MEEN 221- Chapter+17+b

MEEN 221- Chapter+17+b - ENGINEERING MECHANICS - Dynamics...

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Unformatted text preview: ENGINEERING MECHANICS - Dynamics W.F. Riley & L.D. sturges 17—17* A 3-lb collar slides on a frictionless vertical guide as shown. A constant horizontal force of 12 lb is applied to the end of the light inextensible cord that is attached to the collar. If the collar is released from rest when d = 32 in, determine the work done on the collar By the gravitational force as the collar rises 12 in. By the tension in the cord as the- collar rises 12 in. Solution a. weight force W is constant and acts vertically downward. As the 3—lb collar rises 12 in. = 1 ft, the work done by the weight force is U = Fd = —3(1) W = -3 lb'ft b. Initially, the length of cord between the collar and the pulley is L = / 322 + 242 = 40 in. 1 After the collar has risen 12 in., the length of cord between the collar and the pulley is L2 = V 202 + 242 = 31.2410 in. Therefore, the 12—lb force pulls the cord 8.759 in. to the right and does work viz 12(8.759) 105.108 lb'in. = 8.76 lb-ft 849 ENGINEERING MECHANICS - Dynamics W.F. Riley & L.D. Sturges 17-51 The two blocks shown weigh NA = 60 lb and W3 = 40 lb. If the blocks are released from rest from the position shown, determine the velocity of block B after it has moved 10 ft. Solution The length of the cord is constant, = + = L 5A ZSB constant Differentiating the length with respect to time gives = 2 VA VB where VA = 5A and VB = —53 (VA, 5A, and 53 are all positive downward while VB is positive upward). The tension is an internal force — its work will cancel out when the work—energy equations for the two blocks are added together. When block B rises 10 ft, block A will fall a distance of 20 ft, and their weights will do work on the system U = Fd = 60(20) — 40(10) = 800 lb°ft Therefore, the work-energy equation I} + U = T gives f 1 60 2 1 4o 2 + + = ———-——— ————— (0 0) 80° 2 [32.2 ](2VB) + 2 [32.2 ]VB VB = 13.56 ft/s T . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .. Ans. VA = 27.1 ft/s ~L 889 ENGINEERING MECHANICS - Dynamics W.F. Riley & L.D. Sturges 17—52* The pair of blocks shown are connected by a light inextensible bar. Both the horizontal 3kg and vertical guide slots are frictionless. If the blocks are released from rest from the position shown, determine the velocity of the 3-kg block when a. It is at the same level as the 2—kg block. b. It is 150 mm below the 2-kg block. Solution Since the length of the bar is constant, differentiating 2 2 x + y = 2x; + 2y} VB -yVA/x where VA s y (T) and VB = x (-é). Neither NA, Nb, nor the weight of the 2-kg block do work. Only the work done by the weight of the 3-kg block is needed. When A moves down a distance Ay, U = 3(9.81) Ay J Therefore, the work—energy equation T. + U = T gives 1 f 0 + 29.43 Ay [(1/2)(3)v: + (1/2)(2)v:] When y = 0, x = VA = 2.17 m/s ¢ When y = -150 mm, x = 212.4 mm, Ay = 0.39 m, VB 0 + 29.43(o.39) = 1.5V: + (0.7062vA)2 VA = 2.40 m/s l ENGINEERING MECHANICS — Dynamics W.F. Riley & L.D. Sturges 17-55* The two blocks A and B shown weigh NA = 30 lb and NE 20 lb, respectively. The kinetic coefficient of friction between block A and the inclined surface is pk = 0.15 and the horizontal surface supporting block B is smooth. when the blocks are in the position shown, block B is moving to the right with a velocity of 5 ft/s. Determine the distance traveled by block 3 before it comes to rest. Solution The length of the cord is constant, L = 25A + 53 = constant Differentiating the length with respect to time gives = . + 0 25A 53 B = zvh where VA = -5A (VA is positive to the right and EA is positive to the left) and VB = (both are positive to the right). 3 Neither block has any motion normal to the surface. Newtons second law gives +5 2F = man NA - 30(4/5) = o n = 4 b NA 2 l §k= 0.15N = 3.6 lb The tension is an internal force - its work will cancel out when the work-energy equations for the two blocks are added together. When block block B slides to the right a distance b, block A slides up the inclined surface b/2; the work done by the constant weights and friction force is U = FU = —3.6(b/2) — [30(3/5)](b/2) = —lO.80b lb'ft Therefore, the work—energy equation Ti + U = Tf 1 30 2 1 20 2 2 [ 32.2 + 2 [ 32.2 ](5) - 10.80b _ 0 gives b = 0.988 ft ................... . . . . ............ Ans. 893 ENGINEERING MECHANICS - Dynamics W.F. Riley & L.D. Sturges 17—74* A 0.5—kg mass slides on a frictionless rod in a vertical plane as shown. The undeformed length of the spring is £0 = 250 mm, the spring modulus is 5 = 600 N/m, and the distance d = 800 mm. If the slider is released from rest when b = 300 mm, determine the speed of the slider at positions A and B. Solution The normal force N does no work; the spring and weight forces have potentials. The zero of gravitational potential energy will be set at the position where b = 0. In the initial position, b = 0.3 m, the length of the spring is C = 0.5 m, the stretch of the spring is 5 = C — £0 = 0.53 0.25 m, and the initial potential energy is V = V + V = 0 5(9 Bl)(-O 3) + —l— 600 O 25 2 - 17 2785 J igs--.z()(.)—. WhenthesliderisatA,b=0m,£=0.4m,5:0.15m, the final potential energy is 2 vs = o + —:2L-(600)(0.15) = 6.75 J and the work—energy equation Ti + vi + U = Tf + V? gives 1 o + 17.2735 + o = —2-(0.5)v: + 6.75 V When the slider is at B, b the final potential energy is + v5 O.5(9.81)(O.4) + %(600)(0.15)2 = 8.712 J and the work—energy equation 13 + V: + U = Tf + V? gives 0 + 17.2785 + O = -%—(O.5)v: + 8.712 VB = 5.85 m/s . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . ... Ans. 918 ...
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MEEN 221- Chapter+17+b - ENGINEERING MECHANICS - Dynamics...

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