## 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**

**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**

**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