To determine the moment about the $x$-axis ($M_x$) caused by the force $F$ acting at point $A$, we'll need to break the force into its components and then calculate the moment using the moment-arm principle. Here's how we can approach this problem step by step:

Given:

• Force $F = 166 \, \text{N}$
• Angle $\theta = 64^\circ$ (the angle the force makes with the $y$-axis in the $yz$-plane)
• Distance $a = 923 \, \text{mm} = 0.923 \, \text{m}$
• Distance $b = 755 \, \text{mm} = 0.755 \, \text{m}$

Step 1: Resolve the Force into Components

Since the force lies in the $yz$-plane, its components will be along the $y$-axis and the $z$-axis. Using trigonometry, we find:

• $F_y = F \cos(\theta)$
• $F_z = F \sin(\theta)$

Substitute the given values:

$F_y = 166 \cos(64^\circ) = 166 \times 0.43837 \approx 72.77 \, \text{N}$ $F_z = 166 \sin(64^\circ) = 166 \times 0.89879 \approx 149.20 \, \text{N}$

Step 2: Moment Arm and Moment Calculation

The moment about the $x$-axis is caused by the force components $F_y$ and $F_z$ acting at point $A$.

• The moment arm for $F_y$ is the distance along the $z$-axis, which is $a$.
• The moment arm for $F_z$ is the distance along the $y$-axis, which is $b$.

Now, calculate the moment contributions:

• Moment due to $F_y$: $M_{x1} = F_y \times a$
• Moment due to $F_z$: $M_{x2} = F_z \times b$

Substitute the values:

$M_{x1} = 72.77 \, \text{N} \times 0.923 \, \text{m} \approx 67.15 \, \text{N} \cdot \text{m}$ $M_{x2} = 149.20 \, \text{N} \times 0.755 \, \text{m} \approx 112.64 \, \text{N} \cdot \text{m}$

Step 3: Combine the Moments

Since $F_y$ and $F_z$ act in opposite directions relative to the $x$-axis, the total moment about the $x$-axis will be:

$M_x = M_{x2} - M_{x1} = 112.64 \, \text{N} \cdot \text{m} - 67.15 \, \text{N} \cdot \text{m} = 45.49 \, \text{N} \cdot \text{m}$

Thus, the component of the moment about the $x$-axis is approximately:

$\boxed{M_x = 45.5 \, \text{N} \cdot \text{m}}$

Would you like more details or have any questions?

Here are some follow-up questions to explore further:

1. How would the moment change if the force were applied at a different point along the $y$-axis?
2. How do you calculate the total moment when a force has components in all three axes?
3. What effect would reducing the angle $\theta$ have on the moment about the $x$-axis?
4. Can the moment about an axis ever be zero? If so, under what conditions?
5. What is the physical significance of the moment being counterclockwise or clockwise?

Tip: Always use the right-hand rule to determine the direction of the moment: curl your fingers in the direction of the applied force, and your thumb will point in the direction of the moment vector.