We are asked to determine the magnitude of the moment about point $B$ due to force $F_1 = 355 \, \text{N}$. Based on the image, we need to find the moment (or torque) produced by $F_1$ about point $B$.

Key Information:

• Force $F_1 = 355 \, \text{N}$ acts vertically downward.
• Distance from point $A$ to point $B$ is $0.8 \, \text{m} + 0.6 \, \text{m} = 1.4 \, \text{m}$.
• The perpendicular distance between the line of action of $F_1$ and point $B$ is the horizontal distance from $A$ to $B$, which is $1.4 \, \text{m}$.

Moment about point $B$

The formula for the moment $M_B$ about a point is given by: $M_B = F \times d_{\perp}$ where:

• $F$ is the force applied,
• $d_{\perp}$ is the perpendicular distance from the point to the line of action of the force.

Here:

• $F_1 = 355 \, \text{N}$,
• $d_{\perp} = 1.4 \, \text{m}$.

Thus, the moment $M_B$ is: $M_B = 355 \, \text{N} \times 1.4 \, \text{m} = 497 \, \text{N-m}$

Final Answer:

The magnitude of the moment about point $B$ is $497 \, \text{N-m}$.

Would you like more details or have any further questions?

Here are 5 related questions:

1. What is the physical significance of a moment or torque in mechanics?
2. How would the moment change if $F_1$ was applied at an angle?
3. What is the impact of adding force $F_3$ in this problem?
4. How do we calculate moments when forces are applied at different angles?
5. What would be the moment if the distance from $A$ to $B$ was changed?

Tip: The moment arm is the perpendicular distance from the line of action of the force to the point of rotation. Always ensure you're using this distance when calculating moments.