Question

1. the diagram below shows the top view of a door that is 2.0 m wide. two forces are applied to the door as indicated in the diagram. what is the net torque on the door with respect to the hinge?

Explanation:

Step1: Recall torque formula

The formula for torque is $\tau = rF\sin\theta$, where $r$ is the distance from the pivot - point (hinge), $F$ is the force, and $\theta$ is the angle between the position - vector from the pivot to the point of application of the force and the force vector.

Step2: Calculate torque due to first force

For the first force $F_1 = 10\ N$ with $r_1=1.0\ m$ and $\theta_1 = 60^{\circ}$, $\tau_1=r_1F_1\sin\theta_1=(1.0\ m)\times(10\ N)\times\sin60^{\circ}=(1.0\ m)\times(10\ N)\times\frac{\sqrt{3}}{2}=5\sqrt{3}\ N\cdot m$ (counter - clockwise, positive).

Step3: Calculate torque due to second force

For the second force $F_2 = 10\ N$ with $r_2 = 2.0\ m$ and $\theta_2=60^{\circ}$, $\tau_2=r_2F_2\sin\theta_2=(2.0\ m)\times(10\ N)\times\sin60^{\circ}=(2.0\ m)\times(10\ N)\times\frac{\sqrt{3}}{2}=10\sqrt{3}\ N\cdot m$ (clockwise, negative).

Step4: Calculate net torque

The net torque $\tau_{net}=\tau_1-\tau_2=5\sqrt{3}\ N\cdot m- 10\sqrt{3}\ N\cdot m=- 5\sqrt{3}\ N\cdot m\approx - 8.66\ N\cdot m$. The negative sign indicates a clockwise net torque.

Answer:

$-5\sqrt{3}\ N\cdot m\approx - 8.66\ N\cdot m$