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.