Physics (10th Edition)

Published by Wiley
ISBN 10: 1118486897
ISBN 13: 978-1-11848-689-4

Chapter 9 - Rotational Dynamics - Problems - Page 249: 81

Answer

$T_2=4992N$ and $\alpha=6.34rad/s^2$

Work Step by Step

From figure c), according to Newton's 2nd law of motion, $$T_2-mg=ma_y$$ Because the cable pulling the crate is rolled around the smaller pulley, $a_y=l_2\alpha=0.2\alpha$. Therefore, $$T_2-mg=0.2m\alpha$$ The crate's mass $m=451kg$, $g=9.8m/s^2$, so $$T_2-4419.8=90.2\alpha$$ $$T_2-90.2\alpha=4419.8 (1)$$ Both $T_1$ and $T_2$ produce torques with lever arm $l_1$ and $l_2$ respectively. We have $$\sum\tau=T_1l_1-T_2l_2=I\alpha$$ $$T_2l_2+I\alpha=T_1l_1$$ We have $T_1=2150N$, $l_1=0.6m, l_2=0.2m, I=46kg.m^2$ $$0.2T_2+46\alpha=1290 (2)$$ Solving (1) and (2), we get $T_2=4992N$ and $\alpha=6.34rad/s^2$
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