hw21_sol - ENGINEERING MECHANICS - Dynamics W.F. Riley...

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Unformatted text preview: ENGINEERING MECHANICS - Dynamics W.F. Riley & L.D. Sturges 15-4* A 15-kg block of ice slides on a horizontal surface for 20 m before it stops. If the initial speed of the block was 15 m/s, determine The force of friction between the block of r————————QOm——————————j ice and the surface. lSnV Omk The kinetic coefficient of friction Mk between the block of ice and the surface. Solution The free-body diagram includes the weight of the block W = 15(9.81) = 147.15 N, the normal force N and friction force F exerted on the block by the surface. The equations of motion are +6 2F ma : -F X X +T 2F ma N — 147.15 0 y y where ay = 0 since there is no motion in the vertical direction. Therefore N 147.15 N 15a -F Rewriting the x-component of acceleration using the chain—rule of differentiation and integrating 15v dv 2 15V /2 -Fw + C = 1687.5 - Ex where the constant of integration has been chosen so that v = 15 m/s when x = 0. Then, if v = 0 when x = 20 m, F = 84.4 N . . . . . . ... . . . . . ... . . . . . . . . . . . ...... . . . . . . . .... Ans. and the kinetic coefficient of friction is u F/N = 0.573 . . . . . . . . . . . . . . . . . . .... . . . . ... .. Ans. k 429 ENGINEERING MECHANICS - Dynamics W.F. Riley & L.D. Sturges 15-6 A 1500—kg automobile is moving along a level road at a constant speed of 60 km/h. If the automobile accelerates at a constant rate and reaches a speed of 80 km/h in 5 s, determine a. The force required to produce this acceleration. The distance traveled by the automobile during the 5—5 interval that it is accelerating. Solution The free—body diagram includes the weight of the car an the normal force N and pushing force F exerted on the tires by the road. Only the x—component of the equation of motion gives useful information +9 SE} = max: F = 1500a = constant Integrating the acceleration to get the velocity and position gives 1'! 1500V Ft + C1 = Ft + 25,000 H 2 2 Ft /2 + 25,000t + C = Ft /2 + 25,000t 1500 x 2 where the constants of integration have been chosen so that 60 km/h = 16.6667 m/s and x = 0 when t = 0. Then if v = 80 km/h = 22.2222 m/s when t = 5 s, 1500(22.2222) = F(5) + 25,000 F 1667 . . . . . . . . . . . . . .. 833.33(5)2 + 25,000(5) 97.2 m . . . . . . . . . . . . .. 431 ENGINEERING MECHANICS - Dynamics W.F. Riley & L.D. Sturges 15-7* A force of 20 lb is applied to a 25—lb block as shown in Fig. P15—7. Let x = O and V = 0 when t = O and determine the velocity and displacement of the block at t = 5 5 if F The inclined plane supporting the block is smooth. The kinetic coefficient of friction between the inclined plane and the block is uk = 0.25. Solution The free—body diagram includes the weight of the block W, the normal force N and friction force R exerted on the block by the surface, and the applied force F. In terms of coordinates along and normal to the surface, the equations of motion are +2 2F = ma : 2O — 25 sin 200 — R x x +5 2F = ma : N — 25 cos 20° 0 Y Y where ay = 0 since there is no motion in the direction normal to the surface. Therefore N = 23.4923 lb a. If R = 0, then the x—component of the equations of motion gives a = 14.747 ft/s2 = constant and integrating to get the velocity and position gives v = 14.747t + C 14.747t 1 2 2 x = 7.3735t + 02 7.3735t where the constants of integration are both zero since the block starts from rest at x = 0. Then, when t = 5 s V = 73.7 ft/s x = 184.3 ft If R = 0.25N = 5.8731 1b, then 2 a 7.1824 ft/s constant v = 7.1824t + C 7.1824t 3 x 3.5912t2 + c; 3.5912t2 and at t = 5 s, 35.9 ft/s 89.8 ft ....... . . . . . . . . . . ....................... Ans. 432 ENGINEERING MECHANICS — Dynamics W.F. Riley & L.D. Sturges 15-10 A Saturn V rocket has a mass 6 of 2.75(10 ) kg and a thrust of 6 33(10 ) N. Determine The initial vertical acceleration of the rocket. The rocket's velocity 10 s after liftoff. The time required to reach an altitude of 10,000 m. Solution A free-body diagram of the rocket includes its weight W and the thrust T. Only the ybcomponent of the equations of motion gives useful information +T 2F = : T — mg = ma y y 6 a. At lift—off, m = 2.75(1o ) kg, 6 2 33(10 ) N, g = 9.81 m/s , and the initial vertical acceleration of the rocket is 2 a = 2.19 m/s b. Assuming that the thrust and the mass both remain constant, the acceleration is also constant. The first (T = constant) is probably a good assumption but the second (m = constant) is probably not a good assumption unless the time interval is very short. Then, integrating the acceleration to get the velocity of the rocket gives 2 a = 2.190 m/s = constant v = 2.190t m/s (Problem 15-10 continues ...) 435 ENGINEERING MECHANICS — Dynamics W.F. Riley & L.D. Sturges (Problem 15—10 - cont.) where the constant of integration is zero since the rocket starts from rest when t = 0. And when t = 10 s, V = (2.190)(10) = 21.9 m/s ......................... Ans. c. Integrating the velocity to get the position of the rocket gives 2 y = 1.095t m where the constant of integration is again zero since y = 0 when t = 0. Then the rocket will reach an altitude of 10,000 m when 2 10,000 1.095t t 95.6 s (Over such a very long time interval, the rocket will have burned up a large amount of fuel and its weight and mass will have decreased significantly, Therefore, the constant mass assumption is very poor in this case and the answer is not likely to be very accurate.) 436 ENGINEERING MECHANICS - Dynamics W.F. Riley & L.D. Sturges 15—21* Blocks A and B, which weigh 30 lb and 50 lb, respectively, are connected by a rope as shown. The kinetic coefficients of friction pk are 0.35 for block A and 0.15 for block 3. During motion of the blocks down the inclined plane, determine a. The acceleration of block 8. b. The tension in the rope. Solution So long as the rope remains taut, the two blocks will move together and a single free-body diagram may be drawn of the two blocks or separate free—body diagrams may be drawn for each of the two blocks. Choosing to draw separate free—body diagrams results in the equations of motion +2 EEAX 3o sin 350 + T - Eh m a : A Ax + E . - o l FAY V NA 30 cos 35 + E o ‘ 35o — - E FBX 5 Elm T F3 Fe. 0 +1 2F m a : N - 50 cos 35 0 By B. By B where the yhcomponents of acceleration are both zero since there is no motion in the direction normal to the surface and the x—components of acceleration are the same since the blocks move together. Therefore 24. 75 . . NA 5 lb Eh 0 35Nh 8 6011 lb N 40.95 b . . B 8 1 FB 0 15Nfi 6 1436 lb Finally, adding the x—equations together gives 80 sin 35° — 14.7447 = 12.53 ft/s2 E 3.07 lb 450 ENGINEERING MECHANICS - Dynamics W.F. Riley & L.D. Sturges 15—23 Blocks A and B, which weigh 200 lb and 120 lb, respectively, are connected by a rope as shown in Fig. a. During motion of the bodies, Determine the acceleration of block A and the tension in the rope connecting the bodies. Determine the acceleration of block A if block 8 is replaced with a constant force of 120 lb as shown in Fig. b. Solution a. Since the two bodies move in different directions (A moves down and 8 moves up), separate free—body diagrams must be drawn for the two bodies. The equations of equilibrium give +T 2F A Y +T EFBY Measuring the position of both blocks from the ceiling, the length of the rope can be expressed = + L 25A 53 Taking two time derivatives of this relationship and noting that the length of the rope is constant gives +00 = 25A SB 0 h re = — a = — u w e 3A aA nd 33 a3 (5A and 53 are positive downward while aA and a8 are positive upward). Therefore (Problem 15-23 continues ...) 452 ENGINEERING MECHANICS - Dynamics W.F. Riley & L.D. Sturges (Problem 15-23 - cont.) (-a3) = -2(-aA) and combining Eqs. 3 and b gives 200 = — = + 2T 200 + 32.2 3A 240 3A 1.894 ft/s2 T T = 105.9 lb b. If block 3 is replaced with a constant force of 120 lb, then the tension in the cable is just T = 120 lb and Eq. a gives the acceleration of block A 200 2(120) — 200 32.2 a = 6.44 ft/s2 T A 453 ENGINEERING MECHANICS - Dynamics W.F. Riley & L.D. Sturges 15-57 A circular disk rotates in a horizontal plane. A.3-lb block rests on the disk 8 in. from the axis of rotation. The static coefficient of friction between the block and the disk is 0.50. If the disk starts from rest with a constant angular acceleration of 0.5 rad/sz, determine the length of time required for the block to begin to slip. Solution The free-body diagram of the block includes its weight W, a normal force N exerted on the block by the surface, and a friction force exerted on the block by the surface. The friction force is represented in terms of its n— and t—components. The equations of motion are e+ 2F n +T 2F 0 z where the z-component of acceleration is zero since the block has no vertical motion, 2 a 0.5 rad/s = constant w 0.5t rad/s Therefore 3 lb 0.03106 lb 0.01553tz lb 2 2 3 ' (0.03106) + (0.01553: ) _<_ (0.50M)2 = (1.5115)2 and the block will begin to slip when t = 9.83 s 506 ENGINEERING MECHANICS - Dynamics W.F. Riley & L.D. Sturges 15-60 The car shown maintains a constant speed of 100 km/h. At both the bottom and top of the hill, determine the force that the car seat exerts on an BO-kg driver. Solution The free—body diagram of the driver includes his weight W and the normal force N between the driver and the seat. Using coordinates along and normal to the road, the equations of motion are +e 2F : = 80' t mat 0 v +T EFL = ma : N - 80(9.81) = 80a n n 2 where the normal component of acceleration has magnitude v /p and points toward the center of curvature. a. When the car goes over the top of a hump in the road, then = 100 km/h = 27.7778 m/s and p = 90 m. Therefore 2 27.7778 ] = 98.9 N N = 80 [9.81 - 90 is only about 13 percent of the driver's weight! When the car goes over the bottom of a dip in the road, then 2 2 = 100 km/h = 27.7778 m/s and p = 90 m. 2 N = 80 [9.81 + 1237—8] = 1471 N is about 1.87 times the driver's weight! 509 ENGINEERING MECHANICS - Dynamics W.F. Riley & L.D. Sturges 15-61 A highway is designed for traffic moving at 65 mi/h. Along a certain portion of the highway, the radius of a curve is 900 ft. If the curve is banked so that no friction is required to keep cars on the road, determine a. The required angle of banking (angle 9) of the road; b. The minimum coefficient of friction between the tires and the road that would keep traffic from skidding at this speed if the Curve were not banked. Solution The free—body diagram of the car includes its weight W and the normal force N and friction force F exerted on the tires by the road. Using coordinates along and normal to the path of the car in the horizontal plane of motion, the equations of motion are 2 9+ ZFn man F'cos 6 + N sin 6 = m(95.3333 /900) +T SF; ma : N cos 6 — F sin 9 - mg = O z where the normal component of acceleration is an = vz/p acting toward the center of curvature (horizontally to the left), v = 65 mi/h = 95.3333 ft/s, and the z-component of acceleration is zero since the car has no vertical motion. a. If F = 0, then N sin 9 = 10.0983m N cos 6 = 32.2m 6 = tan—1(1O.0983/32.2) O b. If 9 = O , then F = 10.0983m N = 32.2m u = F/N = 10.0983/32.2 510 ...
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hw21_sol - ENGINEERING MECHANICS - Dynamics W.F. Riley...

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