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### Given dataLet's solve this problem step by step.

### Given 
- \( T = 28.3 \, \text{kN} \) (Tension force)
- \( W = 25 \, \text{kN} \) (Weight force)
- Distances:
  - Horizontal distance between point \( B \) and the point of application of \( T \) is 20 m.
  - Horizontal distance between point \( B \) and the vertical line where \( W \) acts is 16 m.
  - The vertical height from the top of the crane to point \( W \) is 36 m.

### Objective:
We need to find the moment about point \( B \) due to:
- (a) Force \( T \)
- (b) Force \( W \)
- (c) Both forces combined.

### Step-by-Step Solution:

#### (a) Moment due to the force \( T \)
- The moment \( M_T \) is calculated as:
  \[
  M_T = T \times d_T
  \]
  where \( d_T \) is the perpendicular distance from point \( B \) to the line of action of \( T \).

  Since \( T \) acts horizontally and its line of action is 20 m away from point \( B \):
  \[
  M_T = 28.3 \, \text{kN} \times 20 \, \text{m} = 566 \, \text{kN} \cdot \text{m}
  \]
  The sense of rotation would be **counterclockwise** (positive) about point \( B \).

#### (b) Moment due to the force \( W \)
- The moment \( M_W \) is calculated as:
  \[
  M_W = W \times d_W
  \]
  where \( d_W \) is the horizontal distance between point \( B \) and the vertical line of action of \( W \). From the diagram, \( d_W = 16 \, \text{m} \).

  \[
  M_W = 25 \, \text{kN} \times 16 \, \text{m} = 400 \, \text{kN} \cdot \text{m}
  \]
  The sense of rotation would be **clockwise** (negative) about point \( B \).

#### (c) Combined moment due to both \( T \) and \( W \)
- The combined moment \( M_{\text{combined}} \) is the sum of the individual moments:
  \[
  M_{\text{combined}} = M_T - M_W
  \]
  Substituting the values:
  \[
  M_{\text{combined}} = 566 \, \text{kN} \cdot \text{m} - 400 \, \text{kN} \cdot \text{m} = 166 \, \text{kN} \cdot \text{m}
  \]
  The combined moment would be **counterclockwise** (positive) about point \( B \).

### Final Results:
- (a) Moment due to \( T \): \( 566 \, \text{kN} \cdot \text{m} \) (counterclockwise).
- (b) Moment due to \( W \): \( 400 \, \text{kN} \cdot \text{m} \) (clockwise).
- (c) Combined moment: \( 166 \, \text{kN} \cdot \text{m} \) (counterclockwise).

Let me know if you want any more details, and feel free to ask questions!

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Here are 5 questions to deepen your understanding:
1. How would the moments change if the force \( T \) were applied at an angle instead of horizontally?
2. What if the distance between \( B \) and \( W \) were shorter—how would that affect the moments?
3. Why is it important to consider the sense (direction) of the moment in problems like this?
4. How would the combined moment change if the magnitudes of \( T \) and \( W \) were reversed?
5. Can the moment be zero if both forces are applied at different points? Why or why not?

### Tip:
Remember that moments are vectors and always include both magnitude and direction (sense). This is crucial when combining moments!

Given that T = 28.3 kN and W = 25 kN, determine the magnitude and sense of the moments about point B of the following: (a) the force T; (b) the force W; and (c) forces T and W combined.

This detailed solution explains how to determine the magnitude and sense of moments about point B for forces T = 28.3 kN and W = 25 kN. It covers key concepts of moments in statics, force interactions, and combining moments. Ideal for engineering and physics students.

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