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Let's break down part (b) of the problem.

### Part (i)
The tension in the wire CD is given as \( 54 \, \text{N} \), and it makes an angle of \( 35^\circ \) with side BC of the sign. The question asks for the vertical component of the tension.

To find the vertical component (\( T_y \)), we use trigonometry:
\[
T_y = T \cdot \sin(35^\circ)
\]
where \( T = 54 \, \text{N} \). Plugging in the values:
\[
T_y = 54 \cdot \sin(35^\circ)
\]
Calculating this will give you the vertical component of the tension.

### Part (ii)
This part asks why the force at point B does not have a moment about point A.

Since the force at B acts vertically and is aligned with point A (directly below B on the same vertical line), the line of action of this force passes through A. For a force to create a moment about a point, it must act at some perpendicular distance from that point. Since the vertical force at B has no horizontal distance from A, it cannot generate a moment about point A.

Do you need more details on any part? 

Here are five related questions to explore further:

1. How would you calculate the horizontal component of the tension in the wire?
2. What role does the weight \( W \) of the sign play in this scenario?
3. How would you find the overall force exerted by the wall on point A?
4. What is the significance of the wire angle in determining stability?
5. How would you calculate the moment created by the weight of the sign about point A?

**Tip:** When analyzing forces and moments, always identify the line of action and check for perpendicular distances to points of interest.

Learn how to solve a statics problem involving a uniform square sign suspended by a wire. This problem explains how to calculate the vertical component of the tension and why the vertical force at point B has no moment about point A. Suitable for students in Grades 10-12.

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