If the coefficient of friction between the wedge and the block is μ, then the maximum acceleration of the wedge for which the block will remain at rest with respect to it is
A
2g(1+μ1−μ)
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B
g(1+μ1−μ)
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C
g(1−μ1+μ)
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D
2g(1−μ1+μ)
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Solution
The correct option is Bg(1+μ1−μ) FBD of block with respect to accelerating frame i.e. wedge is: -
At equilibrium, N=mgcos45∘+masin45∘ .......(1)
& f+mgsin45∘=macos45∘ .......(2) ⇒μN=macos45∘−mgsin45∘
[for maximum acceleration, we have to take limiting case] ⇒μ(mgcos45∘+masin45∘)=macos45∘−mgsin45∘
[from (1)] ⇒a=μgcos45∘+gsin45∘cos45∘−μsin45∘ ⇒a=g(1+μ1−μ)