Find the accelerations a_{1}, a_{2}, a_{3} of the three blocks shown in the following figure if a horizontal force of 10 N is applied on (a) 2 kg block, (b) 3 kg block, (c) 7 kg block. Take g = 10 m/s^{2}.

#### Solution

Given:

μ_{1} = 0.2

μ_{2} = 0.3

μ_{3} = 0.4

Using the free body diagram, we have:

(a) When the 10 N force is applied to the 2 kg block, it experiences maximum frictional force.

Here,

μ_{1}R_{1} = μ_{1} × m_{1}g

μ_{1}R_{1} = μ_{1} × 2g = (0.2) × 20

= 4 N (From the 3 kg block)

Net force experienced by the 2 kg block = 10 − 4 = 6 N

`therefore a_1=6/2=3 "m/s"^2`

But for the 3 kg block (Fig. 3), the frictional force from the 2 kg block, i.e, 4 N, becomes the driving force and the maximum frictional force between the 3 kg and 7 kg blocks.

Thus, we have:

μ_{2} = R_{2} = μ_{2}m_{2}g = (0.3) × 5 kg

= 15 N

Therefore, the 3 kg block cannot move relative to the 7 kg block.

The 3 kg block and the 7 kg block have the same acceleration (a_{2} = a_{3}), which is due to the 4 N force because there is no friction from the floor.

`therefore a_2=a_3=4/10=0.4 "m/s"^2`

(b) When the 10 N force is applied to the 3 kg block, it experiences maximum frictional force of (15 + 4) N, i.e., 19 N, from the 2 kg block and the 7 kg block.

So, it cannot move with respect to them.

As the floor is frictionless, all the three bodies move together.

`therefore a_1=a_2=a_3=10/12=(5/6) "m/s"^2`

(c) Similarly, it can be proved that when the 10 N force is applied to the 7 kg block, all three blocks move together with the same acceleration.

`therefore a_1=a_2=a_3=(5/6)"m/s"^2`