Minimum horizontal force F required to keep the smaller block stationary with respect to the bigger block.
A
(M−m)gμ
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B
(M+m)gμ
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C
Mgμ
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D
mgμ
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Solution
The correct option is B(M+m)gμ Let us suppose, acceleration of system is ′a′, then Fnet=mtotala a=FM+m ...... (1)
We will work from ground frame of reference. F.B.D of smaller block:
On applying ∑F=ma along horizontal, N=ma=mFM+m... (2) [from (1)] Now, on applying ∑F=ma along vertical f−mg=0 [No acceleration along vertical] ⇒f=mg Friction (f)=μN ⇒μN=mg ⇒μmFM+m=mg [from (2)] ⇒F=(M+m)gμ Minimum horizontal force required is (M+m)gμ to keep smaller block stationary with respect to bigger block.