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 $r$ and $F$.

Step2: Calculate torque due to first force

For the force of $F_1 = 10\ N$ at $r_1=1.0\ m$ and $\theta_1 = 60^{\circ}$, $\tau_1=r_1F_1\sin\theta_1$. Substituting the values, we get $\tau_1=(1.0\ m)\times(10\ N)\times\sin60^{\circ}=(1.0\ m)\times(10\ N)\times\frac{\sqrt{3}}{2}=5\sqrt{3}\ N\cdot m$. The direction of this torque is counter - clockwise.

Step3: Calculate torque due to second force

For the force of $F_2 = 10\ N$ at $r_2 = 2.0\ m$ and $\theta_2=60^{\circ}$, $\tau_2=r_2F_2\sin\theta_2$. Substituting the values, we get $\tau_2=(2.0\ m)\times(10\ N)\times\sin60^{\circ}=(2.0\ m)\times(10\ N)\times\frac{\sqrt{3}}{2}=10\sqrt{3}\ N\cdot m$. The direction of this torque is clockwise.

Step4: Calculate net torque

The net torque $\tau_{net}=\tau_2-\tau_1$. Substituting the values of $\tau_1$ and $\tau_2$, we have $\tau_{net}=10\sqrt{3}-5\sqrt{3}=5\sqrt{3}\ N\cdot m\approx8.66\ N\cdot m$. The positive value indicates that the net torque is in the clockwise direction.

Answer:

$5\sqrt{3}\ N\cdot m\approx8.66\ N\cdot m$ (clockwise)