Q. 233.8( 4 Votes )

# Find the accelerations a_{1}, a_{2}, a_{3} 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/s^{2}.

Answer :

The coefficient of friction are given as, 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} × 2*g*

= (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,

a_{1}=6/2=3 m/s^{2}

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:

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.

Due to the 4 N force, the two blocks, 3 kg and 7 kg block have the same acceleration (*a*_{2} = *a*_{3}), which is because there is no friction from the floor.

So *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, 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/s^{2}

(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, a_{1}=a_{2} =a_{3} =5/6 m/s^{2}

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