CHAPTER 2 ENGINEERING MECHANICS YEAR 2012 TWO MARKS • Common Data For Q.1 and 2 Two steel truss members, AC and BC , each having cross sectional area of 100 mm2 , are subjected to a horizontal force F as shown in figure. All the joints are hinged. MCQ 2.1 If F = 1 kN , the magnitude of the vertical reaction force developed at the point B in kN is (A) 0.63 (B) 0.32 (C) 1.26 MCQ 2.2 (D) 1.46 The maximum force F is 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 YEAR 2011 MCQ 2.3 The coefficient of restitution of a perfectly plastic impact is (A) 0 (B) 1 (C) 2 MCQ 2.4 ONE MARK (D) 3 A stone with mass of 0.1 kg is catapulted as shown in the figure. The total force Fx (in N) exerted by the rubber band as a function of distance x GATE Previous Year Solved Paper For Mechanical Engineering Published by: NODIA and COMPANY Visit us at: www.nodia.co.in ISBN: 9788192276250 www.gatehelp.com PAGE 72 ENGINEERING MECHANICS CHAP 2 (in m ) is given by Fx = 300x2 . If the stone is displaced by 0.1 m from the un-stretched position (x = 0) of the rubber band, the energy stored in the rubber band is (A) 0.01 J (B) 0.1 J (C) 1 J (D) 10 J YEAR 2011 MCQ 2.5 TWO MARKS A 1 kg block is resting on a surface with coefficient of friction μ = 0.1. A force of 0.8 N is applied to the block as shown in the figure. The friction force is (A) 0 (B) 0.8 N (C) 0.98 N (D) 1.2 N YEAR 2009 MCQ 2.6 ONE MARK A block weighing 981 N is resting on a horizontal surface. The coefficient of friction between the block and the horizontal surface is μ = 0.2 . A vertical cable attached to the block provides partial support as shown. A man can pull horizontally with a force of 100 N. What will be the tension, T (in N) in the cable if the man is just able to move the block to the right ? (A) 176.2 (B) 196.0 GATE Previous Year Solved Paper For Mechanical Engineering Published by: NODIA and COMPANY Visit us at: www.nodia.co.in ISBN: 9788192276250 www.gatehelp.com CHAP 2 ENGINEERING MECHANICS (C) 481.0 PAGE 73 (D) 981.0 YEAR 2009 MCQ 2.7 TWO MARKS A uniform rigid rod of mass M and length L is hinged at one end as shown in the adjacent figure. A force P is applied at a distance of 2L/3 from the hinge so that the rod swings to the right. The reaction at the hinge is (A) − P (B) 0 (C) P/3 (D) 2P/3 YEAR 2008 MCQ 2.8 ONE MARK A straight rod length L (t), hinged at one end freely extensible at the other end, rotates through an angle θ (t) about the hinge. At time t , L (t) = 1 m, Lo (t) = 1 m/s, θ (t) = π rad and θo(t) = 1 rad/s. The magnitude of the 4 velocity at the other end of the rod is (A) 1 m/s (B) 2 m/s (C) 3 m/s (D) 2 m/s YEAR 2008 MCQ 2.9 TWO MARKS A circular disk of radius R rolls without slipping at a velocity V . The magnitude of the velocity at point P (see figure) is GATE Previous Year Solved Paper For Mechanical Engineering Published by: NODIA and COMPANY Visit us at: www.nodia.co.in ISBN: 9788192276250 www.gatehelp.com PAGE 74 ENGINEERING MECHANICS (A) 3V (C) V/2 MCQ 2.10 CHAP 2 (B) 3 V/2 (D) 2V/ 3 Consider a truss PQR loaded at P with a force F as shown in the figure - The tension in the member QR is (A) 0.5 F (B) 0.63 F (C) 0.73 F (D) 0.87 F YEAR 2007 MCQ 2.11 ONE MARK During inelastic collision of two particles, which one of the following is conserved ? (A) Total linear momentum only (B) Total kinetic energy only (C) Both linear momentum and kinetic energy (D) Neither linear momentum nor kinetic energy YEAR 2007 MCQ 2.12 TWO MARKS A block of mass M is released from point P on a rough inclined plane with inclination angle θ, shown in the figure below. The co-efficient of friction is μ. If μ < tan θ, then the time taken by the block to reach another point Q on the inclined plane, where PQ = s , is (A) 2s g cos θ (tan θ − μ) (B) 2s g cos θ (tan θ + μ) GATE Previous Year Solved Paper For Mechanical Engineering Published by: NODIA and COMPANY Visit us at: www.nodia.co.in ISBN: 9788192276250 www.gatehelp.com CHAP 2 ENGINEERING MECHANICS (C) 2s g sin θ (tan θ − μ) PAGE 75 2s g sin θ (tan θ + μ) (D) YEAR 2006 MCQ 2.13 If a system is in equilibrium and the position of the system depends upon many independent variables, the principles of virtual work states that the partial derivatives of its total potential energy with respect to each of the independent variable must be (A) − 1.0 (B) 0 (C) 1.0 MCQ 2.14 TWO MARKS (D) 3 If point A is in equilibrium under the action of the applied forces, the values of tensions TAB and TAC are respectively (A) 520 N and 300 N (B) 300 N and 520 N (C) 450 N and 150 N (D) 150 N and 450 N YEAR 2005 MCQ 2.15 The time variation of the position of a particle in rectilinear motion is given by x = 2t3 + t2 + 2t . If v is the velocity and a is the acceleration of the particle in consistent units, the motion started with (A) v = 0, a = 0 (B) v = 0, a = 2 (C) v = 2, a = 0 MCQ 2.16 ONE MARK (D) v = 2, a = 2 A simple pendulum of length of 5 m, with a bob of mass 1 kg, is in simple harmonic motion. As it passes through its mean position, the bob has a speed of 5 m/s. The net force on the bob at the mean position is (A) zero (B) 2.5 N (C) 5 N (D) 25 N YEAR 2005 MCQ 2.17 TWO MARKS Two books of mass 1 kg each are kept on a table, one over the other. The GATE Previous Year Solved Paper For Mechanical Engineering Published by: NODIA and COMPANY Visit us at: www.nodia.co.in ISBN: 9788192276250 www.gatehelp.com PAGE 76 ENGINEERING MECHANICS CHAP 2 coefficient of friction on every pair of contacting 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 (A) zero (B) 1.06 N (C) 5.74 N MCQ 2.18 MCQ 2.19 MCQ 2.20 (D) 8.83 N A shell is fired from a cannon. At the instant the shell is just about to leave the barrel, its velocity relative to the barrel is 3 m/s, while the barrel is swinging upwards with a constant angular velocity of 2 rad/s. The magnitude of the absolute velocity of the shell is (A) 3 m/s (B) 4 m/s (C) 5 m/s (D) 7 m/s An elevator (lift) consists of the elevator cage and a counter weight, of mass m each. The cage and the counterweight are connected by chain that passes over a pulley. The pulley is coupled to a motor. It is desired that the elevator should have a maximum stopping time of t seconds from a peak speed v . If the inertias of the pulley and the chain are neglected, the minimum power that the motor must have is (A) 1 mV2 2 2 (B) mV 2t 2 (C) mV t 2 (D) 2mV t A 1 kg mass of clay, moving with a velocity of 10 m/s, strikes a stationary wheel and sticks to it. The solid wheel has a mass of 20 kg and a radius of GATE Previous Year Solved Paper For Mechanical Engineering Published by: NODIA and COMPANY Visit us at: www.nodia.co.in ISBN: 9788192276250 www.gatehelp.com CHAP 2 ENGINEERING MECHANICS PAGE 77 1 m. Assuming that the wheel is set into pure rolling motion, the angular velocity of the wheel immediately after the impact is approximately (A) zero (C) 10 rad/s 3 (B) 1 rad/s 3 (D) 10 rad/s 3 YEAR 2004 MCQ 2.21 ONE MARK The figure shows a pin-jointed plane truss loaded at the point M by hanging a mass of 100 kg. The member LN of the truss is subjected to a load of (A) 0 Newton (B) 490 Newtons in compression (C) 981 Newtons in compression (D) 981 Newtons in tension YEAR 2004 MCQ 2.22 TWO MARKS An ejector mechanism consists of a helical compression spring having a spring constant of k = 981 # 103 N/m . It is pre-compressed by 100 mm from its free state. If it is used to eject a mass of 100 kg held on it, the mass will move up through a distance of GATE Previous Year Solved Paper For Mechanical Engineering Published by: NODIA and COMPANY Visit us at: www.nodia.co.in ISBN: 9788192276250 www.gatehelp.com PAGE 78 MCQ 2.23 MCQ 2.24 ENGINEERING MECHANICS CHAP 2 (A) 100 mm (B) 500 mm (C) 581 mm (D) 1000 mm A rigid body shown in the figure (a) has a mass of 10 kg. It rotates with a uniform angular velocity ‘ω’. A balancing mass of 20 kg is attached as shown in figure (b). The percentage increase in mass moment of inertia as a result of this addition is (A) 25% (B) 50% (C) 100% (D) 200% The figure shows a pair of pin-jointed gripper-tongs holding an object weighting 2000 N. The coefficient of friction (μ) at the gripping surface is 0.1 XX is the line of action of the input force and YY 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 (A) 1000 N (B) 2000 N (C) 2500 N (D) 5000 N GATE Previous Year Solved Paper For Mechanical Engineering Published by: NODIA and COMPANY Visit us at: www.nodia.co.in ISBN: 9788192276250 www.gatehelp.com CHAP 2 ENGINEERING MECHANICS PAGE 79 YEAR 2003 MCQ 2.25 A truss consists of horizontal members (AC,CD, DB and EF) and vertical members (CE and DF) having length l each. The members AE, DE and BF are inclined at 45c to the horizontal. For the uniformly distributed load “p” per unit length on the member EF of the truss shown in figure given below, the force in the member CD is pl 2 (C) 0 (A) MCQ 2.26 ONE MARK (B) pl 2pl (D) 3 A bullet of mass “m ” travels at a very high velocity v (as shown in the figure) and gets embedded inside the block of mass “M ” initially at rest on a rough horizontal floor. The block with the bullet is seen to move a distance “s ” along the floor. Assuming μ to be the coefficient of kinetic friction between the block and the floor and “g ” the acceleration due to gravity what is the velocity v of the bullet ? (A) M + m 2μgs m (C) μ (M + m) 2μgs m (B) M − m 2μgs m (D) M m 2μgs YEAR 2003 TWO MARKS • Common Data For Q.Data for Q. 27 & 28 are given below. Solve the problems and choose correct answers. A reel of mass “m ” and radius of gyration “k ” is rolling down smoothly from rest with one end of the thread wound on it held in the ceiling as depicated in the figure. Consider the thickness of thread and its mass negligible in comparison with the radius “r ” of the hub and the reel mass “m ”. Symbol “g ” represents the acceleration due to gravity. GATE Previous Year Solved Paper For Mechanical Engineering Published by: NODIA and COMPANY Visit us at: www.nodia.co.in ISBN: 9788192276250 www.gatehelp.com PAGE 80 MCQ 2.27 MCQ 2.28 ENGINEERING MECHANICS The linear acceleration of the reel is gr2 (A) 2 (r + k2) grk (C) 2 (r + k2) The tension in the thread is mgr2 (A) 2 (r + k2) mgk2 (C) 2 (r + k2) CHAP 2 gk2 (r2 + k2) mgr2 (D) 2 (r + k2) (B) mgrk (r2 + k2) mg (D) 2 (r + k2) (B) YEAR 2001 MCQ 2.29 MCQ 2.30 ONE MARK A particle P is projected from the earth surface at latitude 45c with escape velocity v = 11.19 km/s . The velocity direction makes an angle α with the local vertical. The particle will escape the earth’s gravitational field (A) only when α = 0 (B) only when α = 45c (C) only when α = 90c (D) irrespective of the value of α The area moment of inertia of a square of size 1 unit about its diagonal is (B) 1 (A) 1 3 4 (C) 1 12 (D) 1 6 GATE Previous Year Solved Paper For Mechanical Engineering Published by: NODIA and COMPANY Visit us at: www.nodia.co.in ISBN: 9788192276250 www.gatehelp.com CHAP 2 ENGINEERING MECHANICS PAGE 81 YEAR 2001 MCQ 2.31 TWO MARKS For the loading on truss shown in the figure, the force in member CD is (A) zero MCQ 2.32 MCQ 2.33 (B) 1 kN (C) 2 kN (D) 1 kN 2 Bodies 1 and 2 shown in the figure have equal mass m . All surfaces are smooth. The value of force P required to prevent sliding of body 2 on body 1 is (A) P = 2 mg (B) P = (C) P = 2 2 mg (D) P = mg 2 mg Mass M slides in a frictionless slot in the horizontal direction and the bob of mass m is hinged to mass M at C , through a rigid massless rod. This system is released from rest with θ = 30c. At the instant when θ = 0c, the velocities of m and M can be determined using the fact that, for the system (i.e., m and M together) GATE Previous Year Solved Paper For Mechanical Engineering Published by: NODIA and COMPANY Visit us at: www.nodia.co.in ISBN: 9788192276250 www.gatehelp.com PAGE 82 ENGINEERING MECHANICS CHAP 2 (A) the linear momentum in x and y directions are conserved but the energy is not conserved. (B) the linear momentum in x and y directions are conserved and the energy is also conserved. (C) the linear momentum in x direction is conserved and the energy is also conserved. (D) the linear momentum in y direction is conserved and the energy is also conserved. ********* GATE Previous Year Solved Paper For Mechanical Engineering Published by: NODIA and COMPANY Visit us at: www.nodia.co.in ISBN: 9788192276250 www.gatehelp.com CHAP 2 ENGINEERING MECHANICS PAGE 83 SOLUTION SOL 2.1 Option (A) is correct. From above figure. Three forces are acting on a common point. Hence by Lami’s Theorem. F = T2 = T1 sin 120c sin 135c sin (105c) T1 F 1 & = = sin 135c sin 105c sin 105c T1 = 0.7320 kN Hence vertical reaction at B , RNT = T1 cos 30c = 0.73205 # cos 30c = 0.634 kN 1 SOL 2.2 Option (B) is correct. From Previous question F = T2 sin 105c sin 120c T2 = sin 120c # F = 0.8965F sin 135 T1 = (0.73205) F T2 > T1 σ = 100 MPa (given) F = σ # A1 Fmax = σmax # A1 and As we know & T2 = 100 # 100 0.8965F = 100 # 100 F = 100 # 100 = 11154.5 N = 11.15 kN 0.8965 GATE Previous Year Solved Paper For Mechanical Engineering Published by: NODIA and COMPANY Visit us at: www.nodia.co.in ISBN: 9788192276250 www.gatehelp.com PAGE 84 SOL 2.3 ENGINEERING MECHANICS CHAP 2 Option (A) is correct. From the Newton’s Law of collision of Elastic bodies. Velocity of separation = e # Velocity of approach (V2 − V1) = e (U1 − U2) Where e is a constant of proportionality & it is called the coefficient of restitution and its value lies between 0 to 1. The coefficient of restitution of a perfectly plastic impact is zero, because all the K.E. will be absorbed during perfectly plastic impact. SOL 2.4 Option (B) is correct. Given : Position of x is, x = 0 to x = 0.1 Fx = 300x2 , The energy stored in the rubber band is equal to work done by the stone. Hence dE = Fx dx Integrating both the sides & put the value of F & limits E #0 dE = #0 0.1 300x2 dx 3 0.1 (0.1) 3 E = 300 :x D = 300 ; = 0.1 Joule 3 E 3 0 SOL 2.5 Option (B) is correct. Given : m = 1 kg , μ = 0.1; From FBD : RN = mg Now static friction force, fS = μRN = μmg = 0.1 # 1 # 9.8 = 0.98 N Applied force F = 0.8 N is less then, the static friction fS = 0.98 N F < fS So, we can say that the friction developed will equal to the applied force F = 0.8 N SOL 2.6 Option (C) is correct. Given : W = 981 N , μ = 0.2 First of all we have to make a FBD of the block Here, RN = Normal reaction force GATE Previous Year Solved Paper For Mechanical Engineering Published by: NODIA and COMPANY Visit us at: www.nodia.co.in ISBN: 9788192276250 www.gatehelp.com CHAP 2 ENGINEERING MECHANICS PAGE 85 T =Tension in string Using the balancing of forces, we have ΣFx = 0 : μRN = 100 N RN = 100 = 100 = 500 N μ 0.2 and ΣFy = 0 or downward forces = upward forces W = T + RN & T = W − RN = 981 − 500 = 481 N SOL 2.7 Option (B) is correct. When rod swings to the right, linear acceleration a and angular acceleration α comes in action. Centre of gravity (G ) acting at the mid-point of the rod. Let R be the reaction at the hinge. ...(i) Linear acceleration a = r.α = L # α = 2a 2 L and about point G , for rotational motion /Μ = IG # α 2 From equation (i) R b L l + P b L l = ML b 2a l 12 L 2 6 R + P = Ma 3 3 ...(ii) a = 3R + P M M GATE Previous Year Solved Paper For Mechanical Engineering G Published by: NODIA and COMPANY Visit us at: www.nodia.co.in ISBN: 9788192276250 www.gatehelp.com PAGE 86 ENGINEERING MECHANICS CHAP 2 By equilibrium of forces in normal direction to the rod From equation (ii) Fm = 0 : P − R = Ma = M b 3R + P l M M / P − R = 3R + P R =0 & SOL 2.8 So, reaction at the hinge is zero. Option (D) is correct. Let : Vt =Tangential Velocity Vr =Relative Velocity V =Resultant Velocity Let rod of length L (t) increases by an amount TL (t). : : Given L (t) = 1 m, L (t)=1 m/sec, θ (t) = π rad, θ (t)=1 rad/sec 4 Time taken by the rod to turn π rad is, 4 π/4 π θ (t) = = sec t = dis tan ce = : 1 4 velocity θ (t) So, increase in length of the rod during this time will be ΔL (t) = L (t) # t = π # 1 = π meter 4 4 π radian. So, increased length after π sec, (New length) Rod turn 4 4 π = a1 + k = 1.785 m 4 Now, tangential velocity, Vt = R.ω = 1.785 # 1 = 1.785 m/sec Radial velocity, : Vr = L (t)=1 m/sec GATE Previous Year Solved Paper For Mechanical Engineering Published by: NODIA and COMPANY Visit us at: www.nodia.co.in ISBN: 9788192276250 ω = θo(t) www.gatehelp.com CHAP 2 ENGINEERING MECHANICS PAGE 87 Therefore, the resultant velocity will be VR = V t2 + V r2 = SOL 2.9 (1.785) 2 + (1) 2 = 2.04 - 2 m/ sec Option (A) is correct. When disc rolling along a straight path, without slipping. The centre of the wheel O moves with some linear velocity and each particle on the wheel rotates with some angular velocity. Thus, the motion of any particular on the periphery of the wheel is a combination of linear and angular velocity. Let wheel rotates with angular velocity=ω rad/sec. ....(i) So, ω =V R Velocity at point P is, VP = ω # PQ From triangle OPQ ...(ii) (OQ) 2 + (OP) 2 − 2OQ # OP # cos (+POQ) PQ = (R) 2 + (R) 2 − 2RR cos 120c = (R) 2 + (R) 2 + (R) 2 = From equation (i), (ii) and (iii) VP = V # 3 R = 3 V R SOL 2.10 3R ...(iii) Option (B) is correct. The forces which are acting on the truss PQR is shown in figure. We draw a perpendicular from the point P , that intersects QR at point S . GATE Previous Year Solved Paper For Mechanical Engineering Published by: NODIA and COMPANY Visit us at: www.nodia.co.in ISBN: 9788192276250 www.gatehelp.com PAGE 88 ENGINEERING MECHANICS CHAP 2 Let PS = QS = a RQ & RR are the reactions acting at point Q & R respectively. Now from the triangle PRS tan 30c = PS & SR = PS = a1 = 3 a = 1.73a SR tan 30c 3 Taking the moment about point R, RQ # (a + 1.73a) = F # 1.73a RQ = 1.73a F = 1.73 F = 0.634 F 2.73a 2.73 From equilibrium of the forces, we have RR + RQ = F RR = F − RQ = F − 0.634 F = 0.366 F To find tension in QR we have to use the method of joint at point Q , and ΣFy = 0 FQP sin 45c = RQ and, ΣFx = 0 FQP = 0.634 F = 0.8966 F 1 2 FQP cos 45c = FQR & FQR = 0.8966 F # 1 = 0.634 F - 0.63 F 2 SOL 2.11 Option (A) is correct. In both elastic & in inelastic collision total linear momentum remains conserved. In the inelastic collision loss in kinetic energy occurs because the coefficient of restitution is less than one and loss in kinetic energy is given by the relation, TK.E. = m1 m2 (u1 − u2) 2 (1 − e2) 2 (m1 + m2) SOL 2.12 Option (A) is correct. First of all we resolve all the force which are acting on the block. GATE Previous Year Solved Paper For Mechanical Engineering Published by: NODIA and COMPANY Visit us at: www.nodia.co.in ISBN: 9788192276250 www.gatehelp.com CHAP 2 ENGINEERING MECHANICS PQ = s μ < tan θ Now from Newton’s second law, Given : F mg sin θ − μN mg sin θ − μmg cos θ g sin θ − μg cos θ = ma = ma = ma =a PAGE 89 where N = Normal fraction force a = Acceleration of block N = mg cos θ a = g cos θ : sin θ − μD = g cos θ (tan θ − μ) cos θ From the Newton;s second law of Motion, s = ut + 1 at2 = 0 + 1 g cos θ (tan θ − μ) t2 2 2 t = ...(i) u=0 2s g cos θ (tan θ − μ) SOL 2.13 Option (B) is correct. If a system of forces acting on a body or system of bodies be in equilibrium and the system has to undergo a small displacement consistent with the geometrical conditions, then the algebraic sum of the virtual works done by all the forces of the system is zero and total potential energy with respect to each of the independent variable must be equal to zero. SOL 2.14 Option (A) is correct. First we solve this problem from Lami’s theorem. Here three forces are given. Now we have to find the angle between these forces Applying Lami’s theorem, we have F = TAB = TAC sin 90c sin 120c sin 150c 600 = TAB = TAC 1 3 /2 1/2 TAB = 600 # 3 = 300 3 . 520 N 2 TAC = 600 = 300 N 2 GATE Previous Year Solved Paper For Mechanical Engineering Published by: NODIA and COMPANY Visit us at: www.nodia.co.in ISBN: 9788192276250 www.gatehelp.com PAGE 90 ENGINEERING MECHANICS CHAP 2 Alternative : Now we using the Resolution of forces. Resolve the TAB & TAC in x & y direction (horizontal & vertical components) We use the Resolution of forces in x & y direction ΣFx = 0 , TAB cos 60c = TAC cos 30c TAB = 3 2= 2 #1 TAC 3 ...(i) ΣFy = 0 , TAB sin 60c + TAC sin 30c = 600 N 3 T + 1 T = 600 N 2 AB 2 AC 3 TAB + TAC = 1200 N TAC = TAB From equation (i) 3 T 3 TAB + AB = 1200 N 3 4TAB = 1200 3 Now, TAB = 1200 3 = 520 N 4 TAC = TAB = 520 = 300 N 3 3 and SOL 2.15 Option (D) is correct. Given ; x = 2t3 + t2 + 2t We know that, v = dx = d (2t3 + t2 + 2t)= 6t2 + 2t + 2 dt dt ...(i) We have to find the velocity & acceleration of particle, when motion stared, So at t = 0 , v =2 Again differentiate equation (i) w.r.t. t 2 a = dv = d x2 = 12t + 2 dt dt At t = 0 , a =2 GATE Previous Year Solved Paper For Mechanical Engineering Published by: NODIA and COMPANY Visit us at: www.nodia.co.in ISBN: 9788192276250 www.gatehelp.com CHAP 2 SOL 2.16 ENGINEERING MECHANICS PAGE 91 Option (A) is correct. We have to make the diagram of simple pendulum Here, We can see easily from the figure that tension in the string is balanced by the weight of the bob and net force at the mean position is always zero. SOL 2.17 Option (D) is correct. Given : m1 = m2 = 1 kg , μ = 0.3 The FBD of the system is shown below : For Book (1) ΣFy = 0 RN1 = mg Then, Friction Force FN1 = μRN1 = μmg From FBD of book second, ΣFx = 0 , F = μRN1 + μRN2 ΣFy = 0 , RN2 = RN1 + mg = mg + mg = 2 mg For slip occurs between the books when ...(i) ...(ii) F $ μRN1 + μRN2$ μmg + μ # 2 mg F $ μ (3 mg)$ 0.3 (3 # 1 # 9.8)$ 8.82 It means the value of F is always greater or equal to the 8.82, for which slip GATE Previous Year Solved Paper For Mechanical Engineering Published by: NODIA and COMPANY Visit us at: www.nodia.co.in ISBN: 9788192276250 www.gatehelp.com PAGE 92 ENGINEERING MECHANICS CHAP 2 occurs between two books. So, F = 8.83 N SOL 2.18 Option (C) is correct. Given : ω = 2 rad/ sec , r = 2 m Tangential velocity of barrel, Resultant velocity of shell, SOL 2.19 Vt = rω = 2 # 2 = 4 m/ sec V = Vr i + Vt j = 3i + 4j V = (3) 2 + (4) 2 = 25 = 5 m/ sec Option (C) is correct. Given : Mass of cage & counter weight = m kg each Peak speed = V Initial velocity of both the cage and counter weight. V1 = V m/ sec Final velocity of both objects V2 = 0 Initial kinetic Energy, E1 = 1 mV2 + 1 mV2 = mV2 2 2 Final kinetic Energy E2 = 1 m (0) 2 + 1 m (0) 2 = 0 2 2 Now, SOL 2.20 Power = Rate of change of K.E. 2 = E1 − E2 = mV t t Option (B) is correct. Given : m1 = 1 kg , V1 = 10 m/ sec , m2 = 20 kg , V2 = Velocity after striking the wheel r = 1 meter Applying the principal of linear momentum on the system dP = 0 & P = constant dt Initial Momentum = Final Momentum m1 # V1 = (m1 + m2) V2 V2 = m1 V1 = 1 # 10 = 10 21 (m1 + m2) 1 + 20 GATE Previous Year Solved Paper For Mechanical Engineering Published by: NODIA and COMPANY Visit us at: www.nodia.co.in ISBN: 9788192276250 www.gatehelp.com CHAP 2 ENGINEERING MECHANICS PAGE 93 Now after the collision the wheel rolling with angular velocity ω. So, V2 = rω & ω = V2 = 10 = 0.476 r 21 # 1 It is nearly equal to 1/3. SOL 2.21 Option (A) is correct. First of all we consider all the forces, which are acting at point L. Now sum all the forces which are acting along x direction, Both are acting in opposite direction FLK = FLM Also summation of all the forces, which are acting along y -direction. Only one forces acting in y -direction FLN = 0 So the member LN is subjected to zero load. SOL 2.22 Option (A) is correct. Given : k = 981 # 103 N/m, xi = 100 mm=0.1 m, m = 100 kg Let, when mass m = 100 kg is put on the spring then spring compressed by x mm. From the conservation of energy : Energy stored in free state = Energy stored after the mass is attach. (K.E.) i = (K.E.) f + (P.E.) f 1 kx 2 = 1 kx2 + mg (x + 0.1) 2 i 2 kx i2 = kx2 + 2mg (x + 0.1) Substitute the values, we get 981 # 103 # (0.1) 2 = (981 # 103 # x2) + [2 # 100 # 9.81 # (x + 0.1)] 103 # 10−2 = 103 x2 + 2 (x + 0.1) 10 = 1000x2 + 2x + 0.2 1000x2 + 2x − 9.8 = 0 Solving above equation, we get x = − 2 ! (2) 2 − 4 # 1000 (− 9.8) = − 2 ! 4 + 39200 = − 2 ! 198 2 # 1000 2000 2000 On taking -ve sign, we get GATE Previous Year Solved Paper For Mechanical Engineering Published by: NODIA and COMPANY Visit us at: www.nodia.co.in ISBN: 9788192276250 www.gatehelp.com PAGE 94 ENGINEERING MECHANICS CHAP 2 x = − 2 − 198 =− 1 , m =− 100 mm 2000 10 (-ve sign shows the compression of the spring) SOL 2.23 Option (B) is correct. Given : First Mass, m1 Balancing Mass, m2 We know the mass moment Where, k Case (I) : When mass of 10 = 10 kg = 20 kg of inertia, I = mk2 = Radius of gyration kg is rotates with uniform angular velocity ‘ω’ I1 = m1 k 12 = 10 # (0.2) 2 = 10 # 0.04 = 0.4 kg m2 k1 = 0.2 m Case (II) : When balancing mass of 20 kg is attached then moment of inertia I2 = 10 # (0.2) 2 + 20 # (0.1) 2 = 0.4 + 0.2 = 0.6 Here k1 = 0.2 m and k2 = 0.1 m Percent increase in mass moment of inertia, I = I2 − I1 # 100 = 0.6 − 0.4 # 100 = 1 # 100 = 50% 2 0.4 I1 SOL 2.24 Option (D) is correct. Given : Weight of object W = 2000 N Coefficient of Friction μ = 0.1 First of all we have to make the FBD of the system. Here, RN = Normal reaction force acting by the pin joint. F = μRN = Friction force In equilibrium condition of all the forces which are acting in y direction. μRN + μRN = 2000 N μRN = 1000 N RN = 1000 = 10000 N 0.1 Taking the moment about the pin, we get 10000 # 150 = F # 300 F = 5000 N GATE Previous Year Solved Paper For Mechanical Engineering Published by: NODIA and COMPANY Visit us at: www.nodia.co.in ISBN: 9788192276250 μ = 0.1 www.gatehelp.com CHAP 2 SOL 2.25 ENGINEERING MECHANICS PAGE 95 Option (A) is correct. Given : AC = CD = DB = EF = CE = DF = l At the member EF uniform distributed load is acting, the U.D.L. is given as “p” per unit length. So, the total load acting on the element EF of length l = Lord per unit length # Total length of element = p # l = pl This force acting at the mid point of EF . From the FBD we get that at A and B reactions are acting because of the roller supports, in the upward direction. In equilibrium condition, Upward force = Downward forces Ra + Rb = pl And take the moment about point A, pl # bl + l l = Rb (l + l + l) 2 ...(i) pl pl # 3 l = Rb # 3l & Rb = 2 2 Substitute the value of Rb in equation (i), we get pl Ra + = pl 2 pl pl pl Ra = pl − = = Rb = 2 2 2 At point A we use the principal of resolution of forces in the y -direction, / Fy = 0 :FAE sin 45c = Ra = pl2 pl pl pl FAE = # 1 = # 2 = 2 2 sin 45c 2 pl 1 = pl And FAC = FAE cos 45c = # 2 2 2 At C , No external force is acting. So, pl FAC = = FCD 2 GATE Previous Year Solved Paper For Mechanical Engineering Published by: NODIA and COMPANY Visit us at: www.nodia.co.in ISBN: 9788192276250 www.gatehelp.com PAGE 96 SOL 2.26 ENGINEERING MECHANICS CHAP 2 Option (A) is correct. Given : Mass of bullet = m Mass of block = M Velocity of bullet = v Coefficient of Kinematic friction = μ Let, Velocity of system (Block + bullet) after striking the bullet = u We have to make the FBD of the box after the bullet strikes, Friction Force (Retardation) = Fr Applying principal of conservation of linear momentum, dP = 0 or P = mV = cons tan t. dt mv = (M + m) u u = mv M+m And, from the FBD the vertical force (reaction force), So, ...(i) RN = (M + m) g Fr = μRN = μ (M + m) g − μ (M + m) g Frictional retardation ...(ii) a = − Fr = =− μg M+m (m + M) Negative sign show the retardation of the system (acceleration in opposite direction). From the Newton’s third law of motion, V f2 = u2 + 2as Vf = Final velocity of system (block + bullet) = 0 2 u + 2as = 0 From equation (ii) u2 =− 2as =− 2 # (− μg) # s = 2μgs Substitute the value of u from equation (i), we get mv 2 a M + m k = 2μgs m2 v2 = 2μgs (M + m) 2 2μgs (M + m) 2 v2 = m2 GATE Previous Year Solved Paper For Mechanical Engineering Published by: NODIA and COMPANY Visit us at: www.nodia.co.in ISBN: 9788192276250 www.gatehelp.com CHAP 2 ENGINEERING MECHANICS v = SOL 2.27 PAGE 97 2μgs # b M + m l = M + m 2μgs m m Option (A) is correct. Given : Mass of real = m Radius of gyration = k We have to make FBD of the system, Where, T =Tension in the thread mg = Weight of the system Real is rolling down. So Angular acceleration (α) comes in the action From FBD, For vertical translation motion, mg − T = ma and for rotational motion, ...(i) ΣMG = IG α T # r = mk2 # a r IG = mk2 , α = a/r 2 ...(ii) T = mk2 # a r From equation (i) & (ii) Substitute the value of T in equation (i), we get 2 mg − mk2 # a = ma r 2 mg = a ;mk2 + mE r gr2 ...(iii) a = 2 k + r2 SOL 2.28 Option (C) is correct. From previous question, T = mg − ma Substitute the value of a from equation (iii), we get mg (k2 + r2) − mgr2 gr2 mgk2 T = mg − m # 2 = = (k + r2) k2 + r2 (k2 + r2) GATE Previous Year Solved Paper For Mechanical Engineering Published by: NODIA and COMPANY Visit us at: www.nodia.co.in ISBN: 9788192276250 www.gatehelp.com PAGE 98 ENGINEERING MECHANICS CHAP 2 SOL 2.29 Option (D) is correct. We know that a particle requires the velocity of 11.2 km/s for escape it from the earth’s gravitational field. The angle α does not effect on it. SOL 2.30 Option (C) is correct. The BD is the diagonal of the square ABCD and CBD = 45c. From the TBCE sin 45c = CE & CE = 1 # sin 45c = 1 unit BC 2 where CE is the height of the triangle TBCD . 3 Now, the area moment of inertia of a triangle about its base BD is bh 12 where b = base of triangle & h = height of triangle So, the triangle TABD are same and required moment of inertia of the square ABCD about its diagonal is, 3 I = 2 # 1 # (BD) # (CE) 3 = 1 # 2 # c 1 m = 1 unit 6 12 12 2 SOL 2.31 Option (A) is correct. The reactions at the hinged support will be in only vertical direction as external loads are vertical. Now, consider the FBD of entire truss. In equilibrium of forces. ...(i) Ra + R f = 1 + 1 = 2 kN Taking moment about point A, we get R f # 3L = 1 # L + 1 # 2L = 3L R f = 1 kN From equation (i), Ra = 2 − 1 = 1 kN GATE Previous Year Solved Paper For Mechanical Engineering Published by: NODIA and COMPANY Visit us at: www.nodia.co.in ISBN: 9788192276250 www.gatehelp.com CHAP 2 ENGINEERING MECHANICS PAGE 99 First consider the FBD of joint A with the direction of forces assumed in the figure. Resolving force vertically, we get Ra = FAB sin 45c FAB = 1 = 2 kN (Compression) sin 45c Resolving forces horizontally FAC FAB cos 45c = 2 # 1 = 1 kN (Tension) 2 Consider the FBD of joint B with known value of force FAB in member AB Resolving forces vertically, FBC = FAB cos 45c = 2 # 1 = 1 kN (Tension) 2 Resolving forces horizontally, FBD = FAB sin 45c = 2 # 1 = 1 kN (Compression) 2 Consider the FBD of joint C with known value of force FBC and FAC Resolving forces vertically, 1 = FBC + FCD sin 45c GATE Previous Year Solved Paper For Mechanical Engineering Published by: NODIA and COMPANY Visit us at: www.nodia.co.in ISBN: 9788192276250 www.gatehelp.com PAGE 104 ENGINEERING MECHANICS CHAP 2 1 = 1 + FCD sin 45c & FCD = 0 SOL 2.32 Option (D) is correct. From the FBD of the system. RN = mg cos 45c All surfaces are smooth, so there is no frictional force at the surfaces. The downward force mg sin 45c is balanced by P cos 45c. mg sin 45c = P cos 45c mg # 1 = P # 1 & P = mg 2 2 SOL 2.33 Option (C) is correct. In this case the motion of mass m is only in x -direction. So, the linear momentum is only in x -direction & it remains conserved. Also from Energy conservation law the energy remains constant i.e. energy is also conserved. ********** GATE Previous Year Solved Paper For Mechanical Engineering Published by: NODIA and COMPANY Visit us at: www.nodia.co.in ISBN: 9788192276250