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Resultant and Equivalent of Force Systems A 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 ‘θ’ with horizontal and line of action), find first their algebraic sum of horizontal and vertical components i.e., ΣX and ΣY, then F y2 2 F3 y3 R x2 x3 ΣY R Rigid θ body O ΣX a1 y a a2 a3 a4 F1 A y1 x x1 y4 F4 x4 Fig. 1.1 ΣY R= ( ΣX ) 2 + ( ΣY ) 2 , tan θ = ...(1.1) ΣX 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 ΣM A ΣMA = ΣF1 . a1 = R . a ; a= ...(1.2) R The force systems are illustrated in A, B & C (graphical methods) and D in space, as follows: 3 4 Engineering Mechanics N = 15 N 20 N 40 N c A B C D E a R 30 N E 15 N e O 0 0 0 0 b a b c 0 e d d Space diagram Force or vector diagram (ii) Unlike II forces kN Gusset 20 = plate Q L – irons kN (angle-irons) 20 = R= Q 20° 46 kN R P=3 = 46 kN 0 kN 20° L ir s ons P=3 (i) Parallelogram 0 kN law of forces R = 46 kN 20° kN P=3 20 Resultant of 0 kN = Q two forces (ii) Triangle of forces a b 0 300 a¢ b N C 0 R c N 110 R= aa¢ e B Block A = O 200 N 250 N Aa 0 d c 0 D 0 0 60° d Force polygon a¢ e (will not close) 100 30° E 200 N R = aa¢ N No. of forces on Block A kN 67 = a R = oe a e 4 25 kN 25 kN 25 kN 3 A C D O 60° B pole 16 m 22 m b 0 47° E 45° 0d x R = 67 kN = 43 m 25 kN 0 c 0 0 c a e b d Space diagram Vector or force diagram See Example 1.3, Fig. 1.3 & 1.4, page 8 (single equivalent tug boat, R) A . Resultant of Coplanar Non Concurrent Forces Resultant and Equivalent of Force Systems 5 N 30° 40 P = 400 N b 180 – b =3 Q = 300 N A B A R g 90° A g R C W B = 2 60° W = 400 N RA RB 00 b a N 180 – g 180 – a B 30° 60° Spherical ball of of forces D Block A C mass 40 kg a R = 500 N rests in a smooth Dr groove A B C W = RA = RB Lami’s Theorem: = = Sin 90° Sin 60° Sin 30° Sin a Sin b Sin g Block A in equilibrium under three forces A A (t) A 160 N kN (C) tie 150 N 18 kN 10 kN B D 45° 27 B E jib B 180 N 200 N 30° 10 kN b R = 150 N 30° 2m a = aa¢ C C D of forces at A E R c a¢ Jib crane d No. of concurrent forces R = Resultant = aa¢ E = Equilibrant = a¢a PLATE 1 B. Equilibrium Under Coplanar Concurrent Forces 6 Engineering Mechanics f RA II t 1 kN 1 kN oc C los D 1.5 kN a RE ing 1 kN B 2.5 lin e 2.8 4m 0 b I J 2.2 O f, g, h, 1.1 4 1.7 0 H 1. d pole c NA 0G 0 0 a i .7 k 0 F 0 c 0e =1 a b E d RA 4m 4m 4m Funicular e b Cl polygon os RE = 3.8 kN Vector diagram or Force ing c lin polygon to find reactions j e d (i) Truss – Space diagram e (with Bow's notation) Superposition of force Maxwell or º polygons at each joint Force diagram [See Ex. 3.1 & Fig. 3.1, page 30] RB = 8 kN (c) B e h 8 (C) = E (D) 3.5 J H 3.5 1.1 f 5.8 3.5 G I R i B 1.7 E RA j A 6.3 F RA = 7 kN 3 kN a 2 kN g (ii) Cantilever Truss – Space diagram Maxwell diagram (with bow's notation) gives also reactions in cantilever trusses [See Ex. 3.3 & Fig. 3.14, page 34] C. Equilibrium Under Coplanar Non Concurrent Forces z z D (x, y) of R = ? 40 kN TA = ? T B = ? TC = ? O R.C.C. mat 4m 20 kN C 30 kN 15 kN G A 2mx y 2m m x 2m 2.5 Column 3m y 2m 2m m loads B 2.5 W = 50 kN (i) Concurrent forces (ii) Parallel forces Resultant and Equivalent of Force Systems 7 z R – C at O = ? z B 50 kN (in x-y plane) Cables TE, TD = ? 30° x 3m RA = ? 1m E 1m 1m 1m 20 3m 20 kN 60 kN kN-m m C 3m (in x-z plane) 1.5 2m 20 kN crate (II to y-axis) f x y m A Forces for f = 0 : 1.5 D Coplanar concurrent at C Non Coplanar concurrent at B y (iii) Space frame-Derrick (iv) Non concurrent, Non parallel forces D. Spatial Force System [Forces in three-dimensions] PLATE 2 Also, ΣMA = (ΣX)y – (ΣY)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 = (ΣX )y – (ΣY ) x ...(1.4) i.e., the moment of the resultant is equal to the algebraic sum of the moments of its components. This is called ‘Varignon’s Theorem’. Example 1.1. Determine the resultant of the forces in Fig. 1.1. Fig. 1.1 → Solution: + ΣX = 150 – 100 sin 45° = 79.29 N → + ↑ ΣY = 50 – 8 – 100 cos 45° = – 28.71 or 28.71 N ↓ R= 79.29 2 + 28.712 = 84.32 N 28.71 tan θ = = 0.3621, θ = 19.9° 79.29 8 Engineering Mechanics Moment of the resultant (at a distance ‘a’ from C) = Moment of the components (about C) R . a = ΣMC + ΣMC = 150 × 1 + 100 × 1 – 50 × 1 – 8 × 1 Fig. 1.1a = 192 N.m ΣMC 192 a= = = 2.27 mm R 84.32 Example 1.2. The resultant (100 kN) of four forces and three of these are shown in Fig. 1.2. Determine the fourth force. Fig. 1.2 → Solution: + ΣX = – 70 cos 60° – 120 cos 60° + 50 cos 50° = – 62.88 or 62.88 kN← + ↑ ΣY = 70 sin 60 – 120 sin 60° – 50 sin 50° F4 = – 81.6 or 81.6 kN ↓ Y4 RX = 100 cos 45° = 70.71 kN q RY = 100 sin 45° = 70.71 kN X4 X4 = 70.71 – (– 62.88) = 133.59 kN → Y4 = 70.71 – (– 81.6) = 152.31 kN ↑ Fig. 1.2a F4 = 133.59 2 + 152.312 = 202.6 kN, Fig. 1.2a 152.31 tan θ = = 1.14, θ = 48.746°. 133.59 Example 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 Systems 9 Boat - 1 Boat - 2 Boat - 3 4 3 60° 16 m 22 m 36 m 30 m 30 m 63 m 66 m Hull of ship 1 1 Boat - 4 Fig. 1.3 Solution: (a) The Varignon’s (French Mathematician) Theorem states that “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. ΣF1. a1 = R.a” 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 ”. (b) To find the resultant ‘R’ due to the four forces acting on the hull of the ship, Fig. 1.4. Y 25 kN SX = 45.18 25 kN F2 5 q = 47.3 F1 4 25 kN F3 21.65 20 3 SY = 48.97 R = 66 kN 60° 15 12.5 22 m 16 m 63 m 66 m 30 m 30 m Fig. 1.4(a) O X 43 m a q SX 22 m SY x 17.68 45° R 25 kN 1 17.68 1 F4 Fig. 1.4 → + ΣX = 12.5 + 15 + 17.68 = 45.18 kN → + ↑ ΣY = – 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) 10 Engineering Mechanics 48.97 θ = tan–1 = 47.3°. 45.18 Let the line of action of R cut the X-axis at x. Taking moments about ‘O’ (and using Varignon’s theorem), ΣM0 = ΣX.y + ΣY.x, y=0 + ΣM0 = 48.97 × x + ΣM0 = 12.5 × 16 + 15 × 22 + 20 × 63 + 25 × 159 – 17.68 × 22 – 17.68 × 129 = 3095.32 kN.m 3095.32 48.97 x = 3095.32, x= = 63.2 m 48.97 From Fig. 1.4, the resultant R will intersect the hull at a distance of (x – a) from O : 22 tan θ = = 1.084, a = 20.3 m, 63.2 – 20.3 ≈ 43 m a i.e., where a single, more powerful tug boat should push, exerting a force of 66.63 kN. See the graphical solution in Plate 1–A, page 5. Example 1.4. Find the resultant of the four forces acting as shown in Fig. 1.5. 70.7 kN 200 kN 50 100 45° 30° 50 1m 173.2 309 70.7 kN 200 kN q 45° 1.5 m R 30° 80 1.5 m a Fig. 1.6(a) 86.6 A SX 1m 30° q A 1m 30° x SY R 80 kN 100 kN 80 kN 100 kN 50 Fig. 1.5 Fig. 1.6 Solution: See Fig. 1.6 → + ΣX = 50 + 173.2 + 86.6 = 309.8 kN → + ↑ ΣY = – 50 + 100 – 50 – 80 = – 80 or 80 kN ↓ Resultant and Equivalent of Force Systems 11 R= Σx 2 + Σy 2 = 309.8 2 + 80 2 = 320 kN, Fig. 1.6(a) 80 θ = tan–1 = 14.48° 309.8 To find the line of action of the resultant R, take moments about A (and applying Varignon’s theorem), ΣMA = R . a = ΣX.y + ΣY.x, y = 0, + ΣMA = 50 × 1.5 – 100 × 1 + 173.2 × 1.5 + 50 × 1 = 284.8 kN ΣM A 284.8 ΣM A 284.8 ∴ a= = = 0.89 m, x= = = 3.56 m R 320 ΣY 80 Note: a = x sin θ = 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 15 kN A C D B 1m 2m 2m 10 kN 20 kN Fig. 1.7 Solution: (a) For an ≡ single force on AB: + ↑ Resultant R = 10 – 25 + 20 – 15 = – 10 or 10 kN ↓ To find its location : ΣMA = R . a + ΣMA = 25 × 1 – 20 × 3 + 15 × 5 = 40 kN.m ΣM A 40 ∴ a= = = 4 m, Fig. 1.8 R 10 Fig. 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: 12 Engineering 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 R = 10 kN 4m D E A 3m 1m (a) R R = 10 kN D A B 3m 10 × 1 = 10 kN.m (b) Fig. 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 E A B (a) 1m R 10 kN A B (b) 10 × 1 = 10 kN.m = M Fig. 1.11 ≡ Force and Moment at B From Fig. 1.7, + ΣMB = 10 × 5 – 25 × 4 + 20 × 2 = – 10 i.e., 10 kN.m Thus, 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 Systems 13 Note: 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 . a Contrariwise, 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 M line of action, the shift being = (i.e., = ‘a’ in Fig. 1.12). Fig. 1.12 F Example 1.6. Replace the system of a force and couple shown in Fig. 1.13 by a single force. 100 N 50 N 0.4 m 50 N x 100 N 100 N Fig. 1.13 Fig. 1.14 Fig. 1.15 Solution: 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 N A A N 0 96 r= 10 0m m 60° F O 50 O mm 60° C 120 N B B Fig. 1.16 Fig. 1.17 Fig. 1.18 14 Engineering Mechanics Solution: Since a force ≡ another force (shifted suitably) and a couple Fig. 1.17, shift ‘x’ to counteract the existing couple: 960 × x = 120 × 0.2 25 x = 0.025 m or 25 mm ; OC = = 50 mm cos 60° The equivalent system is as shown in Fig. 1.18. 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, ΣX = 80 cos 45° = 56.56 N → ΣY = 50 + 80 sin 45° – 150 = – 43.44 or 43.44 N ↓ R= 56.56 2 + 43.44 2 = 71.32 N 43.44 at θ = tan–1 = 37.53° 56.56 + ΣM = – 300 – 80 sin 45° × 3 + 150 × 4 = 130.32 Nm Hence, the equivalent system at A is a ‘force-couple system’ of R and M, Fig. 1.20. 50 N 50 N B 300 Nm 3m 80 N 300 Nm 80 N 2.3 m q R a A 45° q 45° A M R 3m 3m 1m 1m 150 N 150 N Fig. 1.19 Fig. 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, M 130.32 Nm a= = = 1.83 m R 71.32 N a . 183 AB = = = 2.3 m cos θ cos 37.53° 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 Systems 15 is known as the principle of transmissibility of force and it P P B A 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 Pinned C D Roller at A or B, without changing the reactions at the bearing RCH RD support at C or roller support at D. RCY Fig. 1.21 PROBLEMS 1.1. 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 2 Loads in kN 4 2 2 3m 2 1 1 A 3m 3m 3m B Fig. P 1.1 Fig. P 1.2 1.2. For the loaded truss shown in Fig. P 1.2, find the resultant load and where its line of action will intersect AB. [15.46 kN, 75° ; 4.4 m from A] 1.3. Determine the resultant of the forces acting on the dam shown in Fig. P 1.3 and locate its inter- section with the base AB. For safe design, the intersection should occur within the middle-third of the base. Is the design safe ? [137.12 kN, 79.92° ; 2.15 m from B ; Yes] 2m MWL Dam c.g. P = 50 kN TWL Q = 30 kN 2m W = 120 kN A 60° 1m 5m B Base of dam Fig. P 1.3 16 Engineering Mechanics 1.4. (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’. [0.8 N →, Y = 67 m ; 0.8 N →, 53.6 Nm at O] Y 100 N N 0 20 20 0 N 100 N X O 282 N Fig. 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 1.5 Fig. P 1.6 1.6. A 50 N force is applied to a corner plate as shown in Fig. P 1.6. Determine an equivalent force- couple 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 Systems 17 Roof 60° Wall 45° Light 15 N Fig. P 1.7 Fig. P 1.8