# Acceleration of three blocks

Let f1, f2 be the friction forces between 2 and 3 kg blocks and 3 and 7 kg blocks respectively.

Let the normal force on the 2 kg and 3 kg blocks be N1 and N2.

Consider the motion of 2 kg block.

10 – f1 = 2a1

But f1 = u1N1 = 0.2 x 2 = 4 N

Therefore, a1 = 5 – f1/2 = 5 – 4/2 = 3 m/s^2.

Now, the only horizontal forces acting on the 3 kg block are the frictional force f1 due to motion of 2 kg block and frictional force f2. But the maximum value of f2 = u2N2 = 0.3 x 5 = 15 N and only a force greater than this can accelerate the 3 kg block. Thus the force f1 = 4 N cannot move 3 kg block relative to 7 kg block.

But this force f1 = 4 N also acts on the combined mass of 3 kg and 7 kg blocks and since there is no friction between the 7 kg block and floor, both these blocks will move together.

a1 = a3 = f1/10 = 4/10 = 0.4 m/s^2.

When the 10 N force acts on 3 kg block, the maximum frictional force that can arise from the 2 kg and 7 kg blocks is (4+15) 19 N and thus 10 N force applied to it cannot move this block relative to these blocks. But the 10 N force when acts on three blocks with mass 12 kg they will move together as there is no friction between 7 kg block and the floor.

a1 = a2 = a3 = 10/12 = 5/6 m/s^2.

Finally, when 10 N force is applied to 7 kg block, the maximum frictional force that 2 and 3 kg blocks can apply to it is f2 = u2 x 5 = 15 N. Thus it does not move relative to the 2 kg and 3 kg blocks. But there being no friction between the floor and the 7 kg block, all three blocks will move together and

a1 = a2 = a3 = 10/12 = 5/6 m/s^2.