Exercício de Dinamica Aplicada (233)

Text Material Preview

PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, 
reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited 
distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, 
you are using it without permission. 
394 
PROBLEM 12.60 (Continued) 
 
(b) Assume that the block is at rest with respect to the plate. 
 : nm mΣ = + =F a W R a 
 From Part a (above), it then follows that 
 
2
2 298
3
32.2 ft/s
tan ( 50 )
( ) (13 5sin 40 ) ft/sABCD
gφ
ρ φ
− ° = =
− ° 
 or 50 5.752φ − ° = ° 
 and 55.752φ = ° 
 Now 19.29sφ = ° 
 so that sφ φ> 
 The block will slide in the slot and then 
 ,φ φ= k where tan 0.25k k kφ μ μ= = 
 or 14.0362kφ = ° 
 To determine in which direction the block will slide, consider the free-body diagrams for the two 
possible cases. 
 
 Now /plate: n Em m mΣ = + = +F a W R a a 
 From the diagrams it can be concluded that this equation can be satisfied only if the block is sliding 
downward. Then 
 
2
: cos 40 cos sin 40E
x x k
v
F ma W R mφ
ρ
Σ = ° + = ° 
 Now sin kF R φ= 
 Then 
2
cos 40 sin 40
tan
E
k
vF W
W
gφ ρ
° + = ° 
 or 
2
cos40 sin 40E
k
v
F W
g
μ
ρ
 
= − ° + °  
 
 
 1.123 lb= (see the first solution) 
 The block slides downward in the slot and 1.123 lb=F 40° 