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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 N 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.

Physics

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