Get started today!

Good to have you back!
If you've signed in to StudyBlue with Facebook in the past, please do that again.

- StudyBlue
- Wisconsin
- Marquette University
- Civil Engineering
- Civil Engineering 2120
- Bowman
- Dynamics_Part9.pdf

Mike J.

Advertisement

Advertisement

Engineering Mechanics - Dynamics Chapter 15 Solution: W g ⎛ ⎜ ⎝ ⎞ ⎟ ⎠ v A sin θ() Wt− W g ⎛ ⎜ ⎝ ⎞ ⎟ ⎠ 0= t v A sin θ() g = t 0.280 s= W g ⎛ ⎜ ⎝ ⎞ ⎟ ⎠ v A cos θ() 0+ W g ⎛ ⎜ ⎝ ⎞ ⎟ ⎠ v x = v x v A cos θ()= v x 15.59 ft s = Problem 15-3 A block of weight W is given an initial velocity v 0 up a smooth slope of angle θ. Determine the time it will take to travel up the slope before it stops. Given: W 5lb= v 0 10 ft s = θ 45 deg= g 32.2 ft s 2 = Solution: W g ⎛ ⎜ ⎝ ⎞ ⎟ ⎠ v 0 W sin θ()t− 0= t v 0 g sin θ() = t 0.439 s= *Problem 15-4 The baseball has a horizontal speed v 1 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= v 1 35 m s = h 50 m= θ 60 deg= g 9.81 m s 2 = 303 Engineering Mechanics - Dynamics Chapter 15 Solution: Guesses v 2 20 m s = Imp x 1Ns⋅= Imp y 10 N s⋅= Given 1 2 Mv 2 sin θ()() 2 Mgh= M− v 1 Imp x + Mv 2 cos θ()= 0 Imp y + Mv 2 sin θ()= v 2 Imp x Imp y ⎛ ⎜ ⎜ ⎜ ⎝ ⎞ ⎟ ⎟ ⎟ ⎠ Find v 2 Imp x , Imp y ,()= v 2 36.2 m s = Imp x Imp y ⎛ ⎜ ⎝ ⎞ ⎟ ⎠ 21.2 12.5 ⎛ ⎜ ⎝ ⎞ ⎟ ⎠ Ns⋅= Imp x Imp y ⎛ ⎜ ⎝ ⎞ ⎟ ⎠ 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: F 1 0.3 N= t 1 2ms= F 2 0.4 N= t 2 4ms= F 3 0.5 N= t 3 7ms= F 4 0.8 N= t 4 10 ms= F 5 1.2 N= t 5 14 ms= Solution: CONFOR foam: I c 1 2 t 1 F 3 1 2 F 3 F 4 +()t 3 t 1 −()+ 1 2 F 4 t 5 t 3 −()+= I c 6.55 N ms⋅= 304 Engineering Mechanics - Dynamics Chapter 15 Urethane foam: I U 1 2 t 2 F 1 1 2 F 5 F 1 +()t 3 t 2 −()+ 1 2 F 5 F 2 +()t 4 t 3 −()+ 1 2 t 5 t 4 −()F 2 += I U 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 s 2 = Solution: First find the velocity v 1 Guesses v 1 1 m s = t 1s= Given 0 g− 2 ⎛ ⎜ ⎝ ⎞ ⎟ ⎠ t 2 v 1 sin θ()t+= dv 1 cos θ()t= t v 1 ⎛ ⎜ ⎝ ⎞ ⎟ ⎠ Find tv 1 ,()= t 1.85 s= v 1 14.11 m s = Impulse - Momentum 0 Imp+ Mv 1 = Imp M v 1 = 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: T 1a 4lb= t 1a 3s= T 1b 8lb= t 1b 6s= 305 Engineering Mechanics - Dynamics Chapter 15 T 2a 6lb= t 1c 10 s= t 2a 8s= t 2b 10 s= Solution: Impulse is area under curve for hole cavity. I a T 1a t 1a 1 2 T 1a T 1b +()t 1b t 1a −()+ 1 2 T 1b t 1c t 1b −()+= I a 46.00 lb s⋅= For starred cavity: I b T 2a t 2a 1 2 T 2a t 2b t 2a −()+= I b 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 2lb= v 200 ft s = g 32.2 ft s 2 = Solution: I 1 2 90× 10 3 × lb ⎛ ⎜ ⎝ ⎞ ⎟ ⎠ 0.4 10 3− s () = I 18.00 lb s⋅= Δt 0.4 10 3− s×= W− g ⎛ ⎜ ⎝ ⎞ ⎟ ⎠ vI+ WΔt− W g ⎛ ⎜ ⎝ ⎞ ⎟ ⎠ v'= v' v− Ig 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 v 0 when t = 0. If both engines provide a horizontal thrust which varies as shown in the graph, determine the plane’s velocity at time t 1 . Neglect air resistance and the loss of fuel during the motion. Units Used: Mg 10 3 kg= kN 10 3 N= Given: M 250 Mg= v 0 100 m s = t 1 15 s= a 200 kN= b 2 kN s 2 = Solution: Mv 0 0 t 1 tabt 2 + ⌠ ⎮ ⌡ d+ Mv 1 = v 1 v 0 1 M 0 t 1 tabt 2 + ⌠ ⎮ ⌡ d+= v 1 121.00 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 s 2 = 307 Engineering Mechanics - Dynamics Chapter 15 Solution: First find the velocity v A Guesses v A 1 m s = t 1s= Given 0 g− 2 ⎛ ⎜ ⎝ ⎞ ⎟ ⎠ t 2 v A sin θ()t+= dv A cos θ()t= t v A ⎛ ⎜ ⎝ ⎞ ⎟ ⎠ Find tv A ,()= t 1.33 s= v A 13.04 m s = Impulse - Momentum 0 I+ Mv A = IMv A = I 2.61 N s⋅= Problem 15-11 The particle P is acted upon by its weight W and forces F 1 = (ai + btj + ctk) and F 2 = dt 2 i. If the particle originally has a velocity of v 1 = (v 1x i+v 1y j+v 1z k), determine its speed after time t 1 . Given: W 3lb= g 32.2 ft s 2 = v 1x 3 ft s = a 5lb= v 1y 1 ft s = b 2 lb s = v 1z 6 ft s = c 1 lb s = t 1 2s= d 1 lb s 2 = Solution: mv 1 0 t 1 tF 1 F 2 + Wk−() ⌠ ⎮ ⌡ d+ mv 2 = v 2 v 1 1 m 0 t 1 tF 1 F 2 + Wk−() ⌠ ⎮ ⌡ d+= v 2x v 1x g W 0 t 1 tadt 2 + ⌠ ⎮ ⌡ d+= v 2x 138.96 ft s = v 2y v 1y g W 0 t 1 tbt ⌠ ⎮ ⌡ d+= v 2y 43.93 ft s = 308 Engineering Mechanics - Dynamics Chapter 15 v 2z v 1z g W 0 t 1 tct W− ⌠ ⎮ ⌡ d+= v 2z 36.93− ft s = v 2 v 2x 2 v 2y 2 + v 2z 2 += v 2 150.34 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 t 0 , determine the impulse developed by the muscle. Solution: I 0 t 0 tF 0 t T ⎛ ⎜ ⎝ ⎞ ⎟ ⎠ e t− T ⌠ ⎮ ⎮ ⎮ ⌡ d= F 0 t 0 − T−()e t 0 − T TF 0 += IF 0 T 11 t 0 T + ⎛ ⎜ ⎝ ⎞ ⎟ ⎠ e t 0 − 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 t 5 . Units Used: ms 10 3− s= Given: W 175 lb= t 1 25 ms= t 210 ms= 309 Engineering Mechanics - Dynamics Chapter 15 t 2 50 ms= t 3 125 ms= t 4 200 ms= t 5 210 ms= F 2 3.0 lb= F 1 1.5 lb= Solution: Area 1 2 t 1 F 1 F 1 t 2 t 1 −()+ F 1 t 4 t 2 −()+ 1 2 t 5 t 4 −()F 1 + 1 2 F 2 F 1 −()t 4 t 2 −()+= 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 t 1 if it has an initial speed v 0 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 v B relative to A. Given: M 10 kg= v 0 5 m s = F 6N= v B 2 m s = t 1 4s= Solution: Observer A: Mv 0 Ft 1 + Mv 1A = v 1A v 0 F M ⎛ ⎜ ⎝ ⎞ ⎟ ⎠ t 1 += v 1A 7.40 m s = Observer B: Mv 0 v B −()Ft 1 + Mv 1B = v 1B v 0 v B − F M ⎛ ⎜ ⎝ ⎞ ⎟ ⎠ t 1 += v 1B 5.40 m s = Note that v 1A v 1B v B += 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 v 0 , 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 4lb= v 0 10 ft s = a 20 lb= g 32.2 ft s 2 = b 1 s = θ 20 deg= c 1= Solution: Guess t 10 s= Given W g ⎛ ⎜ ⎝ ⎞ ⎟ ⎠ v 0 0 t τabτ c+() ⌠ ⎮ ⌡ d+ W sin θ()t− 0= t Find t()= t 0.069256619− s= *Problem 15-16 If it takes time t 1 for the tugboat of mass m t to increase its speed uniformly to v 1 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 m b . Units Used: Mg 1000 kg= kN 10 3 N= Given: t 1 35 s= m t 50 Mg= v 1 25 km hr = m b 75 Mg= Solution: The barge alone 0 Tt 1 + m b v 1 = T m b v 1 t 1 = T 14.88 kN= The barge and the tug 0 Ft 1 + m t m b +()v 1 = F m t m b +()v 1 t 1 = 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 s 2 = Solution: Guesses v 0 1 ft s = t 1s= Imp 1lbs⋅= Given v 0 cos θ()td= 1− 2 gt 2 v 0 sin θ()t+ 0= Imp W g ⎛ ⎜ ⎝ ⎞ ⎟ ⎠ v 0 = v 0 t Imp ⎛ ⎜ ⎜ ⎜ ⎝ ⎞ ⎟ ⎟ ⎟ ⎠ Find v 0 t, Imp,()= v 0 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 10 3 lb= Given: W 5000 lb= g 32.2 ft s 2 = v 8 ft s = a 3ft= t 1.5 s= b 4ft= 312 Engineering Mechanics - Dynamics Chapter 15 Solution: 0 Wt− 2 b a 2 b 2 + ⎛ ⎜ ⎝ ⎞ ⎟ ⎠ F AB t+ W g ⎛ ⎜ ⎝ ⎞ ⎟ ⎠ v= F AB W g vWt+ ⎛ ⎜ ⎝ ⎞ ⎟ ⎠ a 2 b 2 + 2bt ⎛ ⎜ ⎝ ⎞ ⎟ ⎠ = F AB 3.64 kip= Problem 15-19 The block of mass M is moving downward at speed v 1 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 5kg= v 1 2 m s = h 8m= g 9.81 m s 2 = Solution: Just before impact v 2 v 1 2 2gh+= v 2 12.69 m s = Collision Mv 2 I− 0= IMv 2 = I 63.4 N s⋅= *Problem 15-20 The block of mass M is falling downward at speed v 1 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 5kg= v 1 2 m s = h 8m= Δt 0.9 s= g 9.81 m s 2 = Solution: Just before impact v 2 v 1 2 2gh+= v 2 12.69 m s = Collision Mv 2 FΔt− 0= F Mv 2 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 μ s and μ k 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 s 2 = v 2 m s = μ s 0.3= μ k 0.2= a 300 N s = Solution: Guesses t 1 1s= N C 1N= t 2 1s= Given N C Mgcos θ()− 0= at 1 μ s N C − Mgsin θ()− 0= t 1 t 2 tat Mgsin θ()− μ k N C −() ⌠ ⎮ ⌡ d Mv= t 1 t 2 N C ⎛ ⎜ ⎜ ⎜ ⎝ ⎞ ⎟ ⎟ ⎟ ⎠ Find t 1 t 2 , N C ,()= t 1 1.24 s= t 2 1.93 s= 314 Engineering Mechanics - Dynamics Chapter 15 Problem 15-22 The block of weight W has an initial velocity v 1 in the direction shown. If a force F = {f 1 i + f 2 j} acts on the block for time t, determine the final speed of the block. Neglect friction. Given: W 2lb= a 2ft= f 1 0.5 lb= v 1 10 ft s = b 2ft= f 2 0.2 lb= g 32.2 ft s 2 = c 5ft= t 5s= Solution: θ atan b ca− ⎛ ⎜ ⎝ ⎞ ⎟ ⎠ = Guesses v 2x 1 ft s = v 2y 1 ft s = Given W g ⎛ ⎜ ⎝ ⎞ ⎟ ⎠ v 1 sin θ()− cos θ() ⎛ ⎜ ⎝ ⎞ ⎟ ⎠ f 1 f 2 ⎛ ⎜ ⎝ ⎞ ⎟ ⎠ t+ W g ⎛ ⎜ ⎝ ⎞ ⎟ ⎠ v 2x v 2y ⎛ ⎜ ⎝ ⎞ ⎟ ⎠ = v 2x v 2y ⎛ ⎜ ⎝ ⎞ ⎟ ⎠ Find v 2x v 2y ,()= v 2x v 2y ⎛ ⎜ ⎝ ⎞ ⎟ ⎠ 34.7 24.4 ⎛ ⎜ ⎝ ⎞ ⎟ ⎠ ft s = v 2x v 2y ⎛ ⎜ ⎝ ⎞ ⎟ ⎠ 42.4 ft s = Problem 15-23 The tennis ball has a horizontal speed v 1 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: v 1 15 m s = θ 25 deg= h 10 m= M 180 gm= g 9.81 m s 2 = 315 Engineering Mechanics - Dynamics Chapter 15 Solution: Free flight v 2 sin θ() 2gh= v 2 2gh sin θ() = v 2 33.14 m s = Impulse - momentum M− v 1 I x + Mv 2 cos θ()= I x Mv 2 cos θ() v 1 +()= I x 8.11 N s⋅= 0 I y + Mv 2 sin θ()= I y Mv 2 sin θ()= I y 2.52 N s⋅= II x 2 I y 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 F 1 and F 2 . If these loadings vary in the manner shown on the graph, determine the speed of the block at t = t 3 . Neglect friction and the mass of the pulleys and cords. Given: M 40 kg= v 1.5 m s = t 3 6s= t 2 4s= t 1 2s= P 1 10 N= P 2 20 N= P 3 30 N= P 4 40 N= Solution: The impulses acting on the block are found from the areas under the graph. I 4 P 3 t 2 P 1 t 3 t 2 −()+ ⎡ ⎣ ⎤ ⎦ P 1 t 1 P 2 t 2 t 1 −()+ P 4 t 3 t 2 −()+ ⎡ ⎣ ⎤ ⎦ −= Mv I+ Mv 3 = v 3 v I M += v 3 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: W A 2lb= W B 4lb= t 2s= g 32.2 ft s 2 = Solution: 2s A 2s B + L= v A v B −= Block A 02TW A −()t+ W A g v A = Block B 02TW B −()t+ W B g v A −()= Combining W B W A −()t W B W A + g ⎛ ⎜ ⎝ ⎞ ⎟ ⎠ v A = v A W B W A − W B W A + ⎛ ⎜ ⎝ ⎞ ⎟ ⎠ gt= v B v A −= v A v B ⎛ ⎜ ⎝ ⎞ ⎟ ⎠ 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 μ k . Determine the total time of travel before the package stops sliding. Neglect the size of the package. Given: M 5kg= h 3m= θ 30 deg= g 9.81 m s 2 = μ k 0.2= 317 Engineering Mechanics - Dynamics Chapter 15 Solution: On the slope v 1 2gh= v 1 7.67 m s = t 1 v 1 g sin θ() = t 1 1.56 s= On the flat Mv 1 μ k Mgt 2 − 0= t 2 v 1 μ k g = t 2 3.91 s= tt 1 t 2 += t 5.47 s= Problem 15-27 Block A has weight W A and block B has weight W B . If B is moving downward with a velocity v B0 at t = 0, determine the velocity of A when t = t 1 . Assume that block A slides smoothly. Given: W A 10 lb= W B 3lb= v B0 3 ft s = t 1 1s= g 32.2 ft s 2 = Solution: s A 2s B + L= v A 2− v B = Guess v A1 1 ft s = T 1lb= Given Block A W A g ⎛ ⎜ ⎝ ⎞ ⎟ ⎠ 2v B0 Tt 1 + W A g ⎛ ⎜ ⎝ ⎞ ⎟ ⎠ v A1 = Block B W B − g ⎛ ⎜ ⎝ ⎞ ⎟ ⎠ v B0 2Tt 1 + W B t 1 − W B − g ⎛ ⎜ ⎝ ⎞ ⎟ ⎠ v A1 2 ⎛ ⎜ ⎝ ⎞ ⎟ ⎠ = v A1 T ⎛ ⎜ ⎝ ⎞ ⎟ ⎠ Find v A1 T,()= T 1.40 lb= v A1 10.49 ft s = 318 Engineering Mechanics - Dynamics Chapter 15 *Problem 15-28 Block A has weight W A and block B has weight W B . If B is moving downward with a velocity v B1 at t = 0, determine the velocity of A when t = t 1 . The coefficient of kinetic friction between the horizontal plane and block A is μ k . Given: W A 10 lb= W B 3lb= v B1 3 ft s = μ k 0.15= t 1 1s= g 32.2 ft s 2 = Solution: s A 2s B + L= v A 2− v B = Guess v A2 1 ft s = T 1lb= Given Block A W A g ⎛ ⎜ ⎝ ⎞ ⎟ ⎠ 2v B1 Tt 1 + μ k W A t 1 − W A g ⎛ ⎜ ⎝ ⎞ ⎟ ⎠ v A2 = Block B W B − g ⎛ ⎜ ⎝ ⎞ ⎟ ⎠ v B1 2Tt 1 + W B t 1 − W B − g v A2 2 ⎛ ⎜ ⎝ ⎞ ⎟ ⎠ = v A2 T ⎛ ⎜ ⎝ ⎞ ⎟ ⎠ Find v A2 T,()= T 1.50 lb= v A2 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 10 3 kg= kN 10 3 N= 319 Engineering Mechanics - Dynamics Chapter 15 Given: M 7Mg= t 1 2s= v 40 km hr = t 2 5s= F 1 5kN= t 5s= F 2 15 kN= Solution: The impulse exerted on the plane is equal to the area under the graph. Mv 1 2 F 1 t 1 + 1 2 F 1 F 2 +()t 2 t 1 −()+ Mv 1 = v 1 v 1 2M F 1 t 1 F 1 F 2 +()t 2 t 1 −()+ ⎡ ⎣ ⎤ ⎦ += v 1 16.11 m s = Problem 15-30 The motor pulls on the cable at A with a force F = a + bt 2 . If the crate of weight W is originally at rest at t = 0, determine its speed at time t = t 2 . 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 s 2 = t 2 4s= Solution: 1 2 abt 1 2 +()W− 0= t 1 2Wa− b = t 1 2.00 s= 1 2 t 1 t 2 tabt 2 + ⌠ ⎮ ⌡ d Wt 2 t 1 −()− W g ⎛ ⎜ ⎝ ⎞ ⎟ ⎠ v 2 = v 2 g 2W t 1 t 2 tabt 2 + ⌠ ⎮ ⌡ d gt 2 t 1 −()−= v 2 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 μ s and μ k 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 = t 2 . Originally the tension in the cable is zero. Hint: First determine the force needed to begin moving the log. Given: M 500 kg= t 1 3s= μ s 0.5= T 1 1800 N= μ k 0.4= g 9.81 m s 2 = t 2 5s= Solution: To begin motion we need 2T 1 t 0 2 t 1 2 ⎛ ⎜ ⎜ ⎝ ⎞ ⎟ ⎟ ⎠ μ s Mg= t 0 μ s Mg 2T 1 t 1 = t 0 2.48 s= Impulse - Momentum 0 t 0 t 1 t2T 1 t t 1 ⎛ ⎜ ⎝ ⎞ ⎟ ⎠ 2 ⌠ ⎮ ⎮ ⎮ ⌡ d+ 2T 1 t 2 t 1 −()+ μ k Mg t 2 t 0 −()− Mv 2 = v 2 1 M t 0 t 1 t2T 1 t t 1 ⎛ ⎜ ⎝ ⎞ ⎟ ⎠ 2 ⌠ ⎮ ⎮ ⎮ ⌡ d2T 1 t 2 t 1 −()+ μ k Mg t 2 t 0 −()− ⎡ ⎢ ⎢ ⎢ ⎣ ⎤ ⎥ ⎥ ⎥ ⎦ = v 2 7.65 m s = *Problem 15-32 A railroad car having mass m 1 is coasting with speed v 1 on a horizontal track. At the same time another car having mass m 2 is coasting with speed v 2 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 10 3 kg= kJ 10 3 J= Given: m 1 15 Mg= m 2 12 Mg= 321 Engineering Mechanics - Dynamics Chapter 15 v 1 1.5 m s = v 2 0.75 m s = Solution: m 1 v 1 m 2 v 2 − m 1 m 2 +()v= v m 1 v 1 m 2 v 2 − m 1 m 2 + = v 0.50 m s = T 1 1 2 m 1 v 1 2 1 2 m 2 v 2 2 += T 1 20.25 kJ= T 2 1 2 m 1 m 2 +()v 2 = T 2 3.38 kJ= ΔTT 2 T 1 −= ΔT 16.88− kJ= ΔT− T 1 100 83.33= % loss The energy is dissipated as noise, shock, and heat during the coupling. Problem 15-33 Car A has weight W A and is traveling to the right at speed v A Meanwhile car B of weight W B is traveling at speed v B 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: W A 4500 lb= W B 3000 lb= v A 3 ft s = v B 6 ft s = g 32.2 ft s 2 = Solution: W A g ⎛ ⎜ ⎝ ⎞ ⎟ ⎠ v A W B g ⎛ ⎜ ⎝ ⎞ ⎟ ⎠ v B − W A W B + g ⎛ ⎜ ⎝ ⎞ ⎟ ⎠ v= v W A v A W B v B − W A W B + = v 0.60− ft s = Problem 15-34 The bus B has weight W B and is traveling to the right at speed v B . Meanwhile car A of weight W A is traveling at speed v A 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: W B 15000 lb= v B 5 ft s = W A 3000 lb= v A 4 ft s = g 32.2 ft s 2 = Solution: W B g ⎛ ⎜ ⎝ ⎞ ⎟ ⎠ v B W A g ⎛ ⎜ ⎝ ⎞ ⎟ ⎠ v A − W B W A + g ⎛ ⎜ ⎝ ⎞ ⎟ ⎠ v= v W B v B W A v A − W B W A + = 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 M 1 out the back with a horizontal velocity v bc measured relative to the cart. Determine the final velocity of the cart. Given: M 3kg= h 1.25 m= M 1 0.5 kg= g 9.81 m s 2 = v bc 0.6 m s = Solution: v 1 2gh= MM 1 +()v 1 Mv c M 1 v c v bc −()+= v c v 1 M 1 MM 1 + ⎛ ⎜ ⎝ ⎞ ⎟ ⎠ v bc += v c 5.04 m s = 323 Engineering Mechanics - Dynamics Chapter 15 *Problem 15-36 Two men A and B, each having weight W m , stand on the cart of weight W c . 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: W m 160 lb= W c 200 lb= v 3 ft s = g 32.2 ft s 2 = Solution: m m W m g = m c W c g = (a) A jumps first 0 m m − vv c −()m m m c +()v c1 += v c1 m m v m c 2m m + = v c1 0.923 ft s = And then B jumps m m m c +()v c1 m m − vv c2 −()m c v c2 += v c2 m m vm m m c +()v c1 + m m m c + = v c2 2.26 ft s = (b) Both men jump at the same time 02− m m vv c3 −()m c v c3 += v c3 2m m v 2m m m c + = v c3 1.85 ft s = Problem 15-37 A box of weight W 1 slides from rest down the smooth ramp onto the surface of a cart of weight W 2 . 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: W 1 40 lb= W 2 20 lb= h 15 ft= g 32.2 ft s 2 = Solution: v 1 2gh= W 1 g v 1 W 1 W 2 + g ⎛ ⎜ ⎝ ⎞ ⎟ ⎠ v 2 = v 2 W 1 W 1 W 2 + ⎛ ⎜ ⎝ ⎞ ⎟ ⎠ v 1 = v 2 20.7 ft s = W 1 g ⎛ ⎜ ⎝ ⎞ ⎟ ⎠ v 1 Imp− 0= Imp W 1 g ⎛ ⎜ ⎝ ⎞ ⎟ ⎠ v 1 = Imp 38.6 lb s⋅= Problem 15-38 A boy of weight W 1 walks forward over the surface of the cart of weight W 2 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: W 1 100 lb= W 2 60 lb= v 3 ft s = d 6ft= Solution: 0 W 1 g ⎛ ⎜ ⎝ ⎞ ⎟ ⎠ v c v+() W 2 g ⎛ ⎜ ⎝ ⎞ ⎟ ⎠ v c += v c W 1 W 1 W 2 + − v= v c 1.88− ft s = Assuming that the boy walks the distance d t d v = s c v c t= s c 3.75− ft= Problem 15-39 The barge B has weight W B and supports an automobile weighing W a . 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: W B 30000 lb= W a 3000 lb= d 200 ft= g 32.2 ft s 2 = Solution: m B W B g = m a W a 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. m B v B m a vv B +()+ 0= v B m a − v m B m a + = t d v = s B v B − t= s B m a d m a m B + = s B 18.18 ft= *Problem 15-40 A bullet of weight W 1 traveling at speed v 1 strikes the wooden block of weight W 2 and exits the other side at speed v 2 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 μ k . Given: W 1 0.03 lb= a 3ft= b 4ft= W 2 10 lb= c 5ft= v 1 1300 ft s = d 12 ft= v 2 50 ft s = μ k 0.5= Solution: W 1 g ⎛ ⎜ ⎝ ⎞ ⎟ ⎠ v 1 d c 2 d 2 + ⎛ ⎜ ⎝ ⎞ ⎟ ⎠ W 2 g ⎛ ⎜ ⎝ ⎞ ⎟ ⎠ v B W 1 g ⎛ ⎜ ⎝ ⎞ ⎟ ⎠ v 2 b a 2 b 2 + ⎛ ⎜ ⎝ ⎞ ⎟ ⎠ += 326 Engineering Mechanics - Dynamics Chapter 15 v B W 1 W 2 v 1 d c 2 d 2 + v 2 b a 2 b 2 + − ⎛ ⎜ ⎝ ⎞ ⎟ ⎠ = v B 3.48 ft s = 1 2 W 2 g ⎛ ⎜ ⎝ ⎞ ⎟ ⎠ v B 2 μ k W 2 d− 0= d v B 2 2gμ k = d 0.38 ft= Problem 15-41 A bullet of weight W 1 traveling at v 1 strikes the wooden block of weight W 2 and exits the other side at v 2 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 μ k . Units Used: ms 10 3− s= Given: W 1 0.03 lb= a 3ft= W 2 10 lb= b 4ft= μ k 0.5= c 5ft= Δt 1ms= d 12 ft= v 1 1300 ft s = v 2 50 ft s = Solution: W 1 g v 1 d c 2 d 2 + ⎛ ⎜ ⎝ ⎞ ⎟ ⎠ W 2 g v B W 1 g v 2 b a 2 b 2 + ⎛ ⎜ ⎝ ⎞ ⎟ ⎠ += v B W 1 W 2 v 1 d c 2 d 2 + v 2 b a 2 b 2 + − ⎛ ⎜ ⎝ ⎞ ⎟ ⎠ = v B 3.48 ft s = W 1 − g v 1 c c 2 d 2 + ⎛ ⎜ ⎝ ⎞ ⎟ ⎠ NW 2 −()Δt+ W 1 g v 2 a a 2 b 2 + ⎛ ⎜ ⎝ ⎞ ⎟ ⎠ = N W 1 gΔt v 2 a a 2 b 2 + v 1 c c 2 d 2 + + ⎛ ⎜ ⎝ ⎞ ⎟ ⎠ W 2 += N 503.79 lb= W 2 g ⎛ ⎜ ⎝ ⎞ ⎟ ⎠ v B μ k W 2 t− 0= t v B gμ k = t 0.22 s= 327 Engineering Mechanics - Dynamics Chapter 15 Problem 15-42 The man M has weight W M and jumps onto the boat B which has weight W B . 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 v B away from the pier when he makes the jump, determine the resulting velocity of the man and boat. Given: W M 150 lb= v B 2 ft s = W B 200 lb= g 32.2 ft s 2 = v 3 ft s = Solution: W M g vv B +() W B g v B + W M W B + g ⎛ ⎜ ⎝ ⎞ ⎟ ⎠ v'= v' W M vW M W B +()v B + W M W B + = v' 3.29 ft s = Problem 15-43 The man M has weight W M 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: W M 150 lb= v 3 ft s = v' 2 ft s = g 32.2 ft s 2 = Solution: W M g ⎛ ⎜ ⎝ ⎞ ⎟ ⎠ v W M W B + g ⎛ ⎜ ⎝ ⎞ ⎟ ⎠ v'= W B vv'− v' ⎛ ⎜ ⎝ ⎞ ⎟ ⎠ W M = W B 75.00 lb= 328 Engineering Mechanics - Dynamics Chapter 15 *Problem 15-44 A boy A having weight W A and a girl B having weight W B stand motionless at the ends of the toboggan, which has weight W t . 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: W A 80 lb= W B 65 lb= W t 20 lb= d 4ft= Solution: The center of mass doesn’t move during the motion since there is no friction and therefore no net horizontal force W B dW A W B + W t +()d'= d' W B d W A W B + W t + = d' 1.58 ft= Problem 15-45 The projectile of weight W is fired from ground level with initial velocity v A 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 v 1 , determine the distance between the fragments after they strike the ground. Neglect the size of the gun. Given: W 10 lb= v A 80 ft s = v 1 12 ft s = θ 60 deg= g 32.2 ft s 2 = Solution: At the top vv A cos θ()= 329 Engineering Mechanics - Dynamics Chapter 15 Explosion W g ⎛ ⎜ ⎝ ⎞ ⎟ ⎠ v 0 W 2g ⎛ ⎜ ⎝ ⎞ ⎟ ⎠ v 2x += v 2x 2v= v 2x 80.00 ft s = 0 W 2g ⎛ ⎜ ⎝ ⎞ ⎟ ⎠ v 1 W 2g ⎛ ⎜ ⎝ ⎞ ⎟ ⎠ v 2y −= v 2y v 1 = v 2y 12.00 ft s = Kinematics h v A sin θ()() 2 2g = h 74.53 ft= Guess t 1s= Given 0 g− 2 ⎛ ⎜ ⎝ ⎞ ⎟ ⎠ t 2 v 2y t− h+= t Find t()= t 1.81 s= dv 2x t= d 144.9 ft= Problem 15-46 The projectile of weight W is fired from ground level with an initial velocity v A 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= v A 80 ft s = g 32.2 ft s 2 = d 150 ft= Solution: h v A sin θ()() 2 2g = Guesses v 1 1 ft s = v 2x 1 ft s = v 2y 1 ft s = t 1s= Given W g ⎛ ⎜ ⎝ ⎞ ⎟ ⎠ v A cos θ() W 2g ⎛ ⎜ ⎝ ⎞ ⎟ ⎠ v 2x = 0 W 2g ⎛ ⎜ ⎝ ⎞ ⎟ ⎠ v 1 W 2g ⎛ ⎜ ⎝ ⎞ ⎟ ⎠ v 2y += dv 2x t= 0 h 1 2 gt 2 − v 2y t+= 330 Engineering Mechanics - Dynamics Chapter 15 v 1 v 2x v 2y t ⎛ ⎜ ⎜ ⎜ ⎜ ⎝ ⎞ ⎟ ⎟ ⎟ ⎟ ⎠ Find v 1 v 2x , v 2y , t,()= t 1.87 s= v 1 v 2x v 2y ⎛ ⎜ ⎜ ⎜ ⎝ ⎞ ⎟ ⎟ ⎟ ⎠ 9.56 80.00 9.56− ⎛ ⎜ ⎜ ⎝ ⎞ ⎟ ⎟ ⎠ ft s = v 1 9.56 ft s = v 2x v 2y ⎛ ⎜ ⎝ ⎞ ⎟ ⎠ 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 v rel . If both the car B of mass M B and the jeep A of mass M A 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 10 3 kg= Given: M A 2.5 Mg= v rel 2 m s = M B 1.25 Mg= L 5m= Solution: Guess v A 1 m s = v B 1 m s = Given 0 M A v A M B v B += v A v B − v rel = v A v B ⎛ ⎜ ⎝ ⎞ ⎟ ⎠ Find v A v B ,()= t L v rel = t 2.50 s= v A v B ⎛ ⎜ ⎝ ⎞ ⎟ ⎠ 0.67 1.33− ⎛ ⎜ ⎝ ⎞ ⎟ ⎠ m s = *Problem 15-48 The block of mass M a is held at rest on the smooth inclined plane by the stop block at A. If the bullet of mass M b is traveling at speed v when it becomes embedded in the block of mass M c , determine the distance the block will slide up along the plane before momentarily stopping. 331 Engineering Mechanics - Dynamics Chapter 15 Given: M a 10 kg= v 300 m s = M b 10 gm= θ 30 deg= M c 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 M b v bx M b M a +()v x = M b v cos θ() M b M a +()v x = v x M b v cos θ() M b M a + ⎛ ⎜ ⎝ ⎞ ⎟ ⎠ = v x 0.2595 m s = Conservation of Energy: The datum is set at the block’s initial position. When the block and the embedded bullet are at their highest point they are a distance h above the datum. Their gravitational potential energy is (M a + M b )gh. Applying Eq. 14-21, we have 0 1 2 M a M b +()v x 2 + 0 M a M b +()gh+= h 1 2 v x 2 g ⎛ ⎜ ⎝ ⎞ ⎟ ⎠ = h 3.43 mm= d h sin θ() = d 6.86 mm= Problem 15-49 A tugboat T having mass m T is tied to a barge B having mass m B . If the rope is “elastic” such that it has stiffness k, determine the maximum stretch in the rope during the initial towing. Originally both the tugboat and barge are moving in the same direction with speeds v T1 and v B1 respectively. Neglect the resistance of the water. Units Used: Mg 10 3 kg= kN 10 3 N= 332 Engineering Mechanics - Dynamics Chapter 15 Given: m T 19 Mg= v B1 10 km hr = m B 75 Mg= v T1 15 km hr = k 600 kN m = g 9.81 m s 2 = Solution: At maximum stretch the velocities are the same. Guesses v 2 1 km hr = δ 1m= Given momentum m T v T1 m B v B1 + m T m B +()v 2 = energy 1 2 m T v T1 2 1 2 m B v B1 2 + 1 2 m T m B +()v 2 2 1 2 kδ 2 += v 2 δ ⎛ ⎜ ⎝ ⎞ ⎟ ⎠ Find v 2 δ,()= v 2 11.01 km hr = δ 0.221 m= Problem 15-50 The free-rolling ramp has a weight W r . The crate, whose weight is W c , slides a distance d from rest at A, down the ramp to B. Determine the ramp’s speed when the crate reaches B. Assume that the ramp is smooth, and neglect the mass of the wheels. Given: W r 120 lb= a 3= b 4= W c 80 lb= d 15 ft= g 32.2 ft s 2 = Solution: θ atan a b ⎛ ⎜ ⎝ ⎞ ⎟ ⎠ = Guesses v r 1 ft s = v cr 1 ft s = 333 Engineering Mechanics - Dynamics Chapter 15 Given W c d sin θ() 1 2 W r g ⎛ ⎜ ⎝ ⎞ ⎟ ⎠ v r 2 1 2 W c g ⎛ ⎜ ⎝ ⎞ ⎟ ⎠ v r v cr cos θ()−() 2 v cr sin θ()() 2 + ⎡ ⎣ ⎤ ⎦ += 0 W r g ⎛ ⎜ ⎝ ⎞ ⎟ ⎠ v r W c g ⎛ ⎜ ⎝ ⎞ ⎟ ⎠ v r v cr cos θ()−()+= v r v cr ⎛ ⎜ ⎝ ⎞ ⎟ ⎠ Find v r v cr ,()= v cr 27.9 ft s = v r 8.93 ft s = Problem 15-51 The free-rolling ramp has a weight W r . If the crate, whose weight is W c , is released from rest at A, determine the distance the ramp moves when the crate slides a distance d down the ramp and reaches the bottom B. Given: W r 120 lb= a 3= b 4= W c 80 lb= d 15 ft= g 32.2 ft s 2 = Solution: θ atan a b ⎛ ⎜ ⎝ ⎞ ⎟ ⎠ = Momentum 0 W r g ⎛ ⎜ ⎝ ⎞ ⎟ ⎠ v r W c g ⎛ ⎜ ⎝ ⎞ ⎟ ⎠ v r v cr cos θ()−()+= v r W c W c W r + ⎛ ⎜ ⎝ ⎞ ⎟ ⎠ cos θ()v cr = Integrate s r W c W c W r + ⎛ ⎜ ⎝ ⎞ ⎟ ⎠ cos θ()d= s r 4.80 ft= *Problem 15-52 The boy B jumps off the canoe at A with a velocity v BA relative to the canoe as shown. If he lands in the second canoe C, determine the final speed of both canoes after the motion. Each canoe has a mass M c . The boy’s mass is M B , and the girl D has a mass M D . Both canoes are originally at rest. 334 Engineering Mechanics - Dynamics Chapter 15 Given: M c 40 kg= M B 30 kg= M D 25 kg= v BA 5 m s = θ 30 deg= Solution: Guesses v A 1 m s = v C 1 m s = Given 0 M c v A M B v A v BA cos θ()+()+= M B v A v BA cos θ()+()M c M B + M D +()v C = v A v C ⎛ ⎜ ⎝ ⎞ ⎟ ⎠ Find v A v C ,()= v A v C ⎛ ⎜ ⎝ ⎞ ⎟ ⎠ 1.86− 0.78 ⎛ ⎜ ⎝ ⎞ ⎟ ⎠ m s = Problem 15-53 The free-rolling ramp has a mass M r . A crate of mass M c is released from rest at A and slides down d to point B. If the surface of the ramp is smooth, determine the ramp’s speed when the crate reaches B. Also, what is the velocity of the crate? Given: M r 40 kg= M c 10 kg= d 3.5 m= θ 30 deg= g 9.81 m s 2 = Solution: Guesses v c 1 m s = v r 1 m s = v cr 1 m s = Given 0 M c gdsin θ()+ 1 2 M c v c 2 1 2 M r v r 2 += 335 Engineering Mechanics - Dynamics Chapter 15 v r v cr cos θ()+() 2 v cr sin θ()() 2 + v c 2 = 0 M r v r M c v r v cr cos θ()+()+= v c v r v cr ⎛ ⎜ ⎜ ⎜ ⎝ ⎞ ⎟ ⎟ ⎟ ⎠ Find v c v r , v cr ,()= v cr 6.36 m s = v r v c ⎛ ⎜ ⎝ ⎞ ⎟ ⎠ 1.101− 5.430 ⎛ ⎜ ⎝ ⎞ ⎟ ⎠ m s = Problem 15-54 Blocks A and B have masses m A and m B respectively. They are placed on a smooth surface and the spring connected between them is stretched a distance d. If they are released from rest, determine the speeds of both blocks the instant the spring becomes unstretched. Given: m A 40 kg= d 2m= m B 60 kg= k 180 N m = Solution: Guesses v A 1 m s = v B 1− m s = Given momentum 0 m A v A m B v B += energy 1 2 kd 2 1 2 m A v A 2 1 2 m B v B 2 += v A v B ⎛ ⎜ ⎝ ⎞ ⎟ ⎠ Find v A v B ,()= v A v B ⎛ ⎜ ⎝ ⎞ ⎟ ⎠ 3.29 2.19− ⎛ ⎜ ⎝ ⎞ ⎟ ⎠ m s = Problem 15-55 Block A has a mass M A and is sliding on a rough horizontal surface with a velocity v A1 when it makes a direct collision with block B, which has a mass M B and is originally at rest. If the collision is perfectly elastic, determine the velocity of each block just after collision and the distance between the blocks when they stop sliding. The coefficient of kinetic friction between the blocks and the plane is μ k . Given: M A 3kg= g 9.81 m s 2 = M B 2kg= e 1= v A1 2 m s = μ k 0.3= 336 Engineering Mechanics - Dynamics Chapter 15 Solution: Guesess v A2 3 m s = v B2 5 m s = d 2 1m= Given M A v A1 M A v A2 M B v B2 += ev A1 v B2 v A2 −= d 2 v B2 2 v A2 2 − 2gμ k = v A2 v B2 d 2 ⎛ ⎜ ⎜ ⎜ ⎝ ⎞ ⎟ ⎟ ⎟ ⎠ Find v A2 v B2 , d 2 ,()= v A2 v B2 ⎛ ⎜ ⎝ ⎞ ⎟ ⎠ 0.40 2.40 ⎛ ⎜ ⎝ ⎞ ⎟ ⎠ m s = d 2 0.951 m= *Problem 15-56 Disks A and B have masses M A and M B respectively. If they have the velocities shown, determine their velocities just after direct central impact. Given: M A 2kg= v A1 2 m s = M B 4kg= v B1 5 m s = e 0.4= Solution: Guesses v A2 1 m s = v B2 1 m s = Given M A v A1 M B v B1 − M A v A2 M B v B2 += ev A1 v B1 +()v B2 v A2 −= v A2 v B2 ⎛ ⎜ ⎝ ⎞ ⎟ ⎠ Find v A2 v B2 ,()= v A2 v B2 ⎛ ⎜ ⎝ ⎞ ⎟ ⎠ 4.53− 1.73− ⎛ ⎜ ⎝ ⎞ ⎟ ⎠ m s = Problem 15-57 The three balls each have weight W and have a coefficient of restitution e. If ball A is released from rest and strikes ball B and then ball B strikes ball C, determine the velocity of each ball after the second collision has occurred. The balls slide without friction. Given: W 0.5 lb= r 3ft= 337 Engineering Mechanics - Dynamics Chapter 15 e 0.85= g 32.2 ft s 2 = Solution: v A 2gr= Guesses v A' 1 ft s = v B' 1 ft s = v B'' 1 ft s = v C'' 1 ft s = Given W g ⎛ ⎜ ⎝ ⎞ ⎟ ⎠ v A W g ⎛ ⎜ ⎝ ⎞ ⎟ ⎠ v A' W g ⎛ ⎜ ⎝ ⎞ ⎟ ⎠ v B' += ev A v B' v A' −= W g ⎛ ⎜ ⎝ ⎞ ⎟ ⎠ v B' W g ⎛ ⎜ ⎝ ⎞ ⎟ ⎠ v B'' W g ⎛ ⎜ ⎝ ⎞ ⎟ ⎠ v C'' += ev B' v C'' v B'' −= v A' v B' v B'' v C'' ⎛ ⎜ ⎜ ⎜ ⎜ ⎝ ⎞ ⎟ ⎟ ⎟ ⎟ ⎠ Find v A' v B' , v B'' , v C'' ,()= v A' v B'' v C'' ⎛ ⎜ ⎜ ⎜ ⎝ ⎞ ⎟ ⎟ ⎟ ⎠ 1.04 0.96 11.89 ⎛ ⎜ ⎜ ⎝ ⎞ ⎟ ⎟ ⎠ ft s = Problem 15-58 The ball A of weight W A is thrown so that when it strikes the block B of weight W B it is traveling horizontally at speed v. If the coefficient of restitution between A and B is e, and the coefficient of kinetic friction between the plane and the block is μ k , determine the time before block B stops sliding. Given: W A 1lb= μ k 0.4= W B 10 lb= v 20 ft s = g 32.2 ft s 2 = e 0.6= Solution: Guesses v A2 1 ft s = v B2 1 ft s = t 1s= Given W A g ⎛ ⎜ ⎝ ⎞ ⎟ ⎠ v W A g ⎛ ⎜ ⎝ ⎞ ⎟ ⎠ v A2 W B g ⎛ ⎜ ⎝ ⎞ ⎟ ⎠ v B2 += ev v B2 v A2 −= 338 Engineering Mechanics - Dynamics Chapter 15 W B g ⎛ ⎜ ⎝ ⎞ ⎟ ⎠ v B2 μ k W B t− 0= v A2 v B2 t ⎛ ⎜ ⎜ ⎜ ⎝ ⎞ ⎟ ⎟ ⎟ ⎠ Find v A2 v B2 , t,()= v A2 v B2 ⎛ ⎜ ⎝ ⎞ ⎟ ⎠ 9.09− 2.91 ⎛ ⎜ ⎝ ⎞ ⎟ ⎠ ft s = t 0.23 s= Problem 15-59 The ball A of weight W A is thrown so that when it strikes the block B of weight W B it is traveling horizontally at speed v. If the coefficient of restitution between A and B is e, and the coefficient of kinetic friction between the plane and the block is μ k , determine the distance block B slides before stopping. Given: W A 1lb= μ k 0.4= W B 10 lb= v 20 ft s = g 32.2 ft s 2 = e 0.6= Solution: Guesses v A2 1 ft s = v B2 1 ft s = d 1ft= Given W A g ⎛ ⎜ ⎝ ⎞ ⎟ ⎠ v W A g ⎛ ⎜ ⎝ ⎞ ⎟ ⎠ v A2 W B g ⎛ ⎜ ⎝ ⎞ ⎟ ⎠ v B2 += ev v B2 v A2 −= 1 2 W B g ⎛ ⎜ ⎝ ⎞ ⎟ ⎠ v B2 2 μ k W B d− 0= v A2 v B2 d ⎛ ⎜ ⎜ ⎜ ⎝ ⎞ ⎟ ⎟ ⎟ ⎠ Find v A2 v B2 , d,()= v A2 v B2 ⎛ ⎜ ⎝ ⎞ ⎟ ⎠ 9.09− 2.91 ⎛ ⎜ ⎝ ⎞ ⎟ ⎠ ft s = d 0.33 ft= Problem 15-60 The ball A of weight W A is thrown so that when it strikes the block B of weight W B it is traveling horizontally at speed v. Determine the average normal force exerted between A and B if the impact occurs in time Δt. The coefficient of restitution between A and B is e. Given: W A 1lb= μ k 0.4= 339 Engineering Mechanics - Dynamics Chapter 15 W B 10 lb= v 20 ft s = g 32.2 ft s 2 = e 0.6= Δt 0.02 s= Solution: Guesses v A2 1 ft s = v B2 1 ft s = F N 1lb= Given W A g ⎛ ⎜ ⎝ ⎞ ⎟ ⎠ v W A g ⎛ ⎜ ⎝ ⎞ ⎟ ⎠ v A2 W B g ⎛ ⎜ ⎝ ⎞ ⎟ ⎠ v B2 += ev v B2 v A2 −= W A g ⎛ ⎜ ⎝ ⎞ ⎟ ⎠ vF N Δt− W A g ⎛ ⎜ ⎝ ⎞ ⎟ ⎠ v A2 = v A2 v B2 F N ⎛ ⎜ ⎜ ⎜ ⎝ ⎞ ⎟ ⎟ ⎟ ⎠ Find v A2 v B2 , F N ,()= v A2 v B2 ⎛ ⎜ ⎝ ⎞ ⎟ ⎠ 9.09− 2.91 ⎛ ⎜ ⎝ ⎞ ⎟ ⎠ ft s = F N 45.2 lb= Problem 15-61 The man A has weight W A and jumps from rest from a height h onto a platform P that has weight W P . The platform is mounted on a spring, which has stiffness k. Determine (a) the velocities of A and P just after impact and (b) the maximum compression imparted to the spring by the impact. Assume the coefficient of restitution between the man and the platform is e, and the man holds himself rigid during the motion. Given: W A 175 lb= W P 60 lb= k 200 lb ft = h 8ft= e 0.6= g 32.2 ft s 2 = Solution: m A W A g = m P W P g = δ st W P k = Guesses v A1 1 ft s = v A2 1 ft s = v P2 1− ft s = δ 21 ft= 340 dhanesh_h Mathcad - CombinedMathcads

"StudyBlue is great for studying. I love the study guides, flashcards and quizzes. So extremely helpful for all of my classes!"

Alice , Arizona State University"I'm a student using StudyBlue, and I can 100% say that it helps me so much. Study materials for almost every subject in school are available in StudyBlue. It is so helpful for my education!"

Tim , University of Florida"StudyBlue provides way more features than other studying apps, and thus allows me to learn very quickly!??I actually feel much more comfortable taking my exams after I study with this app. It's amazing!"

Jennifer , Rutgers University"I love flashcards but carrying around physical flashcards is cumbersome and simply outdated. StudyBlue is exactly what I was looking for!"

Justin , LSU