To solve this problem, let's break it down systematically. We'll use Newton's laws and the frictional force formula to calculate the maximum mass $m_C$ that can cause objects A and B to slide together.

Known Information:

• Mass of object A, $m_A = 24 \, \text{kg}$
• Mass of object B, $m_B = 25 \, \text{kg}$
• Coefficient of static friction between A and B, $\mu_s = 0.26$
• Coefficient of kinetic friction between B and the surface, $\mu_k = 0.13$
• Acceleration due to gravity, $g = 9.8 \, \text{m/s}^2$

Step-by-Step Solution:

a) Maximum Mass of $m_C$

The goal is to find the maximum mass of $m_C$ that can cause A and B to slide together. This means we need to determine the static frictional force between A and B and the kinetic frictional force between B and the surface.

1. Static Friction Between A and B: The static friction force $f_{s}$ between A and B can be calculated using the formula: $f_{s} = \mu_s \cdot N_A$ where $N_A$ is the normal force between A and B, which is just the weight of A: $N_A = m_A \cdot g = 24 \cdot 9.8 = 235.2 \, \text{N}$ So, the maximum static friction force is: $f_s = 0.26 \cdot 235.2 = 61.15 \, \text{N}$

2. Kinetic Friction Between B and the Surface: The kinetic friction force $f_k$ between B and the surface is calculated using the formula: $f_k = \mu_k \cdot N_B$ where $N_B$ is the normal force acting on B due to its weight: $N_B = m_B \cdot g = 25 \cdot 9.8 = 245 \, \text{N}$ Therefore, the kinetic friction force is: $f_k = 0.13 \cdot 245 = 31.85 \, \text{N}$

3. Net Force Analysis: For A and B to move together, the static friction force between A and B must balance the force pulling A backward due to $m_C$. The total force required to overcome friction and accelerate both blocks is the net force minus the kinetic friction between B and the surface: $F_{\text{net}} = T - f_k$ where $T$ is the tension in the string due to the mass of $m_C$.

   For the system to move together, the tension $T$ must not exceed the static friction force between A and B, so: $T = m_C \cdot g$ The maximum $T$ is equal to the maximum static friction force between A and B, so: $m_C \cdot g = f_s \implies m_C = \frac{f_s}{g} = \frac{61.15}{9.8} = 6.24 \, \text{kg}$

Thus, the maximum mass of $m_C$ is approximately: $\boxed{6.24 \, \text{kg}}$

b) Frictional Force Between Object B and the Surface:

As calculated earlier, the frictional force between object B and the surface is the kinetic friction force: $\boxed{31.85 \, \text{N}}$

Summary:

• Maximum mass of $m_C$: $6.24 \, \text{kg}$
• Frictional force between B and the surface: $31.85 \, \text{N}$

Would you like further clarification on any step or additional details?

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Here are five related questions for deeper understanding:

1. How would the problem change if the coefficient of static friction between A and B was higher?
2. How do we calculate the acceleration of the system if $m_C$ is greater than the calculated value?
3. What happens to the frictional force if the surface is inclined?
4. How does the mass of the pulley affect the system if it's not massless?
5. What if object A and B had a connecting string?

Tip: Always draw a free-body diagram when dealing with forces; it helps in visualizing the system's forces and motion.