# Find the accelerations a1, a2, a3 of the three blocks shown in figure (6-E8) 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/s2.

The coefficient of friction are given as, μ1 = 0.2
μ2 = 0.3
μ3 = 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,
μ1R1 = μ1 × m1g
μ1R1 = μ1 × 2g

= (0.2) × 20
= 4 N (From the 3 kg block)

So, the net force experienced by the 2 kg block = 10 – 4 = 6 N
Hence,

a1=6/2=3 m/s2

But for the 3 kg block, the frictional force from the 2 kg block becomes the driving force (4N) and the maximum frictional force between the 3 kg and 7 kg blocks.
So, we have:
μ2R2 = μ2m2g = (0.3) × 5 kg
= 15 N

Therefore, the 3 kg block cannot move relative to the 7 kg block.
Due to the 4 N force, the two blocks, 3 kg and 7 kg block have the same acceleration (a2 = a3), which is because there is no friction from the floor.

So a2 = a3 = 4/10 =0.4 m/s2

(b) When the 10 N force is applied to the 3 kg block, it experiences maximum frictional force of (15 + 4) N, that is, 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 will move together.

A = 10/12=5/6 m/s2

(c) Similarly, in this case also, when 10 N force is applied to the 7 kg block, all three blocks will move together with the same acceleration.

Hence, a1=a2 =a3 =5/6 m/s2

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