Determine the resultant moment produced by the forces about point O.

Steps to Solve

1. Calculate the moment due to $F_1$ about point O

The moment due to a force is given by the product of the force and the perpendicular distance from the line of action of the force to the point about which the moment is being calculated.

$F_1$ is 300 lb and acts along the inclined member. The perpendicular distance from O to the line of action of $F_1$ is 6 ft

The moment due to $F_1$ is:

$M_1 = F_1 \times d = 300 \text{ lb} \times 6 \text{ ft} = 1800 \text{ lb.ft}$

The moment $M_1$ is counterclockwise (positive).

2. Calculate the moment due to $F_2$ about point O

The moment due to $F_2$ about point O requires breaking the distance into horizontal and vertical components

The horizontal distance is 6 ft. The perpendicular distance between $F_2$ and point O is $6 \cos(30^\circ)$.

The moment due to $F_2$ is

$M_2 = F_2 \times d = 200 \text{ lb} \times 6\cos(30^\circ) \text{ ft} = 200 \text{ lb} \times 6 \times \frac{\sqrt{3}}{2} \text{ ft} = 600\sqrt{3} \text{ lb.ft} \approx 1039.23 \text{ lb.ft}$

The moment $M_2$ is clockwise (negative).

3. Calculate the resultant moment about point O

The resultant moment is the sum of the individual moments. $M_R = M_1 + M_2 = 1800 \text{ lb.ft} - 1039.23 \text{ lb.ft} = 760.77 \text{ lb.ft}$

The resultant moment is counterclockwise since $M_R$ is positive.