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Question

A force of 100 N is applied on a mass of 3 kg as shown in the figure. The coefficient of friction between the surface and the block is μ=13. The frictional force acting on the block is


A
15 N downwards
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B
25 N upwards
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C
20 N downwards
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D
30 N upwards
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Solution

The correct option is C 20 N downwards
FBD of the block is given below.


Along the horizontal direction,
F cos30=N=10010032

Along vertical direction,
F sin30mg=5030=20 N

So, friction must act in the downward direction to balance 20 N and prevent relative motion between the block and wall.

Limiting friction fmax=μN=13×10032=50 N>20 N

Therefore, friction force will only be 20 N downwards.

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