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: \[ \boxed{\text{(A), (B), and (C)}} \]

| Point | Staff Readings Back side | Staff Readings Fore side | Remarks |
|---|---|---|---|
| P | -2.050 | - | 200.000 |
| Q | 1.050 | 0.95 | Change Point |
| R | - | -1.655 | - |