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## Q. 8.2

THE SWINGING DOOR

GOAL Apply the more general definition of torque.

PROBLEM (a) A man applies a force of $F=3.00 \times 10^{2} \mathrm{~N}$ at an angle of $60.0^{\circ}$ to the door of Figure $8.7 \mathrm{a}, 2.00 \mathrm{~m}$ from well-oiled hinges. Find the torque on the door, choosing the position of the hinges as the axis of rotation. (b) Suppose a wedge is placed $1.50 \mathrm{~m}$ from the hinges on the other side of the door. What minimum force must the wedge exert so that the force applied in part (a) won’t open the door?

STRATEGY Part (a) can be solved by substitution into the general torque equation. In part (b) the hinges, the wedge, and the applied force all exert torques on the door. The door doesn’t open, so the sum of these torques must be zero, a condition that can be used to find the wedge force.

## Verified Solution

(a) Compute the torque due to the applied force exerted at $60.0^{\circ}$.

Substitute into the general torque equation:

\begin{aligned}\tau_{F} &=r F \sin \theta=(2.00 \mathrm{~m})\left(3.00 \times 10^{2} \mathrm{~N}\right) \sin 60.0^{\circ} \\&=(2.00 \mathrm{~m})\left(2.60 \times 10^{2} \mathrm{~N}\right)=5.20 \times 10^{2} \mathrm{~N} \cdot \mathrm{m}\end{aligned}

(b) Calculate the force exerted by the wedge on the other side of the door.

Set the sum of the torques equal to zero:

$\tau_{\text {hinge }}+\tau_{\text {wedge }}+\tau_{F}=0$

The hinge force provides no torque because it acts at the axis $(r=0)$. The wedge force acts at an angle of $90.0^{\circ}$, opposite the upward $260 \mathrm{~N}$ component.

\begin{aligned}&0+F_{\text {wedge }}(1.50 \mathrm{~m}) \sin \left(-90.0^{\circ}\right)+5.20 \times 10^{2} \mathrm{~N} \cdot \mathrm{m}=0 \\&F_{\text {wedge }}=347 \mathrm{~N}\end{aligned}

REMARKS Notice that the angle from the position vector to the wedge force is $-90^{\circ}$. That’s because, starting at the position vector, it’s necessary to go $90^{\circ}$ clockwise (the negative angular direction) to get to the force vector. Measuring the angle that way automatically supplies the correct sign for the torque term and is consistent with the right-hand rule. Alternately, the magnitude of the torque can be found and the correct sign chosen based on physical intuition. Figure $8.7 \mathrm{~b}$ illustrates the fact that the com ponent of the force perpendicular to the lever arm causes the torque.