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

Steps to Solve

1. Resolve F1 into horizontal and vertical components

We need to find the $x$ and $y$ components of $F_1$. Given that $F_1 = 300 \text{ lb}$ and the angle is $30^\circ$, we have: $F_{1x} = F_1 \cos(30^\circ) = 300 \cos(30^\circ) = 300 \cdot \frac{\sqrt{3}}{2} = 150\sqrt{3} \text{ lb}$ $F_{1y} = F_1 \sin(30^\circ) = 300 \sin(30^\circ) = 300 \cdot \frac{1}{2} = 150 \text{ lb}$

2. Calculate the moment due to F2

The force $F_2 = 200 \text{ lb}$ acts horizontally at a distance of $6 \text{ ft}$ from point O. The moment $M_2$ due to $F_2$ is: $M_2 = F_2 \cdot d = 200 \text{ lb} \cdot 6 \text{ ft} = 1200 \text{ lb.ft}$ (clockwise)

3. Calculate the moment due to the x-component of F1

The horizontal component of $F_1$ i.e. $F_{1x} = 150\sqrt{3} \text{ lb}$, acts at a distance of $6 \text{ ft}$ vertically from point O. Therefore, the moment $M_{1x}$ due to $F_{1x}$ is: $M_{1x} = F_{1x} \cdot d = 150\sqrt{3} \text{ lb} \cdot 6 \text{ ft} = 900\sqrt{3} \text{ lb.ft}$ (counter-clockwise)

4. Calculate the moment due to the y-component of F1

The vertical component of $F_1$ i.e. $F_{1y} = 150 \text{ lb}$, acts at a distance of $6\cos(30^\circ) \text{ ft}$ horizontally from point O. Therefore, the moment $M_{1y}$ due to $F_{1y}$ is: $M_{1y} = F_{1y} \cdot d = 150 \text{ lb} \cdot 6\cos(30^\circ) \text{ ft} = 150 \cdot 6 \cdot \frac{\sqrt{3}}{2} = 450\sqrt{3} \text{ lb.ft}$ (clockwise)

5. Calculate the resultant moment

The resultant moment $M_R$ is the sum of all moments about point O, taking clockwise moments as negative and counter-clockwise moments as positive: $M_R = -M_2 + M_{1x} - M_{1y} = -1200 + 900\sqrt{3} - 450\sqrt{3} = -1200 + 450\sqrt{3} \text{ lb.ft}$ $M_R = -1200 + 450\sqrt{3} = -1200 + 450(1.732) = -1200 + 779.4 = -420.6 \text{ lb.ft}$

6. State the final answer with direction

Since the resultant moment is negative, it acts in the clockwise direction. $M_R = -420.6 \text{ lb.ft} \approx 421 \text{ lb.ft}$ (clockwise)