## College Physics (4th Edition)

Published by McGraw-Hill Education

# Review & Synthesis: Chapters 6-8 - Review Exercises - Page 323: 16

#### Answer

The horizontal component of the force exerted on each hinge due to the door is $14.2~N$ The vertical component of the force exerted on each hinge due to the door is $27.4~N$

#### Work Step by Step

Let's consider the torques about a rotation axis at the bottom hinge. If the door's weight exerts a clockwise torque about this axis, then the horizontal force $F_x$ of the top hinge exerts a counterclockwise torque about this axis. Since the system is in equilibrium, the magnitudes of the clockwise and counterclockwise torques must be equal in magnitude. We can find the horizontal force $F_x$ exerted by the top hinge: $\tau_{ccw} = \tau_{cw}$ $r~F_x = (\frac{width}{2})~mg$ $(2.030~m-0.280~m-0.280~m)~F_x = (\frac{0.760~m}{2})(5.60~kg)(9.80~m/s^2)$ $F_x = \frac{(0.380~m)(5.60~kg)(9.80~m/s^2)}{1.47~m}$ $F_x = 14.2~N$ The horizontal force exerted by the top hinge is $14.2~N$. Since the system is in equilibrium, the sum of the horizontal forces in the system must be zero. Therefore, the horizontal force exerted by the bottom hinge is $14.2~N$ in the opposite direction. The horizontal component of the force exerted on each hinge due to the door is $14.2~N$ Since the system is in equilibrium, the sum of the vertical forces in the system must be zero. Then the sum of the vertical forces exerted by each hinge must be equal in magnitude to the door's weight. We can find the vertical force $F_y$ exerted by each hinge: $2F_y = mg$ $F_y = \frac{mg}{2}$ $F_y = \frac{(5.60~kg)(9.80~m/s^2)}{2}$ $F_y = 27.4~N$ The vertical component of the force exerted on each hinge due to the door is $27.4~N$.

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