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Let's analyze the given problem step by step.

### Problem Summary:
We are dealing with a crate of 200 kN resting on a horizontal surface with a coefficient of static friction \( \mu_s = 0.20 \). There are three parts to solve:

1. **Find the smallest angle \( \theta \)** at which the crate will tip over about corner A.
2. **Determine the corresponding value of \( P \)**, the force applied at that tipping angle.
3. **Determine the value of \( P \)** when \( \theta = 60^\circ \), such that the crate will not slide but rather stay in equilibrium.

---

### **1. Smallest Angle \( \theta \) for Tipping:**

For tipping to occur about point A, the moment about point A due to the applied force \( P \) must equal the moment due to the weight of the crate.

- The weight of the crate acts at the center of gravity, which is located at the center of the crate. Since the crate is 1 m tall and 1 m wide (as implied by the dimensions in the figure), the center of gravity is 0.5 m from the base and 0.5 m horizontally from A.
- The force \( P \) is applied at the top of the crate, a distance of 0.5 m from the ground, and it is inclined at an angle \( \theta \) to the horizontal.

We set up the moment equation about point A:

\[
\text{Moment due to } P = P \cdot 0.5 \cdot \sin(\theta)
\]
\[
\text{Moment due to weight} = 200 \, \text{kN} \cdot 0.5
\]

For tipping, these moments must be equal:

\[
P \cdot 0.5 \cdot \sin(\theta) = 200 \cdot 0.5
\]

Simplifying:

\[
P \cdot \sin(\theta) = 200 \, \text{kN}
\]

This is the key equation for tipping. To find the smallest angle \( \theta \), we would need to compute this after finding the relation between \( P \) and \( \theta \) based on static equilibrium conditions.

### **2. Corresponding Value of \( P \) at Tipping Angle:**

From the tipping condition:

\[
P = \frac{200}{\sin(\theta)}
\]

We'll substitute the smallest \( \theta \) (found from part 1) to calculate \( P \).

### **3. Value of \( P \) such that the crate will not slide for \( \theta = 60^\circ \):**

For no sliding to occur, the force \( P \) must not exceed the maximum static friction force, which is given by:

\[
F_{\text{friction}} = \mu_s \cdot N
\]

Where \( N \) is the normal force, which equals the weight of the crate (200 kN) since there's no vertical movement.

\[
F_{\text{friction}} = 0.20 \times 200 = 40 \, \text{kN}
\]

The horizontal component of the applied force \( P \) must be less than or equal to this friction force:

\[
P \cdot \cos(60^\circ) \leq 40 \, \text{kN}
\]

This gives:

\[
P \cdot 0.5 \leq 40
\]

Solving for \( P \):

\[
P \leq 80 \, \text{kN}
\]

Thus, the force \( P \) must be less than or equal to 80 kN for the crate to not slide when \( \theta = 60^\circ \).

---

Would you like to see more detailed steps or clarifications on any of these parts? Here are some questions you could consider:

1. How can we determine the exact tipping angle \( \theta \)?
2. What happens if \( P \) exceeds the static friction limit?
3. How does the height of the force application point affect the tipping moment?
4. Can we adjust the coefficient of friction to allow a larger \( P \) before sliding?
5. What role does the weight distribution of the crate play in these calculations?

**Tip:** Always check both tipping and sliding conditions to ensure full static equilibrium when applying forces to objects on rough surfaces.

Find the smallest angle for tipping a 200 kN crate and corresponding values of P.

This problem involves calculating the smallest angle for which a 200 kN crate tips over, determining the corresponding force P, and finding the value of P to prevent sliding when θ = 60°. Learn about static equilibrium, moments, and static friction conditions. Suitable for undergraduate students.

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