## College Physics (4th Edition)

Published by McGraw-Hill Education

# Chapter 8 - Problems - Page 312: 22

#### Answer

The torque exerted by the rope is $57.4~N \cdot m$.

#### Work Step by Step

The magnitude of the torque can be expressed as $\tau = r~F~sin~\theta$, where $r$ is the displacement between the rotation axis and the point where the force is applied, $F$ is the force, and $\theta$ is the angle between the force vector and the displacement vector $r$. We can assume that the force of gravity pulls down on the door at the center of gravity, that is, at the midpoint. We can find the magnitude of the torque exerted by the door's weight: $\tau = r~ F~sin~\theta$ $\tau = r~ mg~sin~\theta$ $\tau = (0.825~m)(16.8~kg)(9.80~m/s^2)~sin~25^{\circ}$ $\tau = 57.4~N \cdot m$ The magnitude of the torque exerted by the door's weight is $57.4~N \cdot m$. For the door to stay in equilibrium, the rope must exert a torque of equal magnitude. The torque exerted by the rope is $57.4~N \cdot m$.

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