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### Part (a) Maximum mass of piece C (\(m_C\))

We are asked to calculate the maximum mass of \( C \) that can cause both blocks \( A \) and \( B \) to move together without slipping. To solve this, we'll apply Newton's laws and frictional forces.

1. **Forces acting on block A (on top of B):**
   - Weight of \( A \): \( W_A = m_A \cdot g = 24 \, \text{kg} \cdot 9.8 \, \text{m/s}^2 = 235.2 \, \text{N} \)
   - The friction between \( A \) and \( B \) must prevent \( A \) from slipping when \( B \) moves. The static friction force is:
     \[
     f_s = \mu_s \cdot N_A
     \]
     where \( \mu_s = 0.26 \) is the coefficient of static friction and \( N_A = W_A = 235.2 \, \text{N} \).

     Thus,
     \[
     f_s = 0.26 \cdot 235.2 = 61.15 \, \text{N}
     \]

2. **Forces acting on block B (on the surface):**
   - The total weight on \( B \) is the combined mass of \( A \) and \( B \):
     \[
     W_B = (m_A + m_B) \cdot g = (24 + 25) \cdot 9.8 = 49 \cdot 9.8 = 480.2 \, \text{N}
     \]
   - The kinetic friction between \( B \) and the surface is:
     \[
     f_k = \mu_k \cdot N_B = 0.13 \cdot 480.2 = 62.43 \, \text{N}
     \]

3. **For block \( C \):**
   - The force due to the weight of \( C \) is:
     \[
     F_C = m_C \cdot g
     \]
   - For the blocks to move together, the force on \( B \) (caused by \( C \)) must overcome both the kinetic friction between \( B \) and the surface and the static friction between \( A \) and \( B \).
   
   Therefore, the force balance for \( B \) gives:
   \[
   m_C \cdot g = f_s + f_k = 61.15 + 62.43 = 123.58 \, \text{N}
   \]
   \[
   m_C = \frac{123.58}{9.8} = 12.61 \, \text{kg}
   \]

Thus, the maximum mass of \( C \) is **12.61 kg**.

---

### Part (b) Frictional force between object B and the surface

As calculated earlier, the frictional force between \( B \) and the surface is the kinetic friction force:
\[
f_k = 62.43 \, \text{N}
\]

### Summary:
- **(a)** The maximum mass of piece \( C \) is \( 12.61 \, \text{kg} \).
- **(b)** The frictional force between object \( B \) and the surface is \( 62.43 \, \text{N} \).

Would you like more details on any part of this explanation or to clarify the concepts used? 

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Here are some related questions you might find helpful:
1. How does the coefficient of friction affect the force needed to move objects?
2. What would happen if the coefficient of kinetic friction was larger than the static friction?
3. What role does normal force play in determining frictional forces?
4. How does the pulley system affect the motion of the blocks?
5. Could the mass of \( C \) be greater if the surface had less friction?

**Tip:** Static friction keeps objects from moving, and it is generally stronger than kinetic friction, which acts when objects are already in motion.

The system shown in the figure is initially kept at rest. The mass of the object A in the picture is 24 kg and the mass of the object B is 25 kg. Let the coefficient of static friction between A and B be 0.26 and the coefficient of kinetic friction between B and the surface be 0.13. When C is released, what is the maximum mass m_C that C can have to cause A and B to slide together? You can assume that the string is massless and slides smoothly over the pulley. a) Calculate the maximum mass of the piece C. b) Calculate the frictional force between object B and the surface.

Learn how to calculate the maximum mass that causes two blocks to slide together using friction and Newton's laws. The problem involves a pulley system and covers key concepts in classical mechanics, including static and kinetic friction. Suitable for students in Grades 10-12.

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