## College Physics (4th Edition)

Published by McGraw-Hill Education

# Chapter 8 - Problems - Page 311: 20

#### Answer

(a) The net torque is $F~d$ (b) The net torque is $F~d$

#### Work Step by Step

(a) The torque is $\tau = r\times F$, where $r$ is the smallest distance between the force vector and the rotation axis, and $F$ is the perpendicular component of the force vector. The upward force exerts a counterclockwise torque. We can find the magnitude of this torque: $\tau_1 = x_1~F$ The downward force exerts a clockwise torque. We can find the magnitude of this torque: $\tau_2 = x_2~F$ We can find the magnitude of the net torque: $\vert \tau \vert = \vert \tau_2-\tau_1 \vert$ $\vert \tau \vert = \vert x_2~F-x_1~F \vert$ $\vert \tau \vert = \vert F~(x_2-x_1) \vert$ $\vert \tau \vert = F~d$ The net torque is $F~d$ (b) The upward force exerts a counterclockwise torque. We can find the magnitude of this torque: $\tau_1 = x_1~F$ The downward force exerts a clockwise torque. We can find the magnitude of this torque: $\tau_2 = x_2~F$ We can find the magnitude of the net torque: $\vert \tau \vert = \vert \tau_2-\tau_1 \vert$ $\vert \tau \vert = \vert x_2~F-x_1~F \vert$ $\vert \tau \vert = \vert F~(x_2-x_1) \vert$ $\vert \tau \vert = F~d$ The net torque is $F~d$.

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