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 497<br />
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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 />
R1–25. <strong>The</strong> bottle rests at a distance <strong>of</strong> 3 ft from the center<br />
<strong>of</strong> the <strong>horizontal</strong> platform. If the coefficient <strong>of</strong> static friction<br />
between the bottle and the platform is m s = 0.3, determine<br />
the maximum speed that the bottle can attain before<br />
slipping. Assume the angular motion <strong>of</strong> the platform is<br />
slowly increasing.<br />
3 ft<br />
©F b = 0; N - W = 0 N = W<br />
Since the bottle is on the verge <strong>of</strong> slipping, then F f = m s N = 0.3W.<br />
©F n = ma n ; 0.3W = a W<br />
32.2 bay2 3 b<br />
y = 5.38 ft>s<br />
Ans.<br />
R1–26. Work Prob. R1–25 assuming that the platform<br />
starts rotating from rest so that the speed <strong>of</strong> the bottle is<br />
increased at 2 ft>s 2 .<br />
3 ft<br />
Applying Eq. 13–8, we have<br />
©F b = 0; N - W = 0 N = W<br />
Since the bottle is on the verge <strong>of</strong> slipping, then F f = m s N = 0.3W.<br />
©F t = ma t ; 0.3W sin u = a W<br />
32.2 b(2)<br />
©F n = ma n ; 0.3W cos u = a W<br />
32.2 bay2 3 b<br />
[1]<br />
[2]<br />
Solving Eqs. [1] and [2] yields<br />
y = 5.32 ft>s<br />
Ans.<br />
u = 11.95°<br />
497