# What is the moment of force about point O? (Refer to the image below)

$2490.9696 \setminus \setminus \textrm{l b f e e t}$

#### Explanation:

Resolving the force of $600 \setminus l b$ into parallel & vertical components with respect to the axis of given rod/link.

The total torque about the point O in anticlockwise direction,

$= \left(600 \setminus l b\right) \setminus \cos \left({30}^{\setminus} \circ + {20}^{\setminus} \circ\right) \setminus \times \left(0.5 \setminus \setminus \textrm{f e e t}\right) + \left(600 \setminus l b\right) \setminus \sin \left({30}^{\setminus} \circ + {20}^{\setminus} \circ\right) \setminus \times \left(5 \setminus \setminus \textrm{f e e t}\right)$

$= 600 \left(0.5 \setminus \cos {50}^{\setminus} \circ + 5 \setminus \sin {50}^{\setminus} \circ\right) \setminus \setminus \textrm{l b f e e t}$

$= 2490.9696 \setminus \setminus \textrm{l b f e e t}$