Two blocks A and B of masses m$$_A$$ = 1 kg and m$$_B$$ = 3 kg are kept on the table as shown in figure. The coefficients of friction between A and B is 0.2 and between B and the surface of the table is also 0.2. The maximum force F that can be applied on B horizontally, so that the block A does not slide over the block B is:
[Take g = 10 m/s$$^2$$]

To solve for the maximum force $$F$$ such that block $$A$$ does not slide over block $$B$$, we need to analyze the friction limits and the maximum common acceleration they can share.

Here is the structured, step-by-step breakdown of the solution.

1. Identify the Maximum Acceleration of Block A

Block $$A$$ moves forward solely due to the static friction acting between it and block $$B$$.

• Mass of block A: $$m_A = 1\text{ kg}$$
• Coefficient of friction between A and B: $$\mu_1 = 0.2$$
• Coefficient of friction between B and table: $$\mu_2 = 0.2$$
• Total normal force on the table: $$N_{table} = (m_A + m_B) \cdot g = 4\text{ kg} \times 10\text{ m/s}^2 = 40\text{ N}$$
• Correct Option: A (16 N)

The maximum static friction force ($$f_{max, A}$$) that can act on block $$A$$ is:

$$f_{max, A} = \mu_1 \cdot m_A \cdot g$$

$$f_{max, A} = 0.2 \times 1\text{ kg} \times 10\text{ m/s}^2 = 2\text{ N}$$

Using Newton's second law ($$F = m \cdot a$$), the maximum acceleration ($$a_{max}$$) block $$A$$ can achieve without slipping is:

$$a_{max} = \frac{f_{max, A}}{m_A} = \frac{2\text{ N}}{1\text{ kg}} = 2\text{ m/s}^2$$

If the system accelerates faster than $$2\text{ m/s}^2$$, block $$A$$ will begin to slide backward relative to block $$B$$.

2. Analyze the Whole System (A + B)

To find the maximum force $$F$$ required to achieve this common acceleration, we treat both blocks as a single combined system of mass:

$$M_{total} = m_A + m_B = 1\text{ kg} + 3\text{ kg} = 4\text{ kg}$$

Friction from the Table Surface

The table exerts a kinetic/limiting friction force ($$f_{table}$$) opposing the motion of the combined mass:

$$f_{table} = \mu_2 \cdot N_{table} = 0.2 \times 40\text{ N} = 8\text{ N}$$

3. Calculate the Maximum Force ($$F$$)

Using the equation of motion for the entire combined system:

$$F - f_{table} = M_{total} \cdot a_{max}$$

Substitute the values we calculated:

$$F - 8\text{ N} = 4\text{ kg} \times 2\text{ m/s}^2$$

$$F - 8 = 8$$

$$F = 16\text{ N}$$

Final Answer

The maximum horizontal force $$F$$ that can be applied to block $$B$$ without causing block $$A$$ to slide is 16 N.