Two blocks of masses 4 kg and 6 kg connected by a massless string are kept on a
rough surface having coefficient of sliding friction 0.2. Two horizontal forces of 30 N
and 10 N is applied on 4 kg and 6 kg blocks. Find the friction force on both the
blocks and tension in the string. (Consider the string to be taught initially)

To solve this problem, we need to analyze the forces acting on both blocks and how they interact with each other through the tension in the string. Let's break it down step by step.

Identifying the Forces

We have two blocks: Block A (4 kg) and Block B (6 kg). They are subjected to horizontal forces and friction. The applied forces are:

• Force on Block A (4 kg): 30 N
• Force on Block B (6 kg): 10 N

The coefficient of friction (μ) between the blocks and the surface is 0.2. The gravitational force acting on each block can be calculated using the formula:

Weight (W) = mass (m) × gravity (g)

Assuming g = 9.8 m/s², we find:

• Weight of Block A: W_A = 4 kg × 9.8 m/s² = 39.2 N
• Weight of Block B: W_B = 6 kg × 9.8 m/s² = 58.8 N

Calculating the Friction Force

The maximum frictional force (f_max) that can act on each block is given by:

f_max = μ × N

Where N is the normal force, which is equal to the weight of the block in this case since the surface is horizontal.

• Friction force on Block A: f_A = 0.2 × 39.2 N = 7.84 N
• Friction force on Block B: f_B = 0.2 × 58.8 N = 11.76 N

Net Forces and Acceleration

Next, we need to determine the net force acting on each block. The net force (F_net) can be calculated by subtracting the friction force from the applied force:

• For Block A: F_netA = 30 N - 7.84 N = 22.16 N
• For Block B: F_netB = 10 N - 11.76 N = -1.76 N

Since the net force on Block B is negative, it indicates that the friction force is greater than the applied force, meaning Block B does not move. Therefore, Block A will pull Block B through the tension in the string.

Finding the Tension in the String

Since Block B is not moving, we can analyze Block A to find the tension (T) in the string. The net force acting on Block A is also equal to the tension pulling Block B:

F_netA = T - f_A

Substituting the known values:

22.16 N = T - 7.84 N

Solving for T gives:

T = 22.16 N + 7.84 N = 30 N

Summary of Results

To summarize our findings:

• Friction force on Block A: 7.84 N
• Friction force on Block B: 11.76 N
• Tension in the string: 30 N

This analysis shows how the forces interact and how the tension in the string is influenced by the applied forces and friction. Understanding these concepts is crucial in solving problems related to dynamics and forces in physics.