## Moment of a Force

In the below figure, a freely rotating pulley is acted on by two forces. Determine the magnitude for the second force that would make the pulley be in rotational equilibrium.

Hint
The question is asking for
$$F_1=F_2$$$More specifically, the question is asking for the toques/moments to be equal. $$\tau_1=\tau_2$$$
Hint 2
Torque:
$$\tau=F\cdot rsin(\theta)$$$where $$F$$ is the force, $$r$$ is the length of the moment arm, and $$\theta$$ is the angle between the force vector and moment arm. The question is asking for $$F_1=F_2$$$
More specifically, the question is asking for the toques/moments to be equal.
$$\tau_1=\tau_2$$$Remember, to find torque: $$\tau=F\cdot rsin(\theta)$$$
where $$F$$ is the force, $$r$$ is the length of the moment arm, and $$\theta$$ is the angle between the force vector and moment arm.
First, let’s analyze the torque/moment for the first force since there aren’t any unknown variables:
$$\tau_1=(2N)\cdot (0.015m)sin(90^{\circ}) =(2N)(0.015m)(1)=0.03\:N\cdot m$$$Now that both torques are known, let’s solve for $$F_2$$ using the same equation: $$\tau_1=\tau_2=F_2\cdot (0.01m)sin(125^{\circ})=0.03\:N\cdot m$$$
$$F_2=\frac{0.03N\cdot m}{(0.01m)(0.819)}$$$$$F_2=\frac{0.03N\cdot m}{(0.01m)(0.819)}=\frac{0.03N\cdot m}{0.00819m}=3.66\:N$$$
3.66 N