N - W = 0. + F = m a ; N = W. Fs = 0.7W r. Ans. r = 9.32 m

91962_05_R1_p0479-0512 6/5/09 3:53 PM Page 479 R1 1. The ball is thrown horizontally with a speed of 8 m>s. Find the equation of the path, y = f(x), and then determine the ball s velocity and the normal and tangential components of acceleration when t = 0.25 s. y v 8 m/s x Horizontal Motion: The horizontal component of velocity is y x = 8 m>s and the initial horizontal position is (s 0 ) x = 0. : + B s x = (s 0 ) x + (y 0) x t x = 0 + 8t [1] Vertical Motion: The vertical component of initial velocity (y 0) y = 0 and the initial vertical position are (s 0) y = 0. ( + c) s y = (s 0 ) y + (y 0 ) y t + 1 2 (a c) y t 2 y = 0 + 0 + 1 ( -9.81) t2 2 [2] Eliminate t from Eqs. [1] and [2] yields y = -0.0766x 2 The vertical component of velocity when t = 0.25 s is given by (+ c) y y = (y 0) y + (a c ) y t y y = 0 + (-9.81)(0.25) = -2.4525 m>s = 2.4525 m>s T The magnitude and direction angle when t = 0.25 s are y = 2y 2 x + y 2 y = 28 2 + 2.4525 2 = 8.37 m>s u = tan y - 1 y = tan - 1 2.4525 = 17.04 = 17.0 c y x 8 Since the velocity is always directed along the tangent of the path and the acceleration a = 9.81 m>s 2 is directed downward, then tangential and normal components of acceleration are a t = 9.81 sin 17.04 = 2.88 m>s 2 a n = 9.81 cos 17.04 = 9.38 m>s 2 479

91962_05_R1_p0479-0512 6/5/09 3:53 PM Page 480 R1 2. Cartons having a mass of 5 kg are required to move along the assembly line with a constant speed of 8 m>s. Determine the smallest radius of curvature, r, for the conveyor so the cartons do not slip. The coefficients of static and kinetic friction between a carton and the conveyor are ms = 0.7 and mk = 0.5, respectively. + c Fb = m ab ; 8 m/s r N - W = 0 N = W Fs = 0.7W + F = m a ; : n n 0.7W = W 82 a b 9.81 r r = 9.32 m R1 3. small metal particle travels downward through a fluid medium while being subjected to the attraction of a magnetic field such that its position is s = (15t3-3t) mm, where t is in seconds. Determine (a) the particle s displacement from t = 2 s to t = 4 s, and (b) the velocity and acceleration of the particle when t = 5 s. a) s = 15t3-3t t t = 2 s, s1 = 114 mm t t = 4 s, s3 = 948 mm s = 948-114 = 834 mm b) v = ds = 45t2-3 2 = 1122 mm>s = 1.12 m>s dt t=5 a = dv = 90t 2 = 450 mm>s2 = 0.450 m>s2 dt t=5 480

91962_05_R1_p0479-0512 6/5/09 3:53 PM Page 481 *R1 4. The flight path of a jet aircraft as it takes off is defined by the parametric equations x = 1.25t 2 and y = 0.03t 3, where t is the time after take-off, measured in seconds, and x and y are given in meters. If the plane starts to level off at t = 40 s, determine at this instant (a) the horizontal distance it is from the airport, (b) its altitude, (c) its speed, and (d) the magnitude of its acceleration. y x Position: When t = 40 s, its horizontal position is given by x = 1.2540 2 B = 2000 m = 2.00 km and its altitude is given by y = 0.0340 3 B = 1920 m = 1.92 km Velocity: When t = 40 s, the horizontal component of velocity is given by y x = x # = 2.50t t = 40 s = 100 m>s The vertical component of velocity is y y = y # = 0.09t 2 t = 40 s = 144 m>s Thus, the plane s speed at t = 40 s is y y = 2y 2 x + y 2 y = 2100 2 + 144 2 = 175 m>s cceleration: The horizontal component of acceleration is a x = x $ = 2.50 m>s 2 and the vertical component of acceleration is a y = y $ = 0.18t t = 40 s = 7.20 m>s 2 Thus, the magnitude of the plane s acceleration at t = 40 s is a = 2a 2 x + a 2 y = 22.50 2 + 7.20 2 = 7.62 m>s 2 481

91962_05_R1_p0479-0512 6/5/09 3:53 PM Page 482 R1 5. The boy jumps off the flat car at with a velocity of v =4 ft>s relative to the car as shown. If he lands on the second flat car B, determine the final speed of both cars after the motion. Each car has a weight of 80 lb. The boy s weight is 60 lb. Both cars are originally at rest. Neglect the mass of the car s wheels. v 4 ft/s 13 5 12 B Relative Velocity: The horizontal component of the relative velocity of the boy with respect to the car is (y b>) x = 4a 12 b = 3.692 ft>s. Thus, the horizontal 13 component of the velocity of the boy is (y b) x = y + (y b>) x ; + B (y b) x = - y + 3.692 [1] Conservation of Linear Momentum: If we consider the boy and the car as a system, then the impulsive force caused by traction of the shoes is internal to the system. Therefore, they will cancel out. s the result, the linear momentum is conserved along x axis. For car 0 = m b (y b) x + m y ; + B 0 = a 60 32.2 b(y b) x + a 80 32.2 b(-y ) [2] Solving Eqs. [1] and [2] yields y = 1.58 ft>s (y b) x = 2.110 ft>s For car B m b (y b) x = (m b + m B) y B ; + B a 60 60 + 80 b(2.110) = a 32.2 32.2 b y B y B = 0.904 ft>s 482

91962_05_R1_p0479-0512 6/5/09 3:53 PM Page 483 R1-6. The man has a weight of 175 lb and jumps from rest at a height h = 8 ft onto a platform P that has a weight of 60 lb. The platform is mounted on a spring, which has a stiffness k = 200 lb>ft. Determine (a) the velocities of and P just after impact and (b) the maximum compression imparted to the spring by the impact. ssume the coefficient of restitution between the man and the platform is e = 0.6, and the man holds himself rigid during the motion. h Conservation of Energy: The datum is set at the initial position of platform P. When the man falls from a height of 8 ft above the datum, his initial gravitational potential energy is 175(8) = 1400 ft # lb. pplying Eq. 14 21, we have P Conservation of Momentum: T 1 + V 1 = T 2 + V 2 0 + 1400 = 1 2 a 175 32.2 b(y M) 2 1 + 0 (y H) 1 = 22.70 ft>s m M (y M) 1 + m P (y P) 1 = m M (y M) 2 + m p (y p) 2 ( +T) a 175 175 b(22.70) + 0 = a 32.2 32.2 b(y M) 2 + a 60 32.2 b(y p) 2 [1] Coefficient of Restitution: e = (y P) 2 - (y M ) 2 (y M ) 1 - (y p ) 1 ( +T) 0.6 = (y P) 2 - (y P ) 2 22.70-0 [2] Solving Eqs. [1] and [2] yields (y p) 2 = 27.04 ft>s T = 27.0 ft>s T (y M) 2 = 13.4 ft>s T Conservation of Energy: The datum is set at the spring s compressed position. 60 Initially, the spring has been compressed = 0.3 ft and the elastic potential 200 1 energy is. When platform P is at a height of s above the 2 (200) 0.32 B = 9.00 ft # lb datum, its initial gravitational potential energy is 60s. When platform P stops momentary, the spring has been compressed to its maximum and the elastic 1 potential energy at this instant is. pplying 2 (200)(s + 0.3)2 = 100s 2 + 60s + 9 Eq. 14 21, we have T 1 + V 1 = T 2 + V 2 1 2 a 60 32.2 b 27.042 B + 60s + 9.00 = 100s 2 + 60s + 9 s = 2.61 ft 483

91962_05_R1_p0479-0512 6/5/09 3:53 PM Page 484 R1 7. The man has a weight of 100 lb and jumps from rest onto the platform P that has a weight of 60 lb. The platform is mounted on a spring, which has a stiffness k = 200 lb>ft. If the coefficient of restitution between the man and the platform is e = 0.6, and the man holds himself rigid during the motion, determine the required height h of the jump if the maximum compression of the spring is 2 ft. h Conservation of Energy: The datum is set at the initial position of platform P. When the man falls from a height of h above the datum, his initial gravitational potential energy is 100h. pplying Eq. 14 21, we have P Conservation of Momentum: T 1 + V 1 = T 2 + V 2 0 + 100h = 1 2 a 100 32.2 b(y M) 2 1 + 0 (y H ) 1 = 264.4h m M (y M ) 1 + m P (y P ) 1 = m M (y M ) 2 + m P (y P ) 2 ( +T) a 100 100 b(264.4h) + 0 = a 32.2 32.2 b(y M) 2 + a 60 32.2 b(y P) 2 [1] Coefficient of Restitution: e = (y p) 2 - (y M ) 2 (y M ) 1 - (y p ) 1 ( +T) 0.6 = (y p) 2 - (y M ) 2 264.4h - 0 [2] Solving Eqs. [1] and [2] yields (y p ) 2 = 264.4h T (y M ) 2 = 0.4264.4h T Conservation of Energy: The datum is set at the spring s compressed position. 60 Initially, the spring has been compressed = 0.3 ft and the elastic potential 200 1 energy is. Here, the compression of the spring caused by 2 (200) 0.32 B = 9.00 ft # lb impact is (2-0.3) ft = 1.7 ft. When platform P is at a height of 1.7 ft above the datum, its initial gravitational potential energy is 60(1.7) = 102 ft # lb. When platform P stops momentary, the spring has been compressed to its maximum and 1 the elastic potential energy at this instant is. pplying 2 (200)22 B = 400 ft # lb Eq. 14 21, we have T 1 + V 1 = T 2 + V 2 1 2 a 60 32.2 b 264.4hB2 + 102 + 9.00 = 400 h = 4.82 ft 484

91962_05_R1_p0479-0512 6/5/09 3:53 PM Page 485 *R1 8. The baggage truck has a mass of 800 kg and is used to pull each of the 300-kg cars. Determine the tension in the couplings at B and C if the tractive force F on the truck is F = 480 N. What is the speed of the truck when t = 2 s, starting from rest? The car wheels are free to roll. Neglect the mass of the wheels. C B F : + F x = ma x ; 480 = [800 + 2(300)]a a = 0.3429 m>s 2 ( : + ) v = v 0 + a c t v = 0 + 0.3429(2) = 0.686 m>s : + F x = ma x ; T B = 2(300)(0.3429) T B = 205.71 = 206 N : + F x = ma x ; T C = (300)(0.3429) T C = 102.86 = 103 N R1 9. The baggage truck has a mass of 800 kg and is used to pull each of the 300-kg cars. If the tractive force F on the truck is F = 480 N, determine the acceleration of the truck. What is the acceleration of the truck if the coupling at C suddenly fails? The car wheels are free to roll. Neglect the mass of the wheels. C B F : + F x = ma x ; 480 = [800 + 2(300)]a a = 0.3429 = 0.343 m>s 2 : + F x = ma x ; 480 = (800 + 300)a a = 0.436 m>s 2 485

91962_05_R1_p0479-0512 6/5/09 3:53 PM Page 486 R1 10. car travels at 80 ft>s when the brakes are suddenly applied, causing a constant deceleration of 10 ft>s 2. Determine the time required to stop the car and the distance traveled before stopping. a : + b v = v 0 + a c t 0 = 80 + (-10)t t = 8 s a : + b v 2 = v 2 0 + 2a c (s - s 0 ) 0 = (80) 2 + 2(-10)(s - 0) s = 320 ft R1 11. Determine the speed of block B if the end of the cable at C is pulled downward with a speed of 10 ft>s. What is the relative velocity of the block with respect to C? C 10 ft/s 3s B + s C = l 3v B = -v C B 3v B = -(10) v B = -3.33 ft>s = 3.33 ft>s c +TB v B = v C + v B>C -3.33 = 10 + v B>C v B>C = -13.3 ft>s = 13.3 ft>s c 486

91962_05_R1_p0479-0512 6/5/09 3:53 PM Page 487 *R1 12. The skier starts fom rest at and travels down the ramp. If friction and air resistance can be neglected, determine his speed v B when he reaches B. lso, compute the distance s 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 of 70 kg. 50 m B v B 4 m Potential Energy: The datum is set at the lowest point B. When the skier is at point, he is (50-4) = 46 m above the datum. His gravitational potential energy at this position is 70(9.81) (46) = 31588.2 J. s C 30 Conservation of Energy: pplying Eq. 14 21, we have T + V = T B + V B 0 + 31588.2 = 1 2 (70) y2 B y B = 30.04 m>s = 30.0 m>s Kinematics: By considering the vertical motion of the skier, we have ( +T) s y = (s 0 ) y + (y 0 ) y t + 1 2 (a c) y t 2 4 + s sin 30 = 0 + 0 + 1 (9.81) t2 2 [1] By considering the horizontal motion of the skier, we have ; + B s x = (s 0 ) x + y x t s cos 30 = 0 + 30.04 t [2] Solving Eqs. [1] and [2] yields s = 130 m t = 3.753 s 487

91962_05_R1_p0479-0512 6/5/09 3:53 PM Page 488 R1 13. The position of a particle is defined by r = 551cos 2t2i + 41sin 2t2j6 m, where t is in seconds and the arguments for the sine and cosine are given in radians. Determine the magnitudes of the velocity and acceleration of the particle when t = 1 s. lso, prove that the path of the particle is elliptical. Velocity: The velocity expressed in Cartesian vector form can be obtained by applying Eq. 12 7. v = dr dt = {-10 sin 2ri + 8 cos 2rj} m>s When t = 1 s, v = -10 sin 2(1)i + 8 cos 2(1) j = (-9.093i - 3.329j} m>s. Thus, the magnitude of the velocity is y = 2y 2 x + y 2 y = 2(-9.093) 2 + (-3.329) 2 = 9.68 m>s cceleration: The acceleration express in Cartesian vector form can be obtained by applying Eq. 12 9. a = dv dt = {-20 cos 2ri - 16 sin 2rj} m>s2 When t = 1 s, a = -20 cos 2(1) i - 16 sin 2(1) j = {8.323i - 14.549j} m>s 2. Thus, the magnitude of the acceleration is a = 2a 2 x + a 2 y = 28.323 2 + (-14.549) 2 = 16.8 m>s 2 Travelling Path: Here, x = 5 cos 2t and y = 4 sin 2t. Then, x 2 25 = cos2 2t y 2 16 = sin2 2t [1] [2] dding Eqs [1] and [2] yields x 2 25 + y2 16 = cos2 2r + sin 2 2t However, cos 2 2 r + sin 2 2t = 1. Thus, x 2 25 + y2 = 1 (Equation of an Ellipse) 16 (Q.E.D.) 488

91962_05_R1_p0479-0512 6/5/09 3:53 PM Page 489 R1 14. The 5-lb cylinder falls past with a speed v = 10 ft>s onto the platform. Determine the maximum displacement of the platform, caused by the collision. The spring has an unstretched length of 1.75 ft and is originally kept in compression by the 1-ft-long cables attached to the platform. Neglect the mass of the platform and spring and any energy lost during the collision. v 10 ft/s 3 ft Potential Energy: Datum is set at the final position of the platform. When the cylinder is at point, its position is (3 + s) above the datum where s is the maximum displacement of the platform when the cylinder stops momentary. Thus, its gravitational potential energy at this position is 5(3 + s) = (15 + 5s) ft # lb. The 1 and 2 (400) (1.75-1)2 = 112.5 ft # lb initial and final elastic potential energy are 1, respectively. 2 (400) (s + 0.75)2 = 200s 2 + 300s + 112.5 Conservation of Energy: pplying Eq. 14 22, we have 1 ft k 400 lb/ft T + V = T B + V B 1 2 a 5 32.2 b 102 B + (15 + 5s) + 112.5 = 0 + 200s 2 + 300s + 112.5 s = 0.0735 ft R1 15. The block has a mass of 50 kg and rests on the surface of the cart having a mass of 75 kg. If the spring which is attached to the cart and not the block is compressed 0.2 m and the system is released from rest, determine the speed of the block after the spring becomes undeformed. Neglect the mass of the cart s wheels and the spring in the calculation. lso neglect friction. Take k = 300 N>m. k B C T 1 + V 1 = T 2 + V 2 [0 + 0] + 1 2 (300)(0.2)2 = 1 2 (50) v2 b + 1 2 (75) v2 e 12 = 50 v 2 b + 75 v 2 e ( : + ) mv 1 = mv 2 0 + 0 = 50 v b - 75 v e v b = 1.5v e v c = 0.253 m>s ; v b = 0.379 m>s : 489

91962_05_R1_p0479-0512 6/5/09 3:53 PM Page 490 *R1 16. The block has a mass of 50 kg and rests on the surface of the cart having a mass of 75 kg. If the spring which is attached to the cart and not the block is compressed 0.2 m and the system is released from rest, determine the speed of the block with respect to the cart after the spring becomes undeformed. Neglect the mass of the cart s wheels and the spring in the calculation. lso neglect friction. Take k = 300 N>m. k B C T 1 + V 1 = T 2 + V 2 [0 + 0] + 1 2 (300)(0.2)2 = 1 2 (50)v2 b + 1 2 (75) v2 e 12 = 50 v 2 b + 75 v 2 e ( : + ) mv 1 = mv 2 0 + 0 = 50 v b - 75 v e v b = 1.5 v e v c = 0.253 m>s ; v b = 0.379 m>s : v b = v c + v b>c ( : + ) 0.379 = -0.253 + v b>c v b>c = 0.632 m>s : 490

91962_05_R1_p0479-0512 6/5/09 3:53 PM Page 491 R1 17. ball is launched from point at an angle of 30. Determine the maximum and minimum speed v it can have so that it lands in the container. v 30 1 m B C 0.25 m 2.5 m 4 m Min. speed: a : + b s = s 0 + v 0 t 2.5 = 0 + v cos 30 t + c B s = s 0 + v 0 t + 1 2 a c t 2 0.25 = 1 + v sin 30 t - 1 2 (9.81)t2 Solving t = 0.669 s v = v B min = 4.32 m>s Max. speed: a : + b s = s 0 + v 0 t 4 = 0 + v cos 30 t + c B s = s 0 + v 0 t + 1 2 a c t 2 0.25 = 1 + v sin 30 t - 1 (9.81) t2 2 Solving: t = 0.790 s v = (v ) max = 5.85 m>s 491

91962_05_R1_p0479-0512 6/5/09 3:53 PM Page 492 R1 18. t the instant shown, cars and B travel at speeds of 55 mi>h and 40 mi>h, respectively. If B is increasing its speed by 1200 mi>h 2, while maintains its constant speed, determine the velocity and acceleration of B with respect to. Car B moves along a curve having a radius of curvature of 0.5 mi. v B 40 mi/h B 30 v 55 mi/h v B = -40 cos 30 i + 40 sin 30 j = {-34.64i + 20j} mi>h v = {-55i} mi>h v B> = v B - v = (-34.64i + 20j) - (-55i) = {20.36i + 20j} mi>h y B> = 220.36 2 + 20 2 = 28.5 mi>h u = tan -1 a 20 20.36 b = 44.5 a (a B ) n = y2 B r = 402 0.5 = 3200 mi>h2 (a B ) t = 1200 mi>h 2 a B = (3200 sin 30-1200 cos 30 )i + (3200 cos 30 + 1200 sin 30 )j = {560.77i + 3371.28j} mi>h 2 a = 0 a B = a + a B> 560.77i + 3371.28j = 0 + a B> a B> = {560.77i + 3371.28j} mi>h 2 a B> = 2(560.77) 2 + (3371.28) 2 = 3418 mi>h 2 = 3.42 10 3 B mi>h 2 u = tan -1 a 3371.28 560.77 b = 80.6 a 492

91962_05_R1_p0479-0512 6/5/09 3:53 PM Page 493 R1 19. t the instant shown, cars and B travel at speeds of 55 mi>h and 40 mi>h, respectively. If B is decreasing its speed at 1500 mi>h 2 while is increasing its speed at 800 mi>h 2, determine the acceleration of B with respect to. Car B moves along a curve having a radius of curvature of 0.75 mi. v B 40 mi/h B 30 v 55 mi/h (a B ) n = v2 B r = (40)2 0.75 = 2133.33 mi>h2 a B = a + a B> 2133.33 sin 30 i + 2133.33 cos 30 j + 1500 cos 30 i - 1500 sin 30 j = -800i + (a B> ) x i + (a B> ) y j ( : + ) 2133.33 sin 30 + 1500 cos 30 = -800 + (a B> ) x (a B> ) x = 3165.705 : ( + c) 2133.33 cos 30-1500 sin 30 = (a B> ) y (a B> ) y = 1097.521 c (a B> ) = 2(1097.521) 2 + (3165.705) 2 a B> = 3351 mi>h 2 = 3.35 10 3 B mi>h 2 u = tan -1 a 1097.521 3165.705 b = 19.1 a 493

91962_05_R1_p0479-0512 6/5/09 3:54 PM Page 494 *R1 20. Four inelastic cables C are attached to a plate P and hold the 1-ft-long spring 0.25 ft in compression when no weight is on the plate. There is also an undeformed spring nested within this compressed spring. If the block, having a weight of 10 lb, is moving downward at v = 4 ft>s, when it is 2 ft above the plate, determine the maximum compression in each spring after it strikes the plate. Neglect the mass of the plate and springs and any energy lost in the collision. k = 30(12) = 360 lb>ft k = 50(12) = 600 lb>ft ssume both springs compress; 2 ft 0.75 ft k 30 lb/in. v k 50 lb/in. P C 0.5 ft T 1 + V 1 = T 2 + V 2 1 2 ( 10 32.2 )(4)2 + 0 + 1 2 (360)(0.25)2 = 0 + 1 2 (360)(s + 0.25)2 + 1 2 (600)(s - 0.25)2-10(s + 2) 13.73 = 180(s + 0.25) 2 + 300(s - 0.25) 2-10s - 20 (1) 33.73 = 180(s + 0.25) 2 + 300(s - 0.25) 2-10s 480s 2-70s - 3.73 = 0 Choose the positive root; s = 0.1874 ft 6 0.25 ft The nested spring does not deform. Thus Eq. (1) becomes NG! 13.73 = 180(s + 0.25) 2-10s - 20 180 s 2 + 80 s - 22.48 = 0 s = 0.195 ft 494

91962_05_R1_p0479-0512 6/5/09 3:54 PM Page 495 R1 21. Four inelastic cables C are attached to plate P and hold the 1-ft-long spring 0.25 ft in compression when no weight is on the plate. There is also a 0.5-ft-long undeformed spring nested within this compressed spring. Determine the speed v of the 10-lb block when it is 2 ft above the plate, so that after it strikes the plate, it compresses the nested spring, having a stiffness of 50 lb>in., an amount of 0.20 ft. Neglect the mass of the plate and springs and any energy lost in the collision. k = 30(12) = 360 lb>ft 2 ft k 30 lb/in. v k 50 lb/in. P k = 50(12) = 600 lb>ft 0.75 ft C 0.5 ft T 1 + V 1 = T 2 + V 2 1 2 a 10 32.2 bv2 + 1 2 (360)(0.25)2 = 1 2 (360)(0.25 + 0.25 + 0.20)2 + 1 2 (600)(0.20)2-10(2 + 0.25 + 0.20) v = 20.4 ft>s R1 22. The 2-kg spool S fits loosely on the rotating inclined rod for which the coefficient of static friction is m s = 0.2. If the spool is located 0.25 m from, determine the minimum constant speed the spool can have so that it does not slip down the rod. z 0.25 m S 5 4 3 r = 0.25 a 4 5 b = 0.2 m ; + F n = ma n ; N s a 3 5 b - 0.2N sa 4 v2 b = 2a 5 0.2 b + c F b = m a b ; N s a 4 5 b + 0.2N s a 3 5 b - 2(9.81) = 0 N s = 21.3 N v = 0.969 m>s 495

91962_05_R1_p0479-0512 6/5/09 3:54 PM Page 496 R1 23. The 2-kg spool S fits loosely on the rotating inclined rod for which the coefficient of static friction is m s = 0.2. If the spool is located 0.25 m from, determine the maximum constant speed the spool can have so that it does not slip up the rod. z 0.25 m S 5 4 3 r = 0.25 a 4 5 b = 0.2 m ; + F n = ma n ; N s a 3 5 b + 0.2N sa 4 v2 b = 2a 5 0.2 b + c F b = m a b ; N s a 4 5 b - 0.2N s a 3 5 b - 2(9.81) = 0 N s = 28.85 N v = 1.48 m>s *R1 24. The winding drum D draws in the cable at an accelerated rate of 5 m>s 2. Determine the cable tension if the suspended crate has a mass of 800 kg. D s + 2 s B = l a = - 2 a B 5 = - 2 a B a B = -2.5 m>s 2 = 2.5 m>s 2 c + c F y = ma y ; 2T - 800(9.81) = 800(2.5) T = 4924 N = 4.92 kn 496

91962_05_R1_p0479-0512 6/5/09 3:54 PM Page 497 R1 25. The bottle rests at a distance of 3 ft from the center of the horizontal platform. If the coefficient of static friction between the bottle and the platform is m s = 0.3, determine the maximum speed that the bottle can attain before slipping. ssume the angular motion of the platform is slowly increasing. 3 ft F b = 0; N - W = 0 N = W Since the bottle is on the verge of slipping, then F f = m s N = 0.3W. F n = ma n ; 0.3W = a W 32.2 bay2 3 b y = 5.38 ft>s R1 26. Work Prob. R1 25 assuming that the platform starts rotating from rest so that the speed of the bottle is increased at 2 ft>s 2. 3 ft pplying Eq. 13 8, we have F b = 0; N - W = 0 N = W Since the bottle is on the verge of slipping, then F f = m s N = 0.3W. F t = ma t ; 0.3W sin u = a W 32.2 b(2) F n = ma n ; 0.3W cos u = a W 32.2 bay2 3 b [1] [2] Solving Eqs. [1] and [2] yields y = 5.32 ft>s u = 11.95 497

91962_05_R1_p0479-0512 6/5/09 3:54 PM Page 498 R1 27. The 150-lb man lies against the cushion for which the coefficient of static friction is m s = 0.5. Determine the resultant normal and frictional forces the cushion exerts on him if, due to rotation about the z axis, he has a constant speed v = 20 ft>s. Neglect the size of the man. Take u = 60. z 8 ft G u +a F y = m(a n ) y ; N - 150 cos 60 = 150 32.2 a 202 b sin 60 8 N = 277 lb +b F x = m(a n ) x ; -F + 150 sin 60 = 150 32.2 a 202 b cos 60 8 F = 13.4 lb Note: No slipping occurs Since m s N = 138.4 lb 7 13.4 lb *R1 28. The 150-lb man lies against the cushion for which the coefficient of static friction is m s = 0.5. If he rotates about the z axis with a constant speed v = 30 ft>s, determine the smallest angle u of the cushion at which he will begin to slip up the cushion. z 8 ft G u ; + F n = ma n ; 0.5N cos u + N sin u = 150 32.2 a (30)2 8 b + c F b = 0; -150 + N cos u - 0.5 N sin u = 0 N = 150 cos u - 0.5 sin u (0.5 cos u + sin u)150 (cos u - 0.5 sin u) = 150 32.2 a (30)2 8 b 0.5 cos u + sin u = 3.493 79 cos u - 1.746 89 sin u u = 47.5 498

91962_05_R1_p0479-0512 6/5/09 3:54 PM Page 499 R1 29. The motor pulls on the cable at with a force F = (30 + t 2 ) lb, where t is in seconds. If the 34-lb crate is originally at rest on the ground when t = 0, determine its speed when t = 4 s. Neglect the mass of the cable and pulleys. Hint: First find the time needed to begin lifting the crate. t = 2 s for crate to start moving 30 + t 2 = 34 ( + c) mv 1 + L Fdt = mv 2 0 + L 4 2 (30 + t 2 )dt - 34(4-2) = 34 32.2 v 2 [30 t + 1 3 t3 ] 4 2-68 = 34 32.2 v 2 v 2 = 10.1 ft>s R1 30. The motor pulls on the cable at with a force F = (e 2t ) lb, where t is in seconds. If the 34-lb crate is originally at rest on the ground when t = 0, determine the crate s velocity when t = 2 s. Neglect the mass of the cable and pulleys. Hint: First find the time needed to begin lifting the crate. F = e 2t = 34 t = 1.7632 s for crate to start moving ( + c) mv 1 + L F dt = mv 2 0 + L 2 1.7632 e 2 t dt - 34(2-1.7632) = 34 32.2 v 2 1 2 e 2 t 2 2 1.7632-8.0519 = 1.0559 v 2 v 2 = 2.13 ft>s 499

91962_05_R1_p0479-0512 6/5/09 3:54 PM Page 500 R1 31. The collar has a mass of 2 kg and travels along the smooth horizontal rod defined by the equiangular spiral r = (e u ) m, where u is in radians. Determine the tangential force F and the normal force N acting on the collar when if force maintains a constant angular motion u # u = 45, F = 2 rad>s. F r r e u r = e u r # = e u u # r $ = e u (u # ) 2 + e u u # u t u = 45 u # = 2 rad>s u = 0 r = 2.1933 m r = 4.38656 m>s r $ = 8.7731 m>s 2 a r = r $ - r (u # ) 2 = 8.7731-2.1933(2) 2 = 0 a u = r u $ + 2 r # u # = 0 + 2(4.38656)(2) = 17.5462 m>s 2 tan c = r dr du B c = u = 45 = e u >e u = 1 F r = m a r ; -N r cos 45 + F cos 45 = 2(0) F u = m a u ; F sin 45 + N u sin 45 = 2(17.5462) N = 24.8 N F = 24.8 N 500

91962_05_R1_p0479-0512 6/5/09 3:54 PM Page 501 *R1 32. The collar has a mass of 2 kg and travels along the smooth horizontal rod defined by the equiangular spiral r = (e u ) m, where u is in radians. Determine the tangential force F and the normal force N acting on the collar when u if force maintains a constant angular motion u # = 90, F = 2 rad>s. F r r e u r = e u r # = e u u # r $ = e u (u # ) 2 + e u u $ u t u = 90 u # = 2 rad>s u # = 0 r = 4.8105 m r # = 9.6210 m>s r $ = 19.242 m>s 2 a r = r $ - r(u # ) 2 = 19.242-4.8105(2) 2 = 0 a u = r u $ + 2 r # u # = 0 + 2(9.6210)(2) = 38.4838 m>s 2 tan c = r dr du B c = u = 45 = e u >e u = 1 + c F t = m a t ; -N cos 45 + F cos 45 = 2(0) ; + F u = m a u ; F sin 45 + N sin 45 = 2(38.4838) N t = 54.4 N F = 54.4 N 501

91962_05_R1_p0479-0512 6/5/09 3:54 PM Page 502 R1 33. The acceleration of a particle along a straight line is defined by a = (2t - 9) m>s 2, where t is in seconds. When t = 0, s = 1 m and v = 10 m>s. When t = 9 s, determine (a) the particle s position, (b) the total distance traveled, and (c) the velocity. ssume the positive direction is to the right. a = (2t - 9) dv = a dt v t dv = (2t - 9) dt L10 L 0 v - 10 = t 2-9t v = t 2-9t + 10 ds = v dt s t ds = t 2-9t + 10Bdt L1 L 0 s - 1 = 1 3 t3-4.5t 2 + 10t s = 1 3 t3-4.5t 2 + 10t + 1 Note v = 0 at t 2-9t + 10 = 0 t t = 1.298 s, s = 7.127 m t t = 7.702 s, s = -36.627 m t t = 9 s, s = -30.50 m t = 1.298 s and t = 7.702 s a) s = -30.5 m b) s tot = (7.127-1) + 7.127 + 36.627 + (36.627-30.50) = 56.0 m c) v t = 9 = (9) 2-9(9) + 10 = 10 m>s 502

91962_05_R1_p0479-0512 6/5/09 3:55 PM Page 503 R1 34. The 400-kg mine car is hoisted up the incline using the cable and motor M. For a short time, the force in the cable is F = (3200t 2 ) N, where t is in seconds. If the car has an initial velocity v 1 = 2 m>s when t = 0, determine its velocity when t = 2 s. v 1 2 m/s M 17 15 8 +Q F x = ma x ; 3200t 2-400(9.81)a 8 17 b = 400a a = 8t2-4.616 dv = adt y 2 dv = 8t 2-4.616Bdt L2 L 0 v = 14.1 m>s lso, mv 1 + L F dt = mv 2 2 +Q 400(2) + 3200 t 2 dt - 400(9.81)(2-0)a 8 L 17 b = 400v 2 0 800 + 8533.33-3693.18 = 400v 2 v 2 = 14.1 m>s R1 35. The 400-kg mine car is hoisted up the incline using the cable and motor M. For a short time, the force in the cable is F = (3200t 2 ) N, where t is in seconds. If the car has an initial velocity v 1 = 2 m>s at s = 0 and t = 0, determine the distance it moves up the plane when t = 2 s. v 1 2 m/s M 17 15 8 F x = ma x ; 3200t 2-400(9.81)a 8 17 b = 400a a = 8t2-4.616 dv = adt y t dv = 8t 2-4.616Bdt L2 L 0 v = ds dt = 2.667t3-4.616t + 2 s 2 ds = 2.667t 3-4.616t + 2Bdt L2 L 0 s = 5.43 m 503

91962_05_R1_p0479-0512 6/5/09 3:55 PM Page 504 *R1 36. The rocket sled has a mass of 4 Mg and travels v along the smooth horizontal track such that it maintains a T constant power output of 450 kw. Neglect the loss of fuel mass and air resistance, and determine how far the sled must travel to reach a speed of v = 60 m>s starting from rest. : + F x = m a x ; F = m a = ma v dv ds b P = F v = ma v2 dv ds b L P ds = m L v2 dv s v P ds = m v 2 dv L L 0 Ps = m v3 3 s = m v3 3 P 0 s = 4(103 )(60) 3 3(450)(10 3 ) = 640 m R1 37. The collar has a mass of 20 kg and can slide freely on the smooth rod. The attached springs are undeformed when d = 0.5 m. Determine the speed of the collar after the applied force F = 100 N causes it to be displaced so that d = 0.3 m. When d = 0.5 m the collar is at rest. k 15 N/m F 100 N 60 d k 25 N/m T 1 + U 1-2 = T 2 0 + 100 sin 60 (0.5-0.3) + 20(9.81)(0.5-0.3) - 1 2 (15)(0.5-0.3)2-1 2 (25)(0.5-0.3)2 = 1 2 (20)v2 C v C = 2.36 m>s 504

91962_05_R1_p0479-0512 6/5/09 3:55 PM Page 505 R1 38. The collar has a mass of 20 kg and can slide freely on the smooth rod. The attached springs are both compressed 0.4 m when d = 0.5 m. Determine the speed of the collar after the applied force F = 100 N causes it to be displaced so that d = 0.3 m. When d = 0.5 m the collar is at rest. k 15 N/m F 100 N 60 d k 25 N/m T 1 + U 1-2 = T 2 0 + 100 sin 60 (0.5-0.3) + 196.2(0.5-0.3) - [ 1 2 (25)[0.4 + 0.2]2-1 2 (25)(0.4)2 ] - [ 1 2 (15)[0.4-0.2]2-1 2 (15)(0.4)2 ] = 1 2 (20)v2 C v C = 2.34 m>s R1 39. The assembly consists of two blocks and B which have masses of 20 kg and 30 kg, respectively. Determine the speed of each block when B descends 1.5 m. The blocks are released from rest. Neglect the mass of the pulleys and cords. 3 s + s B = l 3 s = - s B B 3 v = - v B T 1 + V 1 = T 2 + V 2 (0 + 0) + (0 + 0) = 1 2 (20)(v ) 2 + 1 2 (30)(-3v ) 2 + 20(9.81)a 1.5 3 b - 30(9.81)(1.5) v = 1.54 m>s v B = 4.62 m>s 505

91962_05_R1_p0479-0512 6/5/09 3:55 PM Page 506 *R1 40. The assembly consists of two blocks and B, which have masses of 20 kg and 30 kg, respectively. Determine the distance B must descend in order for to achieve a speed of 3 m>s starting from rest. 3 s + s B - l 3 s = - s B B 3 v = - v B v B = -9 m>s T 1 + V 1 = T 2 + V 2 (0 + 0) + (0 + 0) = 1 2 (20)(3)2 + 1 2 (30)(-9)2 + 20(9.81)a s B 3 b - 30(9.81)(s B) s B = 5.70 m R1 41. Block, having a mass m, is released from rest, falls a distance h and strikes the plate B having a mass 2 m. If the coefficient of restitution between and B is e, determine the velocity of the plate just after collision. The spring has a stiffness k. Just before impact, the velocity of is B h T 1 + V 1 = T 2 + V 2 k 0 + 0 = 1 2 mv 2 - mgh v = 22gh +TB e = (v B) 2 - (v ) 2 22gh e22gh = (v B ) 2 - (v ) 2 (1) +TB mv 1 = mv 2 m(v ) + 0 = m(v ) 2 + 2m(v B ) 2 (2) Solving Eqs. (1) and (2) for (v B ) 2 yields (v B ) 2 = 1 22gh (1 + e) 3 506

91962_05_R1_p0479-0512 6/5/09 3:55 PM Page 507 R1 42. Block, having a mass of 2 kg, is released from rest, falls a distance h = 0.5 m, and strikes the plate B having a mass of 3 kg. If the coefficient of restitution between and B is e = 0.6, determine the velocity of the block just after collision. The spring has a stiffness k = 30 N>m. B h Just before impact, the velocity of is k T 1 + V 1 = T 2 + V 2 0 + 0 = 1 2 (2)(v ) 2 2-2(9.81)(0.5) (v ) 2 = 22(9.81)(0.5) = 3.132 m>s +TB e = (v B) 3 - (v ) 3 3.132-0 0.6(3.132) = (v B ) 3 - (v ) 3 1.879 = (v B ) 3 - (v ) 3 (1) +TB mv 2 = mv 3 2(3.132) + 0 = 2(v ) 3 + 3(v B ) 3 (2) Solving Eqs. (1) and (2) for (v ) 3 yields (v B ) 3 = 2.00 m>s (v ) 3 = 0.125 m>s R1 43. The cylindrical plug has a weight of 2 lb and it is free to move within the confines of the smooth pipe. The spring has a stiffness k = 14 lb>ft and when no motion occurs the distance d = 0.5 ft. Determine the force of the spring on the plug when the plug is at rest with respect to the pipe.the plug travels in a circle with a constant speed of 15 ft>s, which is caused by the rotation of the pipe about the vertical axis. Neglect the size of the plug. ; + F n = m a n ; F s = 2 32.2 c (15)2 3 - d d F s = ks ; F s = 14(0.5 - d) 3 ft G d k 14 lb/ft Thus, 14(0.5 - d) = 2 32.2 c (15)2 3 - d d (0.5 - d)(3 - d) = 0.9982 1.5-3.5d + d 2 = 0.9982 d 2-3.5d + 0.5018 = 0 Choosing the root 6 0.5 ft d = 0.1498 ft F s = 14(0.5-0.1498) = 4.90 lb 507

91962_05_R1_p0479-0512 6/5/09 3:56 PM Page 508 *R1 44. 20-g bullet is fired horizontally into the 300-g block which rests on the smooth surface. fter the bullet becomes embedded into the block, the block moves to the right 0.3 m before momentarily coming to rest. Determine the speed (v B ) 1 of the bullet. The spring has a stiffness k = 200 N>m and is originally unstretched. (v B ) 1 k 200 N/m fter collision Impact T 1 + U 1-2 = T 2 1 2 (0.320)(v 2) 2-1 2 (200)(0.3)2 = 0 v 2 = 7.50 m>s mv 1 = mv 2 0.02(v B ) 1 + 0 = 0.320(7.50) (v B ) 1 = 120 m>s R1 45. The 20-g bullet is fired horizontally at (v B ) 1 = 1200 m>s into the 300-g block which rests on the smooth surface. Determine the distance the block moves to the right before momentarily coming to rest. The spring has a stiffness k = 200 N>m and is originally unstretched. (v B ) 1 k 200 N/m Impact mv 1 = mv 2 0.02(1200) + 0 = 0.320(v 2 ) v 2 = 75 m>s fter collision; T 1 + U 1-2 = T 2 1 2 (0.320)(75)2-1 2 (200)(x2 ) = 0 x = 3 m 508

91962_05_R1_p0479-0512 6/5/09 3:56 PM Page 509 R1 46. particle of mass m is fired at an angle u 0 with a velocity v 0 in a liquid that develops a drag resistance F = -kv, where k is a constant. Determine the maximum or terminal speed reached by the particle. y Equation of Motion: pplying Eq. 13 7, we have v 0 : + F x = ma x ; -ky x = ma x a x = - k m y x [1] O u 0 x + c F y = ma y ; -mg - ky y = ma y a y = -g - k m y y [2] However,,, and a y = d2 y a y y = dy x = d2 x y x = dx. Substitute these values into dt dt 2 dt dt 2 Eqs. [1] and [2], we have d 2 x dt 2 + k dx m dt = 0 d 2 y dt 2 + k dy m dt = -g The solution for the differential equation, Eq. [3], is in the form of x = C 1 e - k m t + C 2 [3] [4] [5] Thus, x # C 1 k = - m e- k m t [6] However, at t = 0, x = 0 and x # = y 0 cos u 0. Substitute these values into Eqs. [5] and m [6], one obtains C and C 2 = m 1 = -. Substitute C 1 into Eq. [6] k y 0 cos u k y 0 cos u 0 0 and rearrange. This yields x # = e - k m t (y 0 cos u 0 ) The solution for the differential equation, Eq. [4], is in the form of y = C 3 e - k m t + C 4 - mg k t [7] [8] Thus, y # C 3 k = - k m e- m t - mg k [9] However, at t = 0, y = 0 and y # = y 0 sin u 0. Substitute these values into Eq. [8] and [9], one obtains C m and C 4 = m. k ay 0 sin u 0 + mg 3 = - k ay 0 sin u 0 + mg k b k b Substitute C 3 into Eq. [9] and rearrange. This yields y # = e - k m t ay 0 sin u 0 + mg k b - mg k [10] For the particle to achieve terminal speed, t : q and e - k m t : 0. When this happen, from Eqs. [7] and [10], and y y = y # mg y x = x # = 0 = -. Thus, k y max = 2y 2 x + y 2 y = 0 2 mg 2 + a - C k b = mg k 509

91962_05_R1_p0479-0512 6/5/09 3:56 PM Page 510 R1 47. projectile of mass m is fired into a liquid at an y angle u 0 with an initial velocity v 0 as shown. If the liquid develops a friction or drag resistance on the projectile which is proportional to its velocity, i.e., F = -kv, where k is a constant, determine the x and y components of its position at any instant. lso, what is the maximum distance x max that it travels? Equation of Motion: pplying Eq. 13 7, we have O v 0 u 0 x : + F x = ma x ; -ky x = ma x a x = - k m y x [1] + c F y = ma y ; -mg - ky y = ma y a y = -g - k m y y [2] However,,, and a y = d2 y a y y = dy x = d2 x y x = dx. Substituting these values into dt dt 2 dt dt 2 Eqs. [1] and [2], we have d 2 x dt 2 + k dx m dt = 0 d 2 y dt 2 + k dy m dt = -g The solution for the differential equation, Eq. [3], is in the form of x = C 1 e - k m t + C 2 [3] [4] [5] Thus, x # C 1 k = - m e- k m t [6] However, at t = 0, x = 0 and x # = y 0 cos u 0. Substituting these values into Eq. [5] m and [6], one can obtain C and C 2 = m 1 = -. Substituting C k y 0 cos u k y 0 cos u 0 0 1 and C 2 into Eq. [5] and rearrange yields x = m k y 0 cos u 0 1 - e - k m t B When t : q, e - k m t : 0 and x = x max. Then, x max = m k y 0 cos u 0 The solution for the differential equation. Eq. [4], is in the form of y = C 3 e - k m t + C 4 - mg k t [7] Thus, y # C 3 k = - k m e- m t - mg k [8] However, at t = 0, y = 0 and y # = y 0 sin u 0. Substitute these values into Eq. [7] and [8], one can obtain C m and C 4 = m. k ay 0 sin u 0 + mg 3 = - k ay 0 sin u 0 + mg k b k b Substitute C 3 and C 4 into Eq. [7] and rearrange yields y = m k y 0 sin u 0 + mg k k a1 - e- m t b - mg k t 510

91962_05_R1_p0479-0512 6/5/09 3:56 PM Page 511 *R1 48. The position of particles and B are r and r B = 53(t 2 = 53ti + 9t(2 - t)j6 m - 2t + 2)i + 3(t - 2)j6 m, respectively, where t is in seconds. Determine the point where the particles collide and their speeds just before the collision. How long does it take before the collision occurs? When collision occurs, r = r B. 3t = 3(t 2-2t + 2) t 2-3t + 2 = 0 t = 1 s, t = 2 s lso, 9t(2 - t) = 3(t - 2) 3t 2-5t - 2 = 0 The positive root is t = 2 s Thus, t = 2 s x = 3(2) = 6 m y = 9(2)(2-2) = 0 Hence, (6 m, 0) v = dr dt = 3i + (18-18t)j v t = 2 = {3i - 18j} m>s v = 2(3) 2 + (-18) 2 = 18.2 m>s v B = dr B dt = 3(2t - 2)i + 3j v B t = 2 = {6i + 3j} m>s v B = 2(6) 2 + (3) 2 = 6.71 m>s 511

91962_05_R1_p0479-0512 6/5/09 3:56 PM Page 512 R1 49. Determine the speed of the automobile if it has the acceleration shown and is traveling on a road which has a radius of curvature of r = 50 m. lso, what is the automobile s rate of increase in speed? t u 40 a 3 m/s 2 a n = v2 r n 3 sin 40 = v2 50 v = 9.82 m>s a t = 3 cos 40 = 2.30 m>s 2 R1 50. The spring has a stiffness k = 3 lb>ft and an unstretched length of 2 ft. If it is attached to the 5-lb smooth collar and the collar is released from rest at, determine the speed of the collar just before it strikes the end of the rod at B. Neglect the size of the collar. z B 6 ft k 3 lb/ft O 2 ft Potential Energy: Datum is set at point B. The collar is (6-2) = 4 ft above the datum when it is at. Thus, its gravitational potential energy at this point is 5(4) = 20.0 ft # lb. The length of the spring when the collar is at points and B are calculated as l and l OB = 21 2 + 3 2 + 2 2 O = 21 2 + 4 2 + 6 2 = 253 ft = 214 ft, respectively. The initial and final elastic potential energy are 1 1 and, respectively. 2 (3) 214-2B2 = 4.550 ft # lb 2 (3) 253-2B2 = 41.82 ft # lb 4 ft x 3 ft 1 ft 1 ft y Conservation of Energy: pplying Eq. 14 22, we have T + V = T B + V B 0 + 20.0 + 41.82 = 1 2 a 5 32.2 by2 B + 4.550 y B = 27.2 ft>s 512