MEN 201 Problems For Chapter 15 with Solutions, Spring 2012

Engineering Mechanics - Dynamics Chapter 15 Problem 15-1 A block of weight W slides down an inclined plane of angle θ with initial velocity v0. Determine the velocity of the block at time t1 if the coefficient of kinetic friction between the block and the plane is μk. Given: W = 20 lb t1 3 s = θ = 30 deg μk 0.25 = v0 2 ft s = g 32.2 ft s2 = Solution: ( ) mvy1 Σ t1 t2 t Fy' ⌠ ⎮ ⌡ d + mvy2 = 0 + FN t1 − W cos θ( )t1 = 0 FN = W cos θ( ) FN 17.32 lb = ( ) m vx'1 ( ) Σ t1 t2 t Fx' ⌠ ⎮ ⌡ d + m vx'2 ( ) = W g ⎛⎜⎝ ⎞⎟⎠ v0 + W sin θ( )t1 − μk FN t1 W g ⎛⎜⎝ ⎞⎟⎠ v = v W v0 W sin θ( ) t1 g − μk FN t1 g + W = v 29.4 ft s = Problem 15-2 A ball of weight W is thrown in the direction shown with an initial speed vA. Determine the time needed for it to reach its highest point B and the speed at which it is traveling at B. Use the principle of impulse and momentum for the solution. Given: W = 2 lb θ 30 deg = vA 18 ft s = g 32.2 ft s2 = 302 Engineering Mechanics - Dynamics Chapter 15 Solution: W g ⎛⎜⎝ ⎞⎟⎠ vAsin θ( ) − W t W g ⎛⎜⎝ ⎞⎟⎠ 0 = t vAsin θ( ) g = t 0.280 s = W g ⎛⎜⎝ ⎞⎟⎠ vAcos θ( ) + 0 W g ⎛⎜⎝ ⎞⎟⎠ vx = vx = vAcos θ( ) vx 15.59 ft s = Problem 15-3 A block of weight W is given an initial velocity v0 up a smooth slope of angle θ. Determine the time it will take to travel up the slope before it stops. Given: W 5 lb = v0 10 ft s = θ 45 deg = g 32.2 ft s2 = Solution: W g ⎛⎜⎝ ⎞⎟⎠ v0 − W sin θ( )t = 0 t v0 gsin θ( ) = t 0.439 s = *Problem 15-4 The baseball has a horizontal speed v1 when it is struck by the bat B. If it then travels away at an angle θ from the horizontal and reaches a maximum height h, measured from the height of the bat, determine the magnitude of the net impulse of the bat on the ball.The ball has a mass M. Neglect the weight of the ball during the time the bat strikes the ball. Given: M 0.4 kg = v1 35 m s = h 50 m = θ 60 deg = g 9.81 m s2 = 303 Engineering Mechanics - Dynamics Chapter 15 Solution: Guesses v2 20 m s = Impx = 1 N s⋅ Impy 10 N s⋅ = Given 1 2M v2 sin θ( ) ( )2 = M gh −M v1 + Impx = M v2 cos θ( ) 0 + Impy M v2 sin θ( ) = v2 Impx Impy ⎛⎜ ⎜ ⎜⎝ ⎞⎟ ⎟ ⎟⎠ Find v2 Impx , , Impy ( ) = v2 36.2 m s = Impx Impy ⎛ ⎜ ⎝ ⎞ ⎟ ⎠ 21.2 ⎛⎜⎝12.5 ⎞⎟⎠ N s⋅ = Impx Impy ⎛ ⎜ ⎝ ⎞ ⎟ ⎠ 24.7 N s⋅ = Problem 15-5 The choice of a seating material for moving vehicles depends upon its ability to resist shock and vibration. From the data shown in the graphs, determine the impulses created by a falling weight onto a sample of urethane foam and CONFOR foam. Units Used: ms 10 3 − s = Given: F1 = 0.3 N t1 2 ms = F2 = 0.4 N t2 4 ms = F3 = 0.5 N t3 7 ms = F4 = 0.8 N t4 10 ms = F5 = 1.2 N t5 14 ms = Solution: CONFOR foam: Ic 1 2t1 F3 1 2 F3 + F4 ( ) t3 − t1 ( ) + 1 2F4 t5 − t3 ( ) + = Ic 6.55 N ms ⋅ = 304 Engineering Mechanics - Dynamics Chapter 15 Urethane foam: IU 1 2t2 F1 1 2 F5 + F1 ( ) t3 − t2 ( ) + 1 2 F5 + F2 ( ) t4 − t3 ( ) + 1 2 t5 − t4 ( )F2 + = IU 6.05 N ms ⋅ = Problem 15-6 A man hits the golf ball of mass M such that it leaves the tee at angle θ with the horizontal and strikes the ground at the same elevation a distance d away. Determine the impulse of the club C on the ball. Neglect the impulse caused by the ball’s weight while the club is striking the ball. Given: M 50 gm = θ 40 deg = d 20 m = g 9.81 m s2 = Solution: First find the velocity v1 Guesses v1 1 m s = t 1 s = Given 0 −g 2 ⎛⎜⎝ ⎞⎟⎠ t2 + v1 sin θ( )t = d v1 cos θ( )t = t ⎛⎜v1 ⎝ ⎞⎟ ⎠ Find t v1 ( , ) = t = 1.85 s v1 14.11 m s = Impulse - Momentum 0 + Imp = M v1 Imp = M v1 Imp 0.706 N s⋅ = Problem 15-7 A solid-fueled rocket can be made using a fuel grain with either a hole (a), or starred cavity (b), in the cross section. From experiment the engine thrust-time curves (T vs. t) for the same amount of propellant using these geometries are shown. Determine the total impulse in both cases. Given: T1a = 4 lb t1a 3 s = T1b = 8 lb t1b 6 s = 305 Engineering Mechanics - Dynamics Chapter 15 T2a = 6 lb t1c 10 s = t2a = 8 s t2b 10 s = Solution: Impulse is area under curve for hole cavity. Ia T1at1a 1 2 T1a + T1b ( ) t1b − t1a ( ) + 1 2T1b t1c − t1b ( ) + = Ia 46.00 lb s⋅ = For starred cavity: Ib T2at2a 1 2T2a t2b − t2a ( ) + = Ib 54.00 lb s⋅ = *Problem 15-8 During operation the breaker hammer develops on the concrete surface a force which is indicated in the graph. To achieve this the spike S of weight W is fired from rest into the surface at speed v. Determine the speed of the spike just after rebounding. Given: W 2 lb = v 200 ft s = g 32.2 ft s2 = Solution: I 1 2 × 90 × 103 lb ⎛⎜⎝ ⎞⎟⎠ 0.4 10 3 − s ( ) = I = 18.00 lb s⋅ Δt 0.4 10 3 − s × = −W g ⎛⎜⎝ ⎞⎟⎠ v + I − WΔt W g ⎛⎜⎝ ⎞⎟⎠ v' = v' −v I g W ⎛⎜⎝ ⎞⎟⎠ + − gΔt = v' 89.8 ft s = 306 Engineering Mechanics - Dynamics Chapter 15 Problem 15-9 The jet plane has a mass M and a horizontal velocity v0 when t = 0. If both engines provide a horizontal thrust which varies as shown in the graph, determine the plane’s velocity at time t1. Neglect air resistance and the loss of fuel during the motion. Units Used: Mg 103 kg = kN 103 N = Given: M 250 Mg = v0 100 m s = t1 15 s = a 200 kN = b 2 kN s2 = Solution: M v0 0 t1 t a + bt2 ⌠ ⎮⌡ d + M v1 = v1 v0 1 M 0 t1 t a + bt2 ⌠ ⎮⌡ d + = v1 121.00 m s = Problem 15-10 A man kicks the ball of mass M such that it leaves the ground at angle θ with the horizontal and strikes the ground at the same elevation a distance d away. Determine the impulse of his foot F on the ball. Neglect the impulse caused by the ball’s weight while it’s being kicked. Given: M = 200 gm d 15 m = θ = 30 deg g 9.81 m s2 = 307 Engineering Mechanics - Dynamics Chapter 15 Solution: First find the velocity vA Guesses vA 1 m s = t 1 s = Given 0 −g 2 ⎛⎜⎝ ⎞⎟⎠ t2 + vAsin θ( )t = d vAcos θ( )t = t ⎛⎜vA ⎝ ⎞⎟ ⎠ Find t vA ( , ) = t = 1.33 s vA 13.04 m s = Impulse - Momentum 0 + I = M vA I = M vA I 2.61 N s⋅ = Problem 15-11 The particle P is acted upon by its weight W and forces F1 = (ai + btj + ctk) and F2 = dt2i. If the particle originally has a velocity of v1 = (v1xi+v1yj+v1zk), determine its speed after time t1. Given: W = 3 lb g 32.2 ft s2 = v1x 3 ft s = a 5 lb = v1y 1 ft s = b 2 lb s = v1z 6 ft s = c 1 lb s = t1 = 2 s d 1 lb s2 = Solution: mv1 0 t1 t F1 + F2 − W k ( ) ⌠ ⎮⌡ d + = mv2 v2 v1 1 m 0 t1 t F1 + F2 − W k ( ) ⌠ ⎮⌡ d + = v2x v1x g W 0 t1 t a + dt2 ⌠ ⎮⌡ d + = v2x 138.96 ft s = v2y v1y g W 0 t1 bt t ⌠ ⎮⌡ d + = v2y 43.93 ft s = 308 Engineering Mechanics - Dynamics Chapter 15 v2z v1z g W 0 t1 t ct − W ⌠ ⎮⌡ d + = v2z −36.93 ft s = v2 v2x 2 v2y 2 + v2z 2 + = v2 150.34 ft s = *Problem 15-12 The twitch in a muscle of the arm develops a force which can be measured as a function of time as shown in the graph. If the effective contraction of the muscle lasts for a time t0, determine the impulse developed by the muscle. Solution: I 0 t0 t F0 t T ⎛⎜⎝ ⎞⎟⎠ e t − T ⌠⎮ ⎮ ⎮⌡ d = F0 −t0 − T ( ) e t0 − T + T F0 = I F0 T 1 1 t0 ⎛⎜⎝ + T ⎞⎟⎠ e t0 − T − ⎡⎢ ⎢⎣ ⎤⎥ ⎥⎦ = Problem 15-13 From experiments, the time variation of the vertical force on a runner’s foot as he strikes and pushes off the ground is shown in the graph.These results are reported for a 1-lb static load, i.e., in terms of unit weight. If a runner has weight W, determine the approximate vertical impulse he exerts on the ground if the impulse occurs in time t5. Units Used: ms 10 3 − s = Given: W 175 lb = t1 = 25 ms t 210 ms = 309 Engineering Mechanics - Dynamics Chapter 15 t2 = 50 ms t3 125 ms = t4 = 200 ms t5 210 ms = F2 = 3.0 lb F1 1.5 lb = Solution: Area 1 2t1 F1 F1 t2 − t1 ( ) + F1 t4 − t2 ( ) + 1 2 t5 − t4 ( )F1 + 1 2 F2 − F1 ( ) t4 − t2 ( ) + = Imp Area W lb = Imp 70.2 lb s⋅ = Problem 15-14 As indicated by the derivation, the principle of impulse and momentum is valid for observers in any inertial reference frame. Show that this is so, by considering the block of mass M which rests on the smooth surface and is subjected to horizontal force F. If observer A is in a fixed frame x, determine the final speed of the block at time t1 if it has an initial speed v0 measured from the fixed frame. Compare the result with that obtained by an observer B, attached to the x' axis that moves at constant velocity vB relative to A. Given: M = 10 kg v0 5 m s = F 6 N = vB 2 m s = t1 4 s = Solution: Observer A: M v0 + F t1 = M v1A v1A v0 F M ⎛⎜⎝ ⎞⎟⎠ t1 + = v1A 7.40 m s = Observer B: M v0 − vB ( ) + F t1 = M v1B v1B v0 − vB F M ⎛⎜⎝ ⎞⎟⎠ t1 + = v1B 5.40 m s = Note that v1A v1B + vB = Problem 15-15 The cabinet of weight W is subjected to the force F = a(bt+c). If the cabinet is initially moving up the plane with velocity v0, determine how long it will take before the cabinet comes to a stop. F always acts parallel to the plane. Neglect the size of the rollers. 310 Engineering Mechanics - Dynamics Chapter 15 Given: W = 4 lb v0 10 ft s = a = 20 lb g 32.2 ft s2 = b 1 = s θ 20 deg = c 1 = Solution: Guess t = 10 s Given W g ⎛⎜⎝ ⎞⎟⎠ v0 0 t τ a bτ + c ( ) ⌠ ⎮⌡ d + − W sin θ( )t = 0 t = Find t( ) t −0.069256619 s = *Problem 15-16 If it takes time t1 for the tugboat of mass mt to increase its speed uniformly to v1 starting from rest, determine the force of the rope on the tugboat. The propeller provides the propulsion force F which gives the tugboat forward motion, whereas the barge moves freely. Also, determine the force F acting on the tugboat. The barge has mass of mb. Units Used: Mg 1000 kg = kN 103 N = Given: t1 35 s = mt 50 Mg = v1 25 km hr = mb 75 Mg = Solution: The barge alone 0 + T t1 = mb v1 T mb v1 t1 = T 14.88 kN = The barge and the tug 0 + F t1 mt + mb ( )v1 = F mt + mb ( )v1 t1 = F 24.80 kN = 311 Engineering Mechanics - Dynamics Chapter 15 Problem 15-17 When the ball of weight W is fired, it leaves the ground at an angle θ from the horizontal and strikes the ground at the same elevation a distance d away. Determine the impulse given to the ball. Given: W 0.4 lb = d 130 ft = θ 40 deg = g 32.2 ft s2 = Solution: Guesses v0 1 ft s = t = 1 s Imp 1 lb s⋅ = Given v0 cos θ( )t = d −1 2 gt2 + v0 sin θ( )t = 0 Imp W g ⎛⎜⎝ ⎞⎟⎠ v0 = v0 t Imp ⎛⎜ ⎜ ⎜⎝ ⎞⎟ ⎟ ⎟⎠ Find v0 t, , Imp ( ) = v0 65.2 ft s = t = 2.6 s Imp 0.810 lb s⋅ = Problem 15-18 The uniform beam has weight W. Determine the average tension in each of the two cables AB and AC if the beam is given an upward speed v in time t starting from rest. Neglect the mass of the cables. Units Used: kip 103 lb = Given: W = 5000 lb g 32.2 ft s2 = v 8 ft s = a 3 ft = t = 1.5 s b 4 ft = 312 Engineering Mechanics - Dynamics Chapter 15 Solution: 0 − W t 2 b a2 b2 + ⎛ ⎜ ⎝ ⎞ ⎟ ⎠ FABt + W g ⎛⎜⎝ ⎞⎟⎠ v = FAB W g v + W t ⎛⎜⎝ ⎞⎟⎠ a2 b2 + 2bt ⎛ ⎜ ⎝ ⎞ ⎟ ⎠ = FAB 3.64 kip = Problem 15-19 The block of mass M is moving downward at speed v1 when it is a distance h from the sandy surface. Determine the impulse of the sand on the block necessary to stop its motion. Neglect the distance the block dents into the sand and assume the block does not rebound. Neglect the weight of the block during the impact with the sand. Given: M 5 kg = v1 2 m s = h 8 m = g 9.81 m s2 = Solution: Just before impact v2 v1 2 + 2gh = v2 12.69 m s = Collision M v2 − I = 0 I = M v2 I 63.4 N s⋅ = *Problem 15-20 The block of mass M is falling downward at speed v1 when it is a distance h from the sandy surface. Determine the average impulsive force acting on the block by the sand if the motion of the block is stopped in time Δt once the block strikes the sand. Neglect the distance the block dents into the sand and assume the block does not rebound. Neglect the weight of the block during the impact with the sand. 313 Engineering Mechanics - Dynamics Chapter 15 Given: M 5 kg = v1 2 m s = h 8 m = Δt = 0.9 s g 9.81 m s2 = Solution: Just before impact v2 v1 2 + 2gh = v2 12.69 m s = Collision M v2 − FΔt = 0 F M v2 Δt = F 70.5 N = Problem 15-21 A crate of mass M rests against a stop block s, which prevents the crate from moving down the plane. If the coefficients of static and kinetic friction between the plane and the crate are μs and μk respectively, determine the time needed for the force F to give the crate a speed v up the plane. The force always acts parallel to the plane and has a magnitude of F = at. Hint: First determine the time needed to overcome static friction and start the crate moving. Given: M = 50 kg θ = 30 deg g 9.81 m s2 = v 2 m s = μs 0.3 = μk = 0.2 a 300 N s = Solution: Guesses t1 = 1 s NC = 1 N t2 1 s = Given NC − M gcos θ( ) 0 = at1 − μsNC − M gsin θ( ) 0 = t1 t2 t at − M gsin θ( ) − μk NC ( ) ⌠ ⎮ ⌡ d M v = t1 t2 NC ⎛⎜ ⎜ ⎜⎝ ⎞⎟ ⎟ ⎟⎠ Find t1 t2 , , NC ( ) = t1 = 1.24 s t2 1.93 s = 314 Engineering Mechanics - Dynamics Chapter 15 Problem 15-22 The block of weight W has an initial velocity v1 in the direction shown. If a force F = {f1i + f2j} acts on the block for time t, determine the final speed of the block. Neglect friction. Given: W = 2 lb a = 2 ft f1 0.5 lb = v1 10 ft s = b = 2 ft f2 0.2 lb = g 32.2 ft s2 = c = 5 ft t 5 s = Solution: θ atan b c a − ⎛⎜⎝ ⎞⎟⎠ = Guesses v2x 1 ft s = v2y 1 ft s = Given W g ⎛⎜⎝ ⎞⎟⎠ v1 sin θ( ) − ⎛⎜⎝ cos θ( ) ⎞⎟⎠ f1 f2 ⎛ ⎜ ⎝ ⎞ ⎟ ⎠ t + W g ⎛⎜⎝ ⎞⎟⎠ v2x v2y ⎛ ⎜ ⎝ ⎞ ⎟ ⎠ = v2x v2y ⎛ ⎜ ⎝ ⎞ ⎟ ⎠ Find v2x v2y , ( ) = v2x v2y ⎛ ⎜ ⎝ ⎞ ⎟ ⎠ 34.7 ⎛⎜⎝24.4 ⎞⎟⎠ ft s = v2x v2y ⎛ ⎜ ⎝ ⎞ ⎟ ⎠ 42.4 ft s = Problem 15-23 The tennis ball has a horizontal speed v1 when it is struck by the racket. If it then travels away at angle θ from the horizontal and reaches maximum altitude h, measured from the height of the racket, determine the magnitude of the net impulse of the racket on the ball. The ball has mass M. Neglect the weight of the ball during the time the racket strikes the ball. Given: v1 15 m s = θ 25 deg = h 10 m = M 180 gm = g 9.81 m s2 = 315 Engineering Mechanics - Dynamics Chapter 15 Solution: Free flight v2 sin θ( ) 2gh = v2 2gh sin θ( ) = v2 33.14 m s = Impulse - momentum −M v1 + Ix = M v2 cos θ( ) Ix M v2 cos θ( ) + v1 ( ) = Ix 8.11 N s⋅ = 0 + Iy = M v2 sin θ( ) Iy = M v2 sin θ( ) Iy 2.52 N s⋅ = I Ix 2 Iy 2 + = I 8.49 N s⋅ = *Problem 15-24 The slider block of mass M is moving to the right with speed v when it is acted upon by the forces F1 and F2. If these loadings vary in the manner shown on the graph, determine the speed of the block at t = t3. Neglect friction and the mass of the pulleys and cords. Given: M 40 kg = v 1.5 m s = t3 6 s = t2 4 s = t1 2 s = P1 10 N = P2 20 N = P3 30 N = P4 40 N = Solution: The impulses acting on the block are found from the areas under the graph. I 4 P3 t2 P1 t3 − t2 ( ) + ⎡⎣ ⎤⎦ P1 t1 P2 t2 − t1 ( ) + P4 t3 − t2 ( ) + ⎡⎣ ⎤⎦ − = M v + I = M v3 v3 v I + M = v3 12.00 m s = 316 Engineering Mechanics - Dynamics Chapter 15 Problem 15-25 Determine the velocities of blocks A and B at time t after they are released from rest. Neglect the mass of the pulleys and cables. Given: WA 2 lb = WB 4 lb = t 2 s = g 32.2 ft s2 = Solution: 2sA + 2sB L = vA −vB = Block A 0 2T − WA ( )t + WA g vA = Block B 0 2T − WB ( )t + WB g −( vA ) = Combining WB − WA ( )t WB WA + g ⎛⎜⎝ ⎞⎟⎠ vA = vA WB WA − WB WA + ⎛⎜ ⎝ ⎞⎟ ⎠ gt = vB −vA = vA vB ⎛ ⎜ ⎝ ⎞ ⎟ ⎠ 21.47 ⎛⎜⎝−21.47 ⎞⎟⎠ ft s = Problem 15-26 The package of mass M is released from rest at A. It slides down the smooth plane which is inclined at angle θ onto the rough surface having a coefficient of kinetic friction of μk. Determine the total time of travel before the package stops sliding. Neglect the size of the package. Given: M = 5 kg h 3 m = θ = 30 deg g 9.81 m s2 = μk 0.2 = 317 Engineering Mechanics - Dynamics Chapter 15 Solution: On the slope v1 2gh = v1 7.67 m s = t1 v1 gsin θ( ) = t1 1.56 s = On the flat M v1 − μk M gt2 = 0 t2 v1 μk g = t2 3.91 s = t t1 + t2 = t 5.47 s = Problem 15-27 Block A has weight WA and block B has weight WB. If B is moving downward with a velocity vB0 at t = 0, determine the velocity of A when t = t1. Assume that block A slides smoothly. Given: WA 10 lb = WB 3 lb = vB0 3 ft s = t1 1 s = g 32.2 ft s2 = Solution: sA + 2sB = L vA 2 = − vB Guess vA1 1 ft s = T 1 lb = Given Block A WA g ⎛⎜⎝ ⎞⎟⎠ 2vB0 + T t1 WA g ⎛⎜⎝ ⎞⎟⎠ vA1 = Block B WB − g ⎛⎜⎝ ⎞⎟⎠ vB0 + 2T t1 − WBt1 WB − g ⎛⎜⎝ ⎞⎟⎠ vA1 2 ⎛⎜⎝ ⎞⎟⎠ = vA1 ⎛⎜ T ⎝ ⎞⎟ ⎠ Find vA1 T , ( ) = T = 1.40 lb vA1 10.49 ft s = 318 Engineering Mechanics - Dynamics Chapter 15 *Problem 15-28 Block A has weight WA and block B has weight WB. If B is moving downward with a velocity vB1 at t = 0, determine the velocity of A when t = t1. The coefficient of kinetic friction between the horizontal plane and block A is μk. Given: WA 10 lb = WB 3 lb = vB1 3 ft s = μk 0.15 = t1 1 s = g 32.2 ft s2 = Solution: sA + 2sB = L vA 2 = − vB Guess vA2 1 ft s = T 1 lb = Given Block A WA g ⎛⎜⎝ ⎞⎟⎠ 2vB1 + T t1 − μk WAt1 WA g ⎛⎜⎝ ⎞⎟⎠ vA2 = Block B WB − g ⎛⎜⎝ ⎞⎟⎠ vB1 + 2T t1 − WBt1 WB − g vA2 2 ⎛⎜⎝ ⎞⎟⎠ = vA2 ⎛⎜ T ⎝ ⎞⎟ ⎠ Find vA2 T , ( ) = T = 1.50 lb vA2 6.00 ft s = Problem 15-29 A jet plane having a mass M takes off from an aircraft carrier such that the engine thrust varies as shown by the graph. If the carrier is traveling forward with a speed v, determine the plane’s airspeed after time t. Units Used: Mg 103 kg = kN 103 N = 319 Engineering Mechanics - Dynamics Chapter 15 Given: M = 7 Mg t1 2 s = v 40 km hr = t2 5 s = F1 = 5 kN t 5 s = F2 15 kN = Solution: The impulse exerted on the plane is equal to the area under the graph. M v 1 + 2F1 t1 1 2 F1 + F2 ( ) t2 − t1 ( ) + M v1 = v1 v 1 2M F1 t1 F1 + F2 ( ) t2 − t1 ( ) + ⎡⎣ ⎤⎦ + = v1 16.11 m s = Problem 15-30 The motor pulls on the cable at A with a force F = a + bt2. If the crate of weight W is originally at rest at t = 0, determine its speed at time t = t2. Neglect the mass of the cable and pulleys. Hint: First find the time needed to begin lifting the crate. Given: W 17 lb = a 30 lb = b 1 lb s2 = t2 4 s = Solution: 1 2 a ( + bt12 ) − W 0 = t1 2W − a b = t1 2.00 s = 1 2 t1 t2 t a + bt2 ⌠ ⎮ ⌡ d W t2 − t1 ( ) − W g ⎛⎜⎝ ⎞⎟⎠ v2 = v2 g 2W t1 t2 t a + bt2 ⌠ ⎮ ⌡ d g t2 − t1 ( ) − = v2 10.10 ft s = 320 Engineering Mechanics - Dynamics Chapter 15 Problem 15-31 The log has mass M and rests on the ground for which the coefficients of static and kinetic friction are μs and μk respectively. The winch delivers a horizontal towing force T to its cable at A which varies as shown in the graph. Determine the speed of the log when t = t2. Originally the tension in the cable is zero. Hint: First determine the force needed to begin moving the log. Given: M = 500 kg t1 3 s = μs = 0.5 T1 1800 N = μk = 0.4 g 9.81 m s2 = t2 5 s = Solution: To begin motion we need 2T1 t0 2 t1 2 ⎛⎜ ⎜⎝ ⎞⎟ ⎟⎠ = μsM g t0 μsM g 2T1 t1 = t0 2.48 s = Impulse - Momentum 0 t0 t1 t 2T1 t t1 ⎛⎜⎝ ⎞⎟⎠ 2 ⌠⎮ ⎮ ⎮⌡ d + 2T1 t2 − t1 ( ) + μk M g t2 − t0 ( ) − M v2 = v2 1 M t0 t1 t 2T1 t t1 ⎛⎜⎝ ⎞⎟⎠ 2 ⌠⎮ ⎮ ⎮⌡ d 2T1 t2 − t1 ( ) + μk M g t2 − t0 ( ) − ⎡⎢ ⎢ ⎢ ⎣ ⎤⎥ ⎥ ⎥ ⎦ = v2 7.65 m s = *Problem 15-32 A railroad car having mass m1 is coasting with speed v1 on a horizontal track. At the same time another car having mass m2 is coasting with speed v2 in the opposite direction. If the cars meet and couple together, determine the speed of both cars just after the coupling. Find the difference between the total kinetic energy before and after coupling has occurred, and explain qualitatively what happened to this energy. Units used: Mg = 103 kg kJ 103 J = Given: m1 = 15 Mg m2 12 Mg = 321 Engineering Mechanics - Dynamics Chapter 15 v1 1.5 m s = v2 0.75 m s = Solution: m1 v1 − m2 v2 m1 + m2 ( )v = v m1 v1 m2 v2 − m1 + m2 = v 0.50 m s = T1 1 2m1 v1 2 1 2m2 v2 2 + = T1 20.25 kJ = T2 1 2 m1 + m2 ( ) v2 = T2 3.38 kJ = ΔT T2 − T1 = ΔT −16.88 kJ = −ΔT T1 100 = 83.33 % loss The energy is dissipated as noise, shock, and heat during the coupling. Problem 15-33 Car A has weight WA and is traveling to the right at speed vA Meanwhile car B of weight WB is traveling at speed vB to the left. If the cars crash head-on and become entangled, determine their common velocity just after the collision. Assume that the brakes are not applied during collision. Given: WA = 4500 lb WB 3000 lb = vA 3 ft s = vB 6 ft s = g 32.2 ft s2 = Solution: WA g ⎛⎜⎝ ⎞⎟⎠ vA WB g ⎛⎜⎝ ⎞⎟⎠ vB − WA WB + g ⎛⎜⎝ ⎞⎟⎠ v = v WAvA WBvB − WA + WB = v −0.60 ft s = Problem 15-34 The bus B has weight WB and is traveling to the right at speed vB. Meanwhile car A of weight WA is traveling at speed vA to the left. If the vehicles crash head-on and become entangled, determine their common velocity just after the collision. Assume that the vehicles are free to roll during collision. 322 Engineering Mechanics - Dynamics Chapter 15 Given: WB = 15000 lb vB 5 ft s = WA 3000 lb = vA 4 ft s = g 32.2 ft s2 = Solution: WB g ⎛⎜⎝ ⎞⎟⎠ vB WA g ⎛⎜⎝ ⎞⎟⎠ vA − WB WA + g ⎛⎜⎝ ⎞⎟⎠ v = v WBvB WAvA − WB + WA = v 3.50 ft s = Positive means to the right, negative means to the left. Problem 15-35 The cart has mass M and rolls freely down the slope. When it reaches the bottom, a spring loaded gun fires a ball of mass M1 out the back with a horizontal velocity vbc measured relative to the cart. Determine the final velocity of the cart. Given: M = 3 kg h 1.25 m = M1 = 0.5 kg g 9.81 m s2 = vbc 0.6 m s = Solution: v1 2gh = M + M1 ( )v1 M vc M1 vc − vbc ( ) + = vc v1 M1 M M1 + ⎛⎜ ⎝ ⎞⎟ ⎠ vbc + = vc 5.04 m s = 323 Engineering Mechanics - Dynamics Chapter 15 *Problem 15-36 Two men A and B, each having weight Wm, stand on the cart of weight Wc. Each runs with speed v measured relative to the cart. Determine the final speed of the cart if (a) A runs and jumps off, then B runs and jumps off the same end, and (b) both run at the same time and jump off at the same time. Neglect the mass of the wheels and assume the jumps are made horizontally. Given: Wm 160 lb = Wc 200 lb = v 3 ft s = g 32.2 ft s2 = Solution: mm Wm g = mc Wc g = (a) A jumps first 0 −mm v ( − vc ) mm + mc ( )vc1 + = vc1 mmv mc + 2mm = vc1 0.923 ft s = And then B jumps mm + mc ( )vc1 −mm v ( − vc2 ) + mc vc2 = vc2 mmv mm + mc ( )vc1 + mm + mc = vc2 2.26 ft s = (b) Both men jump at the same time 0 2 − mm v ( − vc3 ) + mc vc3 = vc3 2mmv 2mm + mc = vc3 1.85 ft s = Problem 15-37 A box of weight W1 slides from rest down the smooth ramp onto the surface of a cart of weight W2. Determine the speed of the box at the instant it stops sliding on the cart. If someone ties the cart to the ramp at B, determine the horizontal impulse the box will exert at C in order to stop its motion. Neglect friction on the ramp and neglect the size of the box. 324 Engineering Mechanics - Dynamics Chapter 15 Given: W1 = 40 lb W2 = 20 lb h = 15 ft g 32.2 ft s2 = Solution: v1 2gh = W1 g v1 W1 W2 + g ⎛⎜⎝ ⎞⎟⎠ v2 = v2 W1 W1 W2 + ⎛⎜ ⎝ ⎞⎟ ⎠ v1 = v2 20.7 ft s = W1 g ⎛⎜⎝ ⎞⎟⎠ v1 − Imp = 0 Imp W1 g ⎛⎜⎝ ⎞⎟⎠ v1 = Imp 38.6 lb s⋅ = Problem 15-38 A boy of weight W1 walks forward over the surface of the cart of weight W2 with a constant speed v relative to the cart. Determine the cart’s speed and its displacement at the moment he is about to step off. Neglect the mass of the wheels and assume the cart and boy are originally at rest. Given: W1 = 100 lb W2 = 60 lb v 3 ft s = d 6 ft = Solution: 0 W1 g ⎛⎜⎝ ⎞⎟⎠ vc + v ( ) W2 g ⎛⎜⎝ ⎞⎟⎠ vc + = vc W1 W1 + W2 − v = vc −1.88 ft s = Assuming that the boy walks the distance d t d = v sc = vc t sc −3.75 ft = Problem 15-39 The barge B has weight WB and supports an automobile weighing Wa. If the barge is not tied to the pier P and someone drives the automobile to the other side of the barge for unloading, determine how far the barge moves away from the pier. Neglect the resistance of the water. 325 Engineering Mechanics - Dynamics Chapter 15 Given: WB 30000 lb = Wa 3000 lb = d 200 ft = g 32.2 ft s2 = Solution: mB WB g = ma Wa g = v is the velocity of the car relative to the barge. The answer is independent of the acceleration so we will do the problem for a constant speed. mBvB ma v ( + vB ) + = 0 vB −ma v mB + ma = t d = v sB −vB t = sB ma d ma + mB = sB 18.18 ft = *Problem 15-40 A bullet of weight W1 traveling at speed v1 strikes the wooden block of weight W2 and exits the other side at speed v2 as shown. Determine the speed of the block just after the bullet exits the block, and also determine how far the block slides before it stops. The coefficient of kinetic friction between the block and the surface is μk. Given: W1 = 0.03 lb a 3 ft = b = 4 ft W2 10 lb = c 5 ft = v1 1300 ft s = d 12 ft = v2 50 ft s = μk 0.5 = Solution: W1 g ⎛⎜⎝ ⎞⎟⎠ v1 d c2 d2 + ⎛ ⎜ ⎝ ⎞ ⎟ ⎠ W2 g ⎛⎜⎝ ⎞⎟⎠ vB W1 g ⎛⎜⎝ ⎞⎟⎠ v2 b a2 b2 + ⎛ ⎜ ⎝ ⎞ ⎟ ⎠ + = 326 Engineering Mechanics - Dynamics Chapter 15 vB W1 W2 v1 d c2 d2 + v2 b a2 b2 + − ⎛ ⎜ ⎝ ⎞ ⎟ ⎠ = vB 3.48 ft s = 1 2 W2 g ⎛⎜⎝ ⎞⎟⎠ vB 2 − μk W2 d = 0 d vB 2 2gμk = d 0.38 ft = Problem 15-41 A bullet of weight W1 traveling at v1 strikes the wooden block of weight W2 and exits the other side at v2 as shown. Determine the speed of the block just after the bullet exits the block. Also, determine the average normal force on the block if the bullet passes through it in time Δt, and the time the block slides before it stops. The coefficient of kinetic friction between the block and the surface is μk. Units Used: ms 10 3 − s = Given: W1 = 0.03 lb a 3 ft = W2 = 10 lb b 4 ft = μk = 0.5 c 5 ft = Δt = 1 ms d 12 ft = v1 1300 ft s = v2 50 ft s = Solution: W1 g v1 d c2 d2 + ⎛ ⎜ ⎝ ⎞ ⎟ ⎠ W2 g vB W1 g v2 b a2 b2 + ⎛ ⎜ ⎝ ⎞ ⎟ ⎠ + = vB W1 W2 v1 d c2 d2 + v2 b a2 b2 + − ⎛ ⎜ ⎝ ⎞ ⎟ ⎠ = vB 3.48 ft s = W1 − g v1 c c2 d2 + ⎛ ⎜ ⎝ ⎞ ⎟ ⎠ N − W2 ( )Δt + W1 g v2 a a2 b2 + ⎛ ⎜ ⎝ ⎞ ⎟ ⎠ = N W1 gΔt v2 a a2 b2 + v1 c c2 d2 + + ⎛ ⎜ ⎝ ⎞ ⎟ ⎠ + W2 = N 503.79 lb = W2 g ⎛⎜⎝ ⎞⎟⎠ vB − μk W2 t = 0 t vB gμk = t 0.22 s = 327 Engineering Mechanics - Dynamics Chapter 15 Problem 15-42 The man M has weight WM and jumps onto the boat B which has weight WB. If he has a horizontal component of velocity v relative to the boat, just before he enters the boat, and the boat is traveling at speed vB away from the pier when he makes the jump, determine the resulting velocity of the man and boat. Given: WM = 150 lb vB 2 ft s = WB 200 lb = g 32.2 ft s2 = v 3 ft s = Solution: WM g v ( + vB ) WB g vB + WM WB + g ⎛⎜⎝ ⎞⎟⎠ v' = v' WM v WM + WB ( )vB + WM + WB = v' 3.29 ft s = Problem 15-43 The man M has weight WM and jumps onto the boat B which is originally at rest. If he has a horizontal component of velocity v just before he enters the boat, determine the weight of the boat if it has velocity v' once the man enters it. Given: WM 150 lb = v 3 ft s = v' 2 ft s = g 32.2 ft s2 = Solution: WM g ⎛⎜⎝ ⎞⎟⎠ v WM WB + g ⎛⎜⎝ ⎞⎟⎠ v' = WB v − v' v' ⎛⎜⎝ ⎞⎟⎠ WM = WB 75.00 lb = 328 Engineering Mechanics - Dynamics Chapter 15 *Problem 15-44 A boy A having weight WA and a girl B having weight WB stand motionless at the ends of the toboggan, which has weight Wt. If A walks to B and stops, and both walk back together to the original position of A (both positions measured on the toboggan), determine the final position of the toboggan just after the motion stops. Neglect friction. Given: WA 80 lb = WB 65 lb = Wt 20 lb = d 4 ft = Solution: The center of mass doesn’t move during the motion since there is no friction and therefore no net horizontal force WBd WA + WB + Wt ( )d' = d' WBd WA + WB + Wt = d' 1.58 ft = Problem 15-45 The projectile of weight W is fired from ground level with initial velocity vA in the direction shown. When it reaches its highest point B it explodes into two fragments of weight W/2. If one fragment travels vertically upward at speed v1, determine the distance between the fragments after they strike the ground. Neglect the size of the gun. Given: W 10 lb = vA 80 ft s = v1 12 ft s = θ 60 deg = g 32.2 ft s2 = Solution: At the top v vAcos θ( ) = 329 Engineering Mechanics - Dynamics Chapter 15 Explosion W g ⎛⎜⎝ ⎞⎟⎠ v 0 W 2g ⎛⎜⎝ ⎞⎟⎠ v2x + = v2x = 2v v2x 80.00 ft s = 0 W 2g ⎛⎜⎝ ⎞⎟⎠ v1 W 2g ⎛⎜⎝ ⎞⎟⎠ v2y − = v2y = v1 v2y 12.00 ft s = Kinematics h (vAsin θ( ) )2 2g = h = 74.53 ft Guess t 1 s = Given 0 −g 2 ⎛⎜⎝ ⎞⎟⎠ t2 − v2yt + h = t = Find t( ) t 1.81 s = d = v2xt d 144.9 ft = Problem 15-46 The projectile of weight W is fired from ground level with an initial velocity vA in the direction shown. When it reaches its highest point B it explodes into two fragments of weight W/2. If one fragment is seen to travel vertically upward, and after they fall they are a distance d apart, determine the speed of each fragment just after the explosion. Neglect the size of the gun. Given: W = 10 lb θ 60 deg = vA 80 ft s = g 32.2 ft s2 = d 150 ft = Solution: h (vAsin θ( ) )2 2g = Guesses v1 1 ft s = v2x 1 ft s = v2y 1 ft s = t 1 s = Given W g ⎛⎜⎝ ⎞⎟⎠ vAcos θ( ) W 2g ⎛⎜⎝ ⎞⎟⎠ v2x = 0 W 2g ⎛⎜⎝ ⎞⎟⎠ v1 W 2g ⎛⎜⎝ ⎞⎟⎠ v2y + = d = v2xt 0 h 1 − 2gt2 + v2yt = 330 Engineering Mechanics - Dynamics Chapter 15 v1 v2x v2y t ⎛ ⎜ ⎜ ⎜ ⎜ ⎝ ⎞ ⎟ ⎟ ⎟ ⎟ ⎠ Find v1 v2x , , v2y t, ( ) = t 1.87 s = v1 v2x v2y ⎛⎜ ⎜ ⎜⎝ ⎞⎟ ⎟ ⎟⎠ 9.56 80.00 9.56 − ⎛ ⎜ ⎜ ⎝ ⎞ ⎟ ⎟ ⎠ ft s = v1 9.56 ft s = v2x v2y ⎛ ⎜ ⎝ ⎞ ⎟ ⎠ 80.57 ft s = Problem 15-47 The winch on the back of the jeep A is turned on and pulls in the tow rope at speed vrel. If both the car B of mass MB and the jeep A of mass MA are free to roll, determine their velocities at the instant they meet. If the rope is of length L, how long will this take? Units Used: Mg 103 kg = Given: MA = 2.5 Mg vrel 2 m s = MB = 1.25 Mg L 5 m = Solution: Guess vA 1 m s = vB 1 m s = Given 0 MAvA + MBvB = vA − vB vrel = vA vB ⎛ ⎜ ⎝ ⎞ ⎟ ⎠ Find vA vB , ( ) = t L vrel = t 2.50 s = vA vB ⎛ ⎜ ⎝ ⎞ ⎟ ⎠ 0.67 ⎛⎜⎝−1.33 ⎞⎟⎠ m s = *Problem 15-48 The block of mass Ma is held at rest on the smooth inclined plane by the stop block at A. If the bullet of mass Mb is traveling at speed v when it becomes embedded in the block of mass Mc, determine the distance the block will slide up along the plane before momentarily stopping. 331 Engineering Mechanics - Dynamics Chapter 15 Given: Ma = 10 kg v 300 m s = Mb 10 gm = θ 30 deg = Mc 10 kg = Solution: Conservation of Linear Momentum: If we consider the block and the bullet as a system, then from the FBD, the impulsive force F caused by the impact is internal to the system. Therefore, it will cancel out. Also, the weight of the bullet and the block are nonimpulsive forces. As the result, linear momentum is conserved along the x axis Mb vbx Mb + Ma ( )vx = Mb vcos θ( ) Mb + Ma ( )vx = vx Mb v cos θ( ) Mb Ma + ⎛⎜⎝ ⎞⎟⎠ = vx 0.2595 m s = Conservation of Energy: The datum is set at the block’s initial position. When the block and th