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Solution Manual for Engineering Mechanics Dynamics, 13th Edition

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    R1–1. An automobile is traveling with a constant speed along a horizontal circular curve that has a radius If the magnitude of acceleration is determine the speed at which the automobile is traveling. a = 8 ft > s 2 , r = 750 ft. SOLUTION A n s . v = 77.5 ft > s 8 = v 2 750 a = a n = 8 = v 2 r

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    R1–2. SOLUTION a) The maximum friction force between blocks A and B is Thus, both blocks move together. A n s . b) In this case Block A: A n s . Block B : A n s . a B = 3.86 ft > s 2 : + a F x = ma x ; 20(0.3) = 50 32.2 a B a A = 70.8 ft > s 2 : + a F x = ma x ; 20(0.3) = 20 32.2 a A 8 lb 6 F = 50 lb a B = a A = a = 2.76 ft > s 2 : + a F x = ma x ; 6 = 70 32.2 a F max = 0.4(20) = 8 lb 7 6 lb Block B rests on a smooth surface. If the coefficients of friction between A and B are and determine the acceleration of each block if (a) and (b) F = 50 lb. F = 6 lb, m k = 0.3, m s = 0.4 20 lb A B 50 lb F

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    R1–3. The small 2-lb collar starting from rest at A slides down along the smooth rod. During the motion, the collar is acted e r e h w e c r o f a y b n o p u x, y, z are in feet. Determine the collar’s speed when it strikes the wall at B . F = 5 10 i + 6 y j + 2 z k 6 lb, y x z 8 ft 10 ft F 1 ft B 4 ft A SOLUTION A n s . v B = 47.8 ft > s 0 + 2(10 - 1) + L 0 4 10 dx + L 8 0 6 y dy + L 1 10 2 z dz = 1 2 a 2 32.2 b v 2 B T 1 + © L Fds = T 2 r AB = r B - r A = - 4 i + 8 j - 9 k

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    *R1–4. SOLUTION Since Since A n s . a = 2 ( - 0.217) 2 + (0) 2 + (0) 2 = 0.217 m > s 2 a z = z $ = 0 a u = r # u $ + 2 r u = 10(0) + 2(0)(0.1473) = 0 a r = r - r # u 2 = 0 - 10(0.1473) 2 = - 0.217 u = 0 v u = r u # = 1.473 u = 1.473 10 = 0.1473 r = 10 r # = 0 r = 0 v z = - 1.5 sin 10.81° = - 0.2814 m > s v u = 1.5 cos 10.81° = 1.473 m > s v r = 0 v = 1.5 m > s f = tan - 1 a 12 2 p (10) b = 10.81° The automobile travels from a parking deck down along a cylindrical spiral ramp at a constant speed of . If the ramp descends a distance of 12 m for every full revolution, rad, determine the magnitude of the car’s acceleration as it moves along the ramp, . Hint: For part of the solution, note that the tangent to the ramp at any point is at an angle of from the horizontal. Use this to determine the velocity components and , which in turn are used to determine and . z # u # v z v u f = tan - 1 (12 >3 2 p (10) 4 ) = 10.81° r = 10 m u = 2 p v = 1.5 m > s 10 m 12 m

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    R1–5. SOLUTION A n s . ( v R ) 2 = 0.840 m > s 0 + 0 = 0.0015(1400) - 2.5( v R ) 2 : + © m v 1 = © m v 2 A rifle has a mass of 2.5 kg. If it is loosely gripped and a 1.5-g bullet is fired from it with a horizontal muzzle velocity of 1400 m / s, determine the recoil velocity of the rifle just after firing.

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    R1–6. SOLUTION A n s . v 2 = 21.5 ft > s 0 + 150 sin 30°(30) - (0.3)129.9(30) = 1 2 a 150 32.2 b v 2 2 T 1 + © U 1 - 2 = T 2 N C = 129.9 lb +a© F y = 0; N C - 150 cos 30° = 0 If a 150-lb crate is released from rest at A , determine its speed after it slides 30 ft down the plane. The coefficient of kinetic friction between the crate and plane is m k = 0.3. 30 ft 30 ° A

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    R1–7. Th e v a n i s t r a v e lin g at 20 km h wh e n t h e co uplin g o f t h e t r a il e r at A f a il s . If t h e t r a il e r h as a m ass o f 250 k g a nd coasts 45 m b e f o r e co min g to r est, d ete rmin e t h e co n sta n t h o ri zo n ta l f o r ce F c r eate d by r o llin g fri ct i o n whi c h ca u ses t h e t r a il e r to sto p. > A 20 km / h F SOLUTION Ans. : + © F x = ma x ; F = 250(0.3429) = 85.7 N a = - 0.3429 m > s 2 = 0.3429 m > s 2 : 0 = 5.556 2 + 2( a )(45 - 0) a ; + b y 2 = y 2 0 + 2 a c ( s - s 0 ) 20 km > h = 20(10 3 ) 3600 = 5.556 m > s

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    *R1–8. SOLUTION Ans. Ans. y max = 3(10) 2 - 18(10) + 15 = 135 ft > s y max occurs when t = 10 s a max = 6(10) - 18 = 42 ft > s 2 amax occurs at t = 10 s, a = d y dt = 6 t - 18 y = ds dt = 3 t 2 - 18 t + 15 s = t 3 - 9 t 2 + 15 t The position of a particle along a straight line is given by where t is in seconds. Determine its maximum acceleration and maximum velocity during the time interval 0 … t … 10 s. s = 1 t 3 - 9 t 2 + 15 t 2 ft,

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    R1–9. The spool, which has a mass of 4 kg, slides along the rotating rod. At the instant shown, the angular rate of rotation of the rod is and this rotation is increasing at At this same instant, the spool has a velocity of 3 m / s and an acceleration of both measured relative to the rod and directed away from the center O when Determine the radial frictional force and the normal force, both exerted by the rod on the spool at this instant. r = 0.5 m. 1 m > s 2 , u $ = 2 rad > s 2 . u # = 6 rad > s SOLUTION A n s . A n s . N = 2 (148) 2 + (39.24) 2 = 153 N F r = - 68 N N z = 39.24 N © F z = ma z ; N z - 4(9.81) = 0 © F u = ma u ; N u = 4(37) = 148 N © F r = ma r ; F r = 4( - 17) = - 68 N a u = r u $ + 2 r # u # = 0.5(2) + 2(3)(6) = 37 a r = r $ - r u # 2 = 1 - 0.5(6) 2 = - 17 r $ = 1 m > s 2 u $ = 2 rad > s r # = 3 m > s u # = 6 rad > s r = 0.5 m θ •• = 2 rad / s 2 θ • = 6 rad / s v s = 3 m / s a s = 1 m / s 2 r = 0.5 m O

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    R1–10. Packages having a mass of 6 kg slide down a smooth chute and land horizontally with a speed of 3 m / s on the surface of a conveyor belt. If the coefficient of kinetic friction between the belt and a package is determine the time needed to bring the package to rest on the belt if the belt is moving in the same direction as the package with a speed v = 1 m / s. m k = 0.2, SOLUTION A n s . t = 1.02 s ) 3 ( 6 - 0.2(58.86)( t ) = 6(1) ( : + ) m ( v 1 ) x + © L F x dt = m ( v 2 ) x N p = 58.86 N 0 + N p ( t ) - 58.86( t ) = 0 ( +c ) m ( v 1 ) y + © L F y dt = m ( v 2 ) y 1 m / s 3 m / s

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    R1–11. SOLUTION The crate starts moving when From the graph since The time needed for the crate to start moving is Hence, the impulse due to F is equal to the area under the curve from A n s . v 2 = 31.7 m > s 3 8 4 . 4 3 6 = 20 v 2 ( 0 4 1 2 t 2 ) 2 5 2.943 + 200(10 - 5) - 692.292 = 20 v 2 0 + L 5 2.943 200 5 t dt + L 10 5 200 dt - (0.5)196.2(10 - 2.943) = 20 v 2 : + m ( v x ) 1 + © L F dt = m ( v x ) 2 2.943 s … t … 10 s t = 5 200 (117.72) = 2.943 s F = 200 5 t , 0 … t … 5 s F = F r = 0.6(196.2) = 117.72 N A 20-kg block is originally at rest on a horizontal surface for which the coefficient of static friction is and the coefficient of kinetic friction is If a horizontal force F is applied such that it varies with time as shown, determine the speed of the block in 10 s. Hint: First determine the time needed to overcome friction and start the block moving. m k = 0.5. m s = 0.6 F F (N) t (s) 200 5 10 x

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    *R1–12. SOLUTION A n s . x = 0.654 ft = 7.85 in. 0 + 1 2 (20)(12)( x 2 ) = 1 2 a 6 32.2 b (6) 2 + 8(6) T 1 + V 1 = T 2 + V 2 The 6-lb ball is fired from a tube by a spring having a stiffness Determine how far the spring must be compressed to fire the ball from the compressed position to a height of 8 ft, at which point it has a velocity of 6 ft / s. k = 20 lb > in. 8 ft k = 20 lb / in. v = 6 ft / s

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    R1–13. A train car, having a mass of 25 Mg, travels up a 10° incline with a constant speed of 80 km/h. Determine the power required to overcome the force of gravity. SOLUTION A n s . P = 946 kW P = F # v = 25(10 3 )(9.81)(22.22)(sin 10°) v = 80 km > h = 22.22 m > s

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    R1–14. The rocket s led ha s a ma ss of 4 M g and tra v el s from re s t alon g the s mooth horizontal track s uch that it maintain s a con s tant power output of . Ne g lect the lo ss of fuel ma ss and air re s i s tance, and determine how far it mu s t tra v el to reach a s peed of . v = 60 m > s 450 kW SOLUTION Ans. s = 4 (10 3 )(60) 3 3(450)(10 3 ) = 640 m s = m v 3 3 P Ps = m v 3 3 P L s 0 ds = m L v 0 v 2 dv L P ds = m L v 2 dv P = Fv = m a v 2 dv ds b F = m a = m a v dv ds b © F x = m a x ; + : v T

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    ■ R1–15. SOLUTION at Using the Runge–Kutta method: t = 0 s 0 = 0 n = 3 (8 e - 1 + t ) 1 > 2 A projectile, initially at the origin, moves vertically downward along a straight-line path through a fluid medium such that its velocity is defined as where t is in seconds. Plot the position s of the projectile during the first 2 s. Use the Runge-Kutta method to evaluate s with incremental values of h = 0.25 s. v = 3 1 8 e - t + t 2 1 > 2 m > s, t s 0 0 2.01 0.25 3.83 0.50 5.49 0.75 7.03 1.00 8.48 1.25 9.87 1.50 11.2 1.75 12.5 2.00

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    *R1–16. The chain has a mass of 3 kg / m. If the coefficient of kinetic friction between the chain and the plane is determine the velocity at which the end A will pass point B when the chain is released from rest. m k = 0.2, SOLUTION A n s . Also, A n s . v = 4.38 m > s 0 + 2(3)(9.81)(2 sin 40°) - 0.2(45.09)(2) = 1 2 (2)(3)( v 2 ) T 1 + © U 1 - 2 = T 2 v 2 = 4.38 m > s v 2 2 = 0 + 2(4.80)(2) + b v 2 2 = v 2 1 + 2 as a = 4.80 m > s 2 + b© F x = ma x ; 2(3)(9.81) sin 40° - 0.2(45.09) = 2(3) a N C = 45.09 N + a© F y = ma y ; - 2(3)(9.81) cos 40° + N C = 0 2 m A B 40 °

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    R1–17. The motor M pulls in its attached rope with an acceleration Determine the towing force exerted by M on the rope in order to move the 50-kg crate up the inclined plane. The coefficient of kinetic friction between the crate and the plane is Neglect the mass of the pulleys and rope. m k = 0.3. a p = 6 m / s 2 . SOLUTION (1) Kinematics, Taking two time derivatives, yields Thus, Substituting into Eq. (1) and solving, A n s . T = 158 N a C = 6 3 = 2 3 a C = a p 2 s C + ( s C - s p ) = l Q+ © F x = ma x ; 3 T - 0.3(424.79) - 50(9.81) sin 30° = 50 a C N C = 424.79 a+ © F y = ma y ; N C - 50(9.81) cos 30° = 0 aP = 6 m / s 2 P M θ = 30 °

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    R1–18. The drinking fountain is designed such that the nozzle is located from the edge of the basin as shown. Determine the maximum and minimum speed at which water can be ejected from the nozzle so that it does not splash over the sides of the basin at B and C . SOLUTION (1) (2) Substituting Eq. (1) into (2) yields: At point B , . A n s . At point C , . A n s . v A = A 4.905 sin 40°(0.35) 2 sin 2 40°(0.35 cos 40° + 0.05 sin 40°) = 1.76 m > s R = 0.35 m v A = A 4.905 sin 40°(0.1) 2 sin 2 40°(0.1 cos 40° + 0.05 sin 40°) = 0.838 m > s R = 0.1 m v A = A 4.905 sin 40° R 2 sin 2 40°( R cos 40° + 0.05 sin 40°) - 0.05 = v A cos 40° ¢ R v A sin 40° ≤ + 1 2 ( - 9.81) ¢ R v A sin 40° ≤ 2 - 0.05 = 0 + v A cos 40° t + 1 2 ( - 9.81) t 2 ( +c ) s y = A s y B 0 + v y t + 1 2 a c t 2 R = v A sin 40° t t = R v A sin 40° a : + b s x = v x t v A 50 mm A B C 40 ° 250 mm 100 mm

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    R1–19. The 100-kg crate is subjected to the action of two forces, and as shown. If it is originally at rest, determine the distance it slides in order to attain a speed of 6 m / s. The coefficient of kinetic friction between the crate and the surface is m k = 0.2. F 2 = 1.5 kN, F 1 = 800 N SOLUTION A n s . s = 0.933 m s (1928.7) = 1800 0 + 800 cos 30°( s ) - 0.2(867.97)( s ) + 1500 cos 20° ( s ) = 1 2 (100)(6) 2 T 1 + a U 1 - 2 = T 2 N C = 867.97 N + c a F y = 0; N C - 800 sin 30° - 100(9.81) + 1500 sin 20° = 0 F 2 F 1 30 ° 20 °

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    *R1–20. SOLUTION A n s . s = 20.0 ft s = 0 + 12(10) + 1 2 ( - 2)(10) 2 ( : + ) s = s 0 + v 0 t + 1 2 a c t 2 If a particle has an initial velocity to the right, and a constant acceleration of to the left, determine the particle’s displacement in 10 s. Originally s 0 = 0. 2 ft > s 2 v 0 = 12 ft > s

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    R1–21. SOLUTION Ans. h = 0.390 m h = 0 + 8.3328(0.048112) - 1 2 (9.81)(0.048112) 2 ( + c ) s = s 0 + v 0 t + 1 2 a c t 2 t = 0.048112 s 0.75 = 0 + 15.5885 t ( : + ) s = s 0 + v 0 t ( v y ) 2 = 8.3328 m > s ( + c ) e = 0.8 = ( v y ) 2 10.4160 ( v y ) 1 = 10.4160 m > s ( v y ) 1 = 18 sin 30° + 9.81(0.14434) ( + T ) v = v 0 + a c t ( v x ) 1 = ( v x ) 2 = 18 cos 30° = 15.5885 m > s t = 0.14434 s 2.25 = 0 + 18 cos 30° t ( : + ) s = s 0 + v 0 t The ping-pong ball has a mass of 2 g. If it is struck with the velocity shown, determine how high h it rises above the end of the smooth table after the rebound. Take e = 0.8. h 2.25 m 18 m / s 30  0.75 m

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    R1–22. SOLUTION Time to accelerate to 60 m / s, Time to decelerate to a stop, Time to travel at 60 m / s, Total time A n s . t = 10 + 7.5 + 6.25 = 23.8 s t = 6.25 s 5 7 3 = 0 + 60 t ( : + ) s = s 0 + v 0 t 900 - 300 - 225 = 375 m s = 225 m s = 0 + 60(7.5) - 1 2 (8)(7.5 2 ) ( : + ) s = s 0 + v 0 t + 1 2 a c t 2 t = 7.5 s 0 = 60 - 8 t ( : + ) v = v 0 + a c t s = 300 m s = 0 + 0 + 1 2 (6)(10 2 ) ( : + ) s = s 0 + v 0 t + 1 2 a c t 2 t = 10 s 0 6 = 0 + 6 t ( : + ) v = v 0 + a c t A sports car can accelerate at and decelerate at If the maximum speed it can attain is 60 m / s, determine the shortest time it takes to travel 900 m starting from rest and then stopping when s = 900 m. 8 m / s 2 . 6 m / s 2

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    R1–23. A 2-kg particle rests on a smooth horizontal plane and is acted upon by forces and If and when determine the equation which describes the path. y = f 1 x 2 t = 0, v y = 2 m / s v x = 6 m / s, y = 0, x = 0, F y = 3 N. F x = 0 SOLUTION (1) (2) (3) (4) Eliminating t from Eq. (3) and (4) yields: A n s . y = 0.0208 x 2 + 0.333 x ( Parabola ) x = 6 t L x 0 dx = L t 0 6 dt v x = dx dt = 6 L v x 6 dv x = L t 0 0 dt a x = dv x dt = 0 y = 0.75 t 2 + 2 t L y 0 dy = L t 0 (1.5 t + 2) dt v y = dy dt = 1.5 t + 2 L v y 2 dv y = 1.5 L t 0 dt a y = dv y dt = 1.5 : + © F x = ma x ; 0 = 2 a x a x = 0 + c © F y = ma y ; 3 = 2 a y a y = 1.5 m > s 2

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    *R1–24. SOLUTION A n s . N s = 240 lb + c © F n = ma n ; N s - 120 = 120 32.2 c (31.08) 2 30 d r = 3 [1( dy dx ) 2 ] 3 > 2 d 2 y dx 2 3 = (1 + 0) 3 > 2 1 30 = 30 ft d 2 y dx 2 = 1 30 dy dx 2 x = 0 = 1 30 x 2 x = 0 = 0 y = 1 60 x 2 - 15 v B = 31.08 ft > s 0 + (120)(15) = 1 2 a 120 32.2 b v 2 B + 0 T A + V A = T B + V B A skier starts from rest at A (30 ft, 0) and descends the smooth slope, which may be approximated by a parabola. If she has a weight of 120 lb, determine the normal force she exerts on the ground at the instant she arrives at point B . 30 ft 15 ft y B A x y = x 2 – 15 1 __ 60

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    R1–25. SOLUTION Block A : (1) Block B : (2) (3) Block C : (4) Solving Eqs. (1)–(4) for a , A n s . v = 2.68 ft > s v 2 = 0 + 2(1.79)(2 - 0) ( + T ) v 2 = v 2 0 + 2 a c ( s - s 0 ) a = 1.79 ft > s 2 + c © F y = ma y ; T 2 - 6 = 6 32.2 a + c © F y = ma y ; N B - 20 = 0 ; + © F x = ma x ; - T 2 + T 1 - 0.1 N B = 20 32.2 a + T © F y = ma y ; - T 1 + 10 = 10 32.2 a The 20-lb block B rests on the surface of a table for which the coefficient of kinetic friction is Determine the speed of the 10-lb block A after it has moved downward 2 ft from rest. Neglect the mass of the pulleys and cords. m k = 0.1. A 10 lb C 6 lb B 20 lb

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    R1–26. SOLUTION Block A : Block B : Kinematics: Solving, A n s . v A = 26.8 ft > s T v A = 6 + 10.40(2) ( + T ) v A = ( v A ) 0 + a A t a B = - 20.81 ft > s 2 a A = 10.40 ft > s 2 T = 3.38 lb 2 a A = - a B 2 s A + s B = l ; + © F x = ma x ; - T + 0.2(4) = 4 32.2 a B + T © F y = ma y ; 10 - 2 T = 10 32.2 a A At a given instant the 10-lb block A is moving downward with a speed of 6 ft / s. Determine its speed 2 s later. Block B has a weight of 4 lb, and the coefficient of kinetic friction between it and the horizontal plane is Neglect the mass of the pulleys and cord. m k = 0.2. A B

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    R1–27. Two smooth billiard balls A and B have an equal mass of If A strikes B with a velocity of as shown, determine their final velocities just after collision. Ball B is originally at rest and the coefficient of restitution is e = 0.75. 1 v A 2 1 = 2 m / s m = 200 g. SOLUTION (1) (2) Solving Eqs. (1) and (2) For A : For B : Hence A n s . A n s . A n s . ( u A ) 2 = tan - 1 a 0.1915 1.285 b = 8.47° e ( v A ) 2 = 2 ( - 0.1915) 2 + (1.285) 2 = 1.30 m > s ( v B ) 2 = ( v B ) x 2 = 1.34 m > s ; ( v B ) y 2 = 0 ( + c ) m B ( v B ) y 1 = m B ( v B ) y 2 ( v A ) y 2 = 1.285 m > s ( + T ) m A ( v A ) y 1 = m A ( v A ) y 2 ( v B ) x 2 = - 1.3405 m > s ( v A ) x 2 = - 0.1915 m > s 5 7 . 0 = ( v A ) x 2 - ( v B ) x 2 1.532 ( : + ) e = ( v rel ) 2 ( v rel ) 1 - 2(1.532) + 0 = 0.2( v A ) x 2 + 0.2( v B ) x 2 ( : + ) m A ( v A ) x 1 + m B ( v B ) x 1 = m A ( v A ) x 2 + m B ( v B ) x 2 ( v A ) y 1 = - 2 sin 40° = - 1.285 m > s ( v A ) x 1 = - 2 cos 40° = - 1.532 m > s y x 40 ° B A ( v A ) 1

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    *R1–28. SOLUTION A n s . A n s . U 1 - 2 = 14.4 ft # kip U 1 - 2 = (744.4 cos 15°)(20) = 14 380.7 ft # lb N C = 1307.3 lb T = 744.4 lb = 744 lb : + © F x = 0; T cos 15° - 0.55 N C = 0 + c © F y = 0; N C - 1500 + T sin 15° = 0 A crate has a weight of 1500 lb. If it is pulled along the ground at a constant speed for a distance of 20 ft, and the towing cable makes an angle of 15° with the horizontal, determine the tension in the cable and the work done by the towing force. The coefficient of kinetic friction between the crate and the ground is m k = 0.55.

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    R1–29. The collar of negligible size has a mass of 0.25 kg and is attached to a spring having an unstretched length of 100 mm. If the collar is released from rest at A and travels along the smooth guide, determine its speed just before it strikes B . SOLUTION A n s . v B = 10.4 m > s 0 + (0.25)(9.81)(0.6) + 1 2 (150)(0.6 - 0.1) 2 = 1 2 (0.25)( v B ) 2 + 1 2 (150)(0.4 - 0.1) 2 T A + V A = T B + V B 400 mm 200 mm B A k  150 N / m

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    R1–30. Determine the tension developed in the two cords and the acceleration of each block. Neglect the mass of the pulleys and cords. Hint: Since the system consists of two cords, relate the motion of block A to C , and of block B to C . Then, by elimination, relate the motion of A to B . SOLUTION Block A : (1) Block B : (2) Pulley C : (3) Kinematics: Taking the two time derivatives: Also, So that Since (4) Solving Eqs.(1)–(4), A n s . A n s . A n s . A n s . T B = 45.3 N T A = 90.6 N a B = 1.51 m > s 2 a A = 0.755 m > s 2 a B = 2 a A a C ¿ = - a C 2 a C ¿ = a B s C ¿ + ( s C ¿ - s B ) = l ¿ a A = - a C s A + s C = l + c © F y = 0; T A - 2 T B = 0 + c © F y = ma y ; T B - 4(9.81) = 4 a B + T © F y = ma y ; 10(9.81) - T A = 10 a A C B 4 kg A 10 kg 3 m

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