MEEN 221- Chapter+17+b

# MEEN 221- Chapter+17+b - ENGINEERING MECHANICS Dynamics W.F...

This preview shows pages 1–5. Sign up to view the full content.

This preview has intentionally blurred sections. Sign up to view the full version.

View Full Document

This preview has intentionally blurred sections. Sign up to view the full version.

View Full Document
This is the end of the preview. Sign up to access the rest of the document.

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 ...
View Full Document

## This homework help was uploaded on 03/25/2008 for the course MEEN 221 taught by Professor Mcvay during the Spring '08 term at Texas A&M.

### Page1 / 5

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

This preview shows document pages 1 - 5. Sign up to view the full document.

View Full Document
Ask a homework question - tutors are online