A horizontal force of P kN is applied to a homogeneous body of weight 25 kN, as shown in the figure. The coefficient of friction between the body and the floor is 0.3. Which of the following statement(s) is/are correct?

Let the weight of the body be \( W = 25 \, \text{kN} \). The frictional force \( F_f \) acting on the body is given by: \[ F_f = \mu \cdot W, \] where \( \mu = 0.3 \) is the coefficient of friction. Thus: \[ F_f = 0.3 \cdot 25 = 7.5 \, \text{kN}. \]
Step 1: Calculate the torque caused by the force \( P \).
The body will tend to rotate when the horizontal force \( P \) creates a torque about the bottom edge of the body. The distance between the point of application of the force and the point of rotation (the bottom edge) is 2 m. Thus, the torque \( \tau \) produced by the applied force \( P \) is: \[ \tau = P \cdot 2. \]
Step 2: Condition for Overturning.
For the body to overturn, the torque produced by \( P \) must exceed the torque due to the frictional force. The torque due to friction is: \[ \tau_f = F_f \cdot 1 = 7.5 \, \text{kN} \cdot 1 \, \text{m} = 7.5 \, \text{kN} \cdot \text{m}. \] The force \( P \) will cause the body to overturn when: \[ P \cdot 2 \geq 7.5 \Rightarrow P \geq 3.75 \, \text{kN}. \]
Step 3: Condition for Sliding.
For sliding to occur, the applied force \( P \) must overcome the frictional force. Sliding occurs when: \[ P > F_f = 7.5 \, \text{kN}. \]
Step 4: Conclusion.
- The body will overturn when \( P \geq 3.75 \, \text{kN} \).
- The body will slide when \( P > 7.5 \, \text{kN} \).
- For \( P \leq 6 \, \text{kN} \), the body does not move because the frictional force is greater than the applied force.
Thus, the correct answers are (A), (B), and (C).
Final Answer: \[ \boxed{\text{(A), (B), and (C)}} \]
A wedge M and a block N are subjected to forces P and Q as shown in the figure. If force P is sufficiently large, then the block N can be raised. The weights of the wedge and the block are negligible compared to the forces P and Q. The coefficient of friction \( \mu \) along the inclined surface between the wedge and the block is 0.2. All other surfaces are frictionless. The wedge angle is 30°. 
The limiting force \( P \), in terms of \( Q \), required for impending motion of block N to just move it in the upward direction is given as \( P = \alpha Q \). The value of the coefficient \( \alpha \) (round off to one decimal place) is:
Consider a five-digit number PQRST that has distinct digits P, Q, R, S, and T, and satisfies the following conditions:
1. \( P<Q \)
2. \( S>P>T \)
3. \( R<T \)
If integers 1 through 5 are used to construct such a number, the value of P is:



