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To solve the problem of determining the placement \( x \) of the hook at \( B \) such that the force \( P = 4 \, \text{kN} \) creates a maximum moment about point \( O \), we can follow these steps:

### Given Data:
- Force \( P = 4 \, \text{kN} \)
- Length of the crane boom \( L = 9.2 \, \text{m} \)
- Angle \( \theta = 30^\circ \)
- Distance from point \( O \) to the base of the crane (on the ground) = 1 m

### Step 1: Understanding the moment (torque) equation

The moment \( M_O \) about point \( O \) due to the force \( P \) is given by:
\[
M_O = P \times d_{\perp}
\]
where \( d_{\perp} \) is the perpendicular distance from point \( O \) to the line of action of force \( P \). To maximize this moment, we need to find \( x \), the horizontal distance of point \( B \) from \( O \).

### Step 2: Determine \( d_{\perp} \)

From the geometry of the crane:

- The crane boom makes an angle \( \theta = 30^\circ \) with the horizontal.
- The total length of the crane boom is \( L = 9.2 \, \text{m} \).

The perpendicular distance from point \( O \) to the line of action of \( P \), denoted by \( d_{\perp} \), is the vertical component of the boom length. This can be found using trigonometry:
\[
d_{\perp} = L \sin(\theta)
\]
Substitute the given values:
\[
d_{\perp} = 9.2 \times \sin(30^\circ) = 9.2 \times 0.5 = 4.6 \, \text{m}
\]

### Step 3: Relate \( x \) and the geometry

To maximize the moment about \( O \), the force \( P \) should act at the farthest possible distance horizontally from point \( O \). Therefore, we need to account for the geometry of the crane. From the diagram, we see the total horizontal distance \( x \) of point \( B \) is the sum of the distance from point \( O \) to the base (1 m) and the horizontal component of the boom’s length:
\[
x = 1 + L \cos(\theta)
\]
Substitute the values:
\[
x = 1 + 9.2 \times \cos(30^\circ) = 1 + 9.2 \times 0.866 = 1 + 7.97 = 8.97 \, \text{m}
\]

Thus, the placement of the hook at \( B \) to maximize the moment about point \( O \) is approximately **8.97 m**.

### Final Answer:
\[
x = 8.97 \, \text{m}
\]

Would you like further details or clarifications on any part of the solution?

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Here are 5 related questions that expand upon this problem:
1. How would the maximum moment change if the force \( P \) was applied at a different angle?
2. What happens to the moment if the length of the crane boom is increased?
3. How can the moment be minimized instead of maximized?
4. How does the angle \( \theta \) influence the perpendicular distance and the moment?
5. How would the solution change if the force \( P \) was less than 4 kN?

**Tip**: Remember, moments are maximized when forces act as far as possible from the pivot point and perpendicular to the line connecting the pivot to the point of application.

The towline exerts a force of P = 4 kN at the end of the L = 9.2 m crane boom. If θ = 30 degrees, determine the placement x of the hook at B so that this force creates a maximum moment about point O.

Learn how to determine the placement of a hook at point B on a crane boom to maximize the moment about point O. This problem involves a 4 kN force, a crane boom of length 9.2 m, and an angle of 30 degrees. We cover key concepts such as torque, moments, and trigonometry.

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