Mechanics_Ramkumar_Chapter_02_Resultant of Coplanar Forces

,Mechanics: It deals with the effect and analysis of the forces acting on a body or a particle which may be atrest or in the motion.i] Statics : Branch of Mechanics which deals with the study of bodies or particles at rest.ii] Dynamics : Branch of Mechanics which deals with the study of bodies or particles in motion.Body: It is composed of a number of very small particles, held together by cohesion. Body may be rigid ordeformable.Rigid Body: The body which can retain its size and shape under the effect of external forces.Deformable Body: The body which does not retain its size and shape under the effect of external forces.Note: In statics and dynamics, we assume that all the bodies under consideration are rigid.FORCE: It can be defined as any influence which tends to change the state of rest or of motion of a particleor a body. (Unit: N and kN) Force is composed of following three: i) Magnitude of Force (N or kN). ii) Line of action or Direction of Force. iii) Point of Application of Force.Force System: Several forces acting together on a body or particle form a force system. Force SystemProf. Ramkumar's Coplanar Non Coplanar Or SpaceConcurrent Parallel GeneralCoplanar Force System: If all theforces acting on a body or a particle arelying in one plane, it is called as coplanarforce system.Non-coplanar or Space ForceSystem: If all the forces acting on a bodyor a particle do not lie in one plane, it iscalled as space force system.Coplanar Concurrent Force System: If lines of action of all the coplanar forces acting on a body (orparticle) meet at a point, force system is called as coplanar force system. 13

Coplanar Parallel Force System: If several forces acting in a plane are such that they do not meet at apoint even though their lines of action are extended, then they represent parallel force system in a plane. If all the forces act in the same direction, they are called Like otherwise Unlike.Prof. Ramkumar's Like Parallel Forces Unlike Parallel ForcesCoplanar General Force System: If severalforces acting in a plane are such that they do notmeet at a point and are not parallel even though theirlines of action are extended, then they representGeneral force system in a plane. This force system isalso called as Non-concurrent Non-parallel ForceSystem in a plane.Law of Parallelogram: If the two forcesacting at a point be represented inmagnitude and direction by the twoadjacent sides of a parallelogram, drawnfrom one of its angular points, theirresultant is represented by the diagonal ofthe parallelogram passing through thatangular point, in magnitude and direction.Law of Triangle: If the two forces actingat a point are represented in magnitudeand direction by the two sides of a triangletaken in order, their resultant isrepresented by the third side taken inopposite order.Resolution of a Force: Finding the components of a force in two mutually perpendicular directions is calledas resolution of a force. Let Fx and Fy be the components of a force ‘F’ in x and y directions respectively as shown (By Law of Triangle) In ∆ ABC, cos θ = Fx F Fx = Fcos θ and sin θ = Fy F Fy = F sin θExamples: 14

Moment of a Force: When a force is applied on a body, it has the tendency to turn body about some point.The turning tendency of the force about a point is called as moment of the force about that point. Moment of the force about any point is theproduct of force and perpendicular distance from thepoint on the line of action of the force. If the tendency of rotation is clockwise, themoment is called as clockwise. If the tendency of rotation is anti-clockwise, themoment is called as anti-clockwise.Units: N-m, N-mm, kN-m etc.Note: Moment of a force is zero if its line of action passes through moment centre.Prof. Ramkumar'sVARIGNON’S THEOREM: Algebraic sum of moments of a system of coplanar forces about any point isequal to the moment of their resultant about the same point. Mo = MRoDerivation: Let forces P and Q be twoconcurrent forces acting along the linesEX and EY intersecting at E as shown. Let O be any point in the planes ofthe forces. Through O, draw OBCparallel to EX, meeting EY at B. Choosethe scale so that EB represents Q. Withthe same scale, draw EA to representforce P. Complete the parallelogramEACB.Diagonal EC represents R, resultant of forces P and Q. Join OE and OA.Let ∆ represents area of triangle [since one half of parallelogram] ∆ OEC = ∆ OEB + ∆ EBC = ∆ OEB + ∆ ECA ∆ OEC = ∆ OEB + ∆ OEA[since triangle standing on the same base and lying between the same parallels are equal in area]½ ×EC× perpendicular distance from O on EC = ½ × EB × perpendicular distance from O on EB + ½ × EA × perpendicular distance from O on EA.i.e. Moment of the force represented by EC about O = Moment of the force represented by EB about O + Moment of the force represented by EA about Oor Moment of Resultant (R) about O = Moment of Q about O + Moment of P about Oi.e MRO = MOQ + MPO = MOBy the successive application of the above method, Varignon’s theorem can be extended to any number ofconcurrent forces in plane.Note: Varignon’s theorem is also applicable for non-concurrent (parallel and general) force system.Exception: Not applicable for Non-concurrent force system in a plane if resultant is zero.Use of Varignon’s Theorem:(1) Varignon’s theorem is essentially required to be applied for determining the position of resultant force in case of parallel and general force system.(2) This theorem is very useful in finding moment of inclined forces about some point. Using this theorem, complicated geometrical constructions for the determination of perpendicular distances are eliminated. 15

Example: Using Varignon’s theorem, MFA can be easily determined by taking algebraic sum of moment of components of F about A as belowProf. Ramkumar'sMFA = F × d MFA = 100 sin 60 ×10 = 100 × (10 sin60) = 866 Nm ( ) moment of 100 cos 60 about A is zero = 866 Nm( )Principle of Transmissibility of a Force: It statesthat the condition of equilibrium or of motion of a rigidbody will remain unchanged if the point of application ofa force acting on the rigid body is transmitted to act atany other point along its line of action.Note: Not applicable for finding internal forces in body.COUPLE: Two parallel forces of equal magnitude and opposite direction form couple. Couple tends to rotatethe body. Moment of couple M = F × d ……… where, d = arm of couple Consider two parallel forces equal in magnitude and opposite in direction and separated by distance‘d’ as shown in figure. MO = F × 0 + F × d = F × d ( )..........(1) 1 MO = F × 0 + F × d = F × d ( )..........(2) 2 MO 3 = F × d1 + F × d2 = F(d1 + d2 ) = F × d ( ) ..........(3)From (1), (2) and (3), it is clear that moment of couple is fixed and it does not depend upon moment centre.Resolution of a force into a force-couple system: Consider a force F acting on a body at point at A.This is to be replaced by a force and couple at some other point B as shown in figure. Assume two equal and opposite forces at B, each of magnitude F and acting parallel to force F at A.Out of three forces, two forces acting in opposite direction at A and B form a couple of moment M=F×d( )Hence, Force ‘F’ acting at point A can be replaced at ‘B’ (i) by a single force ‘F’ as it is. (ii) and a moment ‘M’ of force ‘F’ at A about B i.e. M = F × d ( ) 16

Prof. Ramkumar'sImportant points related to Resultant: 1) Resultant of parallel and general force systems will be any one of the following (i) Only Resultant Force (ii) Only Resultant Moment (iii) Neither Resultant Force nor Resultant Moment. 2) If resultant of general force system is horizontal, its position can be marked only on y-axis. 3) If resultant of general force system is vertical, its position can be marked only on x- axis. 4) If resultant of general force system is inclined, its position can be marked on both the axes. Sign Convention: For solving the problems of resultant and equilibrium of any force system, following sign convention will be adopted: (i) For determining Fx (algebraic sum of x-components of forces), forces acting towards right ( ) will be taken + ve and forces acting towards left ( ) will be taken ve. (ii) For determining Fy (algebraic sum of y-components of forces) forces acting upwards ( ) will be taken + ve, and forces acting downwards ( ) will be taken ve. (iii) For determining M (algebraic sum of moments of all forces about some point), anti-clock wise ( ) moments will be taken + ve and clockwise ( ) moments will be taken ve. Types of Loads (1) Point Load: It acts over so small distance, that it can be assumed to act at a point. (Also called concentrated Load) (2) Uniformly Distributed Load (u.d.l.): It acts over considerable length of the beam. It may be uniformly distributed over entire length of beam or over part of its length. Intensity of u.d.l. is expressed per unit length. For solving problems, u.d.l. is converted into equivalent point load by multiplying the load intensity with length. Equivalent point load acts at the centre of spread. (3) Uniformly Varying Distributed Load (u.v.d.l.): In this loading, load intensity increases or decreases at a constant rate. This loading is converted into equivalent point load which is equal to area under load diagram. Equivalent point load acts at the centroid of load diagram. Note: While solving problems of mechanics, if direction of any unknown force or reaction is assumed and after solution, unknown is found negative, it indicates that assumed direction of unknown was wrong. 17

-, , ! !Prof. Ramkumar's Find the resultant of five concurrent forces asshown in figure.R = 104.84 kN 11.52°The striker of caram board is being pulled by To move a boat uniformly along a river at afour players as shown in the figure. The players are given speed, a resultant force R = 520 N issitting exactly at the centre of the four sides. required. Two men pulling with forces P and Q byDetermine the resultant of forces in magnitude and means of ropes to do this. The ropes make andirection. (May’08−10 marks) angle of 30 and 40 w.r.t. sides of river. (a) Determine the forces P and Q. (b) If θ1 = 30 , find the value of θ2 such that the force ‘Q’ in the rope is minimum. What is this minimum force ‘Q’. (Jan’04 – 7 marks) R = 29.1N, θ = 45.88° (a) P = 355.7 N, Q = 276.7 N, (b) θ2 = 60 , Qmin = 260 N Find force F4 so as to give the resultant of thesystem of concurrent forces as shown in figure. A force R = 25 N has components Fa, Fb and Fc as shown in figure. If Fc = 20 N find Fa and Fb. (Dec’07−5 marks)F4 = 961.64 N 39.58° Fa = 33.9 N, Fb = 35.03 N Under the action of five forces, followingsystem is in equilibrium. Determine fifthforce. (Dec’04-4marks) 18

&'()*&+ R = (15.91)2 + (7.93)2 Fx = +18 + 22.5 cos 45 − 30 cos 30 = 17.78 N = +7.93 N Fy = +15 + 22.5 sin 45 − 30 sin 30 = +15.91 NNote: Force F5 will be equal and opposite to R. ! Find the resultant of four concurrent forces P1,P2, P3 and P4 as shown in figure.Prof. Ramkumar's tan α = 15.91 α = 63.51° 7.93 17.78 N ! Find resultant of the force system. (Dec’12-4 marks)R = 14.46 kN 5.52° R = 18.83 N α = 57.67° For the system shown, determine− Determine the magnitude and direction of(i) The required value of α if resultant of three forces F1 and F2? When the resultant of givenforces is to be vertical. force system is found to be 800 N along +ve x(ii) The corresponding magnitude of resultant axis. (May’06−5 marks) (Dec’08−5 marks) (i) α = 21.7° (ii)R = 228.64 N ↓ F1 = 527.23 N, F2 = 1161.16 N Five concurrent coplanar forces act on a body For a given concurrent force system, find theas shown in figure. Find the force P and Q such single force which will keep the entire force systemthat the resultant of the five forces is zero. in equilibrium. Hint: Refer Class Work Problem-6. (Dec’09-5 marks)P = 81N, Q = 10.642 N R' = 100.95 N 83.26° 19

-, ! Determine the resultant of the following !systems of parallel forces. Find the single force which is equivalent to four applied forces over vertical bar AB as shown.Prof. Ramkumar'sR = 300 N ( ↑ ) at 9.25 m right of line of action offirst force. Find analytically the resultant of the parallel R = 100kN ← 1m below Aforce system shown in figure. Replace the 1200 N force at C by a force couple system at D. Refer to figure.R = 0, M = 1200 N.m ( ) 1200 N ( ↓ ) and 600 N.m ( ) For the given force system, find equivalent Resolve 15 kN force at ‘A’ into two parallelforce-couple system at A, B and D. components at ‘B’ and ‘C’. (Dec’11-4 marks & May’08−5 marks−old)At A, 45 kN ↑ and 265 kNm At B, 26.25 kN(↓) and at C, 11.25kN(↑)At B, 45 kN ↑ and 355 kNmAt D, 45 kN ↑ and 557.5 kNm 20

\" A 500 N force is applied to the point A ofa L-shaped plate. Find the equivalent forcecouple system at B.Prof. Ramkumar's&'()*&+ 500 N force applied at A will beequivalent to (i) 500 N force at B (as it is) (ii) Moment of 500 N force about B MBF = − 433.01×1 + 250 × 0.5 = − 308.01 Nm = 308.01 Nm ( ) = M# A force of 800 N is acting as shown infigure,(i) Resolve this force into parallelcomponents P and Q acting along aa and bb.(ii) Resolve the same force of 800 N intoparallel components P and Q acting along bband cc&'()*&+ Case (i) Let P and Q be the downward vertical components of given ‘R’Appying Varignon’s TheoremMo = M R o− Q × 5 x = −1600 x Q = 320 N (↓) R = Fy−800 = −P − Q−800 = −P − 320 P = +480 = 480 N (↓)Case (ii) Let P and Q be the downward vertical components of given ‘R’Appying Varignon’s TheoremMo = M R o+ Q × 2x = + 800 × 3x Q = 1200 N (↓)but R = Fy − 800 = − Q − P P = 800 −1200 P = −400 = 400 N (↑)Note: Negative value of ‘P’ indicates that assumed direction of P was wrong. 21

! ! Find the resultant of the parallel force system in Find the resultant of the parallel force systemfigure. shown in figure with reference to point O.Prof. Ramkumar'sR = 1000 N ( ↓ ) at 3.3 m right of O R = 50 N ( ↑ ) at 7.88 m right of O Find the single force which keeps the entire A rigid bar is subjected to a system of parallelforce system in equilibrium. forces as shown. Reduce this system toHint: First find R and at last put it in opp. direction (i) single force (ii) A single force moment systemto find R′ . at B (iii) A single force moment system at C.R′ = 45 kN ↓ at 5.89 m left of A. (i) R = 60 N ( ↓ ), At 0.783 m from 'A' (ii) 60 N ( ↓ ) and 25 Nm ( ) Resolve the force F equal to 900 N acting at B (iii) 60N ( ↓ ) and 23 Nm ( )into (a) parallel components at O and A(b) A couple and force at O. Replace the force system shown by two parallel forces at B and D. Hint: First find resultant of all forces and then solve the problem. (Dec.’04 4 marks)(a) At O, 1800 N ( ↑ ) and at A, 2700 N ( ↓ ) 6.25 kN ( ↓ ) at B and 8.75 kN ( ↓ ) at D(b) 900 N ( ↓ ) and 2700 Nm ( ) # A force of 200 N acting on a bracket as shown\" The resultant of three forces shown in figureand other two forces P and Q acting at A and B is in figure. Determine an equivalent force and couplea couple of magnitude 120 kNm clockwise.Determine the force P and Q. (May’12-5 marks) at A. (Dec.’05 5 marks) Hint: Refer Class Work Problem-7.P = 29 kN (↑) , Q = 16 kN (↓) 22

-, !Determine the resultant of the force system. !Prof. Ramkumar's A member ABC is loaded by distributed load and pure moment as shown in the figure. Find the (i) magnitude and (ii) position along AC of the resultant. (Feb.’02 – 6 marks)R = 25 kN ( → ) at 2.8 m from O upwards R = 90 kN ( ↑ ), x = 23 m right of A Find the resultant of the force system acting on The forces acting on 1 m length of the dam area body OABC, shown in figure. Also find the pointswhere the resultant will cut the x and y axis. What shown in figure. Determine the resultant forceis the distance of resultant from O. acting on the dam. Calculate the point of (Dec’10-5 marks) intersection of the resultant with the base. Hint: Length of dam is not required for solving problem (Jan’04 7 marks)R = 10 kN , θ = 36.87° R = 137.12 kN 79.9°, x = 2.91 m right of Ox = 3.33 m left of O, y = 2.5 m above O, d = 2m Replace the system of forces and couple by a A bracket subjected to forces and couples as single force couple system at A. Refer figure.shown in figure. Find value of M and its direction ifresultant is to pass through; (Feb.’03–4 marks)(1) Point A (2) Point B (3) Point C (4) Point D. (Dec’06−10 marks)(1) 334 Nm (2) 385 Nm(3) 295 Nm (4) 346 Nm 23

\" Find the resultant of four forcestangent to the circle of radius 3 m asshown in figure. Also find its location w.r.t.the centre of circle.Prof. Ramkumar's &'()*&+ Choosing centre of circle O as origin of two axes Fx = 150 − 100 sin 45 = 79.3 N(→) Fy = 50 − 80 − 100 cos 45 = −100.7N = 100.7N (↓)R = (79.3)2 + (100.7)2 = 128.1 Ntan α = 100.7 α = 51.78° 79.3 Mo = −150 × 3 + 50 × 3 − 100× 3 + 80 × 3 = − 360 = 360 Nm ( )Note: All forces are tangential to circlei.e M R will also be oApplying Varignon’s TheoremMo = M R o− 360 = −128.1× d d = 2.81 m where d = Perpendicular distance OP# Find the position and magnitude of theresultant of the force system shown in thefigure. (Dec’00)&'()*&+Fx = 20 kN (→)Fy = −20 − 40 = − 60 kN = 60 kN (↓) R = 202 + 602 = 63.25 kNtan α = 60 20 α = 71.565°Mo = − 20 × 2 − 20 2+ 2 − 40 × 3 − 40 3 = − 253.33 = 253.33 kNm M R is o∴ R should act right of O say at distance x from O Applying Varignon’s Theorem, MO = Fy × x − 253.33 = − 60 × x x = 4.222 m 24

$ A bracket is subjected to a coplanar forcesystem as shown in figure. Determine themagnitude and line of action from A of the singleresultant of the system. If the resultant is to passthrough the point B, what should be themagnitude and direction of applied couple ‘M’ ? (May’04 6marks)Prof. Ramkumar's &'()*&+ Fx = −500 cos 60 = −250 N = 250 N (←) Fy = −400 + 500 sin60 − 400 = −367 = 367 N (↓) R = (250)2 + (367)2 = 444 Ntan α = 367 250 α = 55.74° MA = + 500 sin 60 × 0.5 − 400 × 0.9 + 50 = − 93.494 = 93.494 Nm∴ M R is AHence, R will act right of A as shown Using Varignon’s Theorem, MA = M R A − 93.494 = − 367 × x x = 0.255 m2nd part: R passes through B, M R = 0 B MB = 0 + 400 × 0.5 − 400 × 0.4 + M = 0 M = − 40 = 40 NmNote: Negative value of ‘M’ indicates that assumed direction of M was wrong 25

! ! Find the resultant and its point of application ony–axis for the force system on a triangular plate asshown in figure.R = 20 N ( ← ) at 6 m from A upwardProf. Ramkumar's Replace the force system acting on a bar as Find the resultant of the force system acting onshown in figure by a single force. a body OABC as shown. Also find the points where the resultant will cut the x and y axis. (June’02–5marks) 80 kN ( ↓ ) force at 10 cm right of A. R = 2013.9 N 59.86 , x = 0.3866 m left of O, y = 0.666 m above OReplace the system of forces and couples by a Find the resultant of coplanar force systemsingle force and locate the point on the x-axis given below and locate the same on AB with due consideration to the applied moment.through which the line of action of the resultant (Dec’11-10 marks & June’03 7 marks)passes. (Dec’12-6 marks)R = 21.96N, α = 45.83° ,x = 6.73m right of O R = 510.78 N 66.95º passing through A 26

Find the position (x–axis intercept, y–axisintercept), magnitude and direction of the resultant‘R’ of only the active force acting on the A – framestructure shown in the figure. (Dec.’01–8marks)Note: Resultant of any force system is found onlyfor active forces (Reactions are never considered). R = 134.16 kN, 63.43 x = 6 m right of A, y = 12 m above A\" Replace the four forces acting on the plate asshown by a single force which will keep the forcesystem in equilibrium. Also locate its position onx–axis. Hint: First find R and at last put it in opp.direction to find R′ .Prof. Ramkumar's # Find the resultant of he given force system. R′ = 501.2 N 61.39º, x = 0.527 m left of O R = 0, Resultant Moment = 806.8 kNm$ Replace the loading on the frame by a force % For the figure shown, find resultant force andand moment at point A. moment at point A. (May’09-5 marks) (June’10-5 marks) 27