Doubt & Answer Explanation

I will solve the case when the 10 N force is applied on the 2kg block. Based on my reasoning, you can try to solve the cases when the 10 N force is applied on the 3kg and 7 kg block. If you still have a problem solving the cases when the force is applied on the 3kg and the 7kg block, then in that case, send me your attempt and I will send you the solutions for those cases too.



In this case, the direction of friction forces is as shown.Friction acts on the 2kg block to the left as the 10 N block tends to move it to the right. Friction acts on the 3kg block between the 2-3 kg interface to the right as it tends to move the 3 kg block in the direction of motion of the 2kg block and thus minimize the relative motion between the 2kg and 3kg block. Using similar logic, the direction of the friction forces between 3kg and 7 kg and 7kg and the floor is also shown.

Now the maximum value of f₁ is 0.2 x 2 x 10 = 4 N. Clearly this is less than 10 N. So f₁ = 4N = maximum value.

Now a force of 4N tends to move the 3kg block to the right. The maximum value of f₂ is 0.3 x 3 x 10 = 9 N. Now this maximum value is greater than the force applied on the 3 kg block f₁ which is 4N. Therefore, the maximum value of f₂ DOES NOT act on the 3 kg block. f₂ = 4N acts on it.

Now a force f₂ = 4N acts on the 7 kg block. This causes f₃ to act on the block. The maximum value of  f₃ is 0.5 x 7 x 10 = 35 N. This is much higher than the force 4 N acting on the 7 kg block. 

Thus, the 2 kg block slips on the 3 kg block and the 3kg block and 7 kg block are at rest. The figure below shows these forces.




Therefore, the acceleration of the 2 kg block is 10-4 / 2 = 3 m/s²
The accelerations of the 3kg and 7 kg blocks are zero.