1. Engineering Mechanics Full Material

3.18 | Engineering Mechanics Test 2 1 7. o⇒ΣrF mt =(ma–t()ma=g×–SmOinAg3)S0=ion–θmg Sin θ C ⇒ a= −g Sin q OA AE dw dw dq d w RA (180 × 9.81) N RE Now a = dt = dt × dq = w d q Free Diagram of ACE: or w dw = −g Sin q ΣME R=bAo0d=⇒y7d5Ri6aA.7g×7ra0Nm.2~1o7f–5B17D7N6:5.8 × 0.09 = 0 dq OA ⇒ Free w wdw = − g 45o Sin q dq ⇒ 2.∫83 OA 30∫ o DV 20. B D Dh 1 w2 − 2.832  = + 9.81 Co45o - Cos30o  Bh 2 2 10 cm 5.464 cm w = 2.54 rad/sec Choice (A) BV (180 × 9.81) N B ∑MB = 0 ⇒Σ M B Do=rV0D×V(=1011+452.4N64) + (180 × 9.81 × 10) =0 A Choice (A) 1 8. Since velocity is uniform, therefore acceleration is zero. 6N 10° y x Tangential component of the linear acceleration of point A P 20° (oart)aA = r8A0ara⇒d/s8ec=2 (0.1) × = a 30° Angular acceleration of the system, a = 80 rad/sec µR R Now (at)B = a × rB = 80 × 0.15 = 12 m/s2 (50 × 9.81) N Now wA = VA = 25 = 250 rad/s rA 0.1 Σ+F6x C=oms a1x0⇒o + ΣPFCx o=s020 o –{∵µRa–x =(500}×a9n.d81Σ×FyS=in030o) = 0 or P Cos20o – 0.3 R = 239.341 → (1) NNoowrmwalA c=owmBpo=n2e5n0t rad/s and 6 Sin10o + R – P Sin20o – (50 × 9.81 × Cos30o) = 0 of B = r w2 = 0.15 × 2502 or P Sin20o – R = – 423.743 → (2) = 9375 m/s2 From equation (1) and (2) we get, P = 437.785 N and R = 573.474 N Magnitude of the linear acceleration, a = 93752 +122 ∴ P = 437.785 N ~ 438 N Choice (C) = 9375 m/s2 Choice (D) 19. 21. TΣΣaFFkxy i==ngPRm––oFmmg=en=mts0aaxbout --------- (1) × R = Ia O 1 R ΣM = P × h + F 2 30° or P × h + F × R = mR2 a α at ⇒ P × h + F × R = 1 mRax {∵ ax = R × a} 2 A Dividing the above equation with R/2 we get 30° 2Ph R + 2 F = max  ------- (2) mg Equating right side of equation (1) and (2) we get

Engineering Mechanics Test 2 | 3.19 2Ph + 2F = P − F ∴ Average weight to be pulled = 1 [24.525 + 0] R 2 or 3F = P 1− 2h  = 12.2625 N R  Work done = Average force × distance moved 1 F = 0 when 1 – 2h = 0 or h = R  Choice (C) = 12.2625 × 4 = 3.065 N-m or J Choice (A) R 2 2 6. ΣMhinge = Io × a 2 2. I is the centre of rotation of rod AB. 2 V⇒∴∴∴e locωVωViAtBAAyBB==d××VωiAa1AOAg0BI=r=×raa1m1dB22/I:sme=⇒c/s1 0=ωA×VBA1×I .61.=2  5 mR2 m (OG)2  = 12 [mg × (0.5 + 0.1)] =  +  × a 16 m/s ∴ 0.6 × 9.81 =  2 × 0.12 + 0.62  × a  5  a ⇒ a = 16.17 rad/sec2 Choice (C) 12 27. By the method of joints for equilibrium at point A. 60° i ΣX = ΣY = 0 AΣΣYXt p==o−iSn4Pt+D+S5SC5 Soisn4455+–SS66CSoins4455 = 0 30° = 0 16 AΣΣXYt p==oSi−n6tSSC6inC4o5s4+5S+3 SS3inC4o5s–45R=D 0 ⇒ S6 = S3 = 0 b ∴→ab = 122 +162 = 20 m/s Choice (A) ΣX ==ASS−22t=SpS4oiSn–i3n4S=t52BS–C6 oSs3 4S5in–45S3=C0os⇒45 =0 S3 ΣY S2 = 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 1 ∴ SΣFΣ2rXYo=m==S−S3Σ1=YS–5−eSSqS5iun5Ca4=to5iosS–4n65Sa2+t SpSoi2ninC4t5oAs=450 =0 ( ) ( )U 2 ⇒ S5 = ₋S2 = − k x22 − x12 + F×s = −1 × 730 × 0.32 − 0 + 40 × 9.81× 0.15 FPr=omS5ΣSXine4q5ua–tiSo6nSaint 4p5oi=nt2BS5 Sin45 = 1.414 S5 2 = 26.01 N-m Choice (B) =S1P= S5 Cos45 – S2 Cos45 = 2S5 Cos45 = 1.414 S5 2 4. Initial kinetic energy is zero. Hence the change in kinetic energy, 1 1 ∴ S1 = P Choice (B) 2 2 ∆KE = KE2 – KE1 = mv2 + I ow2 28. As the floor is smooth, there are only vertical reactions at C and D Taking moments at C 1 v2 1  mr 2  w2 2 2  2  ⇒ ∆KE = × 50 × + RD.a – (aa).P = 0 ⇒ RTaDk=inagPm. oments 1 50 × 0.42  v 2 at D 2 4  0.4  ⇒ ∆K .E = × 50 × v2 + × RTRaCCk.=ain+(g1(s−ae−apa)arPaat)ePf=re0e body diagrams for the legs ⇒ Now U = K.E2 – K.E1 50v2 50 ∴ 26.01 = 2 + 4 v2 AD and BC we get (Taking the reactions instead of the force). ⇒ v = 0.833 m/s Choice (C) B 2 5. Weight of the hanging part of the chain = mg A Ye E 4 Xe D C Ye Xe 10 × 9.81 αP (1 – α)P = 4 N = 24.525 N = Maximum weight to be lifted. When the entire hanging portion has been pulled, the weight to be lifted equals zero.

3.20 | Engineering Mechanics Test 2 For the bar AD m = 5 kg, h = 100 mm = 0.1 m: S = 5 mm = 5 × 10–3 m Taking moments around A R = 0.54 N/mm (MaoPm)aen+tsYea(rao/u2n) d– BXe(a/2) = 0  → (1) Energy required to push the nail into the floor = R × S → (2) = 1.032 × 5 = 5.16 N A− dad.(i1ng−(a1))Pan+dY(e2()a/2) + Xe(a/2) = 0  Energy offered by hammer = (KE)impact + (PE)penetration KE during impact 1 where v = aP – (1 − a)P + Ye = 0 = 2 mv 2 2gh ⇒ point E ⇒ YTXehe =e= (1 − 2a)P PE during penetration = (M + m)gS, where M is the ∴ P mass of the nail resultant on is 1 5.16 = 2 m (2gh) + (M + m)gS 2 + Ye2 P2 + 1− 2a 2 P2 ( ) X e Re = = Re = P 1+(1− 2a)2 5.16 =  1 × 5 × 2 × 9.81× 0.1 + (5 × 9.81 × 0.005) +  2 a has the range 0 to 1 ∴ maximum value for Re is at (M × 9.81 × 0.005) Choice (C) When a = 0, Re = P 2 and ⇒ M = 198.77 gms a = 1, Re =P 2  Choice (C) 31. W = 30 N, f = 30o 29. 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 Tcosθ θ R T Pmin mrω2 R φW mg The minimum distance to complete the triangle is the perpendicular from the head of R to the tail of W. T Cos θ = mg and T Sin θ = mr ω2 ∴ Pmin = W Sin j = 30 Sin 30 = 15 N Choice (D) ⇒ Tan q = rw2 ⇒ w = Tan 30 × 9.81 =1.683 rad sec g 2 32. 4m Pα 23 N 90 – φ RW φ ∴ mmr×1 w2 = 23 N, r=1 =4 m Considering the vectors P and W ∴ 4 × 1.6832 23 (90 – f) + 90 + a = 180o ⇒ a = f m = 2.03 kg Choice (C) 2.03× 9.81 ∴ T Cos θ = mg ⇒ T = mg = Cos 30 = 23 N 3 3. + Ve Cos q  4 m/s 8 m/s 30. Choice (A) 6N 2N m v1 − v2 h u2 − u1 e = 0.5 = here v1, u1 are for 6 N ball and v2, u2 are S for 2 N ball. R ⇒ vbv11y––covvn22 s==er0–v.65at⇒×ion(v–2o8=f – 4) = – 6 m/s ⇒ mv1o+m6entum

Engineering Mechanics Test 2 | 3.21 –v6m8W114=×h=ue=641–nv+8+10tvh+.m215e22×mi(um(6/2–sp=+a(8fcmv)ot1=r1)isev61ev=l+1a0+sm.t5i22c)vve22 = 1 ⇒ a = Pr = 10 × 0.05 = 0.005 m W 100 a = 5 mm = 0.005 m ⇒ The distance a is called the coefficient of rolling resis- tance. Choice (B) ⇒ vFv21ro–=mvv21m=+o–1m21e2ntum equation, 3 5. By figure, the shapes are equilateral triangles. By considering moments around E at equilibrium ×ΣM10E)==00⇒ (500 × 40) + (100 × 30) – (RC × 20) + (100 8 1=66=v18+v12 (v1 + 12) 8 = 6v1 + 2v2 – ⇒ RseCc=tio1n20m0eNthod, considering the section ABC. v1 = −16 = – 2 m/s (for e = 1) By 8 100 N ∴ (v1)e=0.5:(v1)e=1 = −0.5 = 0.25  Choice (B) B BD 34. −2 r CD P A 20 60° CE A 500 N 1200 N R By taking the vertical forces. Wa ΣY = 0 ⇒ 1200 – 500 – 100 + CD Sin60 = 0 By applying summation of moments about point A ⇒ CD = 692.82 N ΣMA = 0 ⇒ W × a – P × r = 0 ⇒ W × a = P × r As the value of CD is positive the member is in Tension here, W = 100 N, r = 0.05 m, P = 10 N as per the initial assumption. Choice (B)

Engineering Mechanics Test 3 Number of Questions: 25Time:60 min. Directions for questions 1 to 25: Select the correct alterna- 20 N tive from the given choices. A 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 B 45° 30° C (A) 3351 (B) 3.35 The load in member AB will be approximately (C) 32 (D) 3.2 (A) 21 N (B) 19 N 2. A hollow triangular section is symmetrical about its (C) 17 N (D) 18 N vertical axis. The moment of inertia of the section about the base BC will be 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 A required to just lift 100 N weight? (Take coefficient of friction as 0.25.) Flexible body 0.08 m 0.1 m 100N P B C 80° 0.15 m 0.2 m (A) 132.34 N (B) 121.36 N (C) 70.53 N (D) 141.77 N (A) 1.02 × 10-5 m4 (B) 1.11 × 10-6 m4 (C) 1.02 × 10-6 m4 (D) 1.11 × 10-5 m4 9. The supports which apply force on the body in only 3. The force of friction between two bodies in contact one direction and the direction is always normal to the (A) depends upon the area of their contact contacting surface is known as (B) depends upon the relative velocity between them (A) Fixed support (B) Hinged support (C) is always normal to the surface of their contact (C) Roller support (D) All of these (D) All of the above 1 0. Triangular plate ABC is connected by means of pin at 4. A ball is thrown with a velocity of 12 m/s at an angle of C with another triangular plate CDE as shown in the 60o with the horizontal. How high the ball will rise? figure. The vertical reaction at point D will be (A) 6 m (B) 6.35 m 2 kN 2m (C) 11 m (D) 5.5 m 1m 1m 5. The equation for angular displacement of a particle, BE moving in a circular path of radius 300 m is given by : C 1m θ = 20t + 5t2 – 3t3 where θ is the angular displacement 4 kN at the end of t seconds. The maximum angular velocity of the particle (in rad/s) will be (A) 20.34 (B) 22.78 5m (C) 23.63 (D) 24.39 AD 6. The velocity of piston in a reciprocating pump mecha- (A) 4.5 kN (B) 6 kN nism depends upon (C) 2 kN (D) 4 kN (A) Angular velocity of crank (B) Radius of the crank 11. A beam 5 m long weighing 400 N is suspended in a hor- (C) Length of the connecting rod izontal position by two vertical strings, each of which (D) All of the above can withstand a maximum tension of 450 N only. How far a body of 300 N weight be placed on the beam from 7. Consider a truss ABC loaded at A with a force 20 N as the left end, so that one of the string may just break? shown in figure.

Engineering Mechanics Test 3 | 3.23 (A) 1.81 m (B) 1.43 m (C) 0.834 m (D) 2.12 m 1 2. A figure is shown below. Solve for the force in mem- A ber AB under the actions of the horizontal and vertical B force of 1000 N. θ B AB = 50 cm 1000 N BC = 80 cm 1000 N (A) 25.4o (B) 32.6o (C) 29.2o (D) 34o AC 1 6. A flywheel 2 m in diameter accelerates uniformly from 100 cm rest to 2000 rpm in 25 seconds. 0.6 second after it has started from rest, the linear acceleration of a point on (A) 377.5 N (Tension) the rim of the flywheel (in m/s2) will be (B) 377.5 N (Compression) (A) 24.6 (B) 15.96 (C) 1415.9 N (Tension) (C) 21.34 (D) 26.73 (D) 1415.9 N (Compression) 13. A truss is shown in the figure. Each load is 5 kN and all 1 7. A mass of 2 kg is projected with a speed of 3 m/s up a triangles are equilateral with sides of 4 m. Determine plane inclined 20o with the horizontal as shown in the the force on member GI. 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 BD F H J L m AM 20° CE GI K (A) 35 kN (Compression) (A) 1.2 (B) 2.103 (B) 35 kN (Tension) (C) 1.63 (D) 1.31 (C) 26 kN (Compression) (D) 26 kN (Tension) 1 8. In a device, two equal masses of 100 kg are connected 1 4. A cylinder is shown in the figure. The coefficient of by a very light (negligible mass) tape passing over a friction between the cylinder and wall is 0.25. Will the frictionless pulley as shown in the figure. A mass of 10 180 N force cause the 100 kg cylinder to slip? kg is added to one side, causing that mass to fall and the other to rise. The acceleration (in m/s2) of the masses 180 N will be (100 × 9.81) N N2 F2 F1 N1 m MM (A) No slip (B) Slip will occur (C) Insufficient data (D) None of these (A) 0.49 (B) 0.817 (C) 0.621 (D) 0.467 1 5. Two blocks B and A of mass 40 kg and 13.5 kg respec- tively is kept as shown in the figure. The coefficient of 19. Find the force ‘P’ required to prevent sliding of body 2 friction µ for all surface is 1/3. The value of the angle on body 1. Assume both the bodies have equal mass ‘m’ θ so that the motion of 40 kg block impends down the and all the surfaces are smooth. plane will be

3.24 | Engineering Mechanics Test 3 Body 2 (A) 1588 N (B) 1500 N (C) 520 N (D) 1600 N Body 1 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 P 0.34 seconds. How many revolutions does it make dur- 45° ing the entire time travel? (A) mg Sin45o (B) mg (A) 29.1 rev (B) 20.17 rev (C) 2 mg (D) 2 mg Cos45o (C) 22.34 rev (D) 26.33 rev 2 0. A rope is wound around a 30 kg solid cylinder of radius 24. An eccentric cylinder used in a vibrator weights 198 N 50 cm as shown. Find the speed of its mass centre after and rotates about an axis 5 cm from its geometric cen- it has drop by 2 m from the rest position. tre and perpendicular to the top view as shown in the figure. If the magnitudes of angular velocity and angu- rope lar acceleration are 10 rad/s and 2 rad/s2 in the phase shown, the resultant reaction of the vertical shaft on the cylinder cylinder (in N) and couple applied on the cylinder by the shaft (in N-m) will be ωα (A) 5.11 (B) 7.23 O (C) 6.09 (D) 5.89 G 2 1. A 0.08 N bullet was fired horizontally into a 60 N sand 5 cm 10 cm bag suspended on a rope 1.5 m long as shown in the figure. It was found that the bag with the bullet embed- ded in it swung to a height of 20 mm. Determine speed of the bullet as it entered the bag? (A) 111 and 0.278 (B) 101 and 0.278 (C) 111 and 0.413 (D) 101 and 0.413 θ 2 5. A sphere, rolling with an initial velocity of 12 m/s, starts up a plane inclined 30o with the horizontal as 1.5 m shown in the figure. What is the distance upto which the sphere will roll up the plane? 20 mm yx (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 (A) 10.21 m 30° as shown in the figure. The magnitude of resultant at B (C) 11.86 m (in N) will be (B) 12.31 m AB = 0.6 m (D) 9.63 m BC = 1.2 m AB = 0.6 m C BC = 1.2 m F A B 60° 1200N

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 8. P = Tight side tension 160 × 2 p × 200 W = Slack side tension Power = T × w = 60 ×1000 = 3.35 kW P P e 0.25 × 80 × p  Now, W = e ms q ⇒ 100 = 180  Choice (B) ⇒ P = 141.77 N Choice (D) 2. IBC = BH 3 − bh3 = 0.2 × 0.13 − 0.15 × 0.083 Choice (A) 9. Choice (C) 4 kN 12 12 12 12 Choice (D) D RX 1 0. R ⇒ IBC = 1.0267 × 10–5 m4  2 kN Choice (A) 3. Choice (C) R B ( ) 4. H = u2 Sin2 a = 122 × Sin2 60o = 5.5 m 2g 2 × 9.81 5. θ = 20t + 5t2 – 3t3 dq w = dt = 20 + 10t – 9t2 A For maximum angular velocity, dw = 0 dt dw ∴ dt = 10 – 18t = 0 ΣMA =0 ⇒ 2×1–4×5 ∴t = 10 = 0.556 seconds ⇒ Ry = 4.5 kN  – Ry × 4 = 0 18 1 1. ∴ =wm2a2x .=7820ra+d/1se0c(0.556) – 9(0.556)2 Choice (B) R Choice (D) 6. Vwpi=stonan=gwul[alrCvoesloΦcit+y r Cos θ tan Φ] of crank l = length of connecting rod 5m r = radius of crank  DC A 7. x 300 N 400 N A 20 N x = Distance between the body of weight 300 N and support A (from the left end) x We know that one of the string will just break, when the B 45° 30° C tension will be 450 N (i.e., RA = 450 N) RC + (400 × 2.5) x 1.732 x 4N5o0w×Σ5M=B =0 (5 – x) RB Choice (D) 300 ⇒ x = 0.834 m Choice (C) ∴⇒Σ⇒ M B2RRFN=0AcBoB0×=w==x72F.S0=A31Bi22–nR×.N4C67S58.×3ion224=.=7531o127=2x..9R638B 12. Free body diagram 52.41 29.67 1000 N N AB 1000 N BC N

3.26 | Engineering Mechanics Test 3 Now AΣBFxC=o0s(52.41) – BC Cos(29.67) + 1000 = 0 From equation (1), (2) and (3) Choice (C) ⇒ θ = 29.2o  → (1) and ΣAFBy =0 16. w= wo + at ⇒ Sin(52.41) and wo = 0 and 2p N 2 p × 2000 + BC Sin(29.67) – 1000 = 0 w = 60 = × 60  → (2) ⇒ w = 209.44 rad/sec From equations (1) and (2) we get AB = 377.5 N and BC = 1415.91 N w − wo 209.44 − 0 Now a = t = 25 = 8.4 rad/s2 ∴ Force in member AB is 377.5 N compression  Choice (B) Now velocity after 0.6 seconds 5=.0w4o + at = 0 + (8.4 × 0.6) 13. Taking section passes through JH and GI. w rad/s = 5 kN 5 kN Normal component of acceleration, an = r = 1 × 5.042 ω2 JH J 4m L = 25.4 m/s2 of acceleration, = Tangential component at ra = 1 × 8.4 = 8.4 m/s2 G I 4m K 4m M Total acceleration, a = an2 + at2 15 ⇒ a = 25.44 + 8.42 = 26.753 m s2  Choice (D) Taking moment about point H we get ΣMH = 0 = –(GI) × 2 tan60o – (5 × 4) – (5 × 8) + 15 ×10 17. ⇒ GI = 25.98 kN (Tension) Choice (D) µN 1 4. Since it is unknown whether or not the cylinder slips it Σ∴Σ iΣsMFF nvhoA==tiN1NaLe=ssqp000e11s0uo8t0==u==as=.Nmust2N9Fiis9–5oeb8t1101osnl0a+e–88Nssih–t0stNFs2oNou(F2×l12mm,daF2)–+n,2a,Fe1tdNh9xr1=1(e8iF82+2m=s0µ20s)=oFNyu→i→l13assm2v10tn6e+eaa8d(0s(mn2t1rtfda–N)o)(i+itt3rFni.Fsc)N2T2.eNmv=2qahTa2un×ai,lµshiudxNlrNeeiimFmnb2→1tr1ehuiaNa=uem(nnnm231ds)F8.v=tS01aFhlmia12–unt8eucuF8eissf2ttithnhNbFagee2t, 20° µN N N 19.62 N (2 × 9.81) = 19.62 N (b) (a) From figure (a), N = 19.62 Cos20o = 18.44 N Now V2 =Vo2 + 2as ⇒ a = 0 − (+3)2 = − 4.5 m s2 2 ×1 maximum value of Fn1otorbotataintea.b le is 0.2C5hNoi1c=e 227 Now +fr1o9m.62fig×uSrein(2a)0,oΣ+F(xµ=×m1a8x.44) = 2(4.5) N, the cylinder will (A) ∴ ⇒ µ = 0.124 15. To solve for return speed, refer figure (b) T 1 N1 19.62 Sin 20o – 0.124 (18.44) = 2a 3 N1 ⇒ a = 2.212 m/s2 1 N1 B 1 N2 3 3 Finally V2 = Vo2 + 2as or V2 = 0 + 2(2.212) (1) A ⇒ V = 2.103 m/s2 Choice (B) θ θ 1 8. N1 N2 TT (13.5 × 9.81) = 132.43 N 40 × 9.81 = 392.4 N a From free body diagram of B. 1 1 3 ΣFx = 0 = –392.4 Sin θ + 3 N1 + N2  → (1) a → (2) Mg (M + m)g ΣFrFeye=b0od=yNd2ia–g3ra9m2.4ofCAo.s θ – N1  → (3) → (2) N1 = 132.43 Cos θ  ΣF = T – Mg = Ma → (1) ΣF = Mg + mg – T = (M + m)a

Engineering Mechanics Test 3 | 3.27 From equation (1) and (2) 21. LVe2 t=VV1e=loVceitlyoc(ibtaygo+f bullet before impact m 10 bullet) after impact 2M + 2 ×100 +10 × 9.81= 0.467 m/s2 ( )a= m g ⇒ a = = 2gh = 2 × 9.81× 0.02  Choice (D) = 0.6264 m/s Now momentum before impact = momentum after 19. impact ∴ (V01.0=84×70V.41)2+640m=/[s0 .08 + 60] × 0.6264 Choice (A) ⇒ aP 2 2. ⇒Σ`M BF==06⇒001N200 × 0.6 = F × 1.2 R2 C 2mg 600 N a 45° y A B 60° RH x 45° 1200 N 30° mg RV R R1 ΣFx =Pm=a2x m⇒a–P = 2m(–a) R=H5=19F.6C15osN30o = 600 × Cos 30o ⇒ a=n6d0R0v×=SFinS3i0no3+0o1+2010200 ΣRmFΣrFF1(o–xy=ma==m)emm=gqa/au–CyxaRo⇒t1iso4SnR5ino1(14C)5ooasn4d5o(2=) → (1) = 1500 N ⇒ mg → (2) we ∴ Resultant reaction, R = RH2 + RV2 ⇒ Choice (C) mg get, = 519.6152 +15002 Cos 45o ma = × Sin 45o = 1587.45 N Choice (A) ⇒ a = g tan45o 2 3. From t = 0 to t = 0.5: ⇒ a = g 1 1 Now P = 2 ma ⇒ P = 2mg q1 = 2 (wo + w)t = 2 (0 + 500 60) × 0.5 2 0. ⇒ θF1ro=m2.t0=8304.5retov t = 2.5 s: G KE = 0 (in rest) θ2 = wt = 500 × 2 = 16.67 rev PE = 0 (k = 0) 60 1 From t = 2.5 to rest 1 2m θ3 = 2 (wo + w) t = 1  500 + 0 × 0.34 G 2  60 2 KE = 1 I0ω2 = 1.42 rev = + + 2 Total number of revolutions = 2.0834 + 16.67 + 1.42 θ θ1 θ2 θ3 1 Io w2 = mg(2) = 20.17 rev Choice (B) 2 198 1  mr2 + mr 2  w2 = 30 × 9.81 × 2 2 4. ΣFn = m r w2 = 9.81 × 0.05 × 102 = 101 N 2  2  ⇒ 198 9.81 1 ΣFt = m r a = × 0.05 × 2 = 2.02 N 2 ⇒ [1.5 × 30 × 0.52] × w2 = 30 × 9.81 × 2 1 198 198 2 9.81 9.81 ⇒ w = 10.23 rad/s ΣMo = Io a =  × × 0.152 + × 0.052  VG = w × r = 10.23 × 0.5 = 5.115 m/s Choice (A)   = 0.278 N–m

3.28 | Engineering Mechanics Test 3 Resultant of forces = 1012 + 2.022 = 101.02 N Where, x is the required distance 1 1 2  Choice (B) Initial kinetic energy, k.E1 = 2 mV12 + I o w12 2 5. Now Io = 2 mR2 and V1 = ω1 R 5 F ∴ K.E1 = 1 m V12 + 1 m V12 = 7 m(12)2 2 5 10 30° mg NA Now work done = 0K.–E1270– K.E1 ⇒ –mg Cos30o × x = m(12)2 The initial kinetic Energy tr(aKv.eEl.1)Tdheecorenalysesfortoce final ⇒ x = 10 × 7 ×122 30 o = 11.86 m Kdo.Ees2 = 0 at the top of the that 9.81× Cos work in the component (negative) of the weight W along the plane.  Choice (C) Work done = –[mg Cos30o] × x

Chapter 1 Engineering Mechanics One-mark Questions tan q = AC = 0.5 AB 1 1. A circular object of radius r rolls without slipping θ = 26.56 on a horizontal level floor with the center having Draw FBD of right portion of truss after passing velocity V. The velocity at the point of contact section (1)-(1) 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) Apply equation of equilibrium (εFx = 0) −PAB − PBC cos (26.56) = 0 −PAB − PBC cos (26.56) = 0 (1) εFy = 0 The point of conduct P of the wheel with floor does −10 − PBC sin (26.56) = 0 not slip, which means the point P has zero velocity −10 − 0.44PBC = 0 with respect to point m. PBC = –22.36 N Hence, the correct option is (a). From Equation (1) 2. A two member truss ABC is shown in the figures. The −PAB − (−22.36) × 0.8944 = 0 force (in kN) transmitted in member AB is _____ −PAB + 20 = 0 [2014-S2] 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: (see figure) Solution: (d) Given condition VA = 5 m/s;  VB = 0;  mB = 1000 kg

9.4 | Engineering Mechanics mA = 100 kg;  VB′ = VA′ Solution: (c) VA′ = ? FBD of block 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 Vb′ − VA′ = 0 VB − VA 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 W = 981 N εFy = 0 by [2014-S4] N + T − 918 = 0 N + T = 981 N (a) j = 2m – 3 (b) m = 2j + 1 εFx = 0;  100 − F = 0 100 − µN = 0;  µN = 100 (c) m = 2j − 3 (d) m = 2j − 1 (1) Solution: (c) Condition for statically determinant truss m + r = 2j 100 100 µ 0.2 where m = no. of members of truss N= = = 500 r = no. of reaction at support j = no. of joints in truss From Equation (1) m = 2j − 3 [A ssuming one rollers and N + T − 981 = 0;  500 + T − 981 = 0 one hinged support] T = 418 N Hence, the correct option is (c). Hence, the correct option is (c). 5. The coefficient of restitution of a perfectly plastic 7. The time variation of the position of a particle in impact is [2011] rectilinear motion is given by x = 2t3 + t2 + 2t. If ‘V’ (a) zero (b) 1 is the velocity and ‘a’ acceleration of the particle in (c) 2 (d) infinite consistent units, the motion started with [2005] Solution: (a) (a) V = 0, a = 0 (b) V = 0, a = 2 Coefficient of restitution (CoR) of two colliding (c) V = 2, a = 0 (d) V = 2, a = 2 body is typically a positive real Number between 0 Solution: (d) to 1. Which is ratio of speed after and before impact. Displacement x = 2t3 + t2 + 2t For perfectly plastic body CoR = 0 and for perfectly Differentiating w.r. to t, we get elastic body = 1. Velocity V = 3 × 2 × t2 + 2t + 2 Hence, the correct option is (a). = 6t2 + 2t + 2 (1) 6. A block weighing 981 N is resting on a horizontal Differentiating again, we get surface. The coefficient of friction between the block Acceleration a = 12t + 2 (2) and the horizontal surface is µ = 0.2. A vertical cable at t = 0 attached to the block provides partial support as From Equation (1) V = 2 shown in the figure. A man can pull horizontally with From Equation (2) a = 2 a force of 100 N. What will be the tension, T (in N) in Hence, the correct option is (d). the cable if the man is just able to move the block to 8. The figure shows a pin-jointed plane truss loaded at the point M by hanging a mass of 100 kg. The the right? [2009] member LN of the truss is subjected to a load of [2004] (a) 176.2 (b) 196.0 (c) 481.0 (d) 981.0

Chapter 1  Engineering Mechanics | 9.5 (a) PL (b) PL 2 (c) zero (d) 2PL 2 Solution: (a) (a) zero First find Reaction at support i.e., RA and RB at A and (b) 490 N in compression B. (c) 981 N in compression (d) 981 N in tension Draw FBD of truss. Solution: (a) Use method of section, Pass cutting section which divides truss in two part as shown Apply equation of equilibrium along x- and y-direc- tion εFy = 0 (↑ +ve) RA + RB − PL = 0 PL (1) ⇒ RA +εRFBy == 0 Draw FBD of upper part of section (1)-(1) Taking εMA = 0 [summation of all moment of A = 0] Sign correction Apply equation of equilibrium in y-direction RB × 3L − Pl  2L  = 0 εFy = 0  2  PLN = 0 Hence, the correct option is (a). 3LRB = 3PL2 2 9. A truss consists of horizontal members and vertical PL PL members having length L each. The members AE, ⇒ RB = 2 and RA = 2 DE and BF are inclined at 45o to the horizontal. For the uniformly distributed load P per unit length on To calculate/determine tension in truss member, use the member EF of the truss shown in the figure, the method of join; draw FBD to joint A. force in the member CD is [2003] εFy = 0 PL + PAE sin 45 = 0 2

9.6 | Engineering Mechanics PL 1 PAE =0 where F = Frictional force = mR 2 2 + − u2 2 −PL ⇒ (M + m) = mRS 2 PAE = 2 Again R = (M + m) g εFx = 0 u2 PAC + PAE cos 45 = 0 ⇒ (M + m) 2 = m(M + m) gS PAC  −PL 2 1 =0 ⇒ u2 = 2mgS +  2  2 ⇒ u = 2µgS (1) Again (M u = mV PAC = PL + m) 2 u= mV (2) Draw FBD of joint c So slving this M +m mV = 2µgS M +m εFx = 0 ⇒ V= M+ m 2µgS m PL PCD = PAC = 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 Hence, the correct option is (a). move a distance ‘S’ along the floor. Assuming µ to be the coefficient of kinetic friction between the block 11. A steel wheel of 600 mm diameter on a horizontal and the floor and ‘g’ the acceleration due to gravity, steel rail. It carries a load of 500 N. The coefficient what is the velocity ‘V’ of the bullet? [2003] 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. (a) M +m 2µgs (b) M −m 2µgs m m (c) µ (M + m) 2µgs (d)= M 2µgs m m Solution: (a) Mass of bullet = m Bullet velocity = v Apply equation of equilibrium along x- and y-direc- tion Block mass = m εFy = 0;  N − mg = 0 N − 500 = 0;  N = 500 N Displacement of block From theory of friction = S [After striking by bullet] F = µN = 0.3 × 500 K.E. lost by the block with bullet = work done to = 150 N overcome the frictional force Hence, the correct option is (d). ⇒ (M + m) u2 = F × S 2

Chapter 1  Engineering Mechanics | 9.7 12. The ratio of tension on the tight side to that on the (c) the velocity of both balls is 1 2gh 2 slack side in a flat belt drive is [2000] (a) proportional to the product of coefficient of (d) none of the above Solution: (c) friction and lap angle In a perfectly elastic collision between equal masses of two bodies, velocities exchange on impact. Hence, (b) an exponential function of the product of velocity before impact = velocity immediate after impact coefficient of friction and lap angle (c) proportional to lap angle (d) proportional to the coefficient of friction Solution: (b) VA = 2gh VA′ = final velocity after impact For flat belt VB = 0;   VB′ Form conservation of momentum T1 = tension on tight side M AVA + M BVB T2 = tension on slack side θ = angle of lap of the belt over the pulley µ = coefficient of friction between belt and pulley T1 = eµθ = M AVA′ + M BVB′ T2 ⇒ MAVA = (MA + MB) ∀0 Hence, the correct option is (b). VB 1 2gh 2 13. A car moving with uniform acceleration covers 450 m = in a 5 second interval, and covers 700 m in the next Hence, the correct option is (c). 5 second interval. The acceleration of the car is 15. A wheel of mass m and radius r is in accelerated [1998] rolling motion without slip under a steady axle (a) 7 m/s2 (b) 50 m/s2 torque T. If the coefficient of kinetic friction is m, the (c) 25 m/s2 (d) 10 m/s2 friction force from the ground on the wheel is [1996] Solution: (d) (a) µmg (b) T/r Displacement x = s = ut + 1 at 2 (c) zero (d) none of the above 2 Solution: (a) [u = initial velocity of car, Free body diagram of wheel a = acceleration] 450 = 5u + 1 a (5)2 (1) 2 700 = 5v + 1 a (5)2 (2) 2 Solving Equation (1) and (2), we get v − u = 50 From Equation of equilibrium εFy = 0 [Forces along y-direction] From equation of motion N = mg v − u = at From theory of friction F = µ N ⇒ a = v − u = 50 = 10 m/s2 t 5 F = µ mg Hence, the correct option is (d). Hence, the correct option is (a). 14. A ball A of mass m falls under gravity from a height 16. A stone of mass m at the end of a string of length h and strikes another ball B of mass m, which is l is whirled in a vertical circle at a constant speed. supported at rest on a spring of stiffness k. Assume The tension in the string will be maximum when the perfect elastic impact. Immediately after the impact stone is [1994] [1996] (a) at the top of the circle (a) the velocity of ball A is 1 2gh (b) halfway down from the top 2 (c) quarter-way down from the top (b) the velocity of ball A is zero (d) at the bottom of the circle

9.8 | Engineering Mechanics 18. Instantaneous centre of a body rolling with sliding on Solution: (d) a stationary curved surface lies. [1992] According to Newton’s Second law of motion F = ma (a) at the point of contact εFt = 0 [along tangential direction] mg sin θ = ma (b) on the common normal at the point of contact εFr = 0 [along radial direction] (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] T – mg sin q = mv2 (a) VA and VB are both correct l (b) VA is incorrect but VB is correct (c) VA and VB are both incorrect \\ T= m  v2 + g cos θ  (d) VA is correct but VB is incorrect  l  Solution: (a) Velocity vector at A and at B i.e., VA and VB are where θ = cord makes an angle with always acting tangential to the path at any instant. radical Two-marks Questions l = length of cord Hence, the correct option is (d). 17. The cylinder shown below rolls without slipping. In 1. A block R of mass 100 kg is placed on a block S of which direction does the friction force act? Towards which of the following points is the acceleration of mass 150 kg as shown in the figure. Block R is tied the point of contact A on the cylinder directed? [1993] 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) the mass centre (a) 0.69 (b) the geometric centre (b) 0.88 (c) the point P as marked (c) 0.98 (d) none of the above (d) 1.37 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).

Solution: (d) Chapter 1  Engineering Mechanics | 9.9 Draw FBD of block R Draw FBD of block From equation of equilibrium Apply equation of equilibrium εFy = 0;  N − 200 = 0 εFy = 0 N = 200 N N2 − 981 = 0 N2 = 981 N Static friction is given by Fs = µsN = 0.4 × 200 = 80 N From Friction theory and kinetic friction Fk = µ kN = 0.2 × 200 = 40 N. F2 = µN2 = 0.4 × 981 To start motion of block; P must be greater than Fs F2 = 392.4 N Under static equilibrium Now, draw FBD of block ‘S’ Ws = 150 × 9.81 P − Fs = 0;  10t − 80 = 0 = 1471.5 N ⇒ t= 80 = 8 sec εFy = 0 10 N1 − N2 − Ws = 0;  N1 − 981 − 1471.5 = 0 N1 = 2452.5 N ∴ The block start moving only after when t > 8 sec From friction theory F1 = µN1 During 8 to 10 second of time, = 0.4 × 2452.5 F1 = 981 N According to Newton Second law of motion F = F1 + F2 F = mass × acceleration [total force based on εFx = 0] F = 981 + 392.4 F = ma F = 1373.4 N = 1.3734 kN Hence, the correct option is (d). (P – Fk) = m dV 2. A block weighing 200 N is in contact with a level dt plane whose coefficient of static and kinetic friction are 0.4 and 0.2, respectively. The block is acted upon 10 40) dt 200 V dV by a horizontal force (in newton) P = 10t, where t 9.81 0 denotes the time in seconds. The velocity (in m/s) of (10t the block attained after 10 seconds is _____ [2014-S1] 8 Solution: Static friction ∫ ∫ − = µs = 0.4 → static condition Kinetic friction µk = 0.2 → dynamic/motion condition 10t 2 + 40t 10 = 20.38[V ]V0    2 8 ⇒ [5t 2 + 40t]180 = 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

9.10 | Engineering Mechanics µs = 0.3;  εFx = 0 of dynamic friction between the body and the plane P + 981 sin (10) − F = 0 is 0.5. The body slides down the plane and attains a P + 170.34 − F = 0 (1) velocity of 20 m/s. The distance traveled (in meter) εFy = 0 by the body along the plane is [2014-S3] −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 Solution: (see figure) 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 FBD of mass m m 100 = 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] m = 10 kg Velocity of body = 20 m/s Solution: µk = 0.5 From equation of equilibrium aT = Tangential acceleration εFy = 0 [Normal to surface] aN = radial acceleration/Normal N − mg cos 45 = 0 N − 98.1 cos (45) = 0 acceleration N = 69.36 N α = 12 rad/s2 (Angular acceleration) εFx = 0 [along inclined plane] ω = 4 rad/s (Angular velocity) F − mg sin 45 = 0;  F − 98.1 sin (45) = 0 Tangential acceleration is given by F = 69.36 N aT = rα = 2 × 12 = 24 m/s2 From theory of friction and normal acceleration F = µk N = 0.5 × 69.36 = 34.68 N aN = rω2 = 2 × (4)2 = 32 m/s2 F = 34.68 N Now, Resultant Acceleration By using D’alembert principle εFx = 0 a = aT2 + ar2 = (24)2 + (32)2 mg sin 45 − F = ma a = 40 m/s2. 69.36 − 34.68 = 10 × a a = 3.468 m/s2 5. A body of mass (m) 10 kg is initially stationary on a From equation of motion, displacement 45o inclined plane as shown in figure. The coefficient s = ut + 1 at 2 2 and v2 = µ2 + 2as

Chapter 1  Engineering Mechanics | 9.11 v = 20 m/s Solution: M = 1000 kg u = initial velocity = 0 ω0 = initial angular velocity = 10 rad/sec S= V2 = (20) 2 r = radius of wheel = 0.2 m 20 2 × 3.468 Time = t = 10 sec Mass on each axle (m) s = 57.67 m. 6. An annular disc has a mass m, inner radius R and = 10200 = 500 kg Final angular velocity outer radius 2R. The disc rolls on a flat surface ω0 = 0 From equation of motion, without slipping. If the velocity of the center of mass ω = ω0 + αt [similar to v = µ0 + at] 0 = 10 + α (10) is v, the kinetic energy of the disc is [2014-S3] a = –1 rad/s2 Braking torque m = I2α (a) 9 mV 2 (b) 11 mV 2 Braking torque m = 10 × 1 = 10 N-m. 16 16 (c) 13 mV 2 (d) 15 mV 2 16 16 Solution: (c) 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] Total kinetic energy = 12 mv2 + 12 I ω2 (1) v = rω = Rω = 2Rω w= v = v rad/sec 2R 2R Moment of inertia = I = m2 (R02 − R12 ) (a) 11.25 tension (b) 11.25 compression (c) 13.5 tension (d) 13.5 compression Solution: (a) = m2 (4R2 + R2 ) 5mR2 = 2 From Equation (1) (KE)total = 1 mv2 + 1  5mF 2   v 2 2 2  2   2R  (KE)total = 13mv2 16 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 Pass section (1)-(1), and draw FBD of left portion of truss. 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]

9.12 | Engineering Mechanics tan 2 of the angular acceleration (in rad/s2) of the rod at the 1.5 q = position shown is _____ [2014-S4] q = 53.13 εFy = 0;  −9 − FPS sin (53.13) = 0 Solution: (see figure) −0.8FPS = 9 FPS = –11.25 kN εFx = 0 F PS cos (53.13) + FPQ = 0 (−11.25) × 0.6 + FPQ = 0 FPQ = +6.75 kN Apply equilibrium to whole truss. εFy = 0 ε = Iα [According to Newton Second law] v1 – v2 + 3 – 9 = 0 ⇒ v1 – v2 = 6 εmax R = 0 (1) m = Iα (1) −v1 × 1.5 − 3 × 3 + 9 × 6 = 0 m = 29.43 × 3 = 88.29 Nm I = I0 + Ad 2 –1.5v1 – 9 + 54 = 0 ⇒ v1 = 30 kN = m1l22 + md 2 From Equation (1) = 3 1×282 + 3 × 32 30 – v2 = 6 ⇒ v2 = 24 kN = 43.00 kg m2 From Equation (1) a= m = 88.29 = 2.053 rad/s2 . I 43 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 Pass section (2)-(2), consider right portion of truss. minimum values of the force (in Newton) and the static coefficient of friction µ between the floor and the wardrobe, respectively? [2014-S4] εFy = 0 (a) 490.5 and 0.5 (b) 981 and 0.5 FQS sin (53.13) + 30 − 24 + 3 = 0 (c) 1000.5 and 0.15 (d) 1000.5 and 0.25 FQS = 11.25 kN Hence, the correct option is (a). Solution: (b) 9. A uniform slender rod (8 m length and 3 kg mass) Taking moment at Q = 0 rotates in a vertical plane about a horizontal axis 1 m from its end as shown in the figure. The magnitude εmQ = 0

P × 2 − w × 1 = 0 Chapter 1  Engineering Mechanics | 9.13 Solution: (b) P = 100 × 9.81 = 490.5 N 2 εFy = 0;  N − W = 0 As force F is released, the rod accelerate and N = W = 981 N undergoes pure rotation about hinged point. [weight of an object] From Newton Second law From friction theory εT = Iα [I = moment of Inertia of rod; α = acceleration of rod; εT = external frequency on rod]. F = µN = P;  490.5 = µ × 981 m = 0.5 1 Hence, the correct option is (b). 2 11. A ladder AB of length 5 m and weight (W) 600 N is W = Ia (1) resting against a wall. Assuming frictionless contact at the floor (B) and the wall (A), the magnitude of the I = W 13 [moment of inertia of the g 3 rod about hinged] force P (in newton) required to maintain equilibrium of the ladder is _____ [2014-S4] From Equation (1) Solution: (see figure) W 13 α = W 1 g 3 2 a= 3 g 2 1 For force equilibrium, Applying equation of equilibrium εFy = 0 The inertia force will act upward direction (opposite NB − 600 = 0 to motion) εMA = 0 P × 3 + w × I − NB × 4 = 0 P = 4NB − 2W 1 3 \\ R − W + ∫ pxαdx = 4 × 600 −3 2 × 600 0 = 0 [P = mass/unit length] P = 400 N. R = W − p x2 α =W − p l2 α 2 2 12. A pin jointed uniform rigid rod of weight W and length L is supported horizontally by an external = W − Wgl l22 3 3 2 l 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 R= W 1 − 3 = W 4  4 support is [2013] Hence, the correct option is (b). Common Data for Questions 13 and 14: (a) zero Two steel truss members, AC and BC, each having (b) W/4 (c) W/2 cross-sectional area of 100 mm2 are subjected to a (d) W horizontal force F as shown in figure. All the joints are hinged. [2012]

9.14 | Engineering Mechanics FAC is maximum force 6 = stress = FAC = 0.895F area 100 = 100 MPa F = 100 × 100 0.895 F = 11.17 kN 13. If F = 1 kN, magnitude of the vertical reaction force Hence, the correct option is (b). developed at the point B in kN is 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 (a) 0.63 (b) 0.32 block as shown in the figure. The friction force is [2011] (c) 1.26 (d) 1.46 Solution: (a) Isolate member AC and BC from support, (a) zero (b) 0.8 N (c) 0.89 N (d) 1.2 N Solution: (b) εFx = 0 (→ +ve) FBD of block F − FAC cos 45 − FBC cos 60 = 0 F − 0.707FAC − 0.5FBC = 0 εFy = 0 0.707FAC + 0.5FBC = F (1) N = 9.81 N εFy = 0 From theory of friction FAC sin 45 = FBC sin 60 F = µN = 0.1 × 9.81 = 0.98 N FAC = 1.22FBC (2) The external force applied From Equation (1) = 0.8 N < friction force Hence, Friction force 0.707 (1.22FBC) + 0.5FBC = 1 = external applied force = 0.8 N Hence, the correct option is (b). 0.865FBC + 0.5FBC = 1 16. Consider a truss PQR loaded at P with a force F as 1.365FBC = 1 shown in the figure. The tension in the member QR is [2008] 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 = F sin (120) sin (105) (a) 0.5F (b) 0.63F (c) 0.73F (d) 0 87F FAC = 0.895F

Chapter 1  Engineering Mechanics | 9.15 Solution: (b) (a) 2s − µ) (b) 2s Using method of joint, draw FBD of joint P g cos θ (tan θ g cos θ (tan θ + µ) (c) 2s µ) (d) 2s g sin θ (tan θ − g sin θ (tan θ + µ) Solution: (a) εFx = 0 FBD of mass m FPQ cos 45 = FPR cos 30 0.707FPQ = 0.866FPR (1) 0.866FPR − 0.707FPQ = 0 (2) εFy = 0 F + FPQ sin 45 + FPR sin 30 = 0 From Newton Second law and motion 0.5FPR + 0.707FPQ = −F Solving Equation (1) and (2) EF = ma 0.866FPR − 0.707FPQ = 0 mg sin θ − F = ma (1)    0_._5_F_P_R_+__0_.7_0_7_F__PQ_ _=__−_F_ εFy = 0 [Forces along y-direction] 1.366FPR = −F N − mg cos θ = 0 FPR = –0.732F Draw FBD of joint R mg cos q = N (2) From Equation (1) mg sin θ − µN = ma εFx = 0 a = g (sin θ − µ cos θ) [equation of equilibrium along × direction] s = ut + 1 at 2 2 −FQR − FPR cos 30 = 0 −FQR − (−0.732 F) × 0.866 = 0 [where m = initial velocity FQR = 0.63F of block = 0] Hence, the correct option is (b). From s = 1 at 2 17. A block of mass M is released from point P on a 2 rough inclined plane with inclination angle θ, shown in the figure below. The coefficient of friction is µ. If t= 2s = 2s µ < tan θ, then the time taken by the block to reach a g(sin θ − µ cos θ) another point Q on the inclined plane, where PQ = S, is [2007] = g cos θ[2tasn θ − 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

9.16 | Engineering Mechanics 10 = (1 +20) V1′ Solution: (d) V1′ = 0.476 m/s but V = rω m1 = 1 kg;  m2 = 1 kg FBD of block A w = V1 = 10 = 0.476 rad/sec r 21 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] 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. (a) 1000 N (b) 2000 N (c) 2500 N (d) 5000 N The solid wheel has a mass of 20 kg and a radius of Solution: (d) Draw FBD of weight 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 εFy = 0 [Forces along vertical direction] µR + µR = 2000;  2µR = 2000 (c) 10 rad/s (d) 10/3 rad/s 2 × 0.1 × R = 2000;  R = 10000 N 3 Now, draw FBD of gripper tongs 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′

εmax P = 0 Chapter 1  Engineering Mechanics | 9.17 10000 × 150 = F × 300 (c) Rx = 755.4 N, Ry = 343.4 N F = 5000 N (d) Rx = 755.4 N, Ry = 0 Solution: (d) Hence, the correct option is (d). Draw FBD of bar AB 21. As shown in figure, a person A is standing at the centre tan 125 of a rotating platform facing person B who is riding 275 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] q = θ = 24.45° Draw FBD at B By Lami’s theorem (a) 3 m/s heading East T = HA (b) 3 m/s heading West sin 90 sin (114.45) (c) 8 m/s heading East (d) 13 m/s heading East Solution: (d) = sin3(4135.53.555) Apparent velocity of B w.r.to A = V − (rω) [r = distance of AB] HA = 755.15 N = 8 − 5 (−1) ω = Angular velocity T = 829.55 N = 1 rad/sec = 13 m/s heading east To calculate VA Hence, the correct option is (d). εFy = 0 22. A mass 35 kg is suspended from a weightless bar AB VA + T sin (24.45) − 343.35 which is supported by a cable CB and a pin at A as = 0 shown in figure. The pin reactions at A on the bar AB are [1997] VA + 829.55 × 0.4138 −343.35 = 0 (a) Rx = 343.4 N, Ry = 755.4 N (b) Rx = 343.4 N, Ry = 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

9.18 | Engineering Mechanics VA = ω × IA w= VA = 1 = 2 rad/sec IA 1/2 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. C ollision of 1. Euler’s equation of motion particles B. Stability 2. Minimum kinetic energy Solution: (c) C. Satellite motion 3. Minimum potential energy As in figure hinge ‘B’ is frictionless, no torque is applied to bar CD. So, no angle change occurs. D. Spinning top 4. Impulse-momentum Hence, the correct option is (c). principle 24. A rod of length 1 m is sliding in a corner as shown in 5. C onservation of moment of momentum figure. At an instant when the rod makes an angle of Solution: a-4, b-3, c-1, d-5. 60 degrees with the horizontal plane, the velocity of 26. A spring scale indicates a tension T in the right hand point A on the rod is 1 m/s. The angular velocity of cable of the pulley system shown in figure. Neglecting the rod at this instant is [1996] the mass of the pulleys and ignoring friction between the cable and pulley the mass m is [1995] (a) 2 rad/s (b) 1.5 rad/s (c) 0.5 rad/s (d) 0.75 rad/s Solution: (a) (a) 2T/g (b) T (1 + c4π)/g (c) 4T/g (d) none of the above Solution: (c) VA—velocity of point A acting along a vertical wall FBD of scale = 1 m/s Applying equation of equilibrium εFg = 0 VB—velocity of point B acting along a horizontal T + 2T +T = mg plane = from geometry of triangle ABI 4T = mg IA = OB = l cos θ = 1 m  = OA = l sin θ = 2  m = 4T/g IB  θ = 60°       3 Hence, the correct option is (c). 2 

08. ENGINEERING MECHANICS 1. Forces and Force Systems Ans. (d) : We know that- Given, θ = 90 tan θ = Bsin α 1. Two forces P and Q are acting at an angle . A + Bcos α The resultant force R acts at an angle of with force P, then the value of will be (a) tan −1 Q cosα (b) tan−1 Qsin α P + Q cosα P + Q cos α (c) tan −1 Q cosα (d) tan−1 Qsin α P + Qsin α P + Qsin α HPPSC AE 2018 tan 90o = Bsin α A + B cos α Ans. (b) : Parallelogram Law of Forces–\"If two forces, acting at a point be represented in magnitude and 1 = Bsin α direction by the two adjacent sides of parallelogram, 0 A + Bcos α then their resultant is represented in magnitude and A + Bcos α = 0 direction by the diagonal of the parallelogram passing A = −B cos α through that point\". α = cos −1  − A   B  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. If the angle between the forces is α, then (b) The theorem is applicable for parallel forces. (c) The theorem is not applicable for more than R = P2 + Q2 + 2PQcos α three forces. The direction of resultant will be (d) The theorem is not applicable for less than θ = tan −1  Qsin α  (from Force P) three forces.  + Q cos  (e) The theorem is applicable for coplanar forces.  P α  (CGPCS Polytechnic Lecturer 2017) 2. A rigid body will be in equilibrium under the Ans. (b) : Lami's Theorem–\"If a body is in action of two forces only when: equilibrium under the action of three forces, then each (a) Forces have same magnitude, same line of force is proportional to the sine of the angle between the action and same sense other two forces\". (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: P =Q=R sin α sin γ sin β (a) θ = cos −1  A  (b) θ = cos −1  B  5. The Lami's theorem is applicable only for :  B   A  (a) Coplanar forces (c) θ = cos −1  − B  (d) θ = cos −1  − A  (b) Concurrent forces  A   B  (c) Coplanar and Concurrent forces (d) Any types of the forces  B  (e) θ = tan −1  A  TRB Polytechnic Lecturer 2017 Ans. (c) : The Lami's theorem is applicable only for CGPSC AE 2014- I coplanar and concurrent forces. 391

6. Concurrent forces 2, 3 , 5, 3 and 2 kN act Ans. (a) : We know that, at one of the angular points of a regular Resultant force (R) = P2 + Q2 + 2PQcos θ 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 R2 = P2 + P2 + 2 P2 cosα (P =Q) (c) R = 10 kN; α = 60º = 2P2 + 2P2 cosα (d) R = 20 kN; α = 0º (e) R = 10 kN; α = 90º = 2P2 (1+ cosα ) CGPSC 26th April 1st Shift = 4P2 cos2 (α / 2) Ans. (c) : Resolving all forces horizontally ∑ H = 2cos 0º + 3 cos 30º + 5cos 60º R = 2P cos(α / 2) + 3 cos90º +2cos120º 9. The maximum force which acts on the connecting rod is (a) Force due to gas pressure ( )3× 3  + 5  1  + 3×0 + 2 ×  − 1  2  2   2  (b) Force due to inertia of piston = 2 +  (c) Force due to friction of connecting rod = 5 kN (d) Force due to crank pin Resolving all forces vertically TNPSC AE 2013 ∑V = 2sin 0º + 3 sin 30º +5sin 60º Ans. (a) : Force due to gas pressure is the maximum force acts on the connecting rod. + 3 sin 90º +2sin120º 10. Calculate the resisting torque for static = 0+ 3+5 3+  3 equilibrium in the following figure: 22 3 +  2 × 2  = 8.66 kN The magnitude of the resultant forces R = (∑ H )2 + (∑V )2 = (5)2 + (8.66)2 (a) 100 N-cm (b) 150 N-cm = 9.99 (c) 200 N-cm (d) 300 N-cm ≃ 10kN Ans : (d) : Gujarat PSC AE 2019 Direction of the resultant force tanθ = ∑V = 9.99 =1.732 ∑ H 8.66 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 T = F sin 30o × 30 = 20 × sin 30o × 30 (e) Curvilinear motion with constant speed = 300 N-cm CGPSC 26th April 1st Shift Three forces P, Q and R are acting Ans. (e) : A particle is said to be in equilibrium if the concurrently. The included angles are , and net force acting on the particle is zero. By Newton's 11. . According to Lami's theorem (a) P = Q = R second law, the acceleration of such objects will be sin α sinβ sin γ 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). (b) P = Q = R cosα cosβ cosγ 8. If two of the three concurrent forces in equilibrium have equal magnitude P and angle between them, the magnitude of third force is (c) P = Q = R tan α tanβ tan γ (a) 2P cos (α /2) (b) 2P cos α (c) 2P sin α (d) P cos α (d) sin α = sinβ = sin γ P QR (e) 2P sin (α /2) CGPSC 26th April 1st Shift TNPSC AE 2018 HPPSC AE 2018 392

Ans. (a) : Lami's Theorem–\"If a body is in 15. The angle between two forces P and Q is The resultant of these forces is equilibrium under the action of three forces, then each (a) P2 + Q2 + 2PQsin α force is proportional to the sine of the angle between the other two forces\" (b) P2 + Q2 + 2PQ cos α (c) P2 + Q2 (d) P2 + Q2 − 2PQ cos α UPPSC AE 12.04.2016 Paper-I P =R=Q Ans : (b) sin α sinβ sin γ Parallelogram law of forces:- It states that if two forces, acting simultaneously on a particle, be represented in 12. The force applied on a body of mass 100 kg to produce an acceleration of 5 m/s2 is magnitude and direction by the two adjacent side of a (a) 20 N (b) 100 N parallelogram, then their resultant may be represented in magnitude and direction by the diagonal of a parallelogram which passes through their points of intersection. (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 R = P2 + Q2 + 2PQcos θ 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 tan α = P Qsin θ θ + Qcos AC (in N) is . 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. (a) 200 100 N 17. If a suspended body is struck at the centre of (b)300 percussion, then the pressure on the axis (c) 50 (d)100 passing through the point of suspension will be: RPSC INSP. OF FACTORIES AND BOILER 2016 (a) Maximum (b) Minimum Ans : (d) By Lami's Theorem– (c) Zero (d) Infinity ⇒ 100 = PAC HPPSC W.S. Poly. 2016 sin1200 Sin(900 + 300 ) Ans : (c) If a suspended body is struck at the centre of ⇒ PAC = 100 N percussion, then the pressure on the axis passing through the point of suspension will be zero. 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, The centre of oscillation is termed as centre of R2 = P2 + Q2 + 2PQ cos θ percussion. It is defined as that point at which a blow Given, P = 100 kN, Q = – 75 kN may be struck on a suspended body so that the reaction R2 = (100)2 + (–75)2 + 2 × 100 × (–75) × cos 0o at the support is zero. R2 = 625 18. The effect of a force on a body depends on its : R= 625 (a) Direction (b) Magnitude R = 25 kN (c) Position (d) All of these Hence, the two forces are unlike parallel forces. (OPSC AE. 2016) UKPSC AE 2007 Paper -I 393

Ans : (d) The effect of a force on a body depends on its: Ans : (c) (i) Direction, (ii) Magnitude, (iii) Position (iv) Line of Free body diagram. 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 N and 400 N (d) 300 3 N and 300 3 N TSPSC AEE 2015 Ans : (a) W = F = R sin 1500 sin120 sin 900 horizontal components = 200 cos 30° F = W sin120 FH = 100 3N sin150 Vertical component = 200 sin30° F = 3W Fv = 100 N 22. A 40 mm diameter water jet strikes a hinged 20. What is the thrust at the point 'A' in the post vertical plate of 800 N weight normally at its shown in the figure? surface at its centre of gravity as shown in the UPPSC AE 12.04.2016 Paper-I 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 (a) 0.866 kN (b) 0.5 kN ESE 2018 (c) 1.388 kN (d) 1 kN Ans. (a) : 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 (b) W taking moment about O at equilibrium position 2 F cosθ ⋅ x = W⋅y (d) 2 W F cosθ ⋅ L / 2 = W L sin θ (c) 3W UPPSC AE 12.04.2016 Paper-I cos θ 2 F = W sinθ ρAV2 = W sinθ 103 × π (0.04)2 ×152 = 800 × sinθ 4 sinθ = 0.353 θ = sin−1 (0.353) 394

23. A ball of weight 100N is tied to a smooth wall 25. If the sum of all the forces acting on a moving by a cord making an angle of 30° to the wall. object is zero, the object will (a) continue moving with constant velocity The tension in the cord is (a) 200N (b) 200 N (b) accelerate uniformly 3 (c) change the direction of motion (d) slow down and stop (c) 100N (d) 50 3N UKPSC AE 2012 Paper-I ESE 2017 Ans. (a) : continue moving with constant velocity Ans. (b) : 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. By Lami's theorem (c) P 2, at an angle of 45° with the line of T=W= R action of each force. sin 90° sin120° sin150° (d) Zero T = W = 100 = 200 UKPSC AE 2012 Paper-I sin 90° sin120° 3 / 2 3 Ans. (b) : P 2, at an angle of 45° with the line of T = 200 N action of each force. 3 27. A body subjected to coplanar non-concurrent forces will remain in a state of equilibrium if 24. In a given figure, when angles = , then the (a) ∑Fx = 0 (b) ∑Fy = 0 components (f1 and f2) of force F on either side (c) ∑M = 0 (d) All of the above three are given by an expression: 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 (a) f1 = f2 = F cos α (b) f1 = f2 = Fsin α sin(α + β) sin(α + β) Ans. (a) : ∑Fx = ∑Fy = ∑Fz = 0 (c) f1 = f2 = F cos α (d) f1 = f2 = Fsin α 29. One end of an uniform ladder, of length L and cos(α + β) sin(α) weight W, rests against a rough vertical wall and the other end rests on rough horizontal (e) f1 = f2 = Fsin α ground. The coefficient of friction f is same at sin(β) each end. The inclination of ladder when it is on the point of slipping is (CGPCS Polytechnic Lecturer 2017) Ans. (b) : Using Lami's theorem −1  1 −f 2  tan −1  1 +f 2   2f   2f  (a) tan (b) (c) tan −1  2f  (d) tan −1  2f   +f   −f  1 2 1 2 UKPSC AE 2012 Paper-I f1 f2 F tan −1  1 −f 2  sin α sin β 360° −  2f  = = (α β) Ans. (a) : sin + So, f1 = − Fsin α 30. In the following figure, the tension in the rope sin (α + β ) AC is f2 = − Fsin β ) (α = β) sin (α + β Then, f1 = f2 = Fsin α ) sin(α + β [–ve not considered in answer] 395

(a) 17.32 N (b) 56.60 N 35. Polygon of forces is useful for computing the (c) 169.90 N (d) 113.20 N resultant of (a) concurrent spatial forces UKPSC AE 2012 Paper-I (b) coplanar parallel forces Ans. (c) : (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 From Lamie's theorem (c) unlike parallel (d) concurrent AC = AD = AB UKPSC AE 2007 Paper -I sin120 sin 90 sin150 AD = AB ⇒ AD = 10 Ans. (d) : Concurrent sin 90 sin150 1 0.5 37. The quantity whose dimensions are M2L2T–3 AD = 20 kg could be the product of AC = AD (a) force and velocity (b) mass and power sin120 sin 90 (c) force and pressure (d) force and distance AC = 20 × sin 120 = 20 × 0.860 kg = 20 × 0.866 × 9.81 N UKPSC AE 2007 Paper -I AC = 169.91 N Ans. (b) : Msass and power forces, the 31. When a body is in a state of equilibrium under the action of any force system, the normal 38. In case of concurrent coplanar condition of equilibrium is (a) ∑ H = 0, ∑V = 0, ∑ M = 0 (b) ∑ H = 0, ∑V = 0 stress at a point within the body depends upon (c) ∑ H = 0, ∑ M = 0 (a) elementary area ∆A surrounding the point (d) ∑V = 0, ∑ M = 0 (b) elemental force ∆F acting normal to ∆A UKPSC AE 2007 Paper -I (c) the plane orientation containing the point (d) all the above three Ans. (b) : ∑ H = 0, ∑V = 0 UKPSC AE 2007 Paper -I 39. If a body is in equilibrium then the following is true: Ans. (d) : All the above three (a) There is no force acting on the body (b) Resultant of all forces is zero but the 32. Two forces each equal to P/2 act at right angles. Their effect may be neutralized by a third force moments of forces about any point is not zero acting along their bisector in the opposite (c) The moments of the forces about any point is direction with a magnitude of (a) P (b) P/2 zero, but the resultant of all forces is not zero (d) both (b) and (c) (c) 2P (d) P UKPSC AE 2007 Paper -I 2 Ans. (d) : Both (b) and (c) UKPSC AE 2007 Paper -I 40. Four forces P, 2P, 3 P & 4P act along the sides of Ans. (d) : P a square, taken in order. The resultant force is 2 (a) zero (b) 5P 33. A rigid body is subjected to non-coplanar (c) 2 2P (d) 2P concurrent force system. If the body is to UKPSC AE 2012 Paper-I remain in a state of equilibrium, then Ans. (c) : 2 2P (a) ∑ Fx = ∑ Fy = ∑ Fz = 0 41. According to the Newton’s law of gravitation, (b) ∑ M x = ∑ M y = 0 the force of attraction, between the bodies of (c) ∑ M y = ∑ M z = 0 (d) none of the above masses m1 and m2 situated at a distance ‘d’ UKPSC AE 2007 Paper -I apart, is given by Ans. (a) : ∑ Fx = ∑ Fy = ∑ Fz = 0 (a) F = G m1m22 (b) F = G m12m2 d2 d2 34. The resultant of forces P = -2 i - 3j and Q = 3i - 4j will lie in (quadrants to be reckoned (c) F = G m12m 2 (d) F = G m1m2 2 d2 anticlockwise) quadrant d2 (a) first (b) second (c) third (d) fourth UKPSC AE 2012 Paper-I UKPSC AE 2007 Paper -I Ans. (d) : F = G m1m2 Ans. (d) : Fourth d2 396

2. Moments and Couples 45. The moment of the force about a point is equal to the algebric sum of the component forces 42. Two non-collinear parallel equal forces acting about the same point is known as in opposite direction (a) Tresca theory (a) balance each other (b) Law of Parallelogram (b) constitute a moment (c) Law of tringle (c) constitute a couple (d) Varignon's theorem (d) constitute a moment of couple TNPSC 2019 HPPSC AE 2018 TSPSC AEE 2015 Ans. (c) : Two non-collinear parallel equal force acting in opposite direction constitute a couple. 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 = P × d (clock-wise) (c) Moment (d) Couple HPPSC W.S. Poly. 2016 43. Varignon's theorem of moments states that if a UKPSC AE-2013, Paper-I number of coplanar forces acting on a particle Ans : (d) When trying to turn a key into a lock, the are in equilibrium, then couple is applied. (a) Their algebraic then (b) Their lines of action are at equal distances Couple:- The two equal and opposite forces, whose line (c) The algebraic sum of their moments about of action are different, form a couple. 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 Moment of a couple = P × x. Ans. (d) : Varignon's Theorem– The moment of 47. A 12 kN force produces a moment of 96 kN-m resultant of concurrent forces about any point is equal to then, the moment arm is (a) 2 m (b) 4 m the algebraic sum of the moments of its components about the same point. (c) 6 m (d) 8 m 44. If a force (F) is acting on a rigid body at any TSPSC AEE 2015 point P, then this force (F) can be replaced by: 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 (a) An equal, opposite and parallel force (F) the couple is zero applied at point Q together with a couple (b) the algebraic sum of the force, constituting (b) An equal, parallel and same sense force (F) at the couple, about any point is same point Q together with a couple (c) a couple cannot be balanced by a single force (c) A moment at point Q only (d) An equal, opposite and parallel force (F) only but can be balanced only by a couple of opposite sense at point Q (d) all statements UPRVUNL AE 2016 TSPSC AEE 2015 Ans. (b) : If a force (F) is acting on a rigid body at any Ans. (d) : All statements are correct. point P, then this force (F) can be replaced by an equal, parallel and same sense force (F) at point Q together 49. Varignon’s theorem is related to with a couple. (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 397

51. The dimensions of angular velocity are given 3. The magnitude of limiting friction bears a constant by (a) M0L1T–1 (b) M0L2T–1 ratio to the normal reaction between the mating (c) M0L0T–2 (d) M0L0T–1 Ans. (d) : M0L0T–1 surface. UKPSC AE 2007 Paper -I 4. Limiting friction is independent of the area and shape of contact surface. 52. Opening a Limca bottle is due to 5. Limiting friction depends upon the nature (a) moment (b) couple (roughness or smoothness) of the surface in contact. (c) torque (d) parallel forces 6. At low velocities between sliding surface, the UKPSC AE 2007 Paper -I friction force is practically independent of the Ans. (a) : Moment velocity. However, slight reduction in friction 53. Which of the following statement is correct? occurs when the speed are high. (a) The algebraic sum of forces constituting the 56. If F is the limiting friction and RN is the normal reaction of the surfaces of contact of two couple is zero (b) The algebraic sum of the moments of forces bodies, then the coefficient of friction is constituting the couple about any point is expressed as: same (a) µ = RN/F (b) µ = F/RN (c) A couple cannot be balanced by a single force (c) µ = 2F/RN (d) µ = 0.5F/RN (d) All of the above (e) µ = 0.5RN/F UKPSC AE 2007 Paper -I (CGPCS Polytechnic Lecturer 2017) Ans. (d) : All of the above Ans. (b) : 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 Where, 55. The maximum value of _____, which comes S = Total reaction with the normal reaction. into play, when a body just begins to slide over The ratio of friction force to normal reaction is called the surface of other body, is known as _____. (a) Kinetic friction, Limiting friction coefficient of friction (µ). So, µ = F (b) Dynamic friction, Limiting friction RN (c) Solid friction, Limiting friction tanφ = µ = F , φ → Friction angle RN (d) Boundary friction, Limiting friction The coefficient of rolling resistance is defined (e) Static friction, Limiting friction 57. as the ratio between (CGPCS Polytechnic Lecturer 2017) (a) Rolling resistance to lateral load (b) Lateral load to rolling resistance Ans. (e) : The maximum value of static friction, which (c) Rolling resistance to normal load comes into play, when a body just begins to slide over (d) Normal load to rolling resistance 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 TNPSC AE 2017 external force but there is no relative motion between Ans. (c) : The coefficient of rolling resistance is defined them. Dynamic Friction–The dynamic friction is the as the ratio between Rolling resistance to normal load. frictional force that develops between mating surface when subjected to external forces and there is relative Rolling resistance coefficient (Crr) = FR motion between them. The dynamic friction is also N known as kinematic friction. 58. On a ladder resting on a smooth ground and leaning against rough vertical wall, the force of Laws of Solid Friction [Static or Dynamic] friction acts 1. Friction acts tangential to the surface in contact and (a) upward at its upper end (b) towards the wall at the upper end is in a direction opposite to that in which motion is to impend i.e. take place. (c) towards the wall at lower end 2. Friction force is maximum at the instant of (d) downward at its upper end impending motion. Its variation from zero to BPSC Poly. Lect. 2016 maximum value [Limiting friction] depends upon TNPSC AE 2018 the resultant force tending to cause motion. UKPSC AE 2007 Paper -I 398

Ans : (a) Ans. (c) : Given, W = 1000 N µ = 0.5 P = 100 N for smooth ground µ1 = 0 Frictional force (F) = µN on a ladder resting on a smooth ground and leaning = 0.5 × 1000 (∵ N = W) against rough vertical wall, the force of friction acts towards the wall at the upper end. = 500 N 59. The coefficient of friction is the ratio of Since applied force (100N) is less than limiting friction, (a) the normal reaction to the limiting force of body will be at rest. As long as the body, is at rest friction F = P = 100 N (b) The weight of the body to limiting force of 62. A 13 m long ladder is placed against a smooth friction vertical wall with its lower end 5 m from the (c) the limiting force of friction to the normal reaction wall. What should be the coefficient of friction (d) the weight of the body to the normal reaction between the ladder and the floor so that the TSPSC AEE 2015 ladder remains in equilibrium? (a) 0.29 (b) 0.25 Ans : (c) The coefficient of friction is the ratio of the (c) 0.21 (d) 0.11 limiting force of friction to the normal reaction ESE 2018 Ans. (c) : Using equilibrium equation F = µR. µ= F N2 = µN1 R N1 = W (∵ µ2N2 = 0) Because wall is smooth µ = Coefficient of friction. F = Limiting force of friction Taking moment about A R = Normal reaction. 60. Which statement is wrong in wedge friction? W × 2.5 + µ1N1 × 12 = N1 × 5 µ1 = 0.208 (a) To lift heavy block through small distances µ1 ⇒ 0.21 (b) To lift heavy block through large force (c) To slightly slide one end of the beam relative 63. A body of weight 50 N is kept on a plane inclined at an angle of 30º to the horizontal. It to another end is in limiting equilibrium. The coefficient of (d) Weight of wedge is neglected compared to friction is equal to- weight to be lifted (a) 1 (b) 3 TNPSC AE 2013 3 (d) 3 Ans. (b) : Wedge– (c) 1 5 • Simple machines used to raise heavy loads. 50 3 • Force required to life block is significantly less than block weight. • Friction prevents wedge from sliding out. RPSC AE 2018 • Minimum force D required to raise block. Ans. (a) : Given: 61. A box weight 1000 N is placed on the ground. mg = 50 N The coefficient of friction between the box and the ground is 0.5. When the box is pulled by a θ = 30º 100 N horizontal force, the frictional force For limiting equilibrium mg sin 30º = µ × R developed between the box and the ground at mg sin 30º = µ × mg cos 30º impending motion is µ = tan 30º (a) 50 N (b) 75 N (c) 100 N (d) 500 µ= 1 3 ESE 2018 399

64. A block of mass (m) slides down an inclined α is angle of contact between rope and cylinder plant (coefficient of friction = µ) from the rest α = π radian as in figure. What will be the velocity of block when it reaches the lowest point of plane? 3m = eµπ m 3 = eµπ 1.098 = µπ µ = 0.349 ≈ 0.35 (a) 2gh (1 – µ cot θ) (b) 2gh (1− µ cot θ) 66. The angle of inclination of the plane at which the body begins to move down the plane, is (c) 2ghµ(1− cot θ) (d) 2gh (1− µ sin θ) called– (b) Angle of repose (a) Angle of friction (c) Angle of projection (d) None of these (e) 2gh (µ − tan θ) Vizag Steel (MT) 2017 CGPSC AE 2014- I Ans. (b) : Angle of repose– Minimum angle of inclined Ans. (b) : Kinetic energy at lowest point on inclined plane which causes on object to slide down the plane. plane of the block is 67. Coefficient of friction depends upon ∴ 1 mV 2 = mgh–energy loss due to friction (a) Area of contact only 2 (b) Nature of surface only = mgh − µR × h (c) Both (a) and (b) sin θ (d) None of these Vizag Steel (MT) 2017 Ans. (b) :Coefficient of friction depends upon nature of surface only. = mgh − µ× mg cos θ× h 68. A block weighing 981 N is resting on a sin θ horizontal surface. The coefficient of friction V = 2gh [1− µ cot θ] 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? h is distance covered by block during motion on sin θ inclined plane. 65. What is the minimum coefficient ( ) of friction (a) 176.2N (b) 196.0 N (c) 481.0N (d) 981.0N between the rope and the fixed shaft which will OPSC Civil Services Pre. 2011 prevent the unbalanced cylinder from moving? Ans. (c) : Free body diagram (a) µ = 0.333 (b) µ = 0.350 Since, (c) µ = 0.253 (d) µ = 0.372 N+T=W Ans. (b) : N + T = 981....................(i) APPSC-AE-2019 T1 = eαθ µN = 100 ...................(ii) T2 69. From equation (i) and (ii), 400 N = 100 = 1000 = 500 0.2 2 T = 981 – 500 T = 481N Limiting friction depends upon (a) Materials of the body in contact (b) Weight of the body to be moved

(c) Roughness of surface of contact of the two (a) 0.067 (b) 0.087 bodies (c) 0.098 (d) 0.092 Ans. (b) : UKPSC AE 2012 Paper-I (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 R – 35 g = 0 slide down the plane when the angle of inclination is gradually increased to 30°. The R = 35 g F = µR coefficient of friction between the block and the µ = F = 30 = 0.087 plane is:- (b) 0.578 R 35× 9.8 (a) 0.50 (c) 0.72 (d) 0.866 UKPSC AE-2013, Paper-I 75. Two bodies of mass m and M are hung at the Ans. (b) : From this figure. ends of a rope passing over a frictional pulley. R = W cos 30o µR = W sin 30o The acceleration in which the heavier mass M µ × W cos 30o = W sin 30o comes down is given by the following: (a) g(M + m) (b) g(M + m) M +m M −m (c) gM (d) g×M M +m M −m UKPSC AE 2007 Paper -I µ× 3 = 1 Ans. (a) : g(M + m) 22 M +m 76. The maximum frictional force, which comes into play when a body first begins to slide over µ = 0.5773 the surface of another body, is known as (a) sliding friction (b) limiting friction 71. When the applied force is less than the limiting (c) kinetic friction (d) rolling friction frictional force, the body will:- UKPSC AE 2007 Paper -I (a) Start moving (b) Remain at rest UKPSC AE 2012 Paper-I (c) Slide backward Ans. (b) : Limiting friction (d) Skid 77. Tangent of angle of friction is equal to (a) kinetic friction UKPSC AE-2013, Paper-I (b) limiting friction Ans. (d) : When the applied force is less than the (c) friction force limiting frictional force, the body will remain at rest. (d) coefficient of friction 72. A 44 N block is thrust up a 30° inclined plane UKPSC AE 2007 Paper -I with an initial speed of 5 m/sec. It travels a Ans. (d) : Coefficient of friction distance of 1.5 m before it comes to rest. The 78. A body weight of 200 N is resting on a rough frictional force acting upon it would be horizontal plane, and can be just moved by a (a) 18.3 N (b) 15.3 N (c) 12.3 N (d) 9.3 N force of 80 N applied horizontally. What will UKPSC AE 2012 Paper-I the value of the coefficient of friction? Ans. (b) : 15.3 N (a) 0.4 (b) 0.5 73. Dynamic friction as compared to static friction (c) 0.3 (d) none of the above is UKPSC AE 2007 Paper -I (a) less (b) same Ans. (a) : F = µR (c) more (d) None of the above µ = 80 = 0.4 200 UKPSC AE 2012 Paper-I Ans. (a) : less 79. The coefficient of friction depends upon 74. If a force of 30 N is required to move a mass of (a) speed of the body 35 kg on a flat surface horizontally at a (b) geometrical shape of the body (c) size of the body and nature of contacting constant velocity, what will be the coefficient of surfaces friction? 401

(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 ∫ ∫Ixx = y2.dA and Iyy = x2dA rail is– (b) 5 N Where x and y the Co - ordinate of the differintial (a) 0.5 N elements of area dA. Since this area dA is multiplied by the square of the distance, the moment of inertia is also (c) 15 N (d) 150 N OPSC Civil Services Pre. 2011 Ans. (d) : Given, called the second moment at area. Diameter of wheel (d) = 600 mm 83. For principal axes, the moment of inertia will Radius of wheel (r) = 300 mm be : Load (W) = 500 N Coefficient of rolling resistance (Crr)= 0.3 (a) Ixy = 0 (HPPSC LECT. 2016) Let, Rolling friction force be FR (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) hb3 (b) bh 3 12 3 We know that, (c) bh 3 (d) hb 3 12 3 Crr = Rolling friciton force = Fr W (e) bh 3 Normal force (Load) 36 ∴Fr = Crr × W = 0.3 ×500 Fr = 150 N CGPSC AE 2014- I Ans. (b) : Moment of inertia of a rectangular section 4. Moment of Inertia about its base (x - x). 81. Moment of Inertia of an area dA at a distance x from a reference axis is: (b) ∫x2dA (a) ∫xdA (d) ∫x4dA (c) ∫x3dA (HPPSC LECT. 2016) Ans : (b) Moment of Inertia of an area dA at a distance x from a reference axis is ∫x2dA I x−x = I CG + Ax 2 I C.G. = bh 3 ,A = b× h, x= h 12 2 I x−x = bh 3 + b× h ×  h 2 = bh 3 + bh 3 = bh 3 12  2  12 4 3 The axial moment of inertia of a plane area is the I x−x = bh 3 3 geometrical characteristic of the area defined by the integrals 85. Moment of inertia of a quarter circle (diameter Ixx = ∫y2.dA and Iyy = ∫x2.dA = d) about its straight edge is given by : 82. Moment of inertia of a body does not depend (a) πd 4 (b) πd 4 64 128 on (a) Axis of rotation of the body (c) πd 4 (d) πd 4 256 512 (b) Mass of the body (c) Angular velocity of the body (d) distribution of mass in the body (e) πd 4 32 (KPSC AE 2015) Ans : (c) Moment of Inertia of a body does not depend CGPSC AE 2014- I on angular velocity of the body 402

Ans. (c) : Moment of inertia of an elliptical area about the major axis I xx = π xy3 4 Moment of inertia of an elliptical area about minor axis I yy = π yx3 4 I OX = 1 ×  πd 4  90. Statement (I): Two circular discs of equal masses 4   and thickness made of different materials will  64  have same moment of inertia about their central axes of rotation. πd 4 I OX = 256 Statement (II): Moment of inertia depends upon the distribution of mass within the body. 86. The radius of gyration 'k' for a solid cylinder of ESE 2017 radius 'R' is equal to Ans. (d) : Mass moment of inertia for a uniform disc (a) 2R (b) R / 2 about its axis of rotation (c) 0.6324 R (d) 0.5 R I = MR2 2 TNPSC AE 2017 Ans. (b) : The radius of gyration 'k' for a solid cylinder I = ρAtR2 of radius 'R' is equal to R / 2 . 2 87. The radius of gyration of a disc type fly wheel ∵ Mass & thickness is same, of diameter D is (a) D (b) D/2 ρ1A1t = ρ2A2t (c) D/4 (d) 3 D I1 = ρ1A1tR12 = π ρ1R14 t Ans. (c) : I = AK2 2 2 8 TNPSC AE 2018 I2 = ρ2 A 2 tR 2 = π ρ2R 4 t 2 8 2 2 If ρ1 > ρ2, R1 < R2, I1 > I2 K= I= π D4 Hence statement-I is wrong. A 64 π D2 91. A polar moment of Inertia (I) for hollow shaft 4 with external diameter (D) and internal diameter (d) is given by: K=D (a) 32D4 ( )π D4 − d4 4 πd4 (b) 64 88. What is the unit of moment of inertia of an ( )π D4 − d4 ( )32 D4 − d4 area. (b) kg m (c) (d) (a) kg m2 (d) m4 32 π (c) m3 (e) kg m3 CIL (MT) 2017 IInd Shift CGPSC 26th April 1st Shift (CGPCS Polytechnic Lecturer 2017) Ans. (d) : Units of the mass moment of inertia are kg- m2, gram-cm2. Ans. (c) : Units of the area moment of inertia are m4, mm4. ( )π D4 − d4 Polar moment of inertia of hollow haft = 32 89. Moment of inertia of an elliptical area about 92. Moment of inertia of an area always least with the major axis is (a) πxy3/4 (b) πxy3/3 respect to (a) Bottom-most axis (b) Radius of gyration (c) πx2y3/4 (d) πx2y3 (e) πx2y3/3 (c) Central axis (d) Centroidal axis Vizag Steel (MT) 2017 CGPSC 26th April 1st Shift Ans. (d) : The moment of inertia about any axis passing through centroid is zero so Moment of inertial of an Ans. (a) : 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 403

Ans : (a) : Polar moment of inertia (J) = Ix + Iy = bd 3 + db3 12 12 = 2× 63 + 23 × 6 Given- 12 12 = 40 cm4 94. Radius of gyration of a circular section with IL-L = IT-A diameter D is mR 2 = mR 2 + 1 mL2 (a) D (b) D 2 4 12 2 4 L 2 (c) D (d) D  R  = 3 3 3 APPSC-AE-2019 [L/R] = 3 Ans. (b) : For circular section I = π d 4 , A = π d 2 98. The moment of inertia of a square side (a) 64 4 about an axis through its center of gravity is (a) a4/4 (b) a4/8 π d4 (c) a4/12 (d) a4/36 Radius of gyration = k = I= 64 =d A π d2 4 Vizag Steel (MT) 2017 4 Ans. (c) : 95. Moment of Inertia about the centroidal axis of elliptical quadrant of base 'a' and height 'b' is (a) π ab (a2 + b2 ) (b) π ab (a3 + b3 ) 24 24 (c) π ab (a2 + b2 ) (d) π ab (a3 + b3 ) 16 24 TNPSC AE 2013 ( )Ans. (c) : I = π ab a 2 + b 2 IXX = IYY = a4 16 12 96. When a body of mass moment of Inertia I is 99. Ratio of moment of Inertia of a circular body rotated about that axis with an angular velocity about its x axis to that about y axis is , then the kinetic energy of rotation is (a) 0.5 (b) 1.0 (a) 0.5 I.ω. (b) I.ω. (c) 0.5 I.ω2 (d) I.ω2 (c) 1.5 (d) 2.0 TNPSC AE 2014 TNPSC AE 2013 πd 4 πd 4 Ans. (c) : Kinetic energy of rotation is given as Ans. (b) : I x−x = 64 , I y−y = 64 KE = 1 Iω2 2 97. Length to radius ratio l of a solid cylinder is r 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 πd 4 longitudinal axis 32 I L−L = mR 2 I z−z = I xx + I yy = 2 πd 4 MOI of a uniform solid cylinder about transfer axis then I x−x = 64 =1 I y−y πd 4 I T−A = 1 mR 2 +1 mL2 4 12 64 404

5. Centroid and Center of Gravity 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 100. The C.G. of a solid hemisphere lies on the (b) central radius (c) one-third of the total height above base (a) at distance 3r/2 from the plane base (d) one-half of the total height above base (b) at distance 3r/4 from the plane base three-eighth of the total height above the base (c) at distance 3r/5 from the plane base Ans : (b) HPPSC W.S. Poly. 2016 (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. 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 101. The coordinates of centroid of given geometry (c) 20 mm (d) 10 mm ABCDEFA [DE = 20 mm, EF = 80 mm, FA = TSPSC AEE 2015 100 mm, AB = 20mm] will be given as : (a is Ans : (a) origin) The centre of gravity of a hemisphere is at a distance of 3r/8 from base (a) (60, 20) (b) (25, 60) C.G = 3r/8 (2r = d= 80mm) (c) (25, 65) (d) (65, 25) C.G = 15 mm. (e) (25, 40) CGPSC AE 2014- I 105. The base of triangle is 60 mm and height is 50 mm. The moment of inertia about its base 'BC' is Ans. (c) : The coordinates of centroid is ( x, y) (a) 2,08, 333.33mm4 (b) 6.25,000 mm4 (c) 9,00,000 mm4 (d) 3,00, 000mm4 TSPSC AEE 2015 Ans : (b) x = a1x1 + a 2 x 2 IGG = bh3 a1 + a2 36 = 80× 20 ×10 + 80× 20 × 40 IBC = bh3 80× 20 + 80 × 20 12 x = 25mm 60 × (50)3 y = a1y1 + a 2y 2 = 80 × 20 × 40 + 80 × 20× 90 = 65mm IBC = 12 a1 + a2 80× 20 + 80× 20 IBC = 625000 mm4. So coordinates of centroid (25, 65) 106. The centre of gravity of a plane lamina is not at 102. Shear centre of a semicircular arc is at : its geometrical centre, if it is a (a) 4r/π (b) 3r/π (a) circle (b) square (c) 2r/π (d) r/π (c) rectangle (d) right angled triangle (HPPSC AE 2014) UKPSC AE 2007 Paper -I Ans : (a) Shear centre of a semicircular arc is at 4r/π. Ans. (d) : Right angled triangle 405

107. The eccentricity for the ellipse is ______ 1 and (b) Members of the truss are rigid for hyperbola is ______ 1. (a) equal to, equal to (c) Members of the truss are subjected to bending moments (b) less than, greater than (c) greater than, greater than (d) Members are of uniform cross-section (d) less than, less than TNPSC AE 2013 (e) Less than, equal to Ans. (c) : Members of the truss are not subjected to (CGPCS Polytechnic Lecturer 2017) bending moments. Ans. (b) : The eccentricity for the ellipse is less than 1 111. In a simple truss, if n is the total number of and for hyperbola is greater than 1. joints, the total number of members is equal to 108. Which of the following represents the state of neutral equilibrium? (a) 2n + 3 (b) 2n - 3 (a) Cube resting on one edge (b) A smooth cylinder lying on a curved surface (c) n + 3 (d) n - 3 Gujarat PSC AE 2019 (c) A smooth cylinder lying on a convex surface Ans : (b) : (d) None of the above m = 2n − 3 UKPSC AE 2007 Paper -I Where, m = member of joints Ans. (d) : None of the above n = Total number of joints 109. The coordinate of centroid (x, y) of quarter A rectangular strut is 150 mm wide and 120 circular lamina of radius (R), whose straight 112. mm thick. It carries a load of 180 kN at an edges coincide with the coordinate axis in the first quadrant, are given as: eccentricity of 10 mm in a plane bisecting the (a) (0,4R / 3π) (b) (4R / 3π,0) thickness as shown in the figure (c) (4R / 3π, 8R / 3π) (d) (4R / 3π, 4R / 3π) (e) (8R / 3π, 8R / 3π) CGPSC AE 2014- I Ans. (d) : For circle x = 2r sin α 3α The maximum intensity of stress in the section will be (a) 14 MPa (b) 12 MPa (c) 10 MPa (d) 8 MPa ESE 2019 In case quadrant of a circle- Ans. (a) : Given, 2α = 90o = π radians P = 180 kN 2 b = 150 mm 2r sin ( π / 4) 4r d = 120 mm π 3π x = 3× 4 = 2 × e = 10 mm Resultant normal stress is maximum at the right side The position of centroid with respect to the fiber of the cross section, because the line of action of radii OA and OB will be eccentric axial compressive load is nearer to this fiber. oa ' = 2 × 4r × cos 45o = 4r Maximum intensity of stress 3π 3π = σc + σb ob ' = 2 × 4r cos 45o = 4r σmax = P + M 3π 3π A Z =P + P×e = P + 6P × e A  db2  bd db2 6. Trusses and Beams  6  110. Which one of the following is the wrong = P 1 + 6× e = 180 ×103 1 + 6 ×10  assumption: bd b  120 ×150 150  (a) Members of the truss are pin-connected to each other = 14 MPa 406

113. Choose the correct combination of the stability Joint-C and indeterminacy of the truss given. (a) Statically indeterminate and stable From Lami's theorem (b) Statically determinate and stable F = FAC = FBC (c) Unstable sin 90 sin120 sin150 (d) Statically determinate FAC = 50 × sin120 ⇒ FAC = 43.3 N APPSC-AE-2019 (compressive) Ans. (c&d) : No. of member m = 8 No. of joints, j = 6 Joint-A 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 then, FAB = FAC cos 60 (d) Sufficient but not necessary = 43.3 × cos 60 APPSC-AE-2019 FAB = 21.65 N (Tensile) RA = 37.498 N Ans. (d) : Given condition m + r > 2j 116. In the joint method of plane truss analysis, m > 2j - r value of forces in the member of truss can be found on when joint has: ∴ The member is structurally stable but not (a) Only four unknown force members necessarily have m > 2j - r. It is just sufficient to have (b) Only three unknown force members (c) Not more than two unknown force members m = 2j - r (d) Any number of unknown force members 115. Determine the nature of force in member AB UPRVUNL AE 2016 and AC respectively of the truss shown in figure. A is hinge support and B is roller Ans. (c) : In the joint method of plane truss analysis, support. The direction of reaction at supports value of forces in the member of truss can be found out (RA and RB) is also shown. Load 50N is acting when joint has not more than two unknown force at joint C. 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 (a) Tensile, Compression trusses where as graphical method, joints method and method of sections are used to analysing the trusses. (b) Compression, Compression (c) Tensile, Tensile 118. What will be the axial force in the member EC, (d) Compression, Tensile ED and DC of the plane truss (ABCDE) as shown in figure with end 'A' and 'B' is hinged UPRVUNL AE 2016 to foundation? Ans. (a) : 90º 407

(a) FEC = FED = FDC = 0 sinθ = 3 = 1 (b) FEC = FED = 0; FDC = 1 kN (C) 35 5 (c) FEC = FED = 0; FDC = 1 kN (T) (d) FEC = FED = FDC = 1 kN(T) ∑ M D = 0 ( + ve, - ve) (e) FEC = FED = FDC = 1 kN (C) -FBG cos θ × 3 - FBG sin θ × 6 - 2 × 6 - 8 × 6 = 0 CGPSC AE 2014- I Ans. (c) : Point E- −FBG × 6 × FBG × 6 − 60 = 0 5 5 −2  FBG × 6  = 60  5  FBG = −5 5 kN (compressive) FBG = 11.18 kN (compressive) Point D- 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 F cos 45º = 1 (tension or compression). They cannot take shear force, FDC = F sin 45º bending moment and torsion. = F cos 45º 121. The condition for a truss to be perfect is where = 1 kN (T) m = number of members and J = number of 119. Determine the force in the member BG in the joints given truss (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) (a) 11.18 kN Tension m > 2j - 3 (b) 14.4 kN tension 122. The relation between the number of joints (J) (c) 11.18 kN Compression (d) 14.40 kN compression and number of members (m) in a truss is APPSC-AE-2019 related by : Ans. (c) : Take section 1 - 1 as shown in figure. (a) m = 2J + 3 (b) J = 3m + 3 To find the force in BG, consider the right part of section (1) - (1) and take moment about 'D'. (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 cosθ = 6 = 2 OPSC Civil Services Pre. 2011 35 5 Ans. (a) : In method of section, the section must pass through not more than 3 members. 408

125. If n < (2j - 3), where n is number of members 131. For truss as shown below, the forces in the member AB and AC are 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 (a) Tensile in each (b) Compressive in each j = number of joints (c) Compressive and Tensile respectively (d) Tensile and Compressive respectively 126. Which of the following equilibrium equation UKPSC AE 2012 Paper-I should be satisfied by the joints in truss:- (a) Ans. (c) : Compressive and Tensile respectively (b) ∑ H = 0, ∑V = 0 132. The possible loading in various members framed structure are (c) ∑V = 0, ∑ M = 0 (a) Buckling or shear (d) ∑ H = 0, ∑V = 0 and ∑ M = 0 UKPSC AE-2013, Paper-I (b) Compression or tension Ans. (b) : No bending moment acting on the joints in (c) Shear or tension (d) Bending truss so equilibrium equation should be satisfied by the UPPSC AE 12.04.2016 Paper-I joints in truss will be, ΣH = 0, ΣV = 0 Ans : (b) The possible loading in various members 127. When the number of members ‘n’ in a truss is framed structure are compression or tension. more than 2j-3, where ‘j’ is the number of 133. A truss hinged at one end, supported on rollers joints, the frame is said to be:- at the other, is subjected to horizontal load (a) Perfect truss (b) Imperfect truss only. Its reaction at the hinged end will be: (c) Deficient truss (d) Redundant truss (a) Horizontal (b) Vertical UKPSC AE-2013, Paper-I (c) Resultant of horizontal and vertical Ans. (d) : If (n > 2j – 3), then it become redundant (d) Difference between horizontal and vertical truss. TRB Polytechnic Lecturer 2017 128. Choose the correct relationship between the Ans. (c) : Its reaction at the hinged end will be resultant given statements of Assertion (A) and Reason of horizontal and vertical. (R). Assertion (A) : Only axial forces act in members R = (∆H)2 + (∆V)2 of roof trusses. 134. The force which is not considered in the analysis of the truss is Reason (R) : Truss members are welded together. (a) External applied loads (b) Horizontal forces on joints Code : (c) Vertical forces joints (a) Both (A) & (R) are correct. (R) is the correct (d) Support reactions (e) Weight of the members explanation of (A). (b) Both (A) & (R) are correct. (R) is not the CGPSC 26th April 1st Shift correct explanation of (A). (c) (A) is true, but (R) is false. (d) (A) is false, but (R) is true. Ans. (e) : Assumption for a perfect truss UKPSC AE 2012 Paper-I (i) All the members of truss are straight and connected to each other at their ends by frictionless pins. Ans. (c) : (A) is true, but (R) is false. (ii) All external forces are acting only at pins. 129. The relationship, between number of joints (J), (iii)All the members are assumed to be weightless. (iv) All the members of truss and external forces acting and the number of members (m), in a perfect at pins lies in same plane. truss, is given by (v) Static equilibrium condition is applicable for (a) m = 3j – 2 (b) m = 2j – 3 analysis of perfect truss. (c) m = j – 2 (d) m = 2j – 1 UKPSC AE 2012 Paper-I Ans. (b) : m = 2j – 3 7. Kinematics and Kinetics of Particle 130. In the analysis of truss, the force system acting at each pin (a) is concurrent but not coplanar. 135. The motion of a body moving on a curved path (b) is coplanar and concurrent. is given by a equation x = 4 sin 3t and y = 4 (c) is coplanar and non-concurrent. cos3t. The resultant velocity of the car is (d) does not satisfy rotational equilibrium. (a) 30 m/sec (b) 24 m/sec UKPSC AE 2012 Paper-I (c) 12 m/sec (d) 40 m/sec Ans. (b) : is coplanar and concurrent. TNPSC AE 2014 409

Ans. (c) : x = 4 sin 3t v = dx = 1 5) y = 4 cos 3t dt (4x + velocity component in x & y direction respectively then dv (4x + 5)× 0 −1(4) ux = dx = 4cos3t × 3 a = dt = (4x + 5)2 dt =12 cos 3t −4 vy = dy = −4sin 3t × 3 a = ( 2 ....(1) dt 4x + 5) vy = – 12 sin 3t ∴ t = 12 Resultant velocity - t = 2x2 + 5x 2x2 + 5x = 12 v= u 2 + v 2 2x2 + 5x + 2 = 0 x y (x + 4) (2x – 3) = 0 x = – 4, 2x – 3 = 0 = (12 cos 3t )2 + (−12sin 3t ) 2 = (144) sin 2 3t + cos2 3t x=3 2 = 144 ×1 x = 4 value is substitute in equation (1) v = 12 m / s  = −4 = −4  a  136. The linear velocity of a body rotating at ω rad/s  4× (−4) + 5 2 (11) 2  along a circular path of radius r is = −4 unit 121 (a) ω .r (b) ω/r (c) ω2. r (d) ω2 / r 139. Time variation of the position of the particle in rectilinear motion is given by x = 3t3 + 2t2 + 4t. APPSC AEE 2016 Ans. (a) : We know that, linear velocity, If v is the velocity and a is the acceleration of V = ω .r the particle in consistent units, the motion 137. The midpoint of a rigid link of a mechanism started with moves as a translation along a straight line, (a) v = 0, a = 0 (b) v = 0, a = 4 from rest, with a constant acceleration of 5 m/ (c) v = 4, a = 4 (d) v = 2, a = 4 s2. The distance covered by the said midpoint in (e) v = 4, a = 2 5s of motion is (a) 124.2 m (b) 112.5 m CGPSC 26th April 1st Shift (c) 96.2 m (d) 62.5 m Ans. (c) : Given, x = 3t3 + 2t2 + 4t ESE 2018 velocity (v) = dx = 9t2 + 4t + 4 Ans. (d) : Given, dt ...(1) a = 5 m/s2 u = 0 (rest) t = 5 sec Acceleration (a) = dv = 18t + 4 ...(2) dt Distance covered (s) = ut + 1 at2 ∵ Initially t = 0 so from equation (1) and (2) 2 v = 9t2 + 4t + 4 138. = 0 + 1 × 5× 52 = 62.5 m and at t = 0, v = 4 2 The a = 18t + 4 The functional reaction between time t at t =0, a = 4 distance x, in m, is t = 2x2 + 5x.. 140. When a particle moves with a uniform velocity acceleration in m/s2 at t = 12 s is : along a circular path, then the particle has (a) tangential acceleration only (a) −1 units (b) −4 units (b) normal component of acceleration only 121 1331 (c) centripetal acceleration only (d) both tangential and centripetal acceleration (c) −4 units (d) 4 units TNPSC AE 2013 121 1331 BHEL ET 2019 Ans. (c) : When a particle moves with a uniform velocity along a circular path, then the particle has Ans. (c) : Given- centripetal acceleration only. t = 2x2 + 5x Acceleration a (m/s2) at t = 12 sec. 141. A motorist travelling at a speed of 18 km/hr, t = 2x2 + 5x suddenly applies the brakes and comes to rest after skidding 75 m. The time required for the dt = 4x + 5 car to stop is dx (a) t = 30.25 sec (b) t = 29.94 sec or dx = 1 (c) t = 28.84 sec (d) t = 26.22 sec dt 4x + 5 TNPSC AE 2014 410

Ans. (b) : u = 18 km/hr (c) the reaction on the front wheels increases and on the rear wheels decreases = 5 m/s (d) the reaction on the rear wheels increases and d = 75 m on the front wheels decreases v=0 RPSC Vice Principal ITI 2018 We know that after applies the brake- Ans. (b) : The reaction on the outer wheels increases v2 = u2 – 2ad and on the inner wheels decreases. (5)2 = 2 × a × 75 a = 1 m/s2 146. The mathematical technique for finding the 6 best use of limited resources in an optimum manner is called:- then v = u – a t (a) Linear programming (b) Network analysis u=a×t (c) Queueing theory (d) None of the above 5= 1 ×t 6 UKPSC AE-2013, Paper-I t = 30 sec Ans. (a) : The mathematical technique for finding the best use of limited resources in an optimum manner is called linear programming. 142. The angular motion of a disc is defined by the 147. An object falls from the top of a tower. If relation ( = 3t + t3), where is in radians and t comes down half the height in 2 seconds. Time is in seconds. What will be the angular position taken by the object to reach the ground is:- after 2 seconds? (a) 2.8 s (b) 3.2 s (a) 14 rad (b) 12 rad (c) 4.0 s (d) 4.5 s (c) 18 rad (d) 16 rad UKPSC AE-2013, Paper-I CIL MT 2017 2017 IInd shift Ans. (a) : We assume that the height of tower is h, Ans. (a) : Given angular motion equation θ = 3t + t3 h = ut + 1 gt2 ...(i) Motion covered in 2 seconds = 3 * 2 + 23 = 14 rad. So 22 the angular position after 2 seconds will be 14 rad. u = 0, t = 2s h = 4g 143. The position of a particle in rectilinear motion then, is given by the equation (x = t3 - 2t2 + 10t - 4), v2 = u2 + 2gh 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 v = 2gh = 2g × 4g (c) 15 m/s (d) 30 m/s v=g 8 ...(ii) CIL MT 2017 2017 IInd shift then, Ans. (b) : Position (x) = t3 - 2t2 + 10t - 4 v = u + gt v=g×t Velocity ( v) = dx = 3t2 − 4t +10 t=v=g 8 dt gg At t = 3s V = 3 × 32 - 4 × 3 + 10 = 25 m/s 144. A particle move along a straight line such that t = 2.83 s distance (x) traversal in t seconds is given by x = t2 (t - 4), the acceleration of the particle will 148. Two cars ‘A’ and ‘B’ move at 15m/s in the same direction. Car ‘B’ is 300m ahead of car be given by the equation (b) 3t2 + 2t ‘A’. If car ‘A’ accelerate at 6m/s2 while car ‘B’ (a) 3t3 - 2t (c) 6t - 8 (d) 6t - 4 continues to move with the same velocity, car Ans. (c) : x = t2 (t - 4) TNPSC AE 2018 ‘A’ will overtake car ‘B’ after:- (a) 7.5 s (b) 10 s dx = V = 3t 2 −8t (c) 12 s (d) 15 s dt UKPSC AE-2013, Paper-I d 2x dV Ans. (b) : Two can A & B move initial at 15 m/s dt 2 dt = = a = 6t −8 So initial velocity of car A with respect to car B is zero. 145. A motor car moving at a certain speed takes a So, car A will over take distance 300 m when left turn in a curved path. If the engine rotates accelerate at 6 m/s2 in the same direction as that of wheels, then S = ut + 1 at 2 due to the centrifugal forces 2 (a) the reaction on the inner wheels increases and 300 = 0 × t + 1 × 6 × t2 on the outer wheels decreases 2 (b) the reaction on the outer wheels increases and t = 10 s on the inner wheels decreases 411

149. Two balls are dropped from a common point 8. Kinematics and Kinetics of after an interval of 1 second. If acceleration Rigid Body due to gravity is 10m/s2, separation distance 3 second after the release of the first ball will be:- 152. The tension in the cable supporting a lift is (a) 5 m (b) 15 m more when the lift is (c) 25 m (d) 30 m (a) Moving downwards with uniform velocity. (b) Moving upwards with uniform velocity. UKPSC AE-2013, Paper-I (c) Moving upwards with acceleration. (d) moving downwards with acceleration. Ans. (c) : The first ball has traveled for 3 seconds under UPPSC AE 12.04.2016 Paper-I gravity wit acceleration g and initial velocity u = 0 Ans : (c) The tension in the cable supporting a lift is S1 = 0× 3 + 1 × 10 × (3)2 more when the lift is moving upwards with 2 acceleration. S1=45 m Then second ball has traveled for 2 seconds 153. An object having 10 Kg mass and weights as Then, 9.81 Kg on a spring balance. The value of \"g\" S2 = 0× 2+ 1 ×10× ( 2)2 at that place is (b) 10.m/s2 2 (a) 9.81 m/s2 (d) 98.1 m/s2 (c) 0.981m/s2 S2 = 20 m Then, distance between them will be UJVNL AE 2016 S1 − S2 = 45 − 20 = 25 m Ans : (c) Given, Object mass = 10 kg Spring weight = 9.81 N 150. .......... represents the area under acceleration - W = mg 9.81 = 10 × g time graph. (b) Displacement g = 0.981 m/sec2 (a) Acceleration (d) Change in velocity 154. A rigid body can be replaced by two masses (c) Motion placed at fixed distance apart. The two masses form an equivalent dynamic system, if (select (HPPSC LECT. 2016) the most appropriate answer). Ans : (d) Change in velocity represent the area under acceleration - time graph. 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. a = dv (d) All of the above. dt Ans : (d) (i) The centre of gravity of two masses dv = a dt coincide with that of the body (ii) The sum of the two masses is equal to the total mass ∫ dv =∫ a dt of the body (iii) The sum of the mass moment of inertia of the ∫v2 − v1 = a dt masses about their centre of gravity is equal to the mass moment inertia of the body 151. The speed of a particle moving in circular path 155. If the body falls freely under gravity, then the is 600 rpm. Then, the angular velocity of the gravitational acceleration is taken as particle is (a) + 8.91 m/s2 TSPSC AEE 2015 (b) – 8.91m/s2 (a) 20π rad/ sec (b) 10 π rad/sec (c) + 9.81 m/s2 (d) – 9.81 m/s2 (c) 20 /π rad/sec (d) 10/πrad / sec Ans : (c) If the body falls freely under gravity, then the gravitational acceleration is taken as + 9.81 m/sec2 TSPSC AEE 2015 Ans : (a) Speed. fo a particle = 600 rpm. 156. A car moving with a uniform acceleration Angular velocity = 2πN rad / sec. covers 450 m in 5 sec interval, and covers 700 60 m in next 5 seconds interval. The acceleration = 2π× 600 rad / sec. of the car is: (b) 10 m/sec2 60 (a) 7.5 m/sec2 (d) 20 m/sec2 (c) 12.5 m/sec2 Angular velocity = 20π rad/sec. HPPSC LECT. 2016 412

Ans : (b) S1 = ut + 1 at2 dv = 60t × dt + 40dt 2 Integrating it 450 = 5u + 1 a ×52 ∫ ∫ ∫dv = 60×dt + 40 dt 2 ……. (i) v = 60 t2 + 40t + c1 .....(3) 450 = 5u + 12.5a …….(ii) 2 S2 = ut1 + 1a × t 2 at t = 0, v = 0, ∴ c1 = 0 2 2 Then V = 60t2 + 40t 700 + 450 = 10u + 1 a (10)2 2 2 ds = 60t2 + 40t 1150 = 10u + 50a dt 2 for equation (i) and (ii) Again integrating a = 10m/sec2 ∫ ∫ ∫ds = 60t2 dt + 40t dt 157. How much force will be exerted by the floor of 2 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 60t3 40t 2 s = 6 + 2 + c2 .....(4) (c) 720 N (d) 680 N at t = 0, s = 0, ∴ c2 = 0 CIL MT 2017 2017 IInd shift then s = 60t3 + 40t2 Ans. (c) : Fnet = mg-ma = m(g-a) 62 = 80 (9.81-0.81) = 80×9 = 720 N 158. If velocity of a body change from 50 m/s to 200 at t=5s m/s in 20 seconds, then the acceleration of the s = 60× 53 + 40 ×52 , s = 1750 m 62 body is: (b) 6.5 m/s2 (a) 5.0 m/s2 (d) 7.0 m/s2 160. A car travels from one city to another with the (c) 6.0 m/s2 uniform speed of 40 km/hr for half distance and (e) 7.5 m/s2 with the uniform speed of 60 km/hr for remaining (CGPCS Polytechnic Lecturer 2017) half distance. The average speed of car is: Ans. (e) : v1 = 50 m/s, v2 = 200 m/s, t = 20 s (a) 40 km/hr (b) 45 km/hr Then acceleration (a) is given as (c) 48 km/hr (d) 50 km/hr a = rate change in velocity (e) 42 km/hr a = dv = 200 − 50 = 150 dt 20 20 CGPSC AE 2014- I Ans. (c) : Average speed = Total distance a = 7.5 m / sec2 Total time 159. A particle of mass 1 kg moves in a straight line d+d 2 × 40× 60 under the influence of a force which increases 2 2 60 + 40 linearly with time at the rate of 60N/ s, it being =  d/2 d/2  = = 48 km / hr.   40 N initially. The position of the particle after 40 + 60 a lapse of 5s, if it started from rest at the origin, 161. An elevator has a downward acceleration of 0.1 will be (b) 1500 m g m/s2. What force will be transmitted to the (a) 1250 m floor of elevator by a man of weight 'W' (c) 1750 m (d) 2000 m travelling in the elevator? Ans. (c) : Given, ESE 2019 (a) W (b) W/10 m = 1 kg, dF = 60 N/s (c) 11W/10 (d) 9W/10 UPRVUNL AE 2016 dF = 60 dt dt Ans. (d) : Newton's Second Law of motion Fnet = W - T = ma ∫ ∫dF = 60 dt W - T = W × 0.1× g = 0.1×W g 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) T = W - 01 W = 0.9 W a = 60t + 40 T= 9W dv = 60t + 40 10 dt 413