MEN 201 Problems For Chapter 14 with Solutions, Spring 2012

Engineering Mechanics - Dynamics Chapter 14 Problem 14-1 A woman having a mass M stands in an elevator which has a downward acceleration a starting from rest. Determine the work done by her weight and the work of the normal force which the floor exerts on her when the elevator descends a distance s. Explain why the work of these forces is different. Units Used: kJ 103J = Given: M = 70 kg g 9.81 m s2 = a 4 m s2 = s 6 m = Solution: M g − Np = M a Np M g − M a = Np 406.7 N = UW = M gs UW 4.12 kJ = UNP s = − Np UNP −2.44 kJ = The difference accounts for a change in kinetic energy. Problem 14-2 The crate of weight W has a velocity vA when it is at A. Determine its velocity after it slides down the plane to s = s'. The coefficient of kinetic friction between the crate and the plane is μk. Given: W = 20 lb a 3 = b = 4 vA 12 ft s = s' 6 ft = μk 0.2 = Solution: θ atan a b ⎛⎜⎝ ⎞⎟⎠ = NC = W cos θ( ) F μk NC = Guess v' 1 m s = Given 1 2 W g ⎛⎜⎝ ⎞⎟⎠ vA 2 + W sin θ( )s' − F s' 1 2 W g ⎛⎜⎝ ⎞⎟⎠ v'2 = v' Find v' ( ) = v' 17.72 ft s = 242 Engineering Mechanics - Dynamics Chapter 14 Problem 14-3 The crate of mass M is subjected to a force having a constant direction and a magnitude F, where s is measured in meters. When s = s1, the crate is moving to the right with a speed v1. Determine its speed when s = s2. The coefficient of kinetic friction between the crate and the ground is μk. Given: M = 20 kg F 100 N = s1 = 4 m θ 30 deg = v1 8 m s = a 1 = s2 = 25 m b 1 m 1 − = μk 0.25 = Solution: Equation of motion: Since the crate slides, the friction force developed between the crate and its contact surface is Ff = μkN N + F sin θ( ) − M g = 0 N M g − F sin θ( ) = Principle of work and Energy: The horizontal component of force F which acts in the direction of displacement does positive work, whereas the friction force Ff μk M g − F sin θ( ) ( ) = does negative work since it acts in the opposite direction to that of displacement. The normal reaction N, the vertical component of force F and the weight of the crate do not displace hence do no work. F cos θ( ) − μk N M a = F cos θ( ) μk M g − F sin θ( ) ( ) − M a = a F cos θ( ) μk M g − F sin θ( ) ( ) − M = a 2.503 m s2 = v dv ds = a v2 2 v1 2 2 a s2 − s1 ( ) + = v 2 v1 2 2 a s2 − s1 ( ) + ⎡ ⎢ ⎣ ⎤ ⎥ ⎦ = v 13.004 m s = 243 Engineering Mechanics - Dynamics Chapter 14 *Problem 14-4 The “air spring” A is used to protect the support structure B and prevent damage to the conveyor-belt tensioning weight C in the event of a belt failure D. The force developed by the spring as a function of its deflection is shown by the graph. If the weight is W and it is suspended a height d above the top of the spring, determine the maximum deformation of the spring in the event the conveyor belt fails. Neglect the mass of the pulley and belt. Given: W = 50 lb k 8000 lb ft2 = d 1.5 ft = Solution: T1 + U T2 = 0 W d ( + δ ) + 0 δ x kx2 ⌠ ⎮⌡ d − 0 = Guess δ 1 in = Given W d ( + δ ) k δ3 3 ⎛⎜ ⎝ ⎞⎟ ⎠ − = 0 δ = Find δ( ) δ 3.896 in = Problem 14-5 A car is equipped with a bumper B designed to absorb collisions. The bumper is mounted to the car using pieces of flexible tubing T. Upon collision with a rigid barrier at A, a constant horizontal force F is developed which causes a car deceleration kg (the highest safe deceleration for a passenger without a seatbelt). If the car and passenger have a total mass M and the car is initially coasting with a speed v, determine the magnitude of F needed to stop the car and the deformation x of the bumper tubing. Units Used: Mm 103 kg = kN 103 N = Given: M 1.5 103 kg = v 1.5 m s = k 3 = Solution: The average force needed to decelerate the car is Favg = M kg Favg 44.1 kN = 244 Engineering Mechanics - Dynamics Chapter 14 The deformation is T1 + U12 = T2 1 2M v2 − Favg x 0 = x 1 2 M v2 Favg ⎛ ⎜ ⎝ ⎞ ⎟ ⎠ = x 38.2 mm = Problem 14-6 The crate of mass M is subjected to forces F1 and F2, as shown. If it is originally at rest, determine the distance it slides in order to attain a speed v. The coefficient of kinetic friction between the crate and the surface is μk. Units Used: kN 103 N = Given: M = 100 kg v 6 m s = F1 = 800 N μk 0.2 = F2 1.5 kN = g 9.81 m s2 = θ1 30 deg = θ2 20 deg = Solution: NC F1 sin θ1 ( ) − − M g F2 sin θ2 ( ) + 0 = NC F1 sin θ1 ( ) M g F2 sin θ2 ( ) − + = NC 867.97 N = T1 + U12 T2 = F1 cos θ1 ( )s − μk Nc s F2 cos θ2 ( )s + 1 2M v2 = s M v2 2 F1 cos θ1 ( ) − μk NC F2 cos θ2 ( ) + ( ) = s 0.933 m = Problem 14-7 Design considerations for the bumper B on the train car of mass M require use of a nonlinear spring having the load-deflection characteristics shown in the graph. Select the proper value of k so that the maximum deflection of the spring is limited to a distance d when the car, traveling at speed v, strikes the rigid stop. Neglect the mass of the car wheels. 245 Engineering Mechanics - Dynamics Chapter 14 Units Used: Mg 103 kg = kN 103 N = MN 103 kN = Given: M 5 Mg = d 0.2 m = v 4 m s = Solution: 1 2M v2 0 d x kx2 ⌠ ⎮⌡ d − = 0 1 2M v2 kd3 3 − = 0 k 3M v2 2d3 = k 15 MN m2 = *Problem 14-8 Determine the required height h of the roller coaster so that when it is essentially at rest at the crest of the hill it will reach a speed v when it comes to the bottom. Also, what should be the minimum radius of curvature ρ for the track at B so that the passengers do not experience a normal force greater than kmg? Neglect the size of the car and passengers. Given: v 100 km hr = k 4 = Solution: T1 + U12 T2 = mgh 1 2mv2 = h 1 2 v2 g = h 39.3 m = kmg − mg mv2 ρ = ρ v2 g k − 1 ( ) = ρ 26.2 m = 246 Engineering Mechanics - Dynamics Chapter 14 Problem 14-9 When the driver applies the brakes of a light truck traveling at speed v1 it skids a distance d1 before stopping. How far will the truck skid if it is traveling at speed v2 when the brakes are applied? Given: v1 40 km hr = d1 3 m = v2 80 km hr = Solution: 1 2M v1 2 − μk M gd1 = 0 μk v1 2 2gd1 = μk 2.097 = 1 2M v2 2 − μk M gd2 = 0 d2 v2 2 2μk g = d2 12 m = Problem 14-10 The ball of mass M of negligible size is fired up the vertical circular track using the spring plunger. The plunger keeps the spring compressed a distance δ when x = 0. Determine how far x it must be pulled back and released so that the ball will begin to leave the track when θ = θ1. Given: M 0.5 kg = δ 0.08 m = θ1 135 deg = r 1.5 m = k 500 N m = g 9.81 m s2 = Solution: N = 0 θ θ1 = ΣFn = man N − M gcos θ( ) M v2 r ⎛⎜ ⎝ ⎞⎟ ⎠ = v g − rcos θ( ) = v 3.226 m s = 247 Engineering Mechanics - Dynamics Chapter 14 Guess x 10 mm = Given x δ + δ x k − x ⌠ ⎮⌡ d M gr 1 ( − cos θ( ) ) − 1 = 2M v2 x Find x ( ) = x 178.9 mm = Problem 14-11 The force F, acting in a constant direction on the block of mass M, has a magnitude which varies with the position x of the block. Determine how far the block slides before its velocity becomes v1. When x = 0, the block is moving to the right at speed v0 . The coefficient of kinetic friction between the block and surface is μk.. Given: M = 20 kg c 3 = v1 5 m s = d 4 = v0 2 m s = k 50 N m2 = μk = 0.3 g 9.81 m s2 = Solution: NB − M g c c2 d2 + ⎛ ⎜ ⎝ ⎞ ⎟ ⎠ kx2 − = 0 NB M g c c2 d2 + ⎛ ⎜ ⎝ ⎞ ⎟ ⎠ kx2 + = Guess δ 2 m = Given 1 2M v0 2 d c2 d2 + ⎛ ⎜ ⎝ ⎞ ⎟ ⎠ 0 δ x kx2 ⌠ ⎮⌡ d + − μk M gδ μk 0 δ x c c2 d2 + ⎛ ⎜ ⎝ ⎞ ⎟ ⎠ kx2 ⌠ ⎮ ⎮ ⌡ d − 1 2M v1 2 = δ = Find δ( ) δ 3.413 m = *Problem 14-12 The force F, acting in a constant direction on the block of mass M, has a magnitude which varies with position x of the block. Determine the speed of the block after it slides a distance d1 . When x = 0, the block is moving to the right at v0 .The coefficient of kinetic friction between the block and surface is μk. 248 Engineering Mechanics - Dynamics Chapter 14 Given: M = 20 kg c 3 = d1 = 3 m d 4 = v0 2 m s = k 50 N m2 = μk = 0.3 g 9.81 m s2 = Solution: NB − M g c c2 d2 + ⎛ ⎜ ⎝ ⎞ ⎟ ⎠ kx2 − = 0 NB M g c c2 d2 + ⎛ ⎜ ⎝ ⎞ ⎟ ⎠ kx2 + = Guess v1 2 m s = Given 1 2M v0 2 d c2 d2 + ⎛ ⎜ ⎝ ⎞ ⎟ ⎠ 0 d1 x kx2 ⌠ ⎮⌡ d + − μk M gd1 μk 0 d1 x c c2 d2 + ⎛ ⎜ ⎝ ⎞ ⎟ ⎠ kx2 ⌠ ⎮ ⎮ ⌡ d − 1 2M v1 2 = v1 Find v1 ( ) = v1 3.774 m s = Problem 14-13 As indicated by the derivation, the principle of work and energy 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 if it has an initial speed of v0 and travels a distance d, both directed to the right and measured from the fixed frame. Compare the result with that obtained by an observer B, attached to the x' axis and moving at a constant velocity of vB relative to A. Hint: The distance the block travels will first have to be computed for observer B before applying the principle of work and energy. Given: M 10 kg = F 6 N = v0 5 m s = d 10 m = vB 2 m s = g 9.81 m s2 = 249 Engineering Mechanics - Dynamics Chapter 14 Solution: Observer A: 1 2M v0 2 + F d 1 2M v2 2 = v2 v0 2 2F d M + = v2 6.083 m s = F = M a a F = M a 0.6 m s2 = Guess t 1s = Given d 0 + v0 t 1 + 2at2 = t = Find t( ) t 1.805 s = Observer B: The distance that the block moves as seen by observer B. d' v0 − vB ( )t 1 + 2at2 = d' 6.391 m = 1 2M v0 − vB ( )2 + F d' 1 2M v'2 2 = v'2 v0 − vB ( )2 2F d' M + = v'2 4.083 m s = Notice that v2 v'2 + vB = Problem 14-14 Determine the velocity of the block A of weight WA if the two blocks are released from rest and the block B of weight WB moves a distance d up the incline. The coefficient of kinetic friction between both blocks and the inclined planes is μk. Given: WA 60 lb = WB 40 lb = θ1 60 deg = θ2 30 deg = d 2 ft = μk 0.10 = Solution: L 2sA + sB = 0 2vA + vB = Guesses vA 1 ft s = vB −1 ft s = 250 Engineering Mechanics - Dynamics Chapter 14 Given 0 2vA + vB = WA d 2 ⎛⎜⎝ ⎞⎟⎠ sin θ1 ( ) WBdsin θ2 ( ) − μk WAcos θ1 ( ) d 2 − −μk WBcos θ2 ( )d + ... 1 2g WAvA 2 WBvB 2 + ( ) = vA vB ⎛ ⎜ ⎝ ⎞ ⎟ ⎠ Find vA vB , ( ) = vB −1.543 ft s = vA 0.771 ft s = Problem 14-15 Block A has weight WA and block B has weight WB. Determine the speed of block A after it moves a distance d down the plane, starting from rest. Neglect friction and the mass of the cord and pulleys. Given: WA = 60 lb e 3 = WB = 10 lb f 4 = d = 5 ft g 32.2 ft s2 = Solution: L 2sA + sB = 0 2ΔsA + ΔsB = 0 2vA + vB = 0 WA e e2 f 2 + ⎛ ⎜ ⎝ ⎞ ⎟ ⎠ d + − WB2d 1 2 WA g ⎛⎜⎝ ⎞⎟⎠ vA 2 1 2 WB g ⎛⎜⎝ ⎞⎟⎠ (2vA )2 + = vA 2gd WA + 4WB WA e e2 f 2 + − 2WB ⎛ ⎜ ⎝ ⎞ ⎟ ⎠ = vA 7.178 ft s = *Problem 14-16 The block A of weight WA rests on a surface for which the coefficient of kinetic friction is μk. Determine the distance the cylinder B of weight WB must descend so that A has a speed vA starting from rest. 251 Engineering Mechanics - Dynamics Chapter 14 Given: WA 3 lb = WB 8 lb = μk 0.3 = vA 5 ft s = Solution: L sA + 2sB = Guesses d 1 ft = Given WBd − μk WA2d 1 2g WAvA 2 WB vA 2 ⎛⎜⎝ ⎞⎟⎠ 2 + ⎡ ⎢ ⎣ ⎤ ⎥ ⎦ = d Find d ( ) = d 0.313 ft = Problem 14-17 The block of weight W slides down the inclined plane for which the coefficient of kinetic friction is μk. If it is moving at speed v when it reaches point A, determine the maximum deformation of the spring needed to momentarily arrest the motion. Given: W = 100 lb a 3 m = b = 4 m v 10 ft s = d 10 ft = k 200 lb ft = μk 0.25 = Solution: N b a2 b2 + ⎛ ⎜ ⎝ ⎞ ⎟ ⎠ W = N 80lb = Initial Guess dmax 5 m = 252 Engineering Mechanics - Dynamics Chapter 14 Given 1 2 W g ⎛⎜⎝ ⎞⎟⎠ v2 μk N d ( + dmax ) − 1 2kdmax 2 − W d ( + dmax ) a a2 b2 + ⎛ ⎜ ⎝ ⎞ ⎟ ⎠ + 0 = dmax Find dmax ( ) = dmax 2.56 ft = Problem 14-18 The collar has mass M and rests on the smooth rod. Two springs are attached to it and the ends of the rod as shown. Each spring has an uncompressed length l. If the collar is displaced a distance s = s' and released from rest, determine its velocity at the instant it returns to the point s = 0. Given: M = 20 kg k 50 N m = s' 0.5 m = k' 100 N m = l 1 m = d 0.25 m = Solution: 1 2ks'2 1 + 2k' s'2 1 2M vc 2 = vc k + k' M ⋅s' = vc 1.37 m s = Problem 14-19 The block of mass M is subjected to a force having a constant direction and a magnitude F = k/(a+bx). When x = x1, the block is moving to the left with a speed v1. Determine its speed when x = x2. The coefficient of kinetic friction between the block and the ground is μk . Given: M = 2 kg b 1 m 1 − = x2 = 12 m g 9.81 m s2 = k = 300 N x1 = 4 m θ 30 deg = a = 1 v1 8 m s = μk 0.25 = 253 Engineering Mechanics - Dynamics Chapter 14 Solution: NB − M g k a bx + ⎛⎜⎝ ⎞⎟⎠ sin θ( ) − = 0 NB M g ksin θ( ) a + bx + = U x1 x2 x kcos θ( ) a bx + ⌠⎮ ⎮⌡ d μk x1 x2 x M g ksin θ( ) a + bx + ⌠⎮ ⎮⌡ d − = U 173.177 N m ⋅ = 1 2M v1 2 + U 1 2M v2 2 = v2 v1 2 2U + M = v2 15.401 m s = *Problem 14-20 The motion of a truck is arrested using a bed of loose stones AB and a set of crash barrels BC. If experiments show that the stones provide a rolling resistance Ft per wheel and the crash barrels provide a resistance as shown in the graph, determine the distance x the truck of weight W penetrates the barrels if the truck is coasting at speed v0 when it approaches A. Neglect the size of the truck. Given: Ft = 160 lb d 50 ft = W = 4500 lb k 1000 lb ft3 = v0 60 ft s = g 32.2 ft s2 = Solution: 1 2 W g ⎛⎜⎝ ⎞⎟⎠ v0 2 − 4Ft d k x4 4 − 0 = x 2W v0 2 kg 16Ft d k − ⎛ ⎜ ⎝ ⎞ ⎟ ⎠ 1 4 = x 5.444 ft = Problem 14-21 The crash cushion for a highway barrier consists of a nest of barrels filled with an impact-absorbing material. The barrier stopping force is measured versus the vehicle penetration into the barrier. Determine the distance a car having weight W will penetrate the barrier if it is originally traveling at speed v0 when it strikes the first barrel. 254 Engineering Mechanics - Dynamics Chapter 14 Units Used: kip 103 lb = Given: W 4000 lb = v0 55 ft s = g 32.2 ft s2 = Solution: 1 2 W g ⎛⎜⎝ ⎞⎟⎠ v0 2 − Area 0 = Area 1 2 W g ⎛⎜⎝ ⎞⎟⎠ v0 2 = Area 187.888 kip ft ⋅ = We must produce this much work with the barrels. Assume that 5 ft < x < 15 ft Area (2 ft ) 9 kip ( ) (3 ft ) 18 kip ( ) + x − 5 ft ( ) 27 kip ( ) + = x Area 72 kip ft ⋅ − 27 kip + 5 ft = x = 9.292 ft Check that the assumption is corrrect! Problem 14-22 The collar has a mass M and is supported on the rod having a coefficient of kinetic friction μk. The attached spring has an unstretched length l and a stiffness k. Determine the speed of the collar after the applied force F causes it to be displaced a distance s = s1 from point A. When s = 0 the collar is held at rest. Given: M = 30 kg μk 0.4 = a = 0.5 m θ 45 deg = F = 200 N s1 1.5 m = l 0.2 m = g 9.81 m s2 = k 50 N m = 255 Engineering Mechanics - Dynamics Chapter 14 Solution: Guesses NC = 1 N v 1 m s = Given NC − M g + F sin θ( ) 0 = F cos θ( )s1 − μk NC s1 1 2k a − l ( )2 + 1 2k s1 + a − l ( )2 − 1 2M v2 = NC ⎛⎜ v ⎝ ⎞⎟ ⎠ Find NC v , ( ) = NC = 152.9 N v 1.666 m s = Problem 14-23 The block of weight W is released from rest at A and slides down the smooth circular surface AB. It then continues to slide along the horizontal rough surface until it strikes the spring. Determine how far it compresses the spring before stopping. Given: W = 5 lb μk 0.2 = a = 3 ft θ 90 deg = b = 2 ft k 40 lb ft = Solution: Guess d 1 ft = Given W a μk W b + d ( ) − 1 − 2kd2 0 = d Find d ( ) = d 0.782 ft = *Problem 14-24 The block has a mass M and moves within the smooth vertical slot. If it starts from rest when the attached spring is in the unstretched position at A, determine the constant vertical force F which must be applied to the cord so that the block attains a speed vB when it reaches sB. Neglect the size and mass of the pulley. Hint: The work of F can be determined by finding the difference Δl in cord lengths AC and BC and using UF = F Δl. Given: M = 0.8 kg l 0.4 m = vB 2.5 m s = b 0.3 m = 256 Engineering Mechanics - Dynamics Chapter 14 sB = 0.15 m k 100 N m = Solution: Δl l2 + b2 l ( − sB )2 + b2 − = Guess F 1 N = Given FΔl − M gsB 1 2ksB 2 − 1 2M vB 2 = F Find F ( ) = F 43.9 N = Problem 14-25 The collar has a mass M and is moving at speed v1 when x = 0 and a force of F is applied to it. The direction θ of this force varies such that θ = ax, where θ is clockwise, measured in degrees. Determine the speed of the collar when x = x1. The coefficient of kinetic friction between the collar and the rod is μk. Given: M = 5 kg v1 8 m s = F 60 N = g 9.81 m s2 = μk 0.3 = x1 = 3 m a 10 deg m = Solution: N F sin θ( ) + M g = Guess v 5 m s = Given 1 2M v1 2 0 x1 x F cos ax ( ) ⌠ ⎮⌡ d + μk 0 x1 x F sin ax ( ) + M g ⌠ ⎮⌡ d − 1 2M v2 = v Find v ( ) = v 10.47 m s = 257 Engineering Mechanics - Dynamics Chapter 14 Problem 14-26 Cylinder A has weight WA and block B has weight WB. Determine the distance A must descend from rest before it obtains speed vA. Also, what is the tension in the cord supporting block A? Neglect the mass of the cord and pulleys. Given: WA = 60 lb vA 8 ft s = WB = 10 lb g 32.2 ft s2 = Solution: L 2sA + sB = 0 2vA + vB = System 0 + WAd − WB2d 1 2 WA g ⎛⎜⎝ ⎞⎟⎠ vA 2 1 2 WB g ⎛⎜⎝ ⎞⎟⎠ (2vA )2 + = d WA 4WB + 2g ⎛⎜⎝ ⎞⎟⎠ vA 2 WA − 2WB = d 2.484 ft = Block A alone 0 + WAd − T d 1 2 WA g ⎛⎜⎝ ⎞⎟⎠ vA 2 = T WA WAvA 2 2gd − = T 36lb = Problem 14-27 The conveyor belt delivers crate each of mass M to the ramp at A such that the crate’s velocity is vA, directed down along the ramp. If the coefficient of kinetic friction between each crate and the ramp is μk, determine the speed at which each crate slides off the ramp at B. Assume that no tipping occurs. Given: M 12 kg = vA 2.5 m s = μk 0.3 = g 9.81 m s2 = θ 30 deg = a 3 m = 258 Engineering Mechanics - Dynamics Chapter 14 Solution: Nc M gcos θ( ) = 1 2M vA 2 (M ga )sin θ( ) + − μk Nc a 1 2M vB 2 = vB vA 2 (2ga )sin θ( ) + (2μk g )cos θ( )a − = vB 4.52 m s = *Problem 14-28 When the skier of weight W is at point A he has a speed vA. Determine his speed when he reaches point B on the smooth slope. For this distance the slope follows the cosine curve shown. Also, what is the normal force on his skis at B and his rate of increase in speed? Neglect friction and air resistance. Given: W 150 lb = vA 5 ft s = a 50 ft = b 100 ft = d 35 ft = Solution: y x ( ) a ( )cos π x b ⎛⎜⎝ ⎞⎟⎠ = y' x ( ) x y x ( ) d d = y'' x ( ) x y' x ( ) d d = ρ x ( ) 1 y' x ( )2 ( + ) 3 y'' x ( ) = θB atan y' d ( ) ( ) = ρB ρ d ( ) = Guesses FN = 1 lb v' 1 ft s2 = vB 1 ft s = Given 1 2 W g ⎛⎜⎝ ⎞⎟⎠ vA 2 W y 0 ft ( ) y d ( ) − ( ) + 1 2 W g ⎛⎜⎝ ⎞⎟⎠ vB 2 = FN W cos θB ( ) − W g ⎛⎜⎝ ⎞⎟⎠ vB 2 ρB = −W sin θB ( ) W g ⎛⎜⎝ ⎞⎟⎠ v' = 259 Engineering Mechanics - Dynamics Chapter 14 vB FN v' ⎛⎜ ⎜ ⎜⎝ ⎞⎟ ⎟ ⎟⎠ Find vB FN , , v' ( ) = vB 42.2 ft s = FN = 50.6 lb v' 26.2 ft s2 = Problem 14-29 When the block A of weight W1 is released from rest it lifts the two weights B and C each of weight W2. Determine the maximum distance A will fall before its motion is momentarily stopped. Neglect the weight of the cord and the size of the pulleys. Given: W1 12 lb = W2 15 lb = a 4 ft = Solution: Guess y 10 ft = Given W1 y 2W2 a2 + y2 − a ( ) − = 0 y Find y ( ) = y 3.81 ft = Problem 14-30 The catapulting mechanism is used to propel slider A of mass M to the right along the smooth track. The propelling action is obtained by drawing the pulley attached to rod BC rapidly to the left by means of a piston P. If the piston applies constant force F to rod BC such that it moves it a distance d, determine the speed attained by the slider if it was originally at rest. Neglect the mass of the pulleys, cable, piston, and rod BC. Units Used: kN 103 N = Given: M = 10 kg F = 20 kN d 0.2 m = Solution: 0 + F d 1 = 2M v2 v 2F d M = v 28.284 m s = 260 Engineering Mechanics - Dynamics Chapter 14 Problem 14-31 The collar has mass M and slides along the smooth rod. Two springs are attached to it and the ends of the rod as shown. If each spring has an uncompressed length L and the collar has speed v0 when s = 0, determine the maximum compression of each spring due to the back-and-forth (oscillating) motion of the collar. Given: M = 20 kg a 0.25 m = L = 1 m kA 50 N m = v0 2 m s = kB 100 N m = Solution: 1 2M v0 2 1 2 kA + kB ( ) d2 − = 0 d M kA + kB v0 = d 0.73 m = *Problem 14-32 The cyclist travels to point A, pedaling until he reaches speed vA. He then coasts freely up the curved surface. Determine the normal force he exerts on the surface when he reaches point B. The total mass of the bike and man is M. Neglect friction, the mass of the wheels, and the size of the bicycle. Units Used: kN 103 N = Given: vA 8 m s = M 75 kg = a 4 m = Solution: When y = x 2 y a = y a = 4 y 1 m = 1 2M vA 2 − M g y 1 2M vB 2 = vB vA 2 − 2g y = vB 6.662 m s = Now find the radius of curvature x y + a = 1 2 x dx 1 2 y dy + 0 = 261 Engineering Mechanics - Dynamics Chapter 14 y' y x = − y'' y x x y d d − 2x2 x y = When y = x y' = −1 y'' 1 y = Thus ρ 1 ( + y'2 ) 3 y'' = ρ 8 y = ρ 2.828 m = NB M gcos 45 deg ( ) − M vB 2 ρ ⎛ ⎜ ⎝ ⎞ ⎟ ⎠ = NB M gcos 45 deg ( ) M vB 2 ρ ⎛ ⎜ ⎝ ⎞ ⎟ ⎠ + = NB 1.697 kN = Problem 14-33 The cyclist travels to point A, pedaling until he reaches speed vA. He then coasts freely up the curved surface. Determine how high he reaches up the surface before he comes to a stop. Also, what are the resultant normal force on the surface at this point and his acceleration? The total mass of the bike and man is M. Neglect friction, the mass of the wheels, and the size of the bicycle. Given: vA 4 m s = M = 75 kg a 4 m = Solution: 1 2M vA 2 − M g y = 0 y vA 2 2g = y 0.815 m = x a y − ( )2 = x 1.203 m = y' y x = − θ atan ( y' ) = θ 39.462 deg = NB − M gcos θ( ) = 0 NB = M gcos θ( ) NB 568.03 N = M gsin θ( ) = M at at = gsin θ( ) at 6.235 m s2 = 262 Engineering Mechanics - Dynamics Chapter 14 Problem 14-34 The block of weight W is pressed against the spring so as to compress it a distance δ when it is at A. If the plane is smooth, determine the distance d, measured from the wall, to where the block strikes the ground. Neglect the size of the block. Given: W = 10 lb e 4 ft = δ = 2 ft f 3 ft = k 100 lb ft = g 9.81 m s2 = Solution: θ atan f e ⎛⎜⎝ ⎞⎟⎠ = Guesses vB 1 ft s = t = 1 s d 1 ft = Given 1 2kδ2 − W f 1 2 W g ⎛⎜⎝ ⎞⎟⎠ vB 2 = d = vBcos θ( )t 0 f + vBsin θ( )t g 2 ⎛⎜⎝ ⎞⎟⎠ t2 − = vB t d ⎛⎜ ⎜ ⎜⎝ ⎞⎟ ⎟ ⎟⎠ Find vB t, , d ( ) = vB 33.08 ft s = t = 1.369 s d 36.2 ft = Problem 14-35 The man at the window A wishes to throw a sack of mass M onto the ground. To do this he allows it to swing from rest at B to point C, when he releases the cord at θ = θ1. Determine the speed at which it strikes the ground and the distance R. Given: θ1 30 deg = h 16 m = L 8 m = g 9.81 m s2 = M 30 kg = Solution: 0 M gL cos θ1 ( ) + 1 2M vC 2 = vC 2gL cos θ1 ( ) = vC 11.659 m s = 263 Engineering Mechanics - Dynamics Chapter 14 0 + M gh 1 2M vD 2 = vD 2gh = vD 17.718 m s = Free Flight Guess t = 2 s R 1 m = Given 0 −g 2 ⎛⎜⎝ ⎞⎟⎠ t2 vC sin θ1 ( )t + + h L cos θ1 ( ) − = R vC cos θ1 ( )t L 1 sin θ1 ( ) ( + ) + = t ⎛⎜⎝R ⎞⎟⎠ Find t R ( , ) = t = 2.078 s R 33.0 m = *Problem 14-36 A block of weight W rests on the smooth semicylindrical surface. An elastic cord having a stiffness k is attached to the block at B and to the base of the semicylinder at point C. If the block is released from rest at A(θ = 0°), determine the unstretched length of the cord so the block begins to leave the semicylinder at the instant θ = θ1. Neglect the size of the block. Given: W 2 lb = k 2 lb ft = θ1 45 deg = a 1.5 ft = g 9.81 m s2 = Solution: Guess δ = 1 ft v1 1 ft s = Given W sin θ1 ( ) W g ⎛⎜⎝ ⎞⎟⎠ v1 2 a = 1 2k πa − δ ( )2 1 2k π ( − θ1 )a − δ ⎡⎣ ⎤⎦ 2 − W asin θ1 ( ) − W g ⎛⎜⎝ ⎞⎟⎠ v1 2 2 = v1 ⎛⎜ δ ⎝ ⎞⎟ ⎠ Find v1 δ , ( ) = v1 5.843 ft s = δ 2.77 ft = 264 Engineering Mechanics - Dynamics Chapter 14 Problem 14-37 A rocket of mass m is fired vertically from the surface of the earth, i.e., at r = r1. Assuming that no mass is lost as it travels upward, determine the work it must do against gravity to reach a distance r2. The force of gravity is F = GMem/r2 (Eq. 13-1), where Me is the mass of the earth and r the distance between the rocket and the center of the earth. Solution: F G Me m r2 ⎛ ⎜ ⎝ ⎞ ⎟ ⎠ = U12 r F ⌠ ⎮ ⌡ d = −G Me m r1 r2 r 1 r2 ⌠ ⎮ ⎮ ⌡ d = U12 GMe m 1 r2 1 r1 − ⎛⎜⎝ ⎞⎟⎠ = Problem 14-38 The spring has a stiffness k and an unstretched length l0. As shown, it is confined by the plate and wall using cables so that its length is l. A block of weight W is given a speed vA when it is at A, and it slides down the incline having a coefficient of kinetic friction μk. If it strikes the plate and pushes it forward a distance l1 before stopping, determine its speed at A. Neglect the mass of the plate and spring. Given: W = 4 lb d 3 ft = l0 = 2 ft k 50 lb ft = l = 1.5 ft μk 0.2 = l1 = 0.25 ft a = 3 b 4 = Solution: θ atan a b ⎛⎜⎝ ⎞⎟⎠ = Guess vA 1 ft s = 265 Engineering Mechanics - Dynamics Chapter 14 Given 1 2 W g ⎛⎜⎝ ⎞⎟⎠ vA 2 W sin θ( ) − μk cos θ( ) ( ) d ( + l1 ) + 1 2k l0 − l + l1 ( )2 l0 − l ( )2 − ⎡⎣ ⎤⎦ − 0 = vA Find vA ( ) = vA 5.80 ft s = Problem 14-39 The “flying car” is a ride at an amusement park which consists of a car having wheels that roll along a track mounted inside a rotating drum. By design the car cannot fall off the track, however motion of the car is developed by applying the car’s brake, thereby gripping the car to the track and allowing it to move with a constant speed of the track, vt. If the rider applies the brake when going from B to A and then releases it at the top of the drum, A, so that the car coasts freely down along the track to B (θ = π rad), determine the speed of the car at B and the normal reaction which the drum exerts on the car at B. Neglect friction during the motion from A to B. The rider and car have a total mass M and the center of mass of the car and rider moves along a circular path having a radius r. Units Used: kN 103 N = Given: M 250 kg = r 8 m = vt 3 m s = Solution: 1 2M vt 2 + Mg2r 1 2M vB 2 = vB vt 2 + 4gr = vB 18.0 m s = NB − M g M vB 2 r ⎛ ⎜ ⎝ ⎞ ⎟ ⎠ = NB M g vB 2 r + ⎛ ⎜ ⎝ ⎞ ⎟ ⎠ = NB 12.5 kN = 266 Engineering Mechanics - Dynamics Chapter 14 *Problem 14-40 The skier starts from rest at A and travels down the ramp. If friction and air resistance can be neglected, determine his speed vB when he reaches B. Also, find the distance d to where he strikes the ground at C, if he makes the jump traveling horizontally at B. Neglect the skier’s size. He has a mass M. Given: M = 70 kg h1 50 m = θ = 30 deg h2 4 m = Solution: Guesses vB 1 m s = t = 1 s d 1 m = Given M g h1 − h2 ( ) 1 2M vB 2 = vBt = dcos θ( ) −h2 − dsin θ( ) −1 2 gt2 = vB t d ⎛⎜ ⎜ ⎜⎝ ⎞⎟ ⎟ ⎟⎠ Find vB t, , d ( ) = t = 3.753 s vB 30.0 m s = d 130.2 m = Problem 14-41 A spring having a stiffness k is compressed a distance δ. The stored energy in the spring is used to drive a machine which requires power P. Determine how long the spring can supply energy at the required rate. Units Used: kN 103 N = Given: k 5 kN m = δ = 400 mm P 90 W = Solution: U12 1 = 2kδ2 = P t t 1 2 k δ2 P ⎛⎜ ⎝ ⎞⎟ ⎠ = t 4.44 s = Problem 14-42 Determine the power input for a motor necessary to lift a weight W at a constant rate v. The efficiency of the motor is ε. Given: W = 300 lbf v 5 ft s = ε 0.65 = 267 Engineering Mechanics - Dynamics Chapter 14 Solution: P W v ε = P 4.20 hp = Problem 14-43 An electrically powered train car draws a power P. If the car has weight W and starts from rest, determine the maximum speed it attains in time t. The mechanical efficiency is ε. Given: P = 30 kW W = 40000 lbf t = 30 s ε 0.8 = Solution: εP = F v W g t d v ⎛⎜⎝d ⎞⎟⎠ v = 0 v v v ⌠ ⎮⌡ d 0 t t εP g W ⌠⎮ ⎮⌡ d = v 2εP gt W = v 29.2 ft s = *Problem 14-44 A truck has a weight W and an engine which transmits a power P to all the wheels. Assuming that the wheels do not slip on the ground, determine the angle θ of the largest incline the truck can climb at a constant speed v. Given: W 25000 lbf = v 50 ft s = P 350 hp = Solution: F = W sin θ( ) P W sin θ( )v = θ asin P W v ⎛⎜⎝ ⎞⎟⎠ = θ 8.86 deg = Problem 14-45 An automobile having mass M travels up a slope at constant speed v. If mechanical friction and wind resistance are neglected, determine the power developed by the engine if the automobile has efficiency ε. Units Used: Mg 103 kg = 268 Engineering Mechanics - Dynamics Chapter 14 Given: M = 2 Mg v 100 km hr = θ = 7 deg ε 0.65 = Solution: P = M gsin θ( )v P 66.419 kW = Peng P ε = Peng 102.2 kW = Problem 14-46 The escalator steps move with a constant speed v. If the steps are of height h and length l, determine the power of a motor needed to lift an average mass M per step.There are n steps. Given: M = 150 kg h 125 mm = n = 32 l 250 mm = v 0.6 m s = d nh = Solution: θ atan h l ⎛⎜⎝ ⎞⎟⎠ = P = nMgvsin θ( ) P 12.63 kW = Problem 14-47 If the escalator in Prob. 14−46 is not moving, determine the constant speed at which a man having a mass M must walk up the steps to generate power P—the same amount that is needed to power a standard light bulb. Given: M = 80 kg h 125 mm = n = 32 l 250 mm = v 0.6 m s = P 100 W = 269 Engineering Mechanics - Dynamics Chapter 14 Solution: θ atan h l ⎛⎜⎝ ⎞⎟⎠ = P = F vsin θ( ) v P M gsin θ( ) = v 0.285 m s = *Problem 14-48 An electric streetcar has a weight W and accelerates along a horizontal straight road from rest such that the power is always P. Determine how far it must travel to reach a speed of v. Given: W = 15000 lbf v 40 ft s = P 100 hp = Solution: P = F v W g ⎛⎜⎝ ⎞⎟⎠ av = W g ⎛⎜⎝ ⎞⎟⎠ v2 sc v d ⎛⎜d ⎝ ⎞⎟ ⎠ = Guess d 1 ft = Given 0 d sc P ⌠ ⎮⌡ d 0 v v W g ⎛⎜⎝ ⎞⎟⎠ v2 ⌠⎮ ⎮⌡ d = d Find d ( ) = d 180.8 ft = Problem 14-49 The crate of weight W is given speed v in time t1 starting from rest. If the acceleration is constant, determine the power that must be supplied to the motor when t = t2. The motor has an efficiency ε. Neglect the mass of the pulley and cable. Given: W = 50 lbf t2 2 s = v 10 ft s = ε 0.76 = t1 = 4 s g 32.2 ft s2 = Solution: a v t1 = a 2.5 ft s2 = v2 = at2 v2 5 ft s = 270 Engineering Mechanics - Dynamics Chapter 14 F − W W g ⎛⎜⎝ ⎞⎟⎠ a = F W W g ⎛⎜⎝ ⎞⎟⎠ a + = F 53.882 lbf = P = F v2 P = 0.49 hp Pmotor P ε = Pmotor 0.645 hp = Problem 14-50 A car has a mass M and accelerates along a horizontal straight road from rest such that the power is always a constant amount P. Determine how far it must travel to reach a speed of v. Solution: Power: Since the