3.18 | Engineering Mechanics Test 2 17. C A RA (180 × 9.81) N RE E Free Diagram of ACE: ΣME = 0 ⇒ RA × 0.21 – 1765.8 × 0.09 = 0 ⇒ RA = 756.77 N ~ 757 N Free body diagram of BD: 10 cm ∑MB = 0 BV Bh B Dh DV D 5.464 cm (180 × 9.81) N ΣMB = 0 ⇒ DV × (10 + 5.464) + (180 × 9.81 × 10) = 0 or DV = 1142 N Choice (A) 18. Since velocity is uniform, therefore acceleration is zero. 20° µR R (50 × 9.81) N P 6 N y x 30° 10° ΣFx = max ⇒ ΣFx = 0 {∵ ax = 0} and ΣFy = 0 + 6 Cos 10o + P Cos20o – µR – (50 × 9.81 × Sin 30o ) = 0 or P Cos20o – 0.3 R = 239.341 → (1) and 6 Sin10o + R – P Sin20o – (50 × 9.81 × Cos30o ) = 0 or P Sin20o – R = – 423.743 → (2) From equation (1) and (2) we get, P = 437.785 N and R = 573.474 N ∴ P = 437.785 N ~ 438 N Choice (C) 19. O 30° α 30° mg at A ΣFt = –mg Sin 30o or m(at ) = –mg Sin θ ⇒ m(a × OA) = –mg Sin θ ⇒ a = Sin g OA q − Now a = ddd d dt dt d d wwq w w q q = ×= or Sin d g d OA w w q q − = o o 45 Sin 2.83 30 g d d OA w ww q q − ⇒ = ∫ ∫ 1 2 2 9.81 o o 2.83 = + Co45 -Cos30 2 2 w − w = 2.54 rad/sec Choice (A) 20. A B Tangential component of the linear acceleration of point A (at )A = rA a ⇒ 8 = (0.1) × a or a = 80 rad/sec2 Angular acceleration of the system, a = 80 rad/sec Now (at )B = a × rB = 80 × 0.15 = 12 m/s2 Now wA = 25 250 rad/s 0.1 A A V r = = Now wA = wB = 250 rad/s Normal component of B = r w2 = 0.15 × 2502 = 9375 m/s2 Magnitude of the linear acceleration, a = 2 2 9375 12 + = 9375 m/s2 Choice (D) 21. ΣFx = P – F = max --------- (1) ΣFy = R – mg = 0 Taking moments about R ΣM = P × h + F × R = Ia or P × h + F × R = 1 2 mR2 a ⇒ P × h + F × R = 1 2 mRax {∵ ax = R × a} Dividing the above equation with R/2 we get 2 2 x Ph F ma R + = ------- (2) Equating right side of equation (1) and (2) we get
Engineering Mechanics Test 2 | 3.19 2 2 Ph FPF R + =− or 3F = 2 1 h P R − F = 0 when 1 – 2h R = 0 or h = 2 R Choice (C) 22. I is the centre of rotation of rod AB. ∴ VA = VAO = 12 m/s = VAI ∴ ωAB × AI = 12 ⇒ ωAB × 1.2 = 12 ⇒ ωAB × 10 rad/sec ∴ VB = ωAB × BI = 10 × 1.6 = 16 m/s Velocity diagram: a 12 30° 60° i 16 b 2 2 ab 12 16 → ∴= + = 20 m/s Choice (A) 23. Since the cylinder rolls without slipping, the spring becomes stretched (0.15 × 2) m when the center of the cylinder moves 0.15 m to the right. The work is U = ( ) ( ) 2 2 2 1 1 2 − − +× kx x F s 1 2 730 0.3 0 40 9.81 0.15 2 − = × × − + × × = 26.01 N-m Choice (B) 24. Initial kinetic energy is zero. Hence the change in kinetic energy, ∆KE = KE2 – KE1 = 1 1 2 2 2 2 mv I + ow 2 1 1 2 2 50 2 22 mr KE v w ⇒∆ = × × + 2 2 2 1 50 0.4 . 50 2 4 0.4 v KE v × ⇒∆ = × × + × Now U = K.E2 – K.E1 ∴ 26.01 = 2 50 50 2 2 4 v + v ⇒ v = 0.833 m/s Choice (C) 25. Weight of the hanging part of the chain = mg 4 10 9.81 N 4 × = = 24.525 N = Maximum weight to be lifted. When the entire hanging portion has been pulled, the weight to be lifted equals zero. ∴ Average weight to be pulled = 1 2 [24.525 + 0] = 12.2625 N Work done = Average force × distance moved = 12.2625 × 1 4 = 3.065 N-m or J Choice (A) 26. ΣMhinge = Io × a [mg × (0.5 + 0.1)] = ( ) 2 2 2 5 mR m OG + × a ∴ 0.6 × 9.81 = 2 2 2 0.1 0.6 5 × + × a ⇒ a = 16.17 rad/sec2 Choice (C) 27. By the method of joints for equilibrium at point A. ΣX = ΣY = 0 ΣX = S4 + S5 Cos45 + S6 Cos45 = 0 ΣY = − P + S5 Sin45 – S6 Sin45 = 0 At point D ΣX = − S6 Cos45 + S3 Cos45 = 0 ⇒ S6 = S3 ΣY = S6 Sin45 + S3 Sin45 – RD = 0 At point C ΣX = − S4 – S2 Cos45 – S3 Cos45 = 0 ΣY = S2 Sin45 – S3 Sin45 = 0 ⇒ S2 = S3 ∴ S2 = S3 = S6 At point B ΣX = S1 – S5 Cos45 + S2 Cos45 = 0 ΣY = − S5 Sin45 – S2 Sin45 = 0 ⇒ S5 = ₋S2 ∴ S2 = S3 = − S5 = S6 From ΣY equation at point A P = S5 Sin45 – S6 Sin45 = 2S5 Sin45 = 1.414 S5 From ΣX equation at point B S1 = S5 Cos45 – S2 Cos45 = 2S5 Cos45 = 1.414 S5 = P ∴ S1 = P Choice (B) 28. As the floor is smooth, there are only vertical reactions at C and D Taking moments at C RD.a – (aa).P = 0 ⇒ RD = aP. Taking moments at D RC.a + (a − aa)P = 0 ⇒ RC = (1 − a)P Taking separate free body diagrams for the legs AD and BC we get (Taking the reactions instead of the force). A Ye Xe D C B E Xe Ye αP (1 – α)P
3.20 | Engineering Mechanics Test 2 For the bar AD Taking moments around A (aP)a + Ye (a/2) – Xe (a/2) = 0 → (1) Moments around B − a.(1 − a)P + Ye (a/2) + Xe (a/2) = 0 → (2) Adding (1) and (2) aP – (1 − a)P + Ye = 0 ⇒ Ye = (1 − 2a)P ⇒ Xe = P ∴ The resultant on point E is ( ) 22 2 2 2 R XY P P 1 2 e ee = + = +− a Re = P ( ) 2 1 12 + − a a has the range 0 to 1 ∴ maximum value for Re is at When a = 0, Re = P 2 and a = 1, Re =P 2 Choice (C) 29. Tcosθ θ mrω2 mg T R T Cos θ = mg and T Sin θ = mr ω2 2 Tan 30 9.81 Tan 1.683rad sec 2 r g w q w × ⇒ = ⇒= = 4m 23 N ∴ mr1 w2 = 23 N, r1 = 4 m ∴ m × 4 × 1.6832 = 23 m = 2.03 kg ∴ T Cos θ = mg ⇒ T = mg 2.03 9.81 23 N Cos Cos30 q × = = Choice (A) 30. h R S m m = 5 kg, h = 100 mm = 0.1 m: S = 5 mm = 5 × 10–3 m R = 0.54 N/mm Energy required to push the nail into the floor = R × S = 1.032 × 5 = 5.16 N Energy offered by hammer = (KE)impact + (PE) penetration KE during impact = 1 2 2 mv where v = 2gh PE during penetration = (M + m)gS, where M is the mass of the nail 5.16 = 1 m 2 (2gh) + (M + m)gS 5.16 = 1 5 2 9.81 0.1 2 ××× × + (5 × 9.81 × 0.005) + (M × 9.81 × 0.005) ⇒ M = 198.77 gms Choice (C) 31. W = 30 N, f = 30o By drawing the vector diagram for the forces taking the weight of the block W vertically and the reactive force making an angle of f = 30o with the vertical we get Pmin R W φ The minimum distance to complete the triangle is the perpendicular from the head of R to the tail of W. ∴ Pmin = W Sin j = 30 Sin 30 = 15 N Choice (D) 32. P 90 – φ φ R W α Considering the vectors P and W (90 – f) + 90 + a = 180o ⇒ a = f Choice (C) 33. 6 N 2 N 4 m/s 8 m/s + Ve e = 0.5 = − − v v 1 2 u u 2 1 here v1 , u1 are for 6 N ball and v2 , u2 are for 2 N ball. ⇒ v1 – v2 = 0.5 × (– 8 – 4) = – 6 m/s ⇒ v1 – v2 = –6 ⇒ v2 = v1 + 6 by conservation of momentum
Engineering Mechanics Test 2 | 3.21 m1 u1 + m2 u2 = m1 v1 + m2 v2 6 × 4 + 2 × (– 8) = 6v1 + 2v2 8 = 6v1 + 2(6 + v1 ) – 4 = 8v1 ⇒ v1 = – 0.5 m/s (for e = 0.5) When the impact is elastic e = 1 ⇒ v1 – v2 = – 12 v2 = v1 + 12 From momentum equation, 8 = 6v1 + 2v2 8 = 6v1 + 2 (v1 + 12) – 16 = 8v1 v1 = −16 8 = – 2 m/s (for e = 1) ∴ (v1 ) e=0.5:(v1 )e=1 = − = − 0.5 0.25 2 Choice (B) 34. r P A R W a By applying summation of moments about point A ΣMA = 0 ⇒ W × a – P × r = 0 ⇒ W × a = P × r here, W = 100 N, r = 0.05 m, P = 10 N Pr 10 0.05 0.005 m 100 a W × ⇒= = = a = 5 mm = 0.005 m The distance a is called the coefficient of rolling resistance. Choice (B) 35. By figure, the shapes are equilateral triangles. By considering moments around E at equilibrium ΣME = 0 ⇒ (500 × 40) + (100 × 30) – (RC × 20) + (100 × 10) = 0 ⇒ RC = 1200 N By section method, considering the section ABC. B 20 500 N A 60° BD 1200 N 100 N CD CE By taking the vertical forces. ΣY = 0 ⇒ 1200 – 500 – 100 + CD Sin60 = 0 ⇒ CD = 692.82 N As the value of CD is positive the member is in Tension as per the initial assumption. Choice (B)
Directions for questions 1 to 25: Select the correct alternative from the given choices. 1. A belt wrapped around a pulley 400 mm in diameter has a tension of 800 N on the tight side and a tension of 200 N on the slack side. If the pulley is rotating at 200 rpm then the power being transmitted (in kW) will be (A) 3351 (B) 3.35 (C) 32 (D) 3.2 2. A hollow triangular section is symmetrical about its vertical axis. The moment of inertia of the section about the base BC will be A 0.08 m 0.1 m B C 0.15 m 0.2 m (A) 1.02 × 10-5 m4 (B) 1.11 × 10-6 m4 (C) 1.02 × 10-6 m4 (D) 1.11 × 10-5 m4 3. The force of friction between two bodies in contact (A) depends upon the area of their contact (B) depends upon the relative velocity between them (C) is always normal to the surface of their contact (D) All of the above 4. A ball is thrown with a velocity of 12 m/s at an angle of 60o with the horizontal. How high the ball will rise? (A) 6 m (B) 6.35 m (C) 11 m (D) 5.5 m 5. The equation for angular displacement of a particle, moving in a circular path of radius 300 m is given by : θ = 20t + 5t 2 – 3t 3 where θ is the angular displacement at the end of t seconds. The maximum angular velocity of the particle (in rad/s) will be (A) 20.34 (B) 22.78 (C) 23.63 (D) 24.39 6. The velocity of piston in a reciprocating pump mechanism depends upon (A) Angular velocity of crank (B) Radius of the crank (C) Length of the connecting rod (D) All of the above 7. Consider a truss ABC loaded at A with a force 20 N as shown in figure. 20 N 45° 30° A B C The load in member AB will be approximately (A) 21 N (B) 19 N (C) 17 N (D) 18 N 8. A flexible body is used to lift 100 N from a curved surface is as shown in the figure. What is the force P required to just lift 100 N weight? (Take coefficient of friction as 0.25.) 100N 80° Flexible body P (A) 132.34 N (B) 121.36 N (C) 70.53 N (D) 141.77 N 9. The supports which apply force on the body in only one direction and the direction is always normal to the contacting surface is known as (A) Fixed support (B) Hinged support (C) Roller support (D) All of these 10. Triangular plate ABC is connected by means of pin at C with another triangular plate CDE as shown in the figure. The vertical reaction at point D will be A D B E C 1m 1m 2m 2 kN 1m 5m 4 kN (A) 4.5 kN (B) 6 kN (C) 2 kN (D) 4 kN 11. A beam 5 m long weighing 400 N is suspended in a horizontal position by two vertical strings, each of which can withstand a maximum tension of 450 N only. How far a body of 300 N weight be placed on the beam from the left end, so that one of the string may just break? Engineering Mechanics Test 3 Number of Questions: 25 Time:60 min.
Engineering Mechanics Test 3 | 3.23 (A) 1.81 m (B) 1.43 m (C) 0.834 m (D) 2.12 m 12. A figure is shown below. Solve for the force in member AB under the actions of the horizontal and vertical force of 1000 N. 1000 N 100 cm AB = 50 cm 1000 N BC = 80 cm B A C (A) 377.5 N (Tension) (B) 377.5 N (Compression) (C) 1415.9 N (Tension) (D) 1415.9 N (Compression) 13. A truss is shown in the figure. Each load is 5 kN and all triangles are equilateral with sides of 4 m. Determine the force on member GI. B A C E G I K M D F H J L (A) 35 kN (Compression) (B) 35 kN (Tension) (C) 26 kN (Compression) (D) 26 kN (Tension) 14. A cylinder is shown in the figure. The coefficient of friction between the cylinder and wall is 0.25. Will the 180 N force cause the 100 kg cylinder to slip? (100 × 9.81) N F1 N2 N1 F2 180 N (A) No slip (B) Slip will occur (C) Insufficient data (D) None of these 15. Two blocks B and A of mass 40 kg and 13.5 kg respectively is kept as shown in the figure. The coefficient of friction µ for all surface is 1/3. The value of the angle θ so that the motion of 40 kg block impends down the plane will be A B θ (A) 25.4o (B) 32.6o (C) 29.2o (D) 34o 16. A flywheel 2 m in diameter accelerates uniformly from rest to 2000 rpm in 25 seconds. 0.6 second after it has started from rest, the linear acceleration of a point on the rim of the flywheel (in m/s2 ) will be (A) 24.6 (B) 15.96 (C) 21.34 (D) 26.73 17. A mass of 2 kg is projected with a speed of 3 m/s up a plane inclined 20o with the horizontal as shown in the figure. After travelling 1 m, the mass comes to rest. The speed of the block as the block return to its starting position will be m 20° (A) 1.2 (B) 2.103 (C) 1.63 (D) 1.31 18. In a device, two equal masses of 100 kg are connected by a very light (negligible mass) tape passing over a frictionless pulley as shown in the figure. A mass of 10 kg is added to one side, causing that mass to fall and the other to rise. The acceleration (in m/s2 ) of the masses will be M M m (A) 0.49 (B) 0.817 (C) 0.621 (D) 0.467 19. Find the force ‘P’ required to prevent sliding of body 2 on body 1. Assume both the bodies have equal mass ‘m’ and all the surfaces are smooth.
3.24 | Engineering Mechanics Test 3 45° P Body 1 Body 2 (A) mg Sin45o (B) mg (C) 2 mg (D) 2 mg Cos45o 20. A rope is wound around a 30 kg solid cylinder of radius 50 cm as shown. Find the speed of its mass centre after it has drop by 2 m from the rest position. rope cylinder (A) 5.11 (B) 7.23 (C) 6.09 (D) 5.89 21. A 0.08 N bullet was fired horizontally into a 60 N sand bag suspended on a rope 1.5 m long as shown in the figure. It was found that the bag with the bullet embedded in it swung to a height of 20 mm. Determine speed of the bullet as it entered the bag? 1.5 m θ 20 mm (A) 470.42 m/s (B) 450 m/s (C) 469.8 m/s (D) 474.34 m/s 22. A mass of 1200 N is supported by means of a bell crank as shown in the figure. The magnitude of resultant at B (in N) will be AB = 0.6 m BC = 1.2 m 1200N AB = 0.6 m BC = 1.2 m C F A B 60° (A) 1588 N (B) 1500 N (C) 520 N (D) 1600 N 23. A wheel accelerates uniformly from rest to a speed of 500 rpm in 0.5 seconds. It then rotates at that speed for 2 seconds before decelerating (uniformly) to rest in 0.34 seconds. How many revolutions does it make during the entire time travel? (A) 29.1 rev (B) 20.17 rev (C) 22.34 rev (D) 26.33 rev 24. An eccentric cylinder used in a vibrator weights 198 N and rotates about an axis 5 cm from its geometric centre and perpendicular to the top view as shown in the figure. If the magnitudes of angular velocity and angular acceleration are 10 rad/s and 2 rad/s2 in the phase shown, the resultant reaction of the vertical shaft on the cylinder (in N) and couple applied on the cylinder by the shaft (in N-m) will be G α O 5 cm ω 10 cm (A) 111 and 0.278 (B) 101 and 0.278 (C) 111 and 0.413 (D) 101 and 0.413 25. A sphere, rolling with an initial velocity of 12 m/s, starts up a plane inclined 30o with the horizontal as shown in the figure. What is the distance upto which the sphere will roll up the plane? x 30° y (A) 10.21 m (B) 12.31 m (C) 11.86 m (D) 9.63 m
Engineering Mechanics Test 3 | 3.25 Answer Keys 1. B 2. A 3. C 4. D 5. B 6. D 7. D 8. D 9. C 10. A 11. C 12. B 13. D 14. A 15. C 16. D 17. B 18. D 19. C 20. A 21. A 22. A 23. B 24. B 25. C Hints and Explanations 1. Torque, T = 800 × 0.2 = 160 N Power = T × w = × × p = × 160 2 200 3.35 kW 60 1000 Choice (B) 2. IBC = 3 3 3 3 0.2 0.1 0.15 0.08 12 12 12 12 BH bh × × −= − ⇒ I BC = 1.0267 × 10–5 m4 Choice (A) 3. Choice (C) 4. H = ( ) 2 2 2 2 Sin 12 Sin 60 2 2 9.81 o u g a × = × = 5.5 m Choice (D) 5. θ = 20t + 5t 2 – 3t 3 d dt q w = = 20 + 10t – 9t 2 For maximum angular velocity, 0 d dt w = ∴ d dt w = 10 – 18t = 0 10 0.556 18 ∴= = t seconds ∴ wmax = 20 + 10(0.556) – 9(0.556)2 = 22.78 rad/sec Choice (B) 6. Vpiston = w[l Cos Φ + r Cos θ tan Φ] w = angular velocity of crank l = length of connecting rod r = radius of crank Choice (D) 7. A 20 N 45° x B 30° 1.732 x RB RC C x ΣMB = 0 ⇒ 20 × x = RC × 2.732x Rc = 7.32 N ∴ RB = 20 – 7.32 = 12.68 N Now FAB × Sin45o = RB ⇒ FAB = 12.68 Sin 45o = 17.93 N Choice (D) 8. P = Tight side tension W = Slack side tension Now, 0.25 80 180 100 s P P e e W q p m × × =⇒= ⇒ P = 141.77 N Choice (D) 9. Choice (C) 10. 2 kN 4 kN RX D R A ΣMA = 0 ⇒ 2 × 1 – 4 × 5 – Ry × 4 = 0 ⇒ Ry = 4.5 kN Choice (A) 11. R R 5 m A D C 300 N 400 N B x x = Distance between the body of weight 300 N and support A (from the left end) We know that one of the string will just break, when the tension will be 450 N (i.e., RA = 450 N) Now ΣMB = 0 450 × 5 = 300 (5 – x) + (400 × 2.5) ⇒ x = 0.834 m Choice (C) 12. Free body diagram 52.41 AB 1000 N BC 29.67 1000 N
3.26 | Engineering Mechanics Test 3 Now ΣFx = 0 ⇒ AB Cos(52.41) – BC Cos(29.67) + 1000 = 0 → (1) and ΣFy = 0 ⇒ AB Sin(52.41) + BC Sin(29.67) – 1000 = 0 → (2) From equations (1) and (2) we get AB = 377.5 N and BC = 1415.91 N ∴ Force in member AB is 377.5 N compression Choice (B) 13. Taking section passes through JH and GI. JH 5 kN 5 kN J L G I K M 4 m 4 m 4 m 15 Taking moment about point H we get ΣMH = 0 = –(GI) × 2 tan60o – (5 × 4) – (5 × 8) + 15 ×10 ⇒ GI = 25.98 kN (Tension) Choice (D) 14. Since it is unknown whether or not the cylinder slips it is not possible to F1 = µN1 and F2 = µN2 ΣFh = 0 = F1 – N2 + 180 → (1) ΣFv = 0 = N1 + F2 – 980 → (2) ΣMA = 0 = –180 × 2r + F2 + r + N2 × r → (3) ∴ N1 = 980 – F2 , N2 = 360 – F2 and F1 = 180 – F2 Let us assume F2 is at its maximum value that is 0.25 N2 and solve for N2 , N1 and F1 using equations (1), (2) and (3). Then N2 = 288 N, N1 = 908 N, F1 = 108 N. This means that if F2 assumes its maximum static value then F1 must be 108 N to hold the system in equilibrium. Since the maximum value of F1 obtainable is 0.25 N1 = 227 N, the cylinder will not rotate. Choice (A) 15. A (13.5 × 9.81) = 132.43 N 40 × 9.81 = 392.4 N T B N1 1 3 N1 1 3 N2 1 3 N1 N1 N2 θ θ From free body diagram of B. ΣFx = 0 = –392.4 Sin θ + 1 3 N1 + 1 3 N2 → (1) ΣFy = 0 = N2 – 392.4 Cos θ – N1 → (2) Free body diagram of A. N1 = 132.43 Cos θ → (3) From equation (1), (2) and (3) θ = 29.2o Choice (C) 16. w = wo + at and wo = 0 and w = 2 2 2000 60 60 p p N × × = ⇒ w = 209.44 rad/sec Now a = 209.44 0 25 o t w w− − = = 8.4 rad/s2 Now velocity after 0.6 seconds w = wo + at = 0 + (8.4 × 0.6) = 5.04 rad/s Normal component of acceleration, an = r ω2 = 1 × 5.042 = 25.4 m/s2 Tangential component of acceleration, at = ra = 1 × 8.4 = 8.4 m/s2 Total acceleration, a = 2 2 a a n t + 4 2 2 ⇒= + = a 25.4 8.4 26.753m s Choice (D) 17. (2 × 9.81) = 19.62 N (a) 19.62 N µN µN N 20° N (b) From figure (a), N = 19.62 Cos20o = 18.44 N Now V2 = 2 Vo + 2as ⇒ a = − +( ) = − × 2 2 0 3 4.5 m s 2 1 Now from figure (a), ΣFx = max ∴ +19.62 × Sin20o + (µ × 18.44) = 2(4.5) ⇒ µ = 0.124 To solve for return speed, refer figure (b) 19.62 Sin 20o – 0.124 (18.44) = 2a ⇒ a = 2.212 m/s2 Finally V2 = 2 Vo + 2as or V2 = 0 + 2(2.212) (1) ⇒ V = 2.103 m/s2 Choice (B) 18. a T Mg (M + m)g a T ΣF = T – Mg = Ma → (1) ΣF = Mg + mg – T = (M + m)a → (2)
Engineering Mechanics Test 3 | 3.27 From equation (1) and (2) ( ) 10 9.81 2 2 100 10 m a ga M m = ⇒= × = + × + 0.467 m/s2 Choice (D) 19. a 2mg R2 P mg a 45° 45° R1 y x ΣFx = max ⇒ –P = 2m(–a) ⇒ P = 2ma ΣFx = max ⇒ m(–a) = –R1 Sin45o → (1) ΣFy = may ⇒ R1 Cos45o = mg ⇒ R1 = mg/Cos45o → (2) From equation (1) and (2) we get, ma = Sin 45 Cos 45 o o mg × ⇒ a = g tan45o ⇒ a = g Now P = 2 ma ⇒ P = 2mg Choice (C) 20. 1 KE = 0 (in rest) PE = 0 (k = 0) 2 2 m G G KE = I 1 0 ω2 2 1 2 Io w2 = mg(2) ⇒ 2 1 2 2 2 mr mr + w2 = 30 × 9.81 × 2 1 2 ⇒ [1.5 × 30 × 0.52 ] × w2 = 30 × 9.81 × 2 ⇒ w = 10.23 rad/s VG = w × r = 10.23 × 0.5 = 5.115 m/s Choice (A) 21. Let V1 = Velocity of bullet before impact V2 = Velocity (bag + bullet) after impact = =× × 2 2 9.81 0.02 gh = 0.6264 m/s Now momentum before impact = momentum after impact ∴ (0.08 × V1 ) + 0 = [0.08 + 60] × 0.6264 ⇒ V1 = 470.4264 m/s Choice (A) 22. Σ`MB = 0 ⇒ 1200 × 0.6 = F × 1.2 ⇒ F = 600 N A 1200 N 600 N 60° 30° B RV RH R C RH = F Cos 30o = 600 × Cos 30o = 519.615 N and Rv = F Sin30o + 1200 = 600 × Sin30o + 1200 = 1500 N ∴ Resultant reaction, R = 2 2 R R H V + 2 2 = + 519.615 1500 = 1587.45 N Choice (A) 23. From t = 0 to t = 0.5: 1 ( ) ( ) 1 1 0 500 60 0.5 2 2 o q ww = +=+ × t ⇒ θ1 = 2.0834 rev From t = 0.5 to t = 2.5 s: θ2 = wt = 500 60 × 2 = 16.67 rev From t = 2.5 to rest θ3 = 1 2 (wo + w) t = 1 500 0 2 60 + × 0.34 = 1.42 rev Total number of revolutions θ = θ1 + θ2 + θ3 = 2.0834 + 16.67 + 1.42 = 20.17 rev Choice (B) 24. ΣFn = m r w2 = 198 9.81 × 0.05 × 102 = 101 N ΣFt = m r a = 198 9.81 × 0.05 × 2 = 2.02 N ΣMo = Io a = 1 198 198 2 2 0.15 0.05 2 9.81 9.81 ×× +× = 0.278 N–m
3.28 | Engineering Mechanics Test 3 Resultant of forces = 2 2 101 2.02 + = 101.02 N Choice (B) 25. F NA 30° mg The initial kinetic Energy (K.E1 ) decreases to final K.E2 = 0 at the top of the travel. The only force that does work in the component (negative) of the weight W along the plane. Work done = –[mg Cos30o ] × x Where, x is the required distance Initial kinetic energy, k.E1 = 2 2 1 1 1 1 2 2 mV I + ow Now Io = 2 2 5 mR and V1 = ω1 R ∴ K.E1 = 1 2 m 2 V1 + 1 5 m 2 V1 = 7 10 m(12)2 Now work done = K.E2 – K.E1 ⇒ –mg Cos30o × x = 0 – 7 10 m(12)2 2 7 12 10 9.81 Cos30o x × ⇒ = × × = 11.86 m Choice (C)
Engineering M chanics Chapter 1 One-mark Questions 1. A circular object of radius r rolls without slipping on a horizontal level floor with the center having velocity V. The velocity at the point of contact between the object and the floor is [2014-S1] (a) zero (b) V in the direction of motion (c) V opposite to the direction of motion (d) V vertically upward from the floor Solution: (a) The point of conduct P of the wheel with floor does not slip, which means the point P has zero velocity with respect to point m. Hence, the correct option is (a). 2. A two member truss ABC is shown in the figures. The force (in kN) transmitted in member AB is _____ [2014-S2] Solution: (see figure) tan q = = AC AB 0 5 1 . θ = 26.56 Draw FBD of right portion of truss after passing section (1)-(1) Apply equation of equilibrium (εFx = 0) −PAB − PBC cos (26.56) = 0 −PAB − PBC cos (26.56) = 0 (1) εFy = 0 −10 − PBC sin (26.56) = 0 −10 − 0.44PBC = 0 PBC = –22.36 N From Equation (1) −PAB − (−22.36) × 0.8944 = 0 −PAB + 20 = 0 PBC = 20 N. 3. A mass m1 of 100 kg traveling with a uniform velocity of 5 m/s along a line collides with a stationary mass m2 of 1000 kg. After the collision, both the masses travel together with the same velocity. The coefficient of restitution is [2014-S3] (a) 0.6 (b) 0.1 (c) 0.01 (d) 0 Solution: (d) Given condition VA = 5 m/s; VB = 0; mB = 1000 kg M01_Unit-IX_ME-Gate_C01.indd 3 11/19/2015 5:58:23 PM e
9.4 | Engineering Mechanics mA = 100 kg; VB′ = VA′ VA′ = ? By conservation it momentum gives, mAVA + mBVB = mAVA′ + mBVB′ 100 × 5 + 0 = 100VA′ + 1000 kg VA′ 500 = 1100VA′ ⇒ VA′ = 0.4545 VA′ = VB′ = 0.4545 Coefficient of restitution V V V V b A B A ′ ′ − − = 0 CoR = 0 Hence, the correct option is (d). 4. In a statically determinate plane truss, the number of joints ( j) and the number of members (m) are related by [2014-S4] (a) j = 2m – 3 (b) m = 2j + 1 (c) m = 2j − 3 (d) m = 2j − 1 Solution: (c) Condition for statically determinant truss m + r = 2j where m = no. of members of truss r = no. of reaction at support j = no. of joints in truss m = 2j − 3 [Assuming one rollers and one hinged support] Hence, the correct option is (c). 5. The coefficient of restitution of a perfectly plastic impact is [2011] (a) zero (b) 1 (c) 2 (d) infinite Solution: (a) Coefficient of restitution (CoR) of two colliding body is typically a positive real Number between 0 to 1. Which is ratio of speed after and before impact. For perfectly plastic body CoR = 0 and for perfectly elastic body = 1. Hence, the correct option is (a). 6. A block weighing 981 N is resting on a horizontal surface. The coefficient of friction between the block and the horizontal surface is µ = 0.2. A vertical cable attached to the block provides partial support as shown in the figure. A man can pull horizontally with a force of 100 N. What will be the tension, T (in N) in the cable if the man is just able to move the block to the right? [2009] (a) 176.2 (b) 196.0 (c) 481.0 (d) 981.0 Solution: (c) FBD of block W = 981 N εFy = 0 N + T − 918 = 0 N + T = 981 N (1) εFx = 0; 100 − F = 0 100 − µN = 0; µN = 100 N === 100 100 0 2 500 µ . From Equation (1) N + T − 981 = 0; 500 + T − 981 = 0 T = 418 N Hence, the correct option is (c). 7. The time variation of the position of a particle in rectilinear motion is given by x = 2t 3 + t 2 + 2t. If ‘V’ is the velocity and ‘a’ acceleration of the particle in consistent units, the motion started with [2005] (a) V = 0, a = 0 (b) V = 0, a = 2 (c) V = 2, a = 0 (d) V = 2, a = 2 Solution: (d) Displacement x = 2t 3 + t 2 + 2t Differentiating w.r. to t, we get Velocity V = 3 × 2 × t 2 + 2t + 2 = 6t 2 + 2t + 2 (1) Differentiating again, we get Acceleration a = 12t + 2 (2) at t = 0 From Equation (1) V = 2 From Equation (2) a = 2 Hence, the correct option is (d). 8. The figure shows a pin-jointed plane truss loaded at the point M by hanging a mass of 100 kg. The member LN of the truss is subjected to a load of [2004] M01_Unit-IX_ME-Gate_C01.indd 4 11/19/2015 5:58:23 PM
Chapter 1 Engineering M chanics | 9.5 (a) zero (b) 490 N in compression (c) 981 N in compression (d) 981 N in tension Solution: (a) Use method of section, Pass cutting section which divides truss in two part as shown Draw FBD of upper part of section (1)-(1) Apply equation of equilibrium in y-direction εFy = 0 PLN = 0 Hence, the correct option is (a). 9. A truss consists of horizontal members and vertical members having length L each. The members AE, DE and BF are inclined at 45o to the horizontal. For the uniformly distributed load P per unit length on the member EF of the truss shown in the figure, the force in the member CD is [2003] (a) PL 2 (b) PL (c) zero (d) 2 2 PL Solution: (a) First find Reaction at support i.e., RA and RB at A and B. Draw FBD of truss. Apply equation of equilibrium along x- and y-direction εFy = 0 (↑ +ve) RA + RB − PL = 0 ⇒ RA + RB = PL (1) εFy = 0 Taking εMA = 0 [summation of all moment of A = 0] Sign correction R L Pl L B × − 3 2 2 = 0 3LRB = 3 2 2 PL ⇒ RB = = PL R PL A 2 2 and To calculate/determine tension in truss member, use method of join; draw FBD to joint A. εFy = 0 PL PAE 2 + sin 45 = 0 M01_Unit-IX_ME-Gate_C01.indd 5 11/19/2015 5:58:24 PM e
9.6 | Engineering Mechanics PL PAE 2 1 2 + − = 0 PAE = −PL 2 2 εFx = 0 PAC + PAE cos 45 = 0 P PL AC + − 2 2 1 2 = 0 PAC = PL 2 Draw FBD of joint c εFx = 0 PCD = = P PL AC 2 Hence, the correct option is (a). 10. A bullet of mass ‘m’ travels at a very high velocity ‘V’ (as shown in the figure) and gets embedded inside the block of mass ‘M’ initially at rest on a rough horizontal floor. The block with the bullet is seen to move a distance ‘S’ along the floor. Assuming µ to be the coefficient of kinetic friction between the block and the floor and ‘g’ the acceleration due to gravity, what is the velocity ‘V’ of the bullet? [2003] (a) M m m gs + 2µ (b) M m m gs − 2µ (c) µ µ ( ) M m m gs + 2 (d)= M m 2µgs Solution: (a) Mass of bullet = m Bullet velocity = v Block mass = m Displacement of block = S [After striking by bullet] K.E. lost by the block with bullet = work done to overcome the frictional force ⇒ ( ) M m u + 2 2 = F × S where F = Frictional force = mR ⇒ ( ) M m u + 2 2 = mRS Again R = (M + m) g ⇒ ( ) M m u + 2 2 = + m M( ) m gS ⇒ u2 = 2mgS ⇒ u = 2µgS (1) Again (M + m) u = mV u = + mV M m (2) So slving this mV M m+ = 2µgS ⇒ V = M m+ m 2µgS Hence, the correct option is (a). 11. A steel wheel of 600 mm diameter on a horizontal steel rail. It carries a load of 500 N. The coefficient of rolling resistance is 0.3. The force in Newton, necessary to roll the wheel along the rail is [2000] (a) 0.5 (b) 5 (c) 1.5 (d) 150 Solution: (d) Draw Free body diagram of wheel. Apply equation of equilibrium along x- and y-direction εFy = 0; N − mg = 0 N − 500 = 0; N = 500 N From theory of friction F = µN = 0.3 × 500 = 150 N Hence, the correct option is (d). M01_Unit-IX_ME-Gate_C01.indd 6 11/19/2015 5:58:26 PM
Chapter 1 Engineering Mechanics | 9.7 12. The ratio of tension on the tight side to that on the slack side in a flat belt drive is [2000] (a) proportional to the product of coefficient of friction and lap angle (b) an exponential function of the product of coefficient of friction and lap angle (c) proportional to lap angle (d) proportional to the coefficient of friction Solution: (b) For flat belt T1 = tension on tight side T2 = tension on slack side θ = angle of lap of the belt over the pulley µ = coefficient of friction between belt and pulley T T 1 2 = eµθ Hence, the correct option is (b). 13. A car moving with uniform acceleration covers 450 m in a 5 second interval, and covers 700 m in the next 5 second interval. The acceleration of the car is [1998] (a) 7 m/s2 (b) 50 m/s2 (c) 25 m/s2 (d) 10 m/s2 Solution: (d) Displacement x = = s ut a + t 1 2 2 [u = initial velocity of car, a = acceleration] 450 = + 5 1 2 5 2 u a ( ) (1) 700 = + 5 1 2 5 2 v a ( ) (2) Solving Equation (1) and (2), we get v − u = 50 From equation of motion v − u = at ⇒ a = − = = v u t 50 5 10 m/s 2 Hence, the correct option is (d). 14. A ball A of mass m falls under gravity from a height h and strikes another ball B of mass m, which is supported at rest on a spring of stiffness k. Assume perfect elastic impact. Immediately after the impact [1996] (a) the velocity of ball A is 1 2 2gh (b) the velocity of ball A is zero (c) the velocity of both balls is 1 2 2gh (d) none of the above Solution: (c) In a perfectly elastic collision between equal masses of two bodies, velocities exchange on impact. Hence, velocity before impact = velocity immediate after impact VA = 2gh V ′ A = final velocity after impact VB = 0; V ′ B Form conservation of momentum M V M V A A B B + = M V ′ + M V ′ A A B B ⇒ MAVA = (MA + MB) ∀0 VB = 1 2 2gh Hence, the correct option is (c). 15. A wheel of mass m and radius r is in accelerated rolling motion without slip under a steady axle torque T. If the coefficient of kinetic friction is m, the friction force from the ground on the wheel is[1996] (a) µmg (b) T/r (c) zero (d) none of the above Solution: (a) Free body diagram of wheel From Equation of equilibrium εFy = 0 [Forces along y-direction] N = mg From theory of friction F = µ N F = µ mg Hence, the correct option is (a). 16. A stone of mass m at the end of a string of length l is whirled in a vertical circle at a constant speed. The tension in the string will be maximum when the stone is [1994] (a) at the top of the circle (b) halfway down from the top (c) quarter-way down from the top (d) at the bottom of the circle M01_Unit-IX_ME-Gate_C01.indd 7 11/19/2015 5:58:28 PM
9.8 | Engineering Mechanics Solution: (d) According to Newton’s Second law of motion F = ma εFt = 0 [along tangential direction] mg sin θ = ma εFr = 0 [along radial direction] T – mg sin q = mv l 2 \ T = + m v l g 2 cos θ where θ = cord makes an angle with radical l = length of cord Hence, the correct option is (d). 17. The cylinder shown below rolls without slipping. In which direction does the friction force act? Towards which of the following points is the acceleration of the point of contact A on the cylinder directed? [1993] (a) the mass centre (b) the geometric centre (c) the point P as marked (d) none of the above Solution: (b) According to theory of friction, friction always act opposite to the motion, Here the acceleration at a point of contact will be pass through its geometric center. Hence, the correct option is (b). 18. Instantaneous centre of a body rolling with sliding on a stationary curved surface lies. [1992] (a) at the point of contact (b) on the common normal at the point of contact (c) on the common tangent at the point of contact (d) at the centre of curvature of the stationary surface Solution: (b) 19. A and B are the end-points of a diameter of a disc rolling along a straight line with a counter clock-wise angular velocity as shown in the figure. Referring to the velocity vectors VA and VB shown in the figure [1990] (a) VA and VB are both correct (b) VA is incorrect but VB is correct (c) VA and VB are both incorrect (d) VA is correct but VB is incorrect Solution: (a) Velocity vector at A and at B i.e., VA and VB are always acting tangential to the path at any instant. Two-marks Questions 1. 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 by a massless and in extensible string PQ. If the coefficient of static friction for all surfaces is 0.4, the minimum force F (in kN) needed to move the block S is [2014-S1] (a) 0.69 (b) 0.88 (c) 0.98 (d) 1.37 M01_Unit-IX_ME-Gate_C01.indd 8 11/19/2015 5:58:29 PM
Chapter 1 Engineering Mechanics | 9.9 Solution: (d) Draw FBD of block R Apply equation of equilibrium εFy = 0 N2 − 981 = 0 N2 = 981 N From Friction theory F2 = µN2 = 0.4 × 981 F2 = 392.4 N Now, draw FBD of block ‘S’ Ws = 150 × 9.81 = 1471.5 N εFy = 0 N1 − N2 − Ws = 0; N1 − 981 − 1471.5 = 0 N1 = 2452.5 N From friction theory F1 = µN1 = 0.4 × 2452.5 F1 = 981 N F = F1 + F2 [total force based on εFx = 0] F = 981 + 392.4 F = 1373.4 N = 1.3734 kN Hence, the correct option is (d). 2. A block weighing 200 N is in contact with a level plane whose coefficient of static and kinetic friction are 0.4 and 0.2, respectively. The block is acted upon by a horizontal force (in newton) P = 10t, where t denotes the time in seconds. The velocity (in m/s) of the block attained after 10 seconds is _____ [2014-S1] Solution: Static friction µs = 0.4 → static condition Kinetic friction µk = 0.2 → dynamic/motion condition Draw FBD of block From equation of equilibrium εFy = 0; N − 200 = 0 N = 200 N Static friction is given by Fs = µs N = 0.4 × 200 = 80 N and kinetic friction Fk = µ kN = 0.2 × 200 = 40 N. To start motion of block; P must be greater than Fs Under static equilibrium P − Fs = 0; 10t − 80 = 0 ⇒ t = = 80 10 8sec ∴ The block start moving only after when t > 8 sec During 8 to 10 second of time, According to Newton Second law of motion F = mass × acceleration F = ma (P – Fk) = m dV dt ( ) 10 40 8 10 t d − t ∫ = ∫ 200 9 81 0 . dV V 10 2 40 2 8 10 t + t = 20 38 0 . [V ] V ⇒ [ ] 5 40 2 8 10 t t + = 20.38 [v] ∴ [5 × (10)2 − 40 (10)] − [5 (8)2 + 40 (8)] = 20.38 V V = 4.905 m/s. 3. A truck accelerates up a 10° incline with a crate of 100 kg. Value of static coefficient of friction between the crate and the truck surface is 0.3 The maximum value of acceleration (in m/s2) of the truck such that the crate does not slide down is _____ [2014-S2] Solution: Free body diagram of crate M01_Unit-IX_ME-Gate_C01.indd 9 11/19/2015 5:58:30 PM
9.10 | Engineering Mechanics µs = 0.3; εFx = 0 P + 981 sin (10) − F = 0 P + 170.34 − F = 0 (1) εFy = 0 −mg cos θ + N = 0 N + [−981 cos (10)] = 0 N = 966.09 N From theory of friction F = µN = 0.3 × 966.09 = 289.82 N From Equation (1) P + 170.34 − 289.82 = 0 P = 119.48 N According to Newton Second law of motion F = ma [along inclined plane] 119.48 = ma a = = 119 48 119 48 100 . . m = 1.198 m/s2 a = 1.198 m/s2. 4. A rigid link PQ of length 2 m rotates about the pinned end Q with a constant angular acceleration of 12 rad/s2. When the angular velocity of the link is 4 rad/s, the magnitude of the resultant acceleration (in m/s2) of the end P is _____ [2014-S1] Solution: aT = Tangential acceleration aN = radial acceleration/Normal acceleration α = 12 rad/s2 (Angular acceleration) ω = 4 rad/s (Angular velocity) Tangential acceleration is given by aT = rα = 2 × 12 = 24 m/s2 and normal acceleration aN = rω2 = 2 × (4)2 = 32 m/s2 Now, Resultant Acceleration a = + a a = + T r 2 2 2 2 ( ) 24 ( ) 32 a = 40 m/s2. 5. A body of mass (m) 10 kg is initially stationary on a 45o inclined plane as shown in figure. The coefficient 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 traveled (in meter) by the body along the plane is [2014-S3] Solution: (see figure) FBD of mass m m = 10 kg Velocity of body = 20 m/s µk = 0.5 From equation of equilibrium εFy = 0 [Normal to surface] N − mg cos 45 = 0 N − 98.1 cos (45) = 0 N = 69.36 N εFx = 0 [along inclined plane] F − mg sin 45 = 0; F − 98.1 sin (45) = 0 F = 69.36 N From theory of friction F = µk N = 0.5 × 69.36 = 34.68 N F = 34.68 N By using D’alembert principle εFx = 0 mg sin 45 − F = ma 69.36 − 34.68 = 10 × a a = 3.468 m/s2 From equation of motion, displacement s = + ut at 1 2 2 and v2 = µ2 + 2as M01_Unit-IX_ME-Gate_C01.indd 10 11/19/2015 5:58:30 PM
Chapter 1 Engineering Mechanics | 9.11 v = 20 m/s u = initial velocity = 0 S = = × V 2 20 20 2 2 3 468 ( ) . s = 57.67 m. 6. An annular disc has a mass m, inner radius R and outer radius 2R. The disc rolls on a flat surface without slipping. If the velocity of the center of mass is v, the kinetic energy of the disc is [2014-S3] (a) 9 16 2 mV (b) 11 16 2 mV (c) 13 16 2 mV (d) 15 16 2 mV Solution: (c) Total kinetic energy = + 1 2 1 2 2 2 mv Iω (1) v = rω = Rω = 2Rω w = = v R v 2 2R rad/sec Moment of inertia = = I − m R R 2 0 2 1 2 ( ) = + = m R R mR 2 4 5 2 2 2 2 ( ) From Equation (1) (KE)total = + 1 2 1 2 5 2 2 2 2 2 mv mF v R (KE)total = 13 16 2 mv Hence, the correct option is (c). 7. A four-wheel vehicle of mass 1000 kg moves uniformly in a straight line with the wheels revolving at 10 rad/s. The wheel are identical, each with a radius of 0.2 m, then a constant braking torque is applied to all the wheels and the vehicle experiences a uniform deceleration. For the vehicle to stop in 10 s, the braking torque (in N⋅m) on each wheel is _____ [2014-S4] Solution: M = 1000 kg ω0 = initial angular velocity = 10 rad/sec r = radius of wheel = 0.2 m Time = t = 10 sec Mass on each axle (m) = = 1000 2 500 kg Final angular velocity ω0 = 0 From equation of motion, ω = ω0 + αt [similar to v = µ0 + at] 0 = 10 + α (10) a = –1 rad/s2 Braking torque m = I2α Braking torque m = 10 × 1 = 10 N-m. 8. For the truss shown in the figure, the forces F1 and F2 are 9 kN and 3 kN, respectively. The force (in kN) in the member QS is [2014-S4] (a) 11.25 tension (b) 11.25 compression (c) 13.5 tension (d) 13.5 compression Solution: (a) Pass section (1)-(1), and draw FBD of left portion of truss. M01_Unit-IX_ME-Gate_C01.indd 11 11/19/2015 5:58:32 PM
9.12 | Engineering Mechanics tan q = 2 1 5. q = 53.13 εFy = 0; −9 − FPS sin (53.13) = 0 −0.8FPS = 9 FPS = –11.25 kN εFx = 0 FPS cos (53.13) + FPQ = 0 (−11.25) × 0.6 + FPQ = 0 FPQ = +6.75 kN Apply equilibrium to whole truss. εFy = 0 v1 – v2 + 3 – 9 = 0 ⇒ v1 – v2 = 6 (1) εmax R = 0 −v1 × 1.5 − 3 × 3 + 9 × 6 = 0 –1.5v1 – 9 + 54 = 0 ⇒ v1 = 30 kN From Equation (1) 30 – v2 = 6 ⇒ v2 = 24 kN Pass section (2)-(2), consider right portion of truss. εFy = 0 FQS sin (53.13) + 30 − 24 + 3 = 0 FQS = 11.25 kN Hence, the correct option is (a). 9. A uniform slender rod (8 m length and 3 kg mass) rotates in a vertical plane about a horizontal axis 1 m from its end as shown in the figure. The magnitude of the angular acceleration (in rad/s2) of the rod at the position shown is _____ [2014-S4] Solution: (see figure) ε = Iα [According to Newton Second law] m = Iα (1) m = 29.43 × 3 = 88.29 Nm I = I0 + Ad2 = + ml md 2 2 12 = × + × 3 8 12 3 3 2 2 = 43.00 kg m2 From Equation (1) a = = = m I 88 29 43 2 053 . . . rad/s 2 10. A wardrobe (mass 100 kg, height 4 m, width 2 m, depth 1 m), symmetric about the Y-Y axis, stands on a rough level floor as shown in the figure. A force P is applied at mid-height on the wardrobe so, as to tip it about point Q without slipping. What are the minimum values of the force (in Newton) and the static coefficient of friction µ between the floor and the wardrobe, respectively? [2014-S4] (a) 490.5 and 0.5 (b) 981 and 0.5 (c) 1000.5 and 0.15 (d) 1000.5 and 0.25 Solution: (b) Taking moment at Q = 0 εmQ = 0 M01_Unit-IX_ME-Gate_C01.indd 12 11/19/2015 5:58:33 PM
Chapter 1 Engineering Mechanics | 9.13 P × 2 − w × 1 = 0 P = × = 100 9 81 2 490 5 . . N εFy = 0; N − W = 0 N = W = 981 N [weight of an object] From friction theory F = µN = P; 490.5 = µ × 981 m = 0.5 Hence, the correct option is (b). 11. 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 of the ladder is _____ [2014-S4] Solution: (see figure) Applying equation of equilibrium εFy = 0 NB − 600 = 0 εMA = 0 P × 3 + w × I − NB × 4 = 0 P = 4 2 − 3 N W B = 4 600 2 600 × − × 3 P = 400 N. 12. A pin jointed uniform rigid rod of weight W and length L is supported horizontally by an external force F as shown in the figure below. The force F is suddenly removed. At the instant of force removal, the magnitude of vertical reaction developed at the support is [2013] (a) zero (b) W/4 (c) W/2 (d) W Solution: (b) As force F is released, the rod accelerate and undergoes pure rotation about hinged point. From Newton Second law εT = Iα [I = moment of Inertia of rod; α = acceleration of rod; εT = external frequency on rod]. W 1 2 = Ia (1) I = W g 1 3 3 [moment of inertia of the rod about hinged] From Equation (1) W g 1 3 3 α = W 1 2 a = 3 2 1 g For force equilibrium, The inertia force will act upward direction (opposite to motion) \ R W− + px dx ∫ α 0 1 = 0 [P = mass/unit length] R =− =− W p x W p l 2 2 2 2 α α = − W W gl l l 2 2 3 2 3 R = − W = W 1 3 4 4 Hence, the correct option is (b). Common Data for Questions 13 and 14: Two steel truss members, AC and BC, each having cross-sectional area of 100 mm2 are subjected to a horizontal force F as shown in figure. All the joints are hinged. [2012] M01_Unit-IX_ME-Gate_C01.indd 13 11/19/2015 5:58:35 PM
9.14 | Engineering Mechanics 13. If F = 1 kN, magnitude of the vertical reaction force developed at the point B in kN is (a) 0.63 (b) 0.32 (c) 1.26 (d) 1.46 Solution: (a) Isolate member AC and BC from support, εFx = 0 (→ +ve) F − FAC cos 45 − FBC cos 60 = 0 F − 0.707FAC − 0.5FBC = 0 0.707FAC + 0.5FBC = F (1) εFy = 0 FAC sin 45 = FBC sin 60 FAC = 1.22FBC (2) From Equation (1) 0.707 (1.22FBC) + 0.5FBC = 1 0.865FBC + 0.5FBC = 1 1.365FBC = 1 FBC = 0.732 kN Vertical force at B = FBC sin 60 = 0.732 × sin 60 = 0.634 kN Hence, the correct option is (a). 14. The maximum force F in kN that can be applied at C such that the axial stress in any of the truss members does not exceed 100 MPa is (a) 8.17 (b) 11.15 (c) 14. 14 (d) 22.30 Solution: (b) From Lami’s theorem FAC sin (120) = F sin (105) FAC = 0.895F FAC is maximum force 6 = = stress = area FAC 0 895F 100 . = 100 MPa F = 100 × 100 0.895 F = 11.17 kN Hence, the correct option is (b). 15. A 1 kg block is resting on a surface with coefficient of friction µ = 0.1. A force of 0.8 N is applied to the block as shown in the figure. The friction force is [2011] (a) zero (b) 0.8 N (c) 0.89 N (d) 1.2 N Solution: (b) FBD of block εFy = 0 N = 9.81 N From theory of friction F = µN = 0.1 × 9.81 = 0.98 N The external force applied = 0.8 N < friction force Hence, Friction force = external applied force = 0.8 N Hence, the correct option is (b). 16. Consider a truss PQR loaded at P with a force F as shown in the figure. The tension in the member QR is [2008] (a) 0.5F (b) 0.63F (c) 0.73F (d) 0 87F M01_Unit-IX_ME-Gate_C01.indd 14 11/19/2015 5:58:35 PM
Chapter 1 Engineering M chanics | 9.15 Solution: (b) Using method of joint, draw FBD of joint P εFx = 0 FPQ cos 45 = FPR cos 30 0.707FPQ = 0.866FPR 0.866FPR − 0.707FPQ = 0 (1) εFy = 0 F + FPQ sin 45 + FPR sin 30 = 0 0.5FPR + 0.707FPQ = −F (2) Solving Equation (1) and (2) 0.866FPR − 0.707FPQ = 0 0.5FPR + 0.707FPQ = −F ____________________ 1.366FPR = −F FPR = –0.732F Draw FBD of joint R εFx = 0 [equation of equilibrium along × direction] −FQR − FPR cos 30 = 0 −FQR − (−0.732 F) × 0.866 = 0 FQR = 0.63F Hence, the correct option is (b). 17. A block of mass M is released from point P on a rough inclined plane with inclination angle θ, shown in the figure below. The coefficient of friction is µ. If µ < tan θ, then the time taken by the block to reach another point Q on the inclined plane, where PQ = S, is [2007] (a) 2s g cos ( θ θ tan ) µ− (b) 2s g cos ( θ θ tan ) µ+ (c) 2s g sin ( θ θ tan ) µ− (d) 2s g sin ( θ θ tan ) µ+ Solution: (a) FBD of mass m From Newton Second law and motion EF = ma mg sin θ − F = ma (1) εFy = 0 [Forces along y-direction] N − mg cos θ = 0 mg cos q = N (2) From Equation (1) mg sin θ − µN = ma a = − g (sin θ µ cos ) θ s = + ut at 1 2 2 [where m = initial velocity of block = 0] From s = 1 2 2 at t = = − 2 2 s a s g(sin θ µ cos ) θ = − 2s g cos [ θ θ tan ] µ Hence, the correct option is (a). 18. Two books of mass 1 kg each are kept on a table one over the other. The coefficient of friction on every pair of containing surfaces is 0.3. The lower book is pulled with a horizontal force F. The minimum value of F for which slip occurs between the two books is [2005] (a) zero (b) 1.06 N (c) 5.74 N (d) 8.83 N M01_Unit-IX_ME-Gate_C01.indd 15 11/19/2015 5:58:37 PM e
9.16 | Engineering Mechanics Solution: (d) m1 = 1 kg; m2 = 1 kg FBD of block A Apply Equation of equilibrium εFy = 0; N1 = N2 + 9.81 = 9.81 + 9.81 = 19.62 N From Friction theory F1 = µN1 = 0.3 × 19.62 = 5.886 N εFx = 0 F = F1 + F2 = 5.886 +2.943 = 8.83 N F = 8.83 N Hence, the correct option is (d). 19. A 1 kg mass of clay, moving with a velocity of 10 m/s, strikes a stationery wheel and sticks to it. The solid wheel has a mass of 20 kg and a radius of 1 m. Assuming that the wheel and the ground are both rigid and that the wheel is set into pure rolling motion, the angular velocity of the wheel immediately after the impact is approximately [2005] (a) zero (b) 1/3 rad/s (c) 10 3 rad/s (d) 10/3 rad/s Solution: (b) m1 = 1 kg; m2 = 20 kg V1 = 10 m/s V2 = 0 m/s [initial velocity of solid wheel] V1′ = [final velocity of cloy] V2′ = V1′ = final velocity of solid wheel From conservation of momentum m1V1 + m2V2 = m1V1′ + m2V2′ m1V1 + m2V2 = m1V1′ + m2V1′ (1) (10) + (20) (0) = (m1 + m2) V1′ 10 = (1 +20) V1′ V1′ = 0.476 m/s but V = rω w = = = V r 1 10 21 0. r 476 ad/sec Hence, the correct option is (b). 20. The figure below shown a pair of pin jointed gripper tongs holding an object weighting 2000 N. The coefficient of friction (µ) at the gripping surface is 0.1XX is the line of action of the input force and Y-Y is the line of application of gripping force. If the pin joint is assumed to be frictionless, the magnitude of force F required to hold the weight is [2004] (a) 1000 N (b) 2000 N (c) 2500 N (d) 5000 N Solution: (d) Draw FBD of weight εFy = 0 [Forces along vertical direction] µR + µR = 2000; 2µR = 2000 2 × 0.1 × R = 2000; R = 10000 N Now, draw FBD of gripper tongs M01_Unit-IX_ME-Gate_C01.indd 16 11/19/2015 5:58:37 PM
Chapter 1 Engineering M chanics | 9.17 εmax P = 0 10000 × 150 = F × 300 F = 5000 N Hence, the correct option is (d). 21. As shown in figure, a person A is standing at the centre of a rotating platform facing person B who is riding a bicycle, heading East. The relevant speeds and distances are shown in given figure: person, a bicycle, heading East. At the instant under consideration, what is the apparent velocity of B as seen by A? [1999] (a) 3 m/s heading East (b) 3 m/s heading West (c) 8 m/s heading East (d) 13 m/s heading East Solution: (d) Apparent velocity of B w.r.to A = V − (rω) [r = distance of AB] = 8 − 5 (−1) ω = Angular velocity = 1 rad/sec = 13 m/s heading east Hence, the correct option is (d). 22. A mass 35 kg is suspended from a weightless bar AB which is supported by a cable CB and a pin at A as shown in figure. The pin reactions at A on the bar AB are [1997] (a) Rx = 343.4 N, Ry = 755.4 N (b) Rx = 343.4 N, Ry = 0 (c) Rx = 755.4 N, Ry = 343.4 N (d) Rx = 755.4 N, Ry = 0 Solution: (d) Draw FBD of bar AB tan q = 125 275 θ = 24.45° Draw FBD at B By Lami’s theorem T sin 90 = HA sin (114 45 . ) = 343 35 155 55 . sin ( . ) HA = 755.15 N T = 829.55 N To calculate VA εFy = 0 VA + T sin (24.45) − 343.35 = 0 VA + 829.55 × 0.4138 −343.35 = 0 VA + 343.26 − 343.35 = 0 VA = 0 Answer is HA = 755.15 N VA = 0 Hence, the correct option is (d). 23. AB and CD are two uniform and identical bars of mass 10 kg each, as shown. The hinges at A and B are frictionless. The assembly is released from rest and motion occurs in the vertical plane. At the instant that the hinge B passes the point B, the angle between the two bars will be [1996] (a) 60 degrees (b) 37.4 degrees (c) 30 degrees (d) 45 degrees M01_Unit-IX_ME-Gate_C01.indd 17 11/19/2015 5:58:38 PM e
9.18 | Engineering Mechanics Solution: (c) As in figure hinge ‘B’ is frictionless, no torque is applied to bar CD. So, no angle change occurs. Hence, the correct option is (c). 24. A rod of length 1 m is sliding in a corner as shown in figure. At an instant when the rod makes an angle of 60 degrees with the horizontal plane, the velocity of point A on the rod is 1 m/s. The angular velocity of the rod at this instant is [1996] (a) 2 rad/s (b) 1.5 rad/s (c) 0.5 rad/s (d) 0.75 rad/s Solution: (a) VA—velocity of point A acting along a vertical wall = 1 m/s VB—velocity of point B acting along a horizontal plane = from geometry of triangle ABI IA OB l m IB OA l = = = = = = = ° cos sin θ θ θ 1 2 3 2 60 VA = ω × IA w = = = V I A A 1 1 2 2 / rad/sec Hence, the correct option is (a). 25. Match 4 correct pairs between List-I and List-II. No credit will be given for partially correct matching [1996] List-I List-II A. Collision of particles 1. Euler’s equation of motion B. Stability 2. Minimum kinetic energy C. Satellite motion 3. Minimum potential energy D. Spinning top 4. Impulse-momentum principle 5. Conservation of moment of momentum Solution: a-4, b-3, c-1, d-5. 26. A spring scale indicates a tension T in the right hand cable of the pulley system shown in figure. Neglecting the mass of the pulleys and ignoring friction between the cable and pulley the mass m is [1995] (a) 2T/g (b) T (1 + c4π)/g (c) 4T/g (d) none of the above Solution: (c) FBD of scale Applying equation of equilibrium εFg = 0 T + 2T +T = mg 4T = mg m = 4T/g Hence, the correct option is (c). M01_Unit-IX_ME-Gate_C01.indd 18 11/19/2015 5:58:39 PM
391 08. ENGINEERING MECHANICS 1. Forces and Force Systems 1. Two forces P and Q are acting at an angle . The resultant force R acts at an angle of with force P, then the value of will be (a) 1 Qcos tan P Qcos − α + α (b) 1 Qsin tan P Qcos − α + α (c) 1 Qcos tan P Qsin − α + α (d) 1 Qsin tan P Qsin − α + α HPPSC AE 2018 Ans. (b) : Parallelogram Law of Forces–"If two forces, acting at a point be represented in magnitude and direction by the two adjacent sides of parallelogram, then their resultant is represented in magnitude and direction by the diagonal of the parallelogram passing through that point". If the angle between the forces is α, then 2 2 R P Q 2PQcos = + + α The direction of resultant will be 1 Qsin tan P Qcos − α θ= + α (from Force P) 2. A rigid body will be in equilibrium under the action of two forces only when: (a) Forces have same magnitude, same line of action and same sense (b) Forces have same magnitude and same line of action (c) Forces have same magnitude, same line of action and opposite sense (d) Forces have same magnitude UPRVUNL AE 2016 Ans. (c) : A rigid body will be in equilibrium under the action of two forces only when forces have same magnitude, same line of action and opposite sense. 3. The resultant of two force A and B is perpendicular to A, the angle between the force A and B will be: (a) 1 A cos B − θ = (b) 1 B cos A − θ = (c) 1 B cos A − θ = − (d) 1 A cos B − θ = − (e) 1 B tan A − θ = CGPSC AE 2014- I Ans. (d) : We know thatBsin tan A Bcosα θ = + α Given, θ = 90 o Bsin tan 90 A Bcosα = + α 1 Bsin 0 A Bcosα = + α A Bcos 0 + α = A Bcos = − α 1 A cos B − α = − 4. Pick the odd statement out with regard to Lami's theorem from following: (a) The theorem is applicable only if the body is in equilibrium. (b) The theorem is applicable for parallel forces. (c) The theorem is not applicable for more than three forces. (d) The theorem is not applicable for less than three forces. (e) The theorem is applicable for coplanar forces. (CGPCS Polytechnic Lecturer 2017) Ans. (b) : Lami's Theorem–"If a body is in equilibrium under the action of three forces, then each force is proportional to the sine of the angle between the other two forces". P sinα = Q sin γ = R sin β 5. The Lami's theorem is applicable only for : (a) Coplanar forces (b) Concurrent forces (c) Coplanar and Concurrent forces (d) Any types of the forces TRB Polytechnic Lecturer 2017 Ans. (c) : The Lami's theorem is applicable only for coplanar and concurrent forces.
392 6. Concurrent forces 2, 3 , 5, 3 and 2 kN act at one of the angular points of a regular hexagon towards the remaining five angular points. Determine the magnitude and direction of the resultant force. (a) R = 10 kN; α = 0º (b) R = 12 kN; α = 180º (c) R = 10 kN; α = 60º (d) R = 20 kN; α = 0º (e) R = 10 kN; α = 90º CGPSC 26th April 1st Shift Ans. (c) : Resolving all forces horizontally ∑ = + + H 2cos 0º 3 cos30º 5cos60º + + 3 cos90º 2cos120º = ( ) 3 1 1 2 3 5 3 0 2 2 2 2 + × + + × + × − = 5 kN Resolving all forces vertically ∑ = + + V 2sin 0º 3 sin 30º 5sin 60º + + 3 sin 90º 2sin120º 3 5 3 3 0 3 2 2 2 2 = + + + + × = 8.66 kN The magnitude of the resultant forces 2 2 R H V = ∑ + ∑ ( ) ( ) 2 2 = + (5) (8.66) = 9.99 ≃ 10kN Direction of the resultant force 9.99 tan 1.732 8.66 ∑ = = = ∑ V H θ θ = 59.99 60º ≃ 7. A particle in equilibrium cannot have (a) Linear motion with constant speed (b) Constant velocity (c) Zero acceleration (d) Zero velocity (e) Curvilinear motion with constant speed CGPSC 26th April 1st Shift Ans. (e) : A particle is said to be in equilibrium if the net force acting on the particle is zero. By Newton's second law, the acceleration of such objects will be zero. As a result, either the body is at rest (i.e. zero velocity) or it is moving with a constant velocity (i.e. with constant speed in a straight line). 8. If two of the three concurrent forces in equilibrium have equal magnitude P and angle between them, the magnitude of third force is (a) 2P cos (α /2) (b) 2P cos α (c) 2P sin α (d) P cos α (e) 2P sin (α /2) CGPSC 26th April 1st Shift TNPSC AE 2018 Ans. (a) : We know that, Resultant force (R) = 2 2 P Q 2PQcos + + θ 2 2 2 2 R P P P = + + 2 cosα (P =Q) 2 2 = + 2 2 cos P P α ( ) 2 = + 2 1 cos P α ( ) 2 2 = 4 cos / 2 P α R P = 2 cos / 2 (α ) 9. The maximum force which acts on the connecting rod is (a) Force due to gas pressure (b) Force due to inertia of piston (c) Force due to friction of connecting rod (d) Force due to crank pin TNPSC AE 2013 Ans. (a) : Force due to gas pressure is the maximum force acts on the connecting rod. 10. Calculate the resisting torque for static equilibrium in the following figure: (a) 100 N-cm (b) 150 N-cm (c) 200 N-cm (d) 300 N-cm Gujarat PSC AE 2019 Ans : (d) : T = F sin 30o × 30 = 20 × sin 30o × 30 = 300 N-cm 11. Three forces P, Q and R are acting concurrently. The included angles are , and . According to Lami's theorem (a) P Q R sin sin sin = = α β γ (b) P Q R cos cos cos = = α β γ (c) P Q R tan tan tan = = α β γ (d) sin sin sin P Q R α β γ = = HPPSC AE 2018
393 Ans. (a) : Lami's Theorem–"If a body is in equilibrium under the action of three forces, then each force is proportional to the sine of the angle between the other two forces" P R Q sin sin sin = = α β γ 12. The force applied on a body of mass 100 kg to produce an acceleration of 5 m/s2 is (a) 20 N (b) 100 N (c) 500 N (d) None of these Vizag Steel (MT) 2017 Ans. (c) : F = ma = 100 × 5 = 500 N 13. Two Identical trusses supported a load of 100 N as shown in figure the length of each truss is 0.1, cross sectional area is 200 mm2 Young's modulus E = 200 GPa. The force in the truss AC (in N) is . 100 N (a) 200 (b)300 (c) 50 (d)100 RPSC INSP. OF FACTORIES AND BOILER 2016 Ans : (d) By Lami's Theorem– AC 0 0 0 100 P sin120 Sin(90 30 ) ⇒ = + ⇒ = P 100 N AC 14. Two parallel forces 100 kN and 75 kN act on a body and have resultant of 25 kN. Then, the two forces are (a) Like parallel forces (b) Unlike parallel forces (c) Concurrent forces (d) None of the above Gujarat PSC AE 2019 Ans : (b) : We know that, R2 = P2 + Q2 + 2PQ cos θ Given, P = 100 kN, Q = – 75 kN R2 = (100)2 + (–75)2 + 2 × 100 × (–75) × cos 0o R 2 = 625 R= 625 R = 25 kN Hence, the two forces are unlike parallel forces. 15. The angle between two forces P and Q is The resultant of these forces is (a) 2 2 P Q 2PQsin + + α (b) 2 2 P Q 2PQcos + + α (c) 2 2 P Q+ (d) 2 2 P Q 2PQcos + − α UPPSC AE 12.04.2016 Paper-I Ans : (b) Parallelogram law of forces:- It states that if two forces, acting simultaneously on a particle, be represented in magnitude and direction by the two adjacent side of a parallelogram, then their resultant may be represented in magnitude and direction by the diagonal of a parallelogram which passes through their points of intersection. 2 2 R P Q 2PQcos = + + θ Qsin tan P Qcos θ α = + θ α → Angle make by Resultant force R from P. 16. Which of the following is not a scalar quantity? (a) Mass (b)Volume (c) Time (d)Acceleration (KPSC AE 2015) Ans : (d) Scalar quantity:- Mass, volume, density. Vector quantity:-Displacement, Valocity, acceleration. 17. If a suspended body is struck at the centre of percussion, then the pressure on the axis passing through the point of suspension will be: (a) Maximum (b) Minimum (c) Zero (d) Infinity HPPSC W.S. Poly. 2016 Ans : (c) If a suspended body is struck at the centre of percussion, then the pressure on the axis passing through the point of suspension will be zero. The centre of oscillation is termed as centre of percussion. It is defined as that point at which a blow may be struck on a suspended body so that the reaction at the support is zero. 18. The effect of a force on a body depends on its : (a) Direction (b) Magnitude (c) Position (d) All of these (OPSC AE. 2016) UKPSC AE 2007 Paper -I
394 Ans : (d) The effect of a force on a body depends on its: (i) Direction, (ii) Magnitude, (iii) Position (iv) Line of action. 19. The horizontal and vertical components of a force of 200 N acting on a body at an angle of 30 with the horizonatal is (a) 100 3 and 100 N (b) 200 3 N and 200 N (c) 400 3 N and 400 N (d) 300 3 N and 300 N TSPSC AEE 2015 Ans : (a) horizontal components = 200 cos 30° F 100 3N H = Vertical component = 200 sin30° Fv = 100 N 20. What is the thrust at the point 'A' in the post shown in the figure? UPPSC AE 12.04.2016 Paper-I (a) 0.866 kN (b) 0.5 kN (c) 1.388 kN (d) 1 kN Ans : (a) Thrust at the point A= Vertical component of force 1kN Thrust force = 1 sin 60o Thrust force = 0.866kN 21. A roller of weight W is to be rolled over a wooden block as shown in the figure. The pull F required to just cause the said motion. (a) W 2 (b) W (c) 3W (d) 2 W UPPSC AE 12.04.2016 Paper-I Ans : (c) Free body diagram. 0 0 W F R sin150 sin90 sin120 = = W sin120 F sin150 = F 3W = 22. A 40 mm diameter water jet strikes a hinged vertical plate of 800 N weight normally at its surface at its centre of gravity as shown in the figure below: The angle of deflection is nearly (a) sin–1 0.353 (b) sin–1 0.321 (c) tan–1 0.353 (d) tan–1 0.321 ESE 2018 Ans. (a) : taking moment about O at equilibrium position F cosθ ⋅ x = W⋅y F cosθ ⋅ L / 2 cosθ = L W sin 2 θ F = W sinθ ρAV2 = W sinθ 3 2 2 10 (0.04) 15 4π × × = 800 × sinθ sinθ = 0.353 θ = sin−1 (0.353)
395 23. A ball of weight 100N is tied to a smooth wall by a cord making an angle of 30° to the wall. The tension in the cord is (a) 200N (b) 200 N 3 (c) 100N (d) 50 3N ESE 2017 Ans. (b) : By Lami's theorem T sin90° = W sin120° = R sin150° T sin90° = W sin120° = 100 3 / 2 = 200 3 200 T N 3 = 24. In a given figure, when angles = , then the components (f1 and f2) of force F on either side are given by an expression: (a) 1 2 Fcos f f sin( ) α = = α + β (b) 1 2 Fsin f f sin( ) α = = α + β (c) 1 2 Fcos f f cos( ) α = = α + β (d) 1 2 Fsin f f sin( )α = = α (e) 1 2 Fsin f f sin( )α = = β (CGPCS Polytechnic Lecturer 2017) Ans. (b) : Using Lami's theorem 1 f sinα = 2 f sin β = ( ) F sin 360 ° − α + β So, f1 = ( ) Fsin sin α − α + β f2 = ( ) Fsin sin β − α + β (α = β) Then, f1 = f2 = ( ) Fsin sin α α + β [–ve not considered in answer] 25. If the sum of all the forces acting on a moving object is zero, the object will (a) continue moving with constant velocity (b) accelerate uniformly (c) change the direction of motion (d) slow down and stop UKPSC AE 2012 Paper-I Ans. (a) : continue moving with constant velocity 26. Two equal and mutually perpendicular forces of magnitude ‘P’, are acting at a point. Their resultant force will be (a) P 2, at an angle of 30° with the line of action of any one force. (b) P 2, at an angle of 45° with the line of action of each force. (c) P 2, at an angle of 45° with the line of action of each force. (d) Zero UKPSC AE 2012 Paper-I Ans. (b) : P 2, at an angle of 45° with the line of action of each force. 27. A body subjected to coplanar non-concurrent forces will remain in a state of equilibrium if (a) ∑Fx = 0 (b) ∑Fy = 0 (c) ∑M = 0 (d) All of the above three UKPSC AE 2012 Paper-I Ans. (d) : All of the above three 28. A rigid body is subjected to non-coplanar concurrent force system. If the body is to remain in a state of equilibrium, then (a) ∑Fx = ∑Fy = ∑Fz = 0 (b) ∑Mx = ∑My = 0 (c) ∑My = ∑Mz = 0 (d) None of the above UKPSC AE 2012 Paper-I Ans. (a) : ∑Fx = ∑Fy = ∑Fz = 0 29. One end of an uniform ladder, of length L and weight W, rests against a rough vertical wall and the other end rests on rough horizontal ground. The coefficient of friction f is same at each end. The inclination of ladder when it is on the point of slipping is (a) 2 1 1 f tan 2f − − (b) 2 1 1 f tan 2f − + (c) 1 2 2f tan 1 f − + (d) 1 2 2f tan 1 f − − UKPSC AE 2012 Paper-I Ans. (a) : 2 1 1 f tan 2f − − 30. In the following figure, the tension in the rope AC is
396 (a) 17.32 N (b) 56.60 N (c) 169.90 N (d) 113.20 N UKPSC AE 2012 Paper-I Ans. (c) : From Lamie's theorem AC AD AB sin120 sin90 sin150 = = AD AB AD 10 sin 90 sin150 1 0.5 = ⇒ = AD = 20 kg AC AD sin120 sin90 = AC = 20 × sin 120 = 20 × 0.860 kg = 20 × 0.866 × 9.81 N AC = 169.91 N 31. When a body is in a state of equilibrium under the action of any force system, the normal stress at a point within the body depends upon (a) elementary area ∆A surrounding the point (b) elemental force ∆F acting normal to ∆A (c) the plane orientation containing the point (d) all the above three UKPSC AE 2007 Paper -I Ans. (d) : All the above three 32. Two forces each equal to P/2 act at right angles. Their effect may be neutralized by a third force acting along their bisector in the opposite direction with a magnitude of (a) P (b) P/2 (c) 2P (d) P 2 UKPSC AE 2007 Paper -I Ans. (d) : P 2 33. A rigid body is subjected to non-coplanar concurrent force system. If the body is to remain in a state of equilibrium, then (a) 0 ∑ = ∑ = ∑ = F F F x y z (b) 0 ∑ = ∑ = M M x y (c) 0 ∑ = ∑ = M M y z (d) none of the above UKPSC AE 2007 Paper -I Ans. (a) : 0 ∑ = ∑ = ∑ = F F F x y z 34. The resultant of forces JG G G P = -2 i - 3j and JG G G Q = 3i - 4j will lie in (quadrants to be reckoned anticlockwise) quadrant (a) first (b) second (c) third (d) fourth UKPSC AE 2007 Paper -I Ans. (d) : Fourth 35. Polygon of forces is useful for computing the resultant of (a) concurrent spatial forces (b) coplanar parallel forces (c) coplanar concurrent forces (d) coplanar collinear forces UKPSC AE 2007 Paper -I Ans. (c) : Coplanar concurrent forces 36. If the algebraic sum of all the forces acting on a body is zero, then the body may be in equilibrium provided the forces are (a) parallel (b) like parallel (c) unlike parallel (d) concurrent UKPSC AE 2007 Paper -I Ans. (d) : Concurrent 37. The quantity whose dimensions are M2L 2T –3 could be the product of (a) force and velocity (b) mass and power (c) force and pressure (d) force and distance UKPSC AE 2007 Paper -I Ans. (b) : Msass and power 38. In case of concurrent coplanar forces, the condition of equilibrium is (a) ∑ = ∑ = ∑ = H V M 0, 0, 0 (b) ∑ = ∑ = H V 0, 0 (c) ∑ = ∑ = H M 0, 0 (d) ∑ = ∑ = V M 0, 0 UKPSC AE 2007 Paper -I Ans. (b) : ∑ = ∑ = H V 0, 0 39. If a body is in equilibrium then the following is true: (a) There is no force acting on the body (b) Resultant of all forces is zero but the moments of forces about any point is not zero (c) The moments of the forces about any point is zero, but the resultant of all forces is not zero (d) both (b) and (c) UKPSC AE 2007 Paper -I Ans. (d) : Both (b) and (c) 40. Four forces P, 2P, 3 P & 4P act along the sides of a square, taken in order. The resultant force is (a) zero (b) 5P (c) 2 2P (d) 2P UKPSC AE 2012 Paper-I Ans. (c) : 2 2P 41. According to the Newton’s law of gravitation, the force of attraction, between the bodies of masses m1 and m2 situated at a distance ‘d’ apart, is given by (a) 2 1 2 2 m m F G d = (b) 2 1 2 2 m m F G d = (c) 2 2 1 2 2 m m F G d = (d) 1 2 2 m m F G d = UKPSC AE 2012 Paper-I Ans. (d) : 1 2 2 m m F G d =
397 2. Moments and Couples 42. Two non-collinear parallel equal forces acting in opposite direction (a) balance each other (b) constitute a moment (c) constitute a couple (d) constitute a moment of couple TNPSC 2019 Ans. (c) : Two non-collinear parallel equal force acting in opposite direction constitute a couple. C = P × d (clock-wise) 43. Varignon's theorem of moments states that if a number of coplanar forces acting on a particle are in equilibrium, then (a) Their algebraic then (b) Their lines of action are at equal distances (c) The algebraic sum of their moments about any point is their plane is zero (d) The algebraic sum of their moments about any point is equal to the moments of their resultant forces about the same point Vizag Steel (MT) 2017 Ans. (d) : Varignon's Theorem– The moment of resultant of concurrent forces about any point is equal to the algebraic sum of the moments of its components about the same point. 44. If a force (F) is acting on a rigid body at any point P, then this force (F) can be replaced by: (a) An equal, opposite and parallel force (F) applied at point Q together with a couple (b) An equal, parallel and same sense force (F) at point Q together with a couple (c) A moment at point Q only (d) An equal, opposite and parallel force (F) only at point Q UPRVUNL AE 2016 Ans. (b) : If a force (F) is acting on a rigid body at any point P, then this force (F) can be replaced by an equal, parallel and same sense force (F) at point Q together with a couple. 45. The moment of the force about a point is equal to the algebric sum of the component forces about the same point is known as (a) Tresca theory (b) Law of Parallelogram (c) Law of tringle (d) Varignon's theorem HPPSC AE 2018 TSPSC AEE 2015 Ans. (d) : The moment of the force about a point is equal to the algebric sum of the component forces about the same point is known as Varignon's theorem. 46. When trying to turn a key into a lock, the following is applied : (a) Coplanar force (b) Lever (c) Moment (d) Couple HPPSC W.S. Poly. 2016 UKPSC AE-2013, Paper-I Ans : (d) When trying to turn a key into a lock, the couple is applied. Couple:- The two equal and opposite forces, whose line of action are different, form a couple. Moment of a couple = P × x. 47. A 12 kN force produces a moment of 96 kN-m then, the moment arm is (a) 2 m (b) 4 m (c) 6 m (d) 8 m TSPSC AEE 2015 Ans : (d) Given, F = 12 KN T = 96 KN -m We know that, T = F× d d = 8m 48. Which of the following statement is correct? (a) the algebraic sum of the forces constituting the couple is zero (b) the algebraic sum of the force, constituting the couple, about any point is same (c) a couple cannot be balanced by a single force but can be balanced only by a couple of opposite sense (d) all statements TSPSC AEE 2015 Ans. (d) : All statements are correct. 49. Varignon’s theorem is related to (a) Principle of moments (b) Principle of momentum (c) Principle of force (d) Principle of inertia UKPSC AE 2012 Paper-I Ans. (a) : Principle of moments 50. A rigid body is acted upon by a couple. It undergoes (a) translation (b) plane motion (c) translatory rotation (d) rotation UKPSC AE 2007 Paper -I Ans. (d) : Rotation
398 51. The dimensions of angular velocity are given by (a) M0L 1T –1 (b) M0L 2T –1 (c) M0L 0T –2 (d) M0L 0T –1 UKPSC AE 2007 Paper -I Ans. (d) : M0L 0T –1 52. Opening a Limca bottle is due to (a) moment (b) couple (c) torque (d) parallel forces UKPSC AE 2007 Paper -I Ans. (a) : Moment 53. Which of the following statement is correct? (a) The algebraic sum of forces constituting the couple is zero (b) The algebraic sum of the moments of forces constituting the couple about any point is same (c) A couple cannot be balanced by a single force (d) All of the above UKPSC AE 2007 Paper -I Ans. (d) : All of the above 54. Cycle pedalling is an example of (a) couple (b) moment (c) two equal and opposite forces (d) two unequal parallel forces UKPSC AE 2007 Paper -I Ans. (a) : Couple 3. Friction 55. The maximum value of _____, which comes into play, when a body just begins to slide over the surface of other body, is known as _____. (a) Kinetic friction, Limiting friction (b) Dynamic friction, Limiting friction (c) Solid friction, Limiting friction (d) Boundary friction, Limiting friction (e) Static friction, Limiting friction (CGPCS Polytechnic Lecturer 2017) Ans. (e) : The maximum value of static friction, which comes into play, when a body just begins to slide over the surface of the other body, is known as limiting friction. Static Friction–The static friction is the frictional force that develops between mating surface when subjected to external force but there is no relative motion between them. Dynamic Friction–The dynamic friction is the frictional force that develops between mating surface when subjected to external forces and there is relative motion between them. The dynamic friction is also known as kinematic friction. Laws of Solid Friction [Static or Dynamic] 1. Friction acts tangential to the surface in contact and is in a direction opposite to that in which motion is to impend i.e. take place. 2. Friction force is maximum at the instant of impending motion. Its variation from zero to maximum value [Limiting friction] depends upon the resultant force tending to cause motion. 3. The magnitude of limiting friction bears a constant ratio to the normal reaction between the mating surface. 4. Limiting friction is independent of the area and shape of contact surface. 5. Limiting friction depends upon the nature (roughness or smoothness) of the surface in contact. 6. At low velocities between sliding surface, the friction force is practically independent of the velocity. However, slight reduction in friction occurs when the speed are high. 56. If F is the limiting friction and RN is the normal reaction of the surfaces of contact of two bodies, then the coefficient of friction is expressed as: (a) µ = RN/F (b) µ = F/RN (c) µ = 2F/RN (d) µ = 0.5F/RN (e) µ = 0.5RN/F (CGPCS Polytechnic Lecturer 2017) Ans. (b) : Where, S = Total reaction with the normal reaction. The ratio of friction force to normal reaction is called coefficient of friction (µ). So, µ = N F R tanφ = µ = N F R , φ → Friction angle 57. The coefficient of rolling resistance is defined as the ratio between (a) Rolling resistance to lateral load (b) Lateral load to rolling resistance (c) Rolling resistance to normal load (d) Normal load to rolling resistance TNPSC AE 2017 Ans. (c) : The coefficient of rolling resistance is defined as the ratio between Rolling resistance to normal load. Rolling resistance coefficient (Crr) = FRN 58. On a ladder resting on a smooth ground and leaning against rough vertical wall, the force of friction acts (a) upward at its upper end (b) towards the wall at the upper end (c) towards the wall at lower end (d) downward at its upper end BPSC Poly. Lect. 2016 TNPSC AE 2018 UKPSC AE 2007 Paper -I
399 Ans : (a) for smooth ground µ1 = 0 on a ladder resting on a smooth ground and leaning against rough vertical wall, the force of friction acts towards the wall at the upper end. 59. The coefficient of friction is the ratio of (a) the normal reaction to the limiting force of friction (b) The weight of the body to limiting force of friction (c) the limiting force of friction to the normal reaction (d) the weight of the body to the normal reaction TSPSC AEE 2015 Ans : (c) The coefficient of friction is the ratio of the limiting force of friction to the normal reaction F = µR. F R µ = µ = Coefficient of friction. F = Limiting force of friction R = Normal reaction. 60. Which statement is wrong in wedge friction? (a) To lift heavy block through small distances (b) To lift heavy block through large force (c) To slightly slide one end of the beam relative to another end (d) Weight of wedge is neglected compared to weight to be lifted TNPSC AE 2013 Ans. (b) : Wedge– • Simple machines used to raise heavy loads. • Force required to life block is significantly less than block weight. • Friction prevents wedge from sliding out. • Minimum force D required to raise block. 61. A box weight 1000 N is placed on the ground. The coefficient of friction between the box and the ground is 0.5. When the box is pulled by a 100 N horizontal force, the frictional force developed between the box and the ground at impending motion is (a) 50 N (b) 75 N (c) 100 N (d) 500 ESE 2018 Ans. (c) : Given, W = 1000 N µ = 0.5 P = 100 N Frictional force (F) = µN = 0.5 × 1000 (∵ N = W) = 500 N Since applied force (100N) is less than limiting friction, body will be at rest. As long as the body, is at rest F = P = 100 N 62. A 13 m long ladder is placed against a smooth vertical wall with its lower end 5 m from the wall. What should be the coefficient of friction between the ladder and the floor so that the ladder remains in equilibrium? (a) 0.29 (b) 0.25 (c) 0.21 (d) 0.11 ESE 2018 Ans. (c) : Using equilibrium equation N2 = µN1 N1 = W (∵ µ2N2 = 0) Because wall is smooth Taking moment about A W × 2.5 + µ1N1 × 12 = N1 × 5 µ1 = 0.208 µ1 ⇒ 0.21 63. A body of weight 50 N is kept on a plane inclined at an angle of 30º to the horizontal. It is in limiting equilibrium. The coefficient of friction is equal to- (a) 1 3 (b) 3 (c) 1 50 3 (d) 3 5 RPSC AE 2018 Ans. (a) : Given: mg = 50 N θ = 30º For limiting equilibrium mg sin 30º = µ × R mg sin 30º = µ × mg cos 30º µ = tan 30º 1 3 µ =
400 64. A block of mass (m) slides down an inclined plant (coefficient of friction = µ) from the rest as in figure. What will be the velocity of block when it reaches the lowest point of plane? (a) 2gh (1 – µ cot θ) (b) 2gh 1 cot ( ) − µ θ (c) 2ghµ 1 cot ( ) − θ (d) 2gh 1 sin ( ) − µ θ (e) 2gh tan ( ) µ − θ CGPSC AE 2014- I Ans. (b) : Kinetic energy at lowest point on inclined plane of the block is ∴ 1 2 mV 2 = mgh–energy loss due to friction h mgh R sin = − µ × θ h mgh mg cos sin = − µ× θ× θ V 2gh 1 cot = − µ θ [ ] h sin θ is distance covered by block during motion on inclined plane. 65. What is the minimum coefficient ( ) of friction between the rope and the fixed shaft which will prevent the unbalanced cylinder from moving? (a) µ = 0.333 (b) µ = 0.350 (c) µ = 0.253 (d) µ = 0.372 APPSC-AE-2019 Ans. (b) : 1 2 = T e T αθ α is angle of contact between rope and cylinder α = π radian 3 = m e m µπ 3 = eµπ 1.098 = µπ µ = 0.349 ≈ 0.35 66. The angle of inclination of the plane at which the body begins to move down the plane, is called– (a) Angle of friction (b) Angle of repose (c) Angle of projection (d) None of these Vizag Steel (MT) 2017 Ans. (b) : Angle of repose– Minimum angle of inclined plane which causes on object to slide down the plane. 67. Coefficient of friction depends upon (a) Area of contact only (b) Nature ofsurfaceonly (c) Both (a) and (b) (d) None of these Vizag Steel (MT) 2017 Ans. (b) :Coefficient of friction depends upon nature of surface only. 68. A block weighing 981 N is resting on a horizontal surface. The coefficient of friction between the block and the horizontal surface is 0.2. A vertical cable attached to the block provides partial support as shown. A man can pull horizontally with a force of 100N. What will be the tension 'T' in the cable if the man is just able to move the block to the right? (a) 176.2N (b) 196.0 N (c) 481.0N (d) 981.0N OPSC Civil Services Pre. 2011 Ans. (c) : Free body diagram Since, N + T = W N + T = 981....................(i) µN = 100 ...................(ii) From equation (i) and (ii), N = 100 0.2 = 1000 2 = 500 T = 981 – 500 T 481N = 69. Limiting friction depends upon (a) Materials of the body in contact (b) Weight of the body to be moved
401 (c) Roughness of surface of contact of the two bodies (d) All of the above Gujarat PSC AE 2019 Ans : (d) : Limiting Friction- The maximum value of frictional force, which comes into play when a body just begins to slide over the surface of the other body, is known as limiting friction. Limiting friction depends upon • Materials of the body in contact. • Weight of the body to be moved. • Roughness of surface of contact of the two bodies. 70. A block resting on an inclined plane begins to slide down the plane when the angle of inclination is gradually increased to 30°. The coefficient of friction between the block and the plane is:- (a) 0.50 (b) 0.578 (c) 0.72 (d) 0.866 UKPSC AE-2013, Paper-I Ans. (b) : From this figure. R = W cos 30o µR = W sin 30o µ × W cos 30o = W sin 30o 3 1 × 2 2 µ = µ = 0.5773 71. When the applied force is less than the limiting frictional force, the body will:- (a) Start moving (b) Remain at rest (c) Slide backward (d) Skid UKPSC AE-2013, Paper-I Ans. (d) : When the applied force is less than the limiting frictional force, the body will remain at rest. 72. A 44 N block is thrust up a 30° inclined plane with an initial speed of 5 m/sec. It travels a distance of 1.5 m before it comes to rest. The frictional force acting upon it would be (a) 18.3 N (b) 15.3 N (c) 12.3 N (d) 9.3 N UKPSC AE 2012 Paper-I Ans. (b) : 15.3 N 73. Dynamic friction as compared to static friction is (a) less (b) same (c) more (d) None of the above UKPSC AE 2012 Paper-I Ans. (a) : less 74. If a force of 30 N is required to move a mass of 35 kg on a flat surface horizontally at a constant velocity, what will be the coefficient of friction? (a) 0.067 (b) 0.087 (c) 0.098 (d) 0.092 UKPSC AE 2012 Paper-I Ans. (b) : R – 35 g = 0 R = 35 g F = µR F 30 0.087 R 35 9.8 µ = = = × 75. Two bodies of mass m and M are hung at the ends of a rope passing over a frictional pulley. The acceleration in which the heavier mass M comes down is given by the following: (a) g M m ( ) M m + + (b) g M m ( ) M m + − (c) gM M m+ (d) g M M m × − UKPSC AE 2007 Paper -I Ans. (a) : g M m ( ) M m + + 76. The maximum frictional force, which comes into play when a body first begins to slide over the surface of another body, is known as (a) sliding friction (b) limiting friction (c) kinetic friction (d) rolling friction UKPSC AE 2007 Paper -I UKPSC AE 2012 Paper-I Ans. (b) : Limiting friction 77. Tangent of angle of friction is equal to (a) kinetic friction (b) limiting friction (c) friction force (d) coefficient of friction UKPSC AE 2007 Paper -I Ans. (d) : Coefficient of friction 78. A body weight of 200 N is resting on a rough horizontal plane, and can be just moved by a force of 80 N applied horizontally. What will the value of the coefficient of friction? (a) 0.4 (b) 0.5 (c) 0.3 (d) none of the above UKPSC AE 2007 Paper -I Ans. (a) : F = µR µ = 80 0.4 200 = 79. The coefficient of friction depends upon (a) speed of the body (b) geometrical shape of the body (c) size of the body and nature of contacting surfaces
402 (d) nature of contacting surfaces UKPSC AE 2007 Paper -I Ans. (d) : Nature of contacting surfaces 80. A steel wheel of 600 mm diameter rolls on a horizontal steel rail. It carries a load of 500N. The coefficient of rolling resistance is 0.3 mm. The force necessary to roll the wheel along the rail is– (a) 0.5 N (b) 5 N (c) 15 N (d) 150 N OPSC Civil Services Pre. 2011 Ans. (d) : Given, Diameter of wheel (d) = 600 mm Radius of wheel (r) = 300 mm Load (W) = 500 N Coefficient of rolling resistance (Crr)= 0.3 Let, Rolling friction force be FR We know that, Crr = ( ) Rolling friciton force Fr Normal force Load W = ∴ = × = × F C W 0.3 500 r rr F 150N r = 4. Moment of Inertia 81. Moment of Inertia of an area dA at a distance x from a reference axis is: (a) ∫xdA (b) ∫x 2 dA (c) ∫x 3 dA (d) ∫x 4 dA (HPPSC LECT. 2016) Ans : (b) Moment of Inertia of an area dA at a distance x from a reference axis is ∫x 2 dA The axial moment of inertia of a plane area is the geometrical characteristic of the area defined by the integrals Ixx = ∫y 2 .dA and Iyy = ∫x 2 .dA 82. Moment of inertia of a body does not depend on (a) Axis of rotation of the body (b) Mass of the body (c) Angular velocity of the body (d) distribution of mass in the body (KPSC AE 2015) Ans : (c) Moment of Inertia of a body does not depend on angular velocity of the body 2 2 xx yy I y .dA and I x dA = = ∫ ∫ Where x and y the Co - ordinate of the differintial elements of area dA. Since this area dA is multiplied by the square of the distance, the moment of inertia is also called the second moment at area. 83. For principal axes, the moment of inertia will be : (HPPSC LECT. 2016) (a) Ixy = 0 (b) Ixy = 1 (c) Ixy = ∞ (d) none of these Ans : (b) For principal axes, the moment of Inertia Ixy = 0 84. The moment of inertia of a rectangular section of base (b), height (h) about its base will be : (a) 3 hb 12 (b) 3 bh 3 (c) 3 bh 12 (d) 3 hb 3 (e) 3 bh 36 CGPSC AE 2014- I Ans. (b) : Moment of inertia of a rectangular section about its base (x - x). 2 I I Ax x x CG − = + 3 C.G. bh h I ,A b h, x 12 2 = = × = 3 3 3 3 2 x x bh h bh bh bh I b h 12 2 12 4 3 − = + × × = + = 3 x x bh I 3 − = 85. Moment of inertia of a quarter circle (diameter = d) about its straight edge is given by : (a) 4 d 64 π (b) 4 d 128 π (c) 4 d 256 π (d) 4 d 512 π (e) 4 d 32 π CGPSC AE 2014- I
403 Ans. (c) : 4 OX 1 d I 4 64 π = × 4 OX d I 256 π = 86. The radius of gyration 'k' for a solid cylinder of radius 'R' is equal to (a) 2R (b) R / 2 (c) 0.6324 R (d) 0.5 R TNPSC AE 2017 Ans. (b) : The radius of gyration 'k' for a solid cylinder of radius 'R' is equal to R / 2 . 87. The radius of gyration of a disc type fly wheel of diameter D is (a) D (b) D/2 (c) D/4 (d) 3 D 2 TNPSC AE 2018 Ans. (c) : I = AK2 4 2 D I 64 K A D 4π = = π D K 4 = 88. What is the unit of moment of inertia of an area. (a) kg m2 (b) kg m (c) m3 (d) m4 (e) kg m3 CGPSC 26th April 1st Shift Ans. (d) : Units of the mass moment of inertia are kg- m 2 , gram-cm2 . Units of the area moment of inertia are m4 , mm4 . 89. Moment of inertia of an elliptical area about the major axis is (a) πxy3 /4 (b) πxy3 /3 (c) πx 2 y 3 /4 (d) πx 2 y 3 (e) πx 2 y 3 /3 CGPSC 26th April 1st Shift Ans. (a) : Moment of inertia of an elliptical area about the major axis 3 4 I xy xx = π Moment of inertia of an elliptical area about minor axis 3 4 I yx yy = π 90. Statement (I): Two circular discs of equal masses and thickness made of different materials will have same moment of inertia about their central axes of rotation. Statement (II): Moment of inertia depends upon the distribution of mass within the body. ESE 2017 Ans. (d) : Mass moment of inertia for a uniform disc about its axis of rotation I = 2 MR 2 I = 2 AtR 2 ρ ∵ Mass & thickness is same, ρ1A1t = ρ2A2t I1 = 2 1 1 1 A tR 2 ρ = 4 1 1 R t 8π ρ I2 = 2 2 2 2 A tR 2 ρ = 4 2 2 R t 8π ρ If ρ1 > ρ2, R1 < R2, I1 > I2 Hence statement-I is wrong. 91. A polar moment of Inertia (I) for hollow shaft with external diameter (D) and internal diameter (d) is given by: (a) 4 4 32D πd (b) ( ) 4 4 D d 64 π − (c) ( ) 4 4 D d 32 π − (d) ( ) 4 4 32 D d − π CIL (MT) 2017 IInd Shift (CGPCS Polytechnic Lecturer 2017) Ans. (c) : Polar moment of inertia of hollow haft = ( ) 4 4 D d 32 π − 92. Moment of inertia of an area always least with respect to (a) Bottom-most axis (b) Radius of gyration (c) Central axis (d) Centroidal axis Vizag Steel (MT) 2017 Ans. (d) : The moment of inertia about any axis passing through centroid is zero so Moment of inertial of an area always least with respect to centroidal axis. 93. Polar moment of inertia (Ip), in cm4 , of a rectangular section having width, b = 2 cm and depth, d = 6 cm is : (a) 40 (b) 20 (c) 8 (d) 80 OPSC AEE 2019 Paper-I
404 Ans : (a) : Polar moment of inertia (J) = Ix + Iy = 3 3 12 12 + bd db = 3 3 2 6 2 6 12 12 × × + = 40 cm4 94. Radius of gyration of a circular section with diameter D is (a) 2 D (b) 4 D (c) 3 D (d) 3 D APPSC-AE-2019 Ans. (b) : For circular section 4 2 , 64 4 I d A d π π = = Radius of gyration 4 2 64 4 4 d I d k A d π π = = = = 95. Moment of Inertia about the centroidal axis of elliptical quadrant of base 'a' and height 'b' is (a) 2 2 ( ) 24 ab a b π + (b) 3 3 ( ) 24 ab a b π + (c) 2 2 ( ) 16 ab a b π + (d) 3 3 ( ) 24 ab a b π + TNPSC AE 2013 Ans. (c) : ( ) ab 2 2 I a b 16 π = + 96. When a body of mass moment of Inertia I is rotated about that axis with an angular velocity , then the kinetic energy of rotation is (a) 0.5 I.ω. (b) I.ω. (c) 0.5 I.ω 2 (d) I.ω 2TNPSC AE 2013 Ans. (c) : Kinetic energy of rotation is given as 1 2 KE I 2 = ω 97. Length to radius ratio lr of a solid cylinder is such that the moments of inertia about the longitudinal and transverse axes are equal is (a) 1 (b) 3 (c) 5 (d) 2 TNPSC AE 2014 Ans. (b) : MOI of a uniform solid cylinder about longitudinal axis 2 L L mR I 2 − = MOI of a uniform solid cylinder about transfer axis 2 2 T A 1 1 I mR mL 4 12 − = + Given- IL-L = IT-A 2 2 mR mR 1 2 mL 2 4 12 = + 2 L 3 R = [L / R 3 ] = 98. The moment of inertia of a square side (a) about an axis through its center of gravity is (a) a4 /4 (b) a4 /8 (c) a4 /12 (d) a4 /36 Vizag Steel (MT) 2017 Ans. (c) : 4 X X Y Y a I I 1 2 = = 99. Ratio of moment of Inertia of a circular body about its x axis to that about y axis is (a) 0.5 (b) 1.0 (c) 1.5 (d) 2.0 TNPSC AE 2014 Ans. (b) : 4 x x d I 64 − π = , 4 y y d I 64 − π = 4 z z xx yy d I I I 32 − π = + = then 4 x x 4 y y d I 64 I d 64 − − π = π = 1
405 5. Centroid and Center of Gravity 100. The C.G. of a solid hemisphere lies on the central radius (a) at distance 3r/2 from the plane base (b) at distance 3r/4 from the plane base (c) at distance 3r/5 from the plane base (d) at distance 3r/8 from the plane base TNPSC 2019 Ans. (d) : C.G. of hemisphere is at a distance of 3r 8 from its base measured along the vertical radius. 101. The coordinates of centroid of given geometry ABCDEFA [DE = 20 mm, EF = 80 mm, FA = 100 mm, AB = 20mm] will be given as : (a is origin) (a) (60, 20) (b) (25, 60) (c) (25, 65) (d) (65, 25) (e) (25, 40) CGPSC AE 2014- I Ans. (c) : The coordinates of centroid is (x, y) 1 1 2 2 1 2 a x a x x a a + = + 80 20 10 80 20 40 80 20 80 20 × × + × × = × + × x 25mm = 1 1 2 2 1 2 a y a y 80 20 40 80 20 90 y 65mm a a 80 20 80 20 + × × + × × = = = + × + × So coordinates of centroid (25, 65) 102. Shear centre of a semicircular arc is at : (a) 4r/π (b) 3r/π (c) 2r/π (d) r/π (HPPSC AE 2014) Ans : (a) Shear centre of a semicircular arc is at 4r/π. 103. Centre of gravity of a thin hollow cone lies on the axis at a height of : (a) one-fourth of the total height above base (b) one-third of the total height above base (c) one-half of the total height above base (d) three-eighth of the total height above the base HPPSC W.S. Poly. 2016 Ans : (b) Centre of gravity of a thin hollow cone lies on the axis at a height of one-third of the total height above base. 104. The centre of gravity of a hemisphere of diameter 80 mm form its base diameter is (a) 15 mm (b) 40 mm (c) 20 mm (d) 10 mm TSPSC AEE 2015 Ans : (a) The centre of gravity of a hemisphere is at a distance of 3r/8 from base C.G = 3r/8 (2r = d= 80mm) C.G = 15 mm. 105. The base of triangle is 60 mm and height is 50 mm. The moment of inertia about its base 'BC' is (a) 2,08, 333.33mm4 (b) 6.25,000 mm4 (c) 9,00,000 mm4 (d) 3,00, 000mm4 TSPSC AEE 2015 Ans : (b) IGG = 3 bh 36 3 BC bh I 12 = ( )3 BC 60 50 I 12 × = IBC = 625000 mm4 . 106. The centre of gravity of a plane lamina is not at its geometrical centre, if it is a (a) circle (b) square (c) rectangle (d) right angled triangle UKPSC AE 2007 Paper -I Ans. (d) : Right angled triangle
406 107. The eccentricity for the ellipse is ______ 1 and for hyperbola is ______ 1. (a) equal to, equal to (b) less than, greater than (c) greater than, greater than (d) less than, less than (e) Less than, equal to (CGPCS Polytechnic Lecturer 2017) Ans. (b) : The eccentricity for the ellipse is less than 1 and for hyperbola is greater than 1. 108. Which of the following represents the state of neutral equilibrium? (a) Cube resting on one edge (b) A smooth cylinder lying on a curved surface (c) A smooth cylinder lying on a convex surface (d) None of the above UKPSC AE 2007 Paper -I Ans. (d) : None of the above 109. The coordinate of centroid (x, y) of quarter circular lamina of radius (R), whose straight edges coincide with the coordinate axis in the first quadrant, are given as: (a) (0,4R / 3π) (b) (4R / 3 ,0 π ) (c) (4R / 3 , 8R / 3 π π) (d) (4R / 3 , 4R / 3 π π) (e) (8R / 3 , 8R / 3 π π) CGPSC AE 2014- I Ans. (d) : For circle 2rsin x 3 α = α In case quadrant of a circle- o 2 90 radians 2π α = = 2rsin / 4 ( ) 4r x 2 3 3 4π = = × π π × The position of centroid with respect to the radii OA and OB will be 4r 4r o oa ' 2 cos 45 3 3 = × × = π π 4r 4r o ob ' 2 cos 45 3 3 = × = π π 6. Trusses and Beams 110. Which one of the following is the wrong assumption: (a) Members of the truss are pin-connected to each other (b) Members of the truss are rigid (c) Members of the truss are subjected to bending moments (d) Members are of uniform cross-section TNPSC AE 2013 Ans. (c) : Members of the truss are not subjected to bending moments. 111. In a simple truss, if n is the total number of joints, the total number of members is equal to (a) 2n + 3 (b) 2n - 3 (c) n + 3 (d) n - 3 Gujarat PSC AE 2019 Ans : (b) : m 2n 3 = − Where, m = member of joints n = Total number of joints 112. A rectangular strut is 150 mm wide and 120 mm thick. It carries a load of 180 kN at an eccentricity of 10 mm in a plane bisecting the thickness as shown in the figure The maximum intensity of stress in the section will be (a) 14 MPa (b) 12 MPa (c) 10 MPa (d) 8 MPa ESE 2019 Ans. (a) : Given, P = 180 kN b = 150 mm d = 120 mm e = 10 mm Resultant normal stress is maximum at the right side fiber of the cross section, because the line of action of eccentric axial compressive load is nearer to this fiber. Maximum intensity of stress = σc + σb σmax = P M A Z + = 2 P P e A db 6× + = 2 P 6P e bd db× + = P 6 e 1 bd b × + = 3 180 10 6 10 1 120 150 150 × × + × = 14 MPa
407 113. Choose the correct combination of the stability and indeterminacy of the truss given. (a) Statically indeterminate and stable (b) Statically determinate and stable (c) Unstable (d) Statically determinate APPSC-AE-2019 Ans. (c&d) : No. of member m = 8 No. of joints, j = 6 m < 2j - 3 ∴ The truss is internally unstable but determinate ∴ Both the options (c) and (d) are correct 114. Regarding the ability of a truss, the condition m + r > 2j is (a) Necessary (b) Sufficient (c) Necessary and Sufficient (d) Sufficient but not necessary APPSC-AE-2019 Ans. (d) : Given condition m + r > 2j m > 2j - r ∴ The member is structurally stable but not necessarily have m > 2j - r. It is just sufficient to have m = 2j - r 115. Determine the nature of force in member AB and AC respectively of the truss shown in figure. A is hinge support and B is roller support. The direction of reaction at supports (RA and RB) is also shown. Load 50N is acting at joint C. (a) Tensile, Compression (b) Compression, Compression (c) Tensile, Tensile (d) Compression, Tensile UPRVUNL AE 2016 Ans. (a) : 90º Joint-C From Lami's theorem sin 90 sin120 sin150 F F F AC BC = = 50 sin120 43.3 F F N AC AC = × ⇒ = (compressive) Joint-A then, FAB = FAC cos 60 = 43.3 × cos 60 FAB = 21.65 N (Tensile) RA = 37.498 N 116. In the joint method of plane truss analysis, value of forces in the member of truss can be found on when joint has: (a) Only four unknown force members (b) Only three unknown force members (c) Not more than two unknown force members (d) Any number of unknown force members UPRVUNL AE 2016 Ans. (c) : In the joint method of plane truss analysis, value of forces in the member of truss can be found out when joint has not more than two unknown force members. 117. Which method is not there to analysing the trusses? (a) Graphical Method (b) Analytical Method (c) Method of Joints (d) Method of Sections TNPSC AE 2013 Ans. (b) : Analytical method is not to analysing the trusses where as graphical method, joints method and method of sections are used to analysing the trusses. 118. What will be the axial force in the member EC, ED and DC of the plane truss (ABCDE) as shown in figure with end 'A' and 'B' is hinged to foundation?
408 (a) FEC = FED = FDC = 0 (b) FEC = FED = 0; FDC = 1 kN (C) (c) FEC = FED = 0; FDC = 1 kN (T) (d) FEC = FED = FDC = 1 kN(T) (e) FEC = FED = FDC = 1 kN (C) CGPSC AE 2014- I Ans. (c) : Point EPoint DF cos 45º = 1 FDC = F sin 45º = F cos 45º = 1 kN (T) 119. Determine the force in the member BG in the given truss (a) 11.18 kN Tension (b) 14.4 kN tension (c) 11.18 kN Compression (d) 14.40 kN compression APPSC-AE-2019 Ans. (c) : Take section 1 - 1 as shown in figure. To find the force in BG, consider the right part of section (1) - (1) and take moment about 'D'. 6 2 cos 3 5 5 θ = = 3 1 sin 3 5 5 θ = = 0 ∑ = MD ( + ve, - ve) -FBG cos θ × 3 - FBG sin θ × 6 - 2 × 6 - 8 × 6 = 0 6 6 60 0 5 5 − × × × − = F F BG BG 6 2 60 5 FBG − × = 5 5 kN (compressive) FBG = − 11.18 kN (compressive) FBG = 120. A member in a truss can take (a) axial force and bending moment (b) only axial force (c) only bending moment (d) bending moment and shear force APPSC-AE-2019 Ans. (b) : Truss members take only axial forces (tension or compression). They cannot take shear force, bending moment and torsion. 121. The condition for a truss to be perfect is where m = number of members and J = number of joints (a) m = 2j - 3 (b) m > 2j - 3 (c) m < 2j - 3 (d) m ≥ 2j - 3 TSPSC AEE 2015 Ans. (a) : Statically determinate (Perfect truss) m = 2j - 3 Truss unstable (deficient truss) m < 2j - 3 statically indeterminate (redundant truss) m > 2j - 3 122. The relation between the number of joints (J) and number of members (m) in a truss is related by : (a) m = 2J + 3 (b) J = 3m + 3 (c) m = 2J – 3 (d) J = 3(m–1) OPSC Civil Services Pre. 2011 Ans. (c) : The relation between the number of joints (J) and number of members (m) in a truss is related by– m 2J 3 = − 123. For a perfect frame having 13 members, the number of joints must be : (a) 6 (b) 8 (c) 10 (d) 13 OPSC Civil Services Pre. 2011 Ans. (b) : Given, m = 13 m = 2J – 3 13 = 2J – 3 ⇒ J 8 = 124. In method of section, the section must pass through not more than members. (a) 3 (b) 4 (c) 5 (d) 2 OPSC Civil Services Pre. 2011 Ans. (a) : In method of section, the section must pass through not more than 3 members.
409 125. If n < (2j - 3), where n is number of members and used in a frame structure and j is the number of joints used in the structure, then the frame is called (a) Perfect frame (b) Deficient frame (c) Redundant frame (d) None of the above Gujarat PSC AE 2019 Ans : (b) : n < (2j – 3) (Deficient frame) Where, n = number of members j = number of joints 126. Which of the following equilibrium equation should be satisfied by the joints in truss:- (a) (b) ∑ = ∑ = H V 0, 0 (c) ∑ = ∑ = V M 0, 0 (d) ∑ = ∑ = H V 0, 0 and ∑ = M 0 UKPSC AE-2013, Paper-I Ans. (b) : No bending moment acting on the joints in truss so equilibrium equation should be satisfied by the joints in truss will be, Σ = Σ = H 0, V 0 127. When the number of members ‘n’ in a truss is more than 2j-3, where ‘j’ is the number of joints, the frame is said to be:- (a) Perfect truss (b) Imperfect truss (c) Deficient truss (d) Redundant truss UKPSC AE-2013, Paper-I Ans. (d) : If (n > 2j – 3), then it become redundant truss. 128. Choose the correct relationship between the given statements of Assertion (A) and Reason (R). Assertion (A) : Only axial forces act in members of roof trusses. Reason (R) : Truss members are welded together. Code : (a) Both (A) & (R) are correct. (R) is the correct explanation of (A). (b) Both (A) & (R) are correct. (R) is not the correct explanation of (A). (c) (A) is true, but (R) is false. (d) (A) is false, but (R) is true. UKPSC AE 2012 Paper-I Ans. (c) : (A) is true, but (R) is false. 129. The relationship, between number of joints (J), and the number of members (m), in a perfect truss, is given by (a) m = 3j – 2 (b) m = 2j – 3 (c) m = j – 2 (d) m = 2j – 1 UKPSC AE 2012 Paper-I Ans. (b) : m = 2j – 3 130. In the analysis of truss, the force system acting at each pin (a) is concurrent but not coplanar. (b) is coplanar and concurrent. (c) is coplanar and non-concurrent. (d) does not satisfy rotational equilibrium. UKPSC AE 2012 Paper-I Ans. (b) : is coplanar and concurrent. 131. For truss as shown below, the forces in the member AB and AC are (a) Tensile in each (b) Compressive in each (c) Compressive and Tensile respectively (d) Tensile and Compressive respectively UKPSC AE 2012 Paper-I Ans. (c) : Compressive and Tensile respectively 132. The possible loading in various members framed structure are (a) Buckling or shear (b) Compression or tension (c) Shear or tension (d) Bending UPPSC AE 12.04.2016 Paper-I Ans : (b) The possible loading in various members framed structure are compression or tension. 133. A truss hinged at one end, supported on rollers at the other, is subjected to horizontal load only. Its reaction at the hinged end will be: (a) Horizontal (b) Vertical (c) Resultant of horizontal and vertical (d) Difference between horizontal and vertical TRB Polytechnic Lecturer 2017 Ans. (c) : Its reaction at the hinged end will be resultant of horizontal and vertical. ( ) ( ) 2 2 R H V = ∆ + ∆ 134. The force which is not considered in the analysis of the truss is (a) External applied loads (b) Horizontal forces on joints (c) Vertical forces joints (d) Support reactions (e) Weight of the members CGPSC 26th April 1st Shift Ans. (e) : Assumption for a perfect truss (i) All the members of truss are straight and connected to each other at their ends by frictionless pins. (ii) All external forces are acting only at pins. (iii) All the members are assumed to be weightless. (iv) All the members of truss and external forces acting at pins lies in same plane. (v) Static equilibrium condition is applicable for analysis of perfect truss. 7. Kinematics and Kinetics of Particle 135. The motion of a body moving on a curved path is given by a equation x = 4 sin 3t and y = 4 cos3t. The resultant velocity of the car is (a) 30 m/sec (b) 24 m/sec (c) 12 m/sec (d) 40 m/sec TNPSC AE 2014
410 Ans. (c) : x = 4 sin 3t y = 4 cos 3t velocity component in x & y direction respectively then x dx u 4cos3t 3 dt = = × =12 cos 3t y dy v 4sin 3t 3 dt = = − × vy = – 12 sin 3t Resultant velocity - 2 2 x y v u v = + ( ) ( ) 2 2 = + − 12cos 3t 12sin 3t ( ) 2 2 = + 144 sin 3t cos 3t = × 144 1 v 12 m / s = 136. The linear velocity of a body rotating at ω rad/s along a circular path of radius r is (a) ω.r (b) ω/r (c) 2 ω .r (d) 2 ω / r APPSC AEE 2016 Ans. (a) : We know that, linear velocity, V = ω.r 137. The midpoint of a rigid link of a mechanism moves as a translation along a straight line, from rest, with a constant acceleration of 5 m/ s2 . The distance covered by the said midpoint in 5s of motion is (a) 124.2 m (b) 112.5 m (c) 96.2 m (d) 62.5 m ESE 2018 Ans. (d) : Given, a = 5 m/s2 u = 0 (rest) t = 5 sec Distance covered (s) = 1 2 ut at 2 + = 1 2 0 5 5 2 + × × = 62.5 m 138. The functional reaction between time t and distance x, in m, is t = 2x2 + 5x.. The acceleration in m/s2 at t = 12 s is : (a) 1 units 121 − (b) 4 units 1331 − (c) 4 units 121 − (d) 4 units 1331 BHEL ET 2019 Ans. (c) : Given- t = 2x2 + 5x Acceleration a (m/s2 ) at t = 12 sec. t = 2x2 + 5x dt 4x 5 dx = + or dx 1 dt 4x 5 = + ( ) dx 1 v dt 4x 5 = = + ( ) ( ) ( ) dv 4x 5 0 1 4 a dt 4x 5 2 + × − = = + ( ) 2 4 a 4x 5 − = + ....(1) ∴ t = 12 t = 2x2 + 5x 2x2 + 5x = 12 2x2 + 5x + 2 = 0 (x + 4) (2x – 3) = 0 x = – 4, 2x – 3 = 0 3 x 2 = x = 4 value is substitute in equation (1) ( ) ( ) 2 2 4 4 a 4 4 5 11 − − = = × − + 4 unit 121 − = 139. Time variation of the position of the particle in rectilinear motion is given by x = 3t3 + 2t2 + 4t. If v is the velocity and a is the acceleration of the particle in consistent units, the motion started with (a) v = 0, a = 0 (b) v = 0, a = 4 (c) v = 4, a = 4 (d) v = 2, a = 4 (e) v = 4, a = 2 CGPSC 26th April 1st Shift Ans. (c) : Given, x = 3t3 + 2t2 + 4t dx 2 velocity (v) = 9 4 4 dt = + + t t ...(1) dv Acceleration (a) = 18 4 dt = +t ...(2) ∵ Initially t = 0 so from equation (1) and (2) v = 9t2 + 4t + 4 at t = 0, v = 4 a = 18t + 4 at t =0, a = 4 140. When a particle moves with a uniform velocity along a circular path, then the particle has (a) tangential acceleration only (b) normal component of acceleration only (c) centripetal acceleration only (d) both tangential and centripetal acceleration TNPSC AE 2013 Ans. (c) : When a particle moves with a uniform velocity along a circular path, then the particle has centripetal acceleration only. 141. A motorist travelling at a speed of 18 km/hr, suddenly applies the brakes and comes to rest after skidding 75 m. The time required for the car to stop is (a) t = 30.25 sec (b) t = 29.94 sec (c) t = 28.84 sec (d) t = 26.22 sec TNPSC AE 2014
411 Ans. (b) : u = 18 km/hr = 5 m/s d = 75 m v = 0 We know that after applies the brake- v 2 = u 2 – 2ad (5)2 = 2 × a × 75 1 2 a m / s 6 = then v = u – a t u = a × t 5 = 1 6 × t t 30 sec = 142. The angular motion of a disc is defined by the relation ( = 3t + t3 ), where is in radians and t is in seconds. What will be the angular position after 2 seconds? (a) 14 rad (b) 12 rad (c) 18 rad (d) 16 rad CIL MT 2017 2017 IInd shift Ans. (a) : Given angular motion equation θ = 3t + t3 Motion covered in 2 seconds = 3 * 2 + 23 = 14 rad. So the angular position after 2 seconds will be 14 rad. 143. The position of a particle in rectilinear motion is given by the equation (x = t3 - 2t2 + 10t - 4), where x is in meters and t is in seconds. What will the velocity of the particle at 3s? (a) 20 m/s (b) 25 m/s (c) 15 m/s (d) 30 m/s CIL MT 2017 2017 IInd shift Ans. (b) : Position (x) = t3 - 2t2 + 10t - 4 Velocity ( ) dx 2 v 3t 4t 10 dt = = − + At t = 3s V = 3 × 32 - 4 × 3 + 10 = 25 m/s 144. A particle move along a straight line such that distance (x) traversal in t seconds is given by x = t2 (t - 4), the acceleration of the particle will be given by the equation (a) 3t3 - 2t (b) 3t2 + 2t (c) 6t - 8 (d) 6t - 4 TNPSC AE 2018 Ans. (c) : x = t2 (t - 4) dx 2 V 3t 8t dt = = − G 2 2 d x dV a 6t 8 dt dt = = = − G 145. A motor car moving at a certain speed takes a left turn in a curved path. If the engine rotates in the same direction as that of wheels, then due to the centrifugal forces (a) the reaction on the inner wheels increases and on the outer wheels decreases (b) the reaction on the outer wheels increases and on the inner wheels decreases (c) the reaction on the front wheels increases and on the rear wheels decreases (d) the reaction on the rear wheels increases and on the front wheels decreases RPSC Vice Principal ITI 2018 Ans. (b) : The reaction on the outer wheels increases and on the inner wheels decreases. 146. The mathematical technique for finding the best use of limited resources in an optimum manner is called:- (a) Linear programming (b) Network analysis (c) Queueing theory (d) None of the above UKPSC AE-2013, Paper-I Ans. (a) : The mathematical technique for finding the best use of limited resources in an optimum manner is called linear programming. 147. An object falls from the top of a tower. If comes down half the height in 2 seconds. Time taken by the object to reach the ground is:- (a) 2.8 s (b) 3.2 s (c) 4.0 s (d) 4.5 s UKPSC AE-2013, Paper-I Ans. (a) : We assume that the height of tower is h, h 1 2 ut gt 2 2 = + u = 0, t = 2s h 4g = ...(i) then, v2 = u2 + 2gh v 2gh 2g 4g = = × v g 8 = ...(ii) then, v = u + gt v = g × t v g 8 t g g = = t 2.83 s = 148. Two cars ‘A’ and ‘B’ move at 15m/s in the same direction. Car ‘B’ is 300m ahead of car ‘A’. If car ‘A’ accelerate at 6m/s2 while car ‘B’ continues to move with the same velocity, car ‘A’ will overtake car ‘B’ after:- (a) 7.5 s (b) 10 s (c) 12 s (d) 15 s UKPSC AE-2013, Paper-I Ans. (b) : Two can A & B move initial at 15 m/s So initial velocity of car A with respect to car B is zero. So, car A will over take distance 300 m when accelerate at 6 m/s2 1 2 S ut at 2 = + 1 2 300 0 t 6 t 2 = × + × × t 10 s =
412 149. Two balls are dropped from a common point after an interval of 1 second. If acceleration due to gravity is 10m/s2 , separation distance 3 second after the release of the first ball will be:- (a) 5 m (b) 15 m (c) 25 m (d) 30 m UKPSC AE-2013, Paper-I Ans. (c) : The first ball has traveled for 3 seconds under gravity wit acceleration g and initial velocity u = 0 ( )2 1 1 S 0 3 10 3 2 = × + × × S1=45 m Then second ball has traveled for 2 seconds Then, ( )2 2 1 S 0 2 10 2 2 = × + × × S2 = 20 m Then, distance between them will be S S 45 20 25m 1 2 − = − = 150. .......... represents the area under acceleration - time graph. (a) Acceleration (b) Displacement (c) Motion (d) Change in velocity (HPPSC LECT. 2016) Ans : (d) Change in velocity represent the area under acceleration - time graph. dv a dt = dv a dt = dv a dt = ∫ ∫ v v a dt 2 1 − = ∫ 151. The speed of a particle moving in circular path is 600 rpm. Then, the angular velocity of the particle is (a) 20π rad/ sec (b) 10 π rad/sec (c) 20 /π rad/sec (d) 10/πrad / sec TSPSC AEE 2015 Ans : (a) Speed. fo a particle = 600 rpm. Angular velocity = 2 N rad /sec. 60π = 2 600 rad /sec. 60 π× Angular velocity = 20π rad/sec. 8. Kinematics and Kinetics of Rigid Body 152. The tension in the cable supporting a lift is more when the lift is (a) Moving downwards with uniform velocity. (b) Moving upwards with uniform velocity. (c) Moving upwards with acceleration. (d) moving downwards with acceleration. UPPSC AE 12.04.2016 Paper-I Ans : (c) The tension in the cable supporting a lift is more when the lift is moving upwards with acceleration. 153. An object having 10 Kg mass and weights as 9.81 Kg on a spring balance. The value of "g" at that place is (a) 9.81 m/s2 (b) 10.m/s2 (c) 0.981m/s2 (d) 98.1 m/s2 UJVNL AE 2016 Ans : (c) Given, Object mass = 10 kg Spring weight = 9.81 N W = mg 9.81 = 10 × g g = 0.981 m/sec2 154. A rigid body can be replaced by two masses placed at fixed distance apart. The two masses form an equivalent dynamic system, if (select the most appropriate answer). UPPSC AE 12.04.2016 Paper-I (a) The sum of the two masses is equal to the total mass of the body. (b) The centre of gravity of two masses coincide with that of the body (c) The sum of the mass moment of inertia of the masses about their centre of gravity is equal to the mass moment inertia of the body. (d) All of the above. Ans : (d) (i) The centre of gravity of two masses coincide with that of the body (ii) The sum of the two masses is equal to the total mass of the body (iii) The sum of the mass moment of inertia of the masses about their centre of gravity is equal to the mass moment inertia of the body 155. If the body falls freely under gravity, then the gravitational acceleration is taken as TSPSC AEE 2015 (a) + 8.91 m/s2 (b) – 8.91m/s2 (c) + 9.81 m/s2 (d) – 9.81 m/s2 Ans : (c) If the body falls freely under gravity, then the gravitational acceleration is taken as + 9.81 m/sec2 156. A car moving with a uniform acceleration covers 450 m in 5 sec interval, and covers 700 m in next 5 seconds interval. The acceleration of the car is: (a) 7.5 m/sec2 (b) 10 m/sec2 (c) 12.5 m/sec2 (d) 20 m/sec2 HPPSC LECT. 2016
413 Ans : (b) S1 = ut + 1 2 at2 450 = 5u + 1 2 a ×5 2 450 = 5u + 12.5a ……. (i) S2 = ut1 + 1 2 a × 2 2 t 700 + 450 = 10u + ( ) 1 2 a 10 2 1150 = 10u + 50a …….(ii) for equation (i) and (ii) a = 10m/sec2 157. How much force will be exerted by the floor of the lift on a passenger of 80 kg mass when lift is accelerating downward at 0.81 m/s2 ? (a) 740 N (b) 700 N (c) 720 N (d) 680 N CIL MT 2017 2017 IInd shift Ans. (c) : Fnet = mg-ma = m(g-a) = 80 (9.81-0.81) = 80×9 = 720 N 158. If velocity of a body change from 50 m/s to 200 m/s in 20 seconds, then the acceleration of the body is: (a) 5.0 m/s2 (b) 6.5 m/s2 (c) 6.0 m/s2 (d) 7.0 m/s2 (e) 7.5 m/s2 (CGPCS Polytechnic Lecturer 2017) Ans. (e) : v1 = 50 m/s, v2 = 200 m/s, t = 20 s Then acceleration (a) is given as a = rate change in velocity a = dv dt = 200 50 20− = 150 20 2 a 7.5 m /sec = 159. A particle of mass 1 kg moves in a straight line under the influence of a force which increases linearly with time at the rate of 60N/ s, it being 40 N initially. The position of the particle after a lapse of 5s, if it started from rest at the origin, will be (a) 1250 m (b) 1500 m (c) 1750 m (d) 2000 m ESE 2019 Ans. (c) : Given, m = 1 kg, dF dt = 60 N/s dF = 60 dt dF ∫ = 60 dt ∫ F = 60 t + c .....(1) at t = 0 F = 40 c = 40 [from (1)] ∴ F = 60t + 40 .....(2) Using Newton's second law – F = ma = 1 × a = a From equation (2) a = 60t + 40 dv dt = 60t + 40 dv = 60t × dt + 40dt Integrating it dv ∫ = 60 dt 40 dt × + ∫ ∫ v = 2 1 t 60 40t c 2 + + .....(3) at t = 0, v = 0, ∴ c1 = 0 Then V = 2 60t 40t 2 + ds dt = 2 60t 40t 2 + Again integrating 2 60t ds dt 40t dt 2 = + ∫ ∫ ∫ s = 3 2 2 60t 40t c 6 2 + + .....(4) at t = 0, s = 0, ∴ c2 = 0 then s = 3 2 60t 40t 6 2 + at t = 5 s s = 3 2 60 5 40 5 6 2 × × + , s = 1750 m 160. A car travels from one city to another with the uniform speed of 40 km/hr for half distance and with the uniform speed of 60 km/hr for remaining half distance. The average speed of car is: (a) 40 km/hr (b) 45 km/hr (c) 48 km/hr (d) 50 km/hr (e) 42 km/hr CGPSC AE 2014- I Ans. (c) : Total distance Average speed = Total time d d 2 2 = d / 2 d / 2 40 60 + + 2 40 60 48 km / hr. 60 40 × × = = + 161. An elevator has a downward acceleration of 0.1 g m/s2 . What force will be transmitted to the floor of elevator by a man of weight 'W' travelling in the elevator? (a) W (b) W/10 (c) 11W/10 (d) 9W/10 UPRVUNL AE 2016 Ans. (d) : Newton's Second Law of motion Fnet = W - T = ma W - T = 0.1 0.1 W g W g × × = × T = W - 01 W = 0.9 W 9 10 T W =