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`y2F2Resultant and Equivalent of Force SystemsA force on a body produces translation while a couple produces rotation. The forces acting on a body may be in the same plane (coplanar) or in different planes (non-coplanar) or in space ; see plate 1 and plate 2. The coplanar forces may be parallel (like or unlike) or concurrent. If a number of forces are acting on a body, Fig. 1.1, to find their resultant R (in magnitude, direction `' with horizontal and line of action), find first their algebraic sum of horizontal and vertical components i.e., X and Y, thenF3y3Rx2x3 YR  X yRigid body a1 a3 F1 x1 y1 a A a2 a4 xOy4F4 x4Fig. 1.1R=( X ) 2 + ( Y ) 2 ,tan  =Y X...(1.1)To find the line of action of R (at `a' w.r.t. a point A or a line), find the algebraic sum of the moments of all the forces about A, then MA = F1 . a1 = R . a ; a=M A R...(1.2)The force systems are illustrated in A, B &amp; C (graphical methods) and D in space, as follows:3420 N A B N = 15 N 40 N C D EEngineering Mechanicsc a R 0 a 0 b 30 N E 0 c 0 15 N 0 e d e b d Force or vector diagram OSpace diagram (ii) Unlike II forcesGusset plate20 = Q L ­ irons(angle-irons)kNQR = 46 kN 20° L ir s ons P=3 0 kN=R= 20° 46 kN P=3 0 kN (i) Parallelogram law of forces20kNResultant of two forcesR = 46 kN kN 20° 20 P=3 = 0 kN Q(ii) Triangle of forces0 B 200 N Aa 0 0 a¢ 100 NkN300 b N Block Aa C c 0 a¢N 110 R= aa¢ =bR O d ce250 N 0 D d 200 N0 60° e E 30°Force polygon (will not close) R = aa¢No. of forces on Block A= 67a R = oe e 4 a 25 kN 25 kN 25 kN 3 D C A 60° B 16 m 22 m 0 47° 0d 45° E x 0 = 43 m 25 kN c 0 0 a bbO polecR = 67 kN eSpace diagramd Vector or force diagramSee Example 1.3, Fig. 1.3 &amp; 1.4, page 8 (single equivalent tug boat, R) A . Resultant of Coplanar Non Concurrent ForcesResultant and Equivalent of Force SystemsN540Q = 300 N BA g a C 180 ­ g 180 ­ a C R = 500 N a B C A = = Sin a Sin b Sin g B W RA RBRBA b Block AgR90° = 2 60° W = 400 N 00 N30° 60° of forces Spherical ball of mass 40 kg rests in a smooth Dr groove W = RA = RB Sin 90° Sin 60° Sin 30°Lami's Theorem:Block A in equilibrium under three forcesA(t)AkN10 kN BkNtie45°(C)18150 N D E 180 N a E R a¢ bA B160 Njib272m30°10 kNCD of forces at AJib craneC30°B200 N R = 150 N = aa¢ cd No. of concurrent forces R = Resultant = aa¢ E = Equilibrant = a¢aPLATE 1 B. Equilibrium Under Coplanar Concurrent ForcesA=3P = 400 Nb180 ­ b30°D6f 1 kN C 2.52.2Engineering Mechanics1 kN D 1.5 kN4RA a b c d eII t1 kNB 2.8REoclosinglineRA=1.7 kNA1.7 0 H 0G 0 0 F 0 a b 4m 4m Cl os ing lin eI 1.1J1.4m 0 dO pole a b c df, g, h, i0 c 0e E 4m RE = 3.8 kNFunicular polygonVector diagram or Force polygon to find reactionsj(i) Truss ­ Space diagram (with Bow's notation)e Superposition of force Maxwell or º polygons at each joint Force diagram[See Ex. 3.1 &amp; Fig. 3.1, page 30]RB = 8 kN B8(c) e (C) = E (D)3.5h1.1J3.5 H3.5GI 1.7f i E 3 kN a RARA = 7 kNA6.3(ii) Cantilever Truss ­ Space diagram (with bow's notation)[See Ex. 3.3 &amp; Fig. 3.14, page 34] C. Equilibrium Under Coplanar Non Concurrent Forces5.8RBF 2 kNj gMaxwell diagram gives also reactions in cantilever trussesDz TA = ? T B = ? TC = ?z 40 kN O (x, y) of R = ? R.C.C. mat 20 kN 30 kN 15 kN4m C A G2mxy2m 3m2.5 m2my2mB W = 50 kN2mm x 2.5 Column loads(i) Concurrent forces(ii) Parallel forcesResultant and Equivalent of Force Systemsz R ­ C at O = ? z B Cables 3m Em 1.5750 kN (in x-y plane) TE, TD = ? RA = ? C 20 kN crate y x Forces for f = 0 : Coplanar concurrent at C Non Coplanar concurrent at B (iv) Non concurrent, Non parallel forces 30° x 1m 1m 20 1m 1m 20 kN kN-m 60 kN (in x-z plane) (II to y-axis)3m 2m A3m1.5mfD y(iii) Space frame-DerrickD. Spatial Force System [Forces in three-dimensions] PLATE 2Also,MA = (X)y ­ (Y)x...(1.3)This Eqn. can be solved for the ratio of y to x corresponding to any point `O' along the line of action of R as indicated by the dashed line, that is the force can be applied at any point on its line of action. This is called the principle of tansmissibility of force. From Eqs. (1.2) and (1.3), R . a = (X )y ­ (Y ) x This is called `Varignon's Theorem'. Example 1.1. Determine the resultant of the forces in Fig. 1.1. ...(1.4) i.e., the moment of the resultant is equal to the algebraic sum of the moments of its components.Fig. 1.1Solution: + X = 150 ­ 100 sin 45° = 79.29 N  +  Y = 50 ­ 8 ­ 100 cos 45° = ­ 28.71 or 28.71 N  R= tan  =79.29 2 + 28.712 = 84.32 N28.71 = 0.3621, 79.29 = 19.9°8Moment of the resultant (at a distance `a' from C) = Moment of the components (about C) R . a = MC + MC = 150 × 1 + 100 × 1 ­ 50 × 1 ­ 8 × 1 = 192 N.m a=Engineering MechanicsFig. 1.1aMC 192 = = 2.27 mm R 84.32Example 1.2. The resultant (100 kN) of four forces and three of these are shown in Fig. 1.2. Determine the fourth force.Fig. 1.2Solution:+ X = ­ 70 cos 60° ­ 120 cos 60° + 50 cos 50°= ­ 62.88 = ­ 81.6 or 62.88 kN or 81.6 kN F4 Y4 q X4+  Y = 70 sin 60 ­ 120 sin 60° ­ 50 sin 50° RX = 100 cos 45° = 70.71 kN RY = 100 sin 45° = 70.71 kN X4 = 70.71 ­ (­ 62.88) = 133.59 kN  Y4 = 70.71 ­ (­ 81.6) = 152.31 kN  F4 = tan  =133.59 2 + 152.312 = 202.6 kN, Fig. 1.2aFig. 1.2a152.31 = 1.14,  = 48.746°. 133.59Example 1.3. (a) State `Varignon's Theorem'. (b) Four tugboats exert 25 kN each (as shown in Fig. 1.3) to bring an ocean liner to the pier. Determine the point on the hull where a single, more powerful tugboat should push to produce the same effect as the original four boats.Resultant and Equivalent of Force SystemsBoat - 1 4 3 60° 16 m 22 m 36 m 63 m 66 m 30 m 30 m Boat - 2 Boat - 39Hull of ship 11 Boat - 4Fig. 1.3Solution: (a) The Varignon's (French Mathematician) Theorem states that &quot;if a number of coplanar forces acting on a body, the algebraic sum of the moments of all the forces about any point is equal to the moment of their resultant about the same points, i.e. F1. a1 = R.a&quot; or simply ``the moment of a force about any point equals the algebraic sum of the moments of its components about the same point, i.e. F . a = X . x + Y . y, where F =x 2 + y 2 &quot;.(b) To find the resultant `R' due to the four forces acting on the hull of the ship, Fig. 1.4.Y 25 kN F1 60° 12.5 O 16 m 25 kN F2 5 4 3 15 22 m 63 m 43 m 66 m a SY x R q SX 22 m 45° 30 m 30 m X SX = 45.1821.652025 kN F3 SY = 48.97q = 47.3 R = 66 kNFig. 1.4(a)17.6825 kN 1 1 F4Fig. 1.4+ X = 12.5 + 15 + 17.68 = 45.18 kN +  Y = ­ 21.65 ­ 20 ­ 25 + 17.68 = ­ 48.97 or 48.97 kN  R=45.18 2 + 48.97 2 = 66.63 kN, Fig. 1.4(a)17.6810 = tan­1Engineering Mechanics48.97 = 47.3°. 45.18Let the line of action of R cut the X-axis at x. Taking moments about `O' (and using Varignon's theorem), M0 = X.y + Y.x, + M0 = 48.97 × x y=0+ M0 = 12.5 × 16 + 15 × 22 + 20 × 63 + 25 × 159 ­ 17.68 × 22 ­ 17.68 × 129 = 3095.32 kN.m 48.97 x = 3095.32, x=3095.32 = 63.2 m 48.97From Fig. 1.4, the resultant R will intersect the hull at a distance of (x ­ a) from O : tan  =22 = 1.084, aa = 20.3 m,63.2 ­ 20.3  43 mi.e., where a single, more powerful tug boat should push, exerting a force of 66.63 kN. See the graphical solution in Plate 1­A, page 5. Example 1.4. Find the resultant of the four forces acting as shown in Fig. 1.5.50 10070.7 kN 200 kN45° 50 70.7 kN 45° 200 kN 30° a A A 1m 1m 30° 100 kN 80 kN 50 x 1m30° 173.2 q 1.5 m 80 R 3091.5 mFig. 1.6(a)86.6 30° SY 100 kN q SX R80 kNFig. 1.5Fig. 1.6Solution: See Fig. 1.6+ X = 50 + 173.2 + 86.6 = 309.8 kN +  Y = ­ 50 + 100 ­ 50 ­ 80 = ­ 80 or 80 kN Resultant and Equivalent of Force SystemsR=x 2 + y 2 = 309.8 2 + 80 2 = 320 kN, Fig. 1.6(a)11 = tan­180 = 14.48° 309.8To find the line of action of the resultant R, take moments about A (and applying Varignon's theorem), MA = R . a = X.y + Y.x, +  y = 0, MA = 50 × 1.5 ­ 100 × 1 + 173.2 × 1.5 + 50 × 1 = 284.8 kN a=M A 284.8 = = 0.89 m, R 320x=M A 284.8 = = 3.56 m 80 YNote: a = x sin  = 3.56 sin 14.48° = 0.89 m Example 1.5. A rigid bar AB is subjected to a system of parallel forces as shown in Fig. 1.7. Reduce the given system of forces to an equivalent (a) single force, (b) force and moment at A, (c) force and moment at D, and (d) force and moment at B.25 kN A 1m C 2m D 2m 15 kNB10 kN20 kNFig. 1.7Solution: (a) For an  single force on AB: +  Resultant R = 10 ­ 25 + 20 ­ 15 = ­ 10 or 10 kN  To find its location : MA = R . a +  MA = 25 × 1 ­ 20 × 3 + 15 × 5 = 40 kN.m a=M A 40 = = 4 m, Fig. 1.8 R 10Fig. 1.8  Single Force on AB To find  force and couple at any point other than E, introduce equal and opposite forces R at the required point, which will not affect the equilibrium:12Engineering Mechanics(b) A force R at E may be treated as another force at A and a couple as shown in Fig. 1.9(b)Fig. 1.9.  Force and Moment at A(c) A force at E  another force at D and a couple as shown in Fig. 1.10(b)R 4m A 3m R R = 10 kN A 3m D 10 × 1 = 10 kN.m (b) B D 1m E (a) R = 10 kNFig. 1.10  Force and Moment at D(d) A force at E  another force at B and a couple as shown in Fig. 1.11(b)R = 10 kN R 4m A (a) E 1m R 10 kN A (b) 10 × 1 = 10 kN.m = M B BFig. 1.11  Force and Moment at BFrom Fig. 1.7, +MB = 10 × 5 ­ 25 × 4 + 20 × 2 = ­ 10 i.e., 10 kN.mThus, the equivalent system at any point other than E is the same resultant force R and a moment which is equal to the algebraic sum of the moments of all the forces acting on the bar AB about that point.Resultant and Equivalent of Force SystemsNote: A single force F (acting at A in Fig. 1.12) can be looked upon as a force of the same magnitude and direction shifted parallel to itself (by AB = a) accompanied by a couple of moment (M) = force × shift (= F . a) i.e., F at A = F at B + F . a13Contrariwise, a force, F and a couple of moment M (= F . a) acting at point (B) is equivalent to a single force (F) of the same magnitude and direction shifted parallel to its original line of action, the shift being =M (i.e., = `a' in Fig. 1.12). FFig. 1.12Example 1.6. Replace the system of a force and couple shown in Fig. 1.13 by a single force.100 N 50 N0.4 m 50 N x 100 N 100 NFig. 1.13Fig. 1.14Fig. 1.15Solution: Since a force  another force (shifted suitably) and a couple, Fig. 1.14, shift `x' to counteract the existing couple : 100 × x = 50 × 0.4 x = 0.2 m or 1.8 m from the support The equivalent system is as shown in Fig. 1.15. Example 1.7. Replace the system of a force and couple shown in Fig. 1.16 by a single force on the line AB.120 Nr= 10 0mA96AO50 mmO C 60°B120 NBFig. 1.16Fig. 1.17Fig. 1.18Fm60°0N14shift `x' to counteract the existing couple: 960 × x = 120 × 0.2 x = 0.025 m or 25 mm ; OC = The equivalent system is as shown in Fig. 1.18.Engineering MechanicsSolution: Since a force  another force (shifted suitably) and a couple Fig. 1.17,25 = 50 mm cos 60°Example 1.8. An angle is subjected to a force-couple system as shown in Fig. 1.19. Reduce the system (a) to an equivalent system at A, (b) as a single resultant anywhere. Solution: (a) At A, X = 80 cos 45° = 56.56 N  Y = 50 + 80 sin 45° ­ 150 = ­ 43.44 or 43.44 N  R= at +56.56 2 + 43.44 2 = 71.32 N = tan­143.44 = 37.53° 56.56M = ­ 300 ­ 80 sin 45° × 3 + 150 × 4 = 130.32 NmHence, the equivalent system at A is a `force-couple system' of R and M, Fig. 1.20.50 N50 NB3m300 Nm 80 N q R 45° 3m300 Nm80 N2.3 m a q M RA 3m45° 1m 150 NA1m 150 NFig. 1.19Fig. 1.20(b) The force-couple system at A can be reduced to a single resultant R acting at B, Fig. 1.20, in the same direction at a distance, a=130.32 Nm M = = 1.83 m R 71.32 N a 183 . = = 2.3 m cos  cos 37.53°AB =Principle of Transmissibility. The point of application of a force may be transmitted along its Line of Action to another point, without changing its effect on any rigid body. ThisResultant and Equivalent of Force Systemsis known as the principle of transmissibility of force and it enables to regard a force acting on a rigid body as a Sliding Vector. In Fig. 1.21. the force P on the body can be applied at A or B, without changing the reactions at the bearing support at C or roller support at D.P C Pinned RCH RCY RD B A D P15RollerFig. 1.21PROBLEMS1.1. Six forces are acting along the sides of a regular hexagon of side 100 mm as shown in Fig. P 1.1. Find the resultant and its distance from A. [185. 2N, 57.32° , 276.5 mm]2 4 2A 1 3m 22Loads in kN 2 3m 13m3mBFig. P 1.1 intersect AB.Fig. P 1.2 [15.46 kN, 75° ; 4.4 m from A]1.2. For the loaded truss shown in Fig. P 1.2, find the resultant load and where its line of action will 1.3. Determine the resultant of the forces acting on the dam shown in Fig. P 1.3 and locate its intersection with the base AB. For safe design, the intersection should occur within the middle-third of the base. Is the design safe ?2m MWL[137.12 kN,79.92° ; 2.15 m from B ; Yes]DamP = 50 kN 2m Ac.g. TWL Q = 30 kN W = 120 kN 5m Base of dam 60° B 1mFig. P 1.3161.4.Engineering Mechanics(a) Determine completely the resultant of the five forces shown in Fig. P 1.4. The squares are 100 mm × 100 mm. (b) Give an equivalent system at `O'.Y[0.8 N , Y = 67 m ; 0.8 N , 53.6 Nmat O]20 0100NN200N100NX O 282 NFig. P 1.4 1.5. A force of 500 N is acting at A in Fig. P 1.5 produces a moment of 1200 Nm about O. Find the y-intercept of the force. Give an equivalent system with the same force acting at `O'. [3 m ; 500 N and 1200 Nm at O]Fig. P 1.5Fig. P 1.61.6. A 50 N force is applied to a corner plate as shown in Fig. P 1.6. Determine an equivalent forcecouple system at A. [50 N, 3.08 Nm] 1.7. Find the single resultant of the forces acting on the pulley in Fig. P 1.7. Give its intercept with the axes. [1020 N ], (0.1, 0.02) m] 1.8. An electric light fixture weighing 15 N is hung from two strings from the roof and wall as shown in Fig. P 1.8. Find the tension in the strings. [7.8, 11 N]Resultant and Equivalent of Force SystemsRoof 60° Wall 45°17Light 15 NFig. P 1.7Fig. P 1.8`

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