## Physics (10th Edition)

Published by Wiley

# Chapter 9 - Rotational Dynamics - Problems - Page 245: 27

#### Answer

a) $T=1.6\times10^4N$ b) $P=2.17\times10^4N$

#### Work Step by Step

1) The net force is zero: $$\sum F_x=P_x-T_x=0 (1)$$ $$\sum F_y=P_y-T_y-W_1-W_2=0$$ $$P_y-T_y=8400 (2)$$ 2) The net torque is zero: We choose the axis of rotation to be at P, so $P$ creates no torque. The length of the boom is $L$, and the center of gravity of the boom is at the midpoint of the boom. We have $$\sum\tau=-\tau_{W_1}-\tau_{W_2}+\tau_{T}=0$$ $$-W_1(0.5L)\cos48-W_2L\cos48+TL\sin(48-32)=0$$ $$-W_1(0.5\cos48)-W_2\cos48+T\sin(16)=0$$ $$T=\frac{W_1(0.5\cos48)+W_2\cos48}{\sin16}$$ We have $W_1=3600N$ and $W_2=4800N$, so $T=1.6\times10^4N$ (1): $P_x=T\sin58=1.36\times10^4N$ (2): $P_y=T\cos58+8400N=1.69\times10^4N$ So $P=\sqrt{P_x^2+P_y^2}=2.17\times10^4N$

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