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Question

In the arrangement shown tension in the string connecting 4 kg and 6 kg masses is :
293557_e0e83ae472a04926b3fb986d47f00a75.png

A
8N
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B
2N
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C
6N
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D
4N
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Solution

The correct option is A 8N
Let us consider the forces acting on the system of the three blocks together.
i. 20N (to the right)
ii. Weight = 12g (downwards)
iii. Normal force = 12g (upwards)
iv. Friction = f (to the left)
The maximum friction that can act = μmg=0.2×12g=24N
Since the maximum friction is greater than the force required to stop motion, the system of three bodies will not accelerate.
Deciding tension and friction acting on each body:
We know that tension is an elastic force. This means that unless there is a deformation (extension) in the string, tension will not act. Therefore, first friction will act to resist motion, and only if maximum friction is in sufficient to stop motion, the string gets deformed and tension will begin to act.
The forces acting on the 2kg block are:

i. 20 N (to the right)
ii. Weight = 2g (downwards)
ii. Normal force = 2g (upwards)
iii. Tension = T1 (to the left)
iv. Friction = f1(to the left)

Max friction = μmg=0.2×2g=4N
Since 4N < 20N, in order for the block to remain at rest, T1=204=16N
The forces acting on the 4 kg block are:

i. T1=16N (to the right)
ii. Weight = 4g (downwards)
ii. Normal force = 4g (upwards)
iii. Tension = T2 (to the left)
iv. Friction = f2(to the left)

Max friction = μmg=0.2×4g=8N
Since 8N < 16N, in order for the block to remain at rest, T2=168=8N

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