax Quarter circular arc y Circular arc a0 a 2 a x y a a  a cosθ  2a sinθ  2 q y x pr 0 2r π rr y • CG π r 2r 2r y 2π π x x y a r 2r sinα/2 0 α a/2 x a/2 (Continued)

3.44  |  Part III  •  Unit 1  •  Engineering Mechanics Description Shape L xc yc Rectangle y h b/2 2 bh b 2 h/2 c b x y Square ca a2 a a Parallelogram ax 2 2 Triangle y a ab sin a b + acosα a sinα Semi circle b 2 2 Quarter circle c Sector of a circle Quarter ellipse a x y h bh a + b h 23 3 ax πR  2 0 4R  b 2 3π y 4R 3p R •c ox y πr 2 4R  4R  xc c 2 3π 3π • yc Rx y a x R2a 2 R sinα 0 3α a• xc y b xc • yc πa  b 4a  4b  3 3π 3π a x

Chapter 3  •  Friction, Centre of Gravity, Moment of Inertia  |  3.45 y y2 = kx Quarter parabola xc • b πa  b 3a 3b General spandrel x 35 5 yc a y y = kxn xc c b ab 3a 3b yc • 34 4 oa x Description Figure x lX lY Rectangle y ab3 ba3 12 12 Circle b/2 πr  4 πr  4 b/2 4 4 a/2 a/2 y rx b π ab3 π ba3 4 4 Ellipse a Triangle y bh3 hb3 Quadrant Circle 36 36 hC x h/3 0.0549 r4 0.0549 r4 b 4r 3p r C x 4r 3π

3.46  |  Part III  •  Unit 1  •  Engineering Mechanics Centroid of Solids P G• If dm is an elemental mass in a body of mass M and xG, yG are the coordinates of the center of gravity of the body from the reference axes y-axis and x-axis respectively, then: XG = ∫ xdm = ∫ xdm , yG = ∫ ydm = ∫ ydm CO D ∫ dm M ∫ dm M (A) OG = 5 cm (B) OG = 3 OP (C) CO = 10 cm 8 Let us consider a right circular solid cone whose centre of gravity is to be determined. Let the diameter of the base of (D) OD = 2 × OG the right circular solid cone be 2R and its height H as shown in the following figure. Solution: The centre of gravity of a hollow hemisphere with respect Since the cone is symmetric about the VX axis, its centre to the x-axis would lie on an perpendicular axis along which of gravity will lie on this axis. The cone can be imagined the homogeneous hemisphere is symmetrical. to be consisting of an infinite number of circular discs with different radii, parallel to the base. Since G is the centre of gravity, then the hemisphere should be symmetrical along OP, i.e., CO = OD. V It can also be deciphered that CO = OD = radius of the y hemisphere = OP = 10 cm. C xD H Now OG will be equal to R/2, where R is the radius of the dy hollow hemisphere. FE \\ OG = 0.5 OP = 5 cm It can be written OP = CO = OD = 2 OG, hence the option (B) is NOT correct. AXB NOTE 2R Option B would be right if the hemisphere had been a homogeneous solid hemisphere. Consider one such disc of radius x, thickness dy and at a depth y from the vertex of the cone, i.e., from V. Mass Moment of Inertia From the geometry of the above figure, The Moment of Inertia of an element of mass is the product of the mass of the element and the square of the distance of x = y or X = yR the element from the axis. R H H The mass moment of inertia of the body with respect to Volume of disc = p X 2dy = p y2R2 dy Cartesian frame xyz is given by: H2 If r is the density of the material making up the cone, y2R2 ∫ ∫I xx = ( y2 + z2 ) dm = ( y2 + z2) rdv then dm = r p H2 dy v (x2 + z2) dm = (x2 + z2) rdv H p y3R2 ∫ ∫IYY = H2 dy v 0 dy ∫ ∫∫\\ yG H r [ ]ydm 3 3 ∫ ∫I zz = (x + y2) dm = (x2 + y2) rdv, where, IXX, IYY and = 0 = 4 y H = 4 H v dm 0 p y2R2 IZZ are the axial moments of inertia of mass with respect to r H2 the x-, y- and z-axes respectively. \\ C entroid or centre of gravity of a right circular cone For thin plates essentially in the x-y plane, the following 3 relations hold. 4 is situated at a distance of H from its vertex V and ∫Ixx = y2dm lies on its axis VX. ∫IYY = x2dm ∫ ∫I zz = z2dm = (x2 + y2) dm Example 16:  In the homogenous hollow hemisphere, shown in the following figure, OP = 10 cm = the radius of the I zz = I XX + IYY hemisphere. The points P, G and O lie on a straight line that is perpendicular to the base CD. If G is the centroid of the hollow Izz is also called the polar moment of inertia. hemisphere, then which one of the following statements is not correct?

Chapter 3  •  Friction, Centre of Gravity, Moment of Inertia  |  3.47 Mass Moment of Inertia and IYY = 1 m 2 Radius of Gyration 12 I xx = Kx2m Radius of gyration about a centroidal axis perpendicular I yy = K y2m I zz = Kz2m to a uniform thin rod of the length , mass m and a small cross section is given by Kx = I xx K y = 12 m The mass moment of inertia about the longitudinal and Ky = IYY transverse axes passing through the centre of mass of a rec- m tangular prism (block) of cross section (axb), uniform den- 1 I zz sity r and length  is given by I xx = 12 m ( 2 + b2 ). m Kz = The parallel-axis theorem for the mass moment of iner- IYY = 1 m (a2 + b2 ) tia states that the mass moment of inertia with respect to 12 any axis is equal to the moment of inertia of the mass with respect to a parallel axis through the centre of mass plus I zz = 1 m (a2 + 2) the product of the mass and the square of the perpendicular 12 distance between the axes. In the above case, if the three axes were chosen through a Mathematically IAB = IG + md2 corner instead of centre of mass, the results are: For a thin plate, I xx = 1 m ( 2 + b2) I xx(mass) = rt I xx(area) 3 IYY (mass) = rt IYY (area) IYY = 1 m (a 2 + b2) 3 I zz(mass) = rt I zz(area) I zz = 1 m ( a2 + 2) Where t is the uniform thickness and r is the mass of the 3 thin plate. For a right circular cylinder of radius R, length or height  I zz = I xx + IYY and mass m, the mass moment of inertia about the centroi- The mass moment of inertia about a centroidal axis per- dal x-axis is given by pendicular to a uniform thin rod of length , mass m and small cross section is given by I xx = m ⎡ R4 + 2⎤ ⎢ 4 ⎥ ⎣ 12 ⎦ Solid Body Centroid Mass moment of inertia z Solid hemisphere G xG = yG = 0 IXX IYY IZZ 2 mR2 R 3 5 O x ZG = 8 R = = = y Solid sphere z IXX = IYY = IZZ = 2 mR2 5 xG = yG = zG = 0 GR x Ky = 2 R O 5 y (Continued)

3.48  |  Part III  •  Unit 1  •  Engineering Mechanics Solid Body Centroid Mass moment of inertia z Solid cylinder R xG = yG = 0 IXX = IYY = 1 mR2 + 1 mL2 4 3 LG L zG = 2 IZZ = 1 mR2 O 2 y x Rectangular block (cuboid) a z b xG = yG = 0 Ixx = 1 ma2 + 1 mL2 12 3 LG zG = L Iyy = 1 mb2 + 1 mL2 2 12 3 Izz = 1 m ( a2 + b2 ) 12 O x y Slender rod (thin cylinder) z O Ixx = 0 L zG = L mL2 G 2 3 IYY = IZZ = y yG = zG = 0 x z IXX = IYY = mR2 Solid disk 4 R y O xG = yG = zG = 0 Izz = mR2 2 x Kz = r 2

Chapter 3  •  Friction, Centre of Gravity, Moment of Inertia  |  3.49 Exercises Practice Problems 1 Direction for question 4: A locomotive of weight W is at rest.   Select the correct alternative from the given choices. 4. The reactions at A and B are 1. A belt supports two weights W1 and W2 over a pulley CP as shown in the figure. If W1 = 2000 N, the minimum weight W2 to keep W1 in equilibrium (assume that the • b pulley is locked and m = 0.25) is: • b A • OB Aa aB RA W RB (A) W N (B) 2WN 2 (C) 32WN (D) 3WN T1 T2 Direction for question 5: When it is pulling a wagon, the W1 W2 draw bar pull P is just equal to the total friction at the points 2000 N of contact, A and B. (B) 812.8 N (A) 911.9 N (D) 715.5 N 5. The new magnitudes of the vertical reactions at A and (C) 913 N B respectively are: (A) Wa - Pb , Wa + Pb (B) W2 , W 2a 2a 2a 2.   25 cm 50 cm (C) W2 , W (D) W2 , 23W B 3 A• C 6. A four wheel vehicle with passengers has a mass of • 2000 kg passengers. The road, on which the vehicle is r = 0.25 m moving, is inclined at an angle q with the horizontal. If 45° the coefficient of static friction between tyres and the road is 0.3, the maximum inclination q at which the vehicle can still climb is: q 0.25 m 1 m 0.5 • A rotating wheel is braked by a belt AB attached to the • lever ABC hinged at B. The coefficient of friction be- q mg tween the belt and the wheel is 0.5. The braking mo- (A) 18° (B) 16.7° (C) 15° (D) 17.2° ment exerted by the vertical weight W = 200 N is: 7. A weight W of 2000 N is to be raised by a system of (A) 98.23 Nm (B) 95.96 Nm pulleys as shown in the following figure. (C) 95.00 Nm (D) 93.24 Nm 3. A screw jack has square threaded screw of 5 cm diam- • +dv eter and 1 cm pitch. The coefficient of friction at the screw thread is 0.15. The force required at the end of a P 70 cm long handle to raise a load of 1000 N and the force required, at the end of the same handle to raise the • same load, if the screw jack is considered to be an ideal machine, respectively, are: −d y W 2000 N 2 (A) 7.702 N and 2.123 N (B) 7.702 N and 2.273 N (C) 8.162 N and 1.850 N (D) 8.162 N and 1.798 N

3.50  |  Part III  •  Unit 1  •  Engineering Mechanics The value of the force P which can hold the system in 8. The values of the normal reaction R and the limiting friction F, respectively are: equilibrium is: (A) 500 cos q and 500m cos q (A) 5000 N (B) 1000 N (B) 400 cos q and 400m cos q (C) 600 cos q and 600m cos q (C) 2000 N (D) 1500 N (D) 600 cos q and 500m cos q Direction for questions 8, 9 and 10: A weight of 600 N just 9. The inclination of the plane q is: starts moving down a rough inclined plane supported by a force of 200 N acting parallel to the plane and it is at the point (A) 30° (B) 25.6° of moving up the plane when pulled by a force of 300 N paral- lel to the plane. (C) 24.6° (D) 32.1° q P = 200 N 1 0. The coefficient of friction is: •F (A) 0.092 (B) 0.1124 (C) 0.1510 (D) 0.2130 R q Practice Problems 2 4. A body of weight 50 N is kept on a plane inclined at an Direction for questions 1 to 10:  Select the correct alterna- angle of 30° to the horizontal. It is in limiting equilib- tive from the given choices. rium. The co-efficient friction is the equal to: (A) 1 (B) 3 1. The block shown in figure below is kept in equilibrium and prevented from sliding down by applying a force of 3 600 N. The co-efficient of friction is 3 . The weight of (C) 1 (D) 3 the block would be: 5 50 3 5 5. A man of weight 60 N stands on the middle rung of a lad- der of weight 15 N. The co-efficient of friction between 600 N contacting surfaces is 0.25. The reaction at the floor is: (A) 80 N (B) 73.25 N (C) 85.6 N (D) 72.75 N 6. Determine the effort required at the end of an arm 50 cm 30° long to lift a load of 5 kN by means of a simple screw (A) 4000 N (B) 2500 N jack with screw threads of pitch 1 cm if the efficiency (C) 3000 N (D) 5000 N at this load is 45%. (A) 40.8 N (B) 43.6 N 2. Mention the statements which are governing the laws (C) 44.8 N (D) 35.36 N of friction between dry surfaces. 7. Determine the effort needed if the jack in above question  (i) The friction force is independent on the velocity of is converted into a differential screw jack with internal sliding. threads of pitch 7 mm and efficiency of operation is 30%. (ii) The friction force is proportional to the normal (A) 15.9 N (B) 19.8 N force across surface of contact (C) 17.2 N (D) 18 N (iii) The friction force is dependent on the materials of 8. A wooden block is being split by a 20° wedge with a force of 70 N applied horizontally as shown. Taking the co- the contacting surfaces. efficient of friction between wood and the wedge as 0.4  (iv) The friction force is independent of the area of estimate the vertical force tending to split the wood apart. contact. (A) 2, 3, 4 (B) 1 and 3 (C) 2 and 4 (D) 1, 2, 3 and 4 3. The limiting friction between two bodies in contact is 70 N independent of 20° Wedge (A) Nature of surfaces in contact; (B) The area of surfaces in contact; (C) Normal reaction between the surfaces. (D) All of the above.

Chapter 3  •  Friction, Centre of Gravity, Moment of Inertia  |  3.51 (A) -54.24 N, 54 N (B) -65 N, 64 N (A) 1985 N (B) 1723 N (C) -48 N, 46 N (D) -56 N, 54 N (C) 1630 N (D) 1874 N 9. A screw thread of screw jack has a mean diameter of 10. Efficiency of the screw jack in problem above is: 10 cm and a pitch of 1.25 cm. The co-efficient of fric- tion between the screw and its nut housing is 0.25. The (A) 12% (B) 13.7% force F that must be applied at the end of a 50 cm lever arm to raise a mass 6000 kg, is: (C) 15% (D) 16.4% Previous Years’ Questions 1. An elevator (lift) consists of the elevator cage and T a counter weight, of mass m each. The cage and the counterweight are connected by a chain that passes G 100 N over a pulley. The pulley is coupled to a motor. It is desired that the elevator should have a maximum μ = 0⋅2 stopping time of t seconds from a peak speed v. If the inertia of the pulley and the chain are neglected, the minimum power that the motor must have is: [2005] Pulley (A) 176.2 (B) 196.0 (C) 481.0 (D) 981.0 Chain 4. A 1 kg block is resting on a surface with coefficient of friction m = 0.1. A force of 0.8 N is applied to the block as shown in the figure. The friction force is: [2011] vm 0⋅8 N 1 kg Cage mv (A) 0 (B) 0.8 N (C) 0.98 N (D) 1.2 N Counter weight 5. A block R of mass 100 kg is placed on a block S of mass 150 kg as shown in the figure. Block R is tied to the wall (A) 12 mv2 (B) mv2 by a massless and inextensible string PQ. If the coeffi- 2t cient of static friction for all surfaces is 0.4, the minimum force F (in kN) needed to move the block S is: [2014] (C) mv2 (D) 2mtv2 PQ t R 2. If a system is in equilibrium and the position of the S system depends upon many independent variables, the principle of virtual work states that the partial derivatives of its potential energy with respect to each F of the independent variable must be: [2006] (A) -1.0 (B) 0 (C) 1.0 (D) ∞ 3. A block weighing 981 N is resting on a horizon- (A) 0.69 (B) 0.88 (C) 0.98 (D) 1.37 tal surface. The coefficient of friction between the block and the horizontal surface is m = 0.2. A verti- 6. A block weighing 200 N is in contact with a level cal cable attached to the block provides partial sup- plane whose coefficients of static and kinetic friction port as shown. A man can pull horizontally with a are 0.4 and 0.2 respectively. The block is acted upon force of 100 N. What will be the tension, T (in N) by a horizontal force (in newton) P = 10t, where t denotes the time in seconds. The velocity (in m/s) of in the cable if the man is just able to move the block the block attained after 10 seconds is: ____ [2014] to the right? [2009]

3.52  |  Part III  •  Unit 1  •  Engineering Mechanics 7. A body of mass (M) 10 kg is initially stationary on a (A) Solid Cylinder (B) Rimmed wheel (C) Solid sphere (D) Solid cube 45° inclined plane as shown in figure. The coefficient O′ of dynamic friction between the body and the plane is 0.5.? The body slides down the plane and attains a velocity of 20 m/s. The distance travelled (in meter) by the body along the plane is _______ [2014] O O′ M 45° O   10. The value of moment of inertia of the section shown in the figure about the axis-XX is: [2015] 8. A wardrobe (mass 100 kg, height 4 m, width 2 m, 60 depth 1 m), symmetric about the Y-Y axis, stands on 30 a rough level floor as shown in the figure. A force P is applied at mid-height on the wardrobe so as to tip 45 15 120 it about point Q without slipping. What are the mini- All dimensions 15 are in mm mum values of the force (in newton) and the static X coefficient of friction m between the floor and the X wardrobe, respectively? [2014] 45 Y 30 2m 4m (A) 8.5050 × 106 mm4 (B) 6.8850 × 106 mm4 P (C) 7.7625 × 106 mm4 (D) 8.5725 × 106 mm4 Q 1 1. A block of mass m rests on an inclined plane and is Y attached by a string to the wall as shown in the figure. The coefficient of static friction between the plane and the block (A) 490.5 and 0.5 is 0.25. The string can withstand a maximum force of 20 N. (B) 981 and 0.5 The maximum value of the mass (m) for which the string (C) 1000.5 and 0.15 will not break and the block will be in static equilibrium is (D) 1000.5 and 0.25 _______ kg.[2016] Take cos q = 0.8 and sin q = 0.6 Acceleration due to gravity g = 10 m/s2 9. For the same material and the mass, which of the fol- lowing configurations of flywheel will have maximum mass moment of inertia about the axis of rotation OO′ passing through the center of gravity. [2015] θ O O′ O O′    1 2. The figure shows cross-section of a beam subjected to bending. The area moment of inertia (in mm4) of this cross-section about its base is ________. [2016]

Chapter 3  •  Friction, Centre of Gravity, Moment of Inertia  |  3.53 10 the angular momentum of the system of particles about G R4 is (The quantity ri indicates second derivative of ρi with respect to time and likewise for ri).[2016] All dimensions R4 are in mm mi System boundary 8 ρ 10 G 10 ri 1 3. A system of particles in motion has mass center G r as shown in the figure. The particle i has mass mi and its position with respect to a fixed point O is given by the O position vector ri. The position of the particle with respect to G is given by the vector ρi. The time rate of change of (A) Σiri × mi ri (B) Σiri × miri (C) Σiri × miri (D) Σiri × miri Answer Keys Exercises Practice Problems 1 1. A 2. B 3. B 4. A 5. A 6. B 7. B 8. C 9. C 10. A 9. B 10. B Practice Problems 2 4. A 9. B 10. B 1. C 2. A 3. B 4. B 5. D 6. D 7. A 8. A 13. B Previous Years’ Questions 1. C 2. B 3. C 5. D 6.  4.8 to 5 7.  56 to 59 8. A 11. 5 12.  1873 to 1879

Chapter 4 Rectilinear Motion CHAPTER HIGHLIGHTS ☞☞ Dynamics ☞☞ Freely Falling Body ☞☞ Dynamics of a Particle ☞☞ Rectilinear Motion: Displacement, Distance, ☞☞ Work and Energy velocity and Acceleration ☞☞ Law of Conservation of Energy ☞☞ Impact ☞☞ Motion at a Uniform Acceleration ☞☞ Elastic Impact ☞☞ Plastic or Inelastic Impact ☞☞ Vertical Motion Under Gravity ☞☞ Coefficient of Restitution ☞☞ Kinetics of a Particle ☞☞ Differential Equation of Rectilinear Motion ☞☞ M otion of a Particle Acted Upon by a Constant Force Dynamics Rectilinear Motion: Displacement, Distance, Velocity and Acceleration Dynamics is the branch of mechanics dealing with the motion of a particle or a system of particles under the action 1. Displacement and distance: of a force. Dynamics is broadly divided into two categories: AB 1. Kinematics 2. Kinetics xx x Kinematics is the study of motion of a body without any Let the particle be at the position A at any point of time reference to the forces or other factors which causes the t. Let the position of the particle be at B at time t + dt motion. Kinematics relates displacement, velocity and (dt > 0). Then the particle is said to move from A to acceleration of a particle of system of particles. B. The change in position is the displacement x. It is the shortest distance between A and B. Distance is the Kinetics studies the force which causes the motion. length of the path described by the particle from point It relates the force and the mass of a body and hence the A to point B. motion of the body. So in fact, the motion of a particle or body is largely covered and interpreted by Kinematics and R Kinetics. x Types of Motion z The rate of change of position is motion. The type of motion P yQ is explained by the type of path traced by it. If the path traced is a straight line, the motion is said to be rectilinear Let a body start from a point P and move towards a motion or translation. point Q and then turn and reach a point R. During this course of motion, the total displacement is denoted by If the path traced by the motion (or path traversed by x. The distance traversed is given by y + z. the particle) is a curve, it is known as curvilinear motion. When the curve becomes a circle, then it is known as cir- NOTE cular motion. When the motion of a particle is considered along a line The two types of motion, i.e., rectilinear and curvilinear segment, both distance and displacement are the same in motions, explained above can be together termed as the gen- magnitudes. eral plane motion.