479 Horizontal Motion: The horizontal component of velocity ... - Wuala
479 Horizontal Motion: The horizontal component of velocity ... - Wuala
479 Horizontal Motion: The horizontal component of velocity ... - Wuala
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91962_05_R1_p0<strong>479</strong>-0512 6/5/09 3:54 PM Page 496<br />
© 2010 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently<br />
exist. No portion <strong>of</strong> this material may be reproduced, in any form or by any means, without permission in writing from the publisher.<br />
R1–23. <strong>The</strong> 2-kg spool S fits loosely on the rotating inclined<br />
rod for which the coefficient <strong>of</strong> static friction is m s = 0.2. If<br />
the spool is located 0.25 m from A, determine the maximum<br />
constant speed the spool can have so that it does not slip up<br />
the rod.<br />
z<br />
0.25 m<br />
S<br />
5<br />
4<br />
3<br />
A<br />
r = 0.25 a 4 5 b = 0.2 m<br />
; + ©F n = ma n ; N s a 3 5 b + 0.2N sa 4 v2<br />
b = 2a<br />
5 0.2 b<br />
+ c ©F b = m a b ; N s a 4 5 b - 0.2N s a 3 5 b - 2(9.81) = 0<br />
N s = 28.85 N<br />
v = 1.48 m>s<br />
Ans.<br />
*R1–24. <strong>The</strong> winding drum D draws in the cable at an<br />
accelerated rate <strong>of</strong> 5 m>s 2 . Determine the cable tension if<br />
the suspended crate has a mass <strong>of</strong> 800 kg.<br />
D<br />
s A + 2 s B = l<br />
a A =<br />
- 2 a B<br />
5 = - 2 a B<br />
a B = -2.5 m>s 2 = 2.5 m>s 2<br />
c<br />
+ c ©F y = ma y ; 2T - 800(9.81) = 800(2.5)<br />
T = 4924 N = 4.92 kN<br />
Ans.<br />
496
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