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Question

Two blocks A of 6 kg and B of 4 kg are placed in contact with each other as shown in figure.

There is no friction between A and ground and between both the blocks. The coefficient of friction between B and ground is 0.5. A horizontal force F is applied on A. Find the minimum and maximum values of F, which can be applied so that both the blocks can move combinely without any relative motion between them.

A
10 N, 50 N
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B
12 N, 50 N
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C
12 N, 75 N
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D
None of these
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Solution

The correct option is C 12 N, 75 N
F N sin37° = 6a F(3N5)= 6a (i)


N2 = 4gN cos37° = 40 4N5 (ii)
N sin37° f = 4a (iii)
From Eqs. (i) and (iii):
F f = 10a
F μN2 = 10a (iv) F μ(404N5] = 10a (v)
Put the value of μ from Eq. (i) in (v) and also put the value of μ to get a=5F6042
Now to start motion : a>0
F>12 N

So the minimum force F to just start the motion is 12 N.
Now maximum F will be when N2 just becomes zero.
Then from Eq. (ii): N = 50 N.
From Eqs. (i) and (iv), we get F = 75 N (by putting N2 = 0)
If we apply F>75 N, then B will start sliding up on A.

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