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Question

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+μ1μ)
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B
g(1+μ1μ)
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C
g(1μ1+μ)
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D
2g(1μ1+μ)
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Solution

The correct option is B g(1+μ1μ)
FBD of block with respect to accelerating frame i.e. wedge is: -


At equilibrium,
N=mgcos45+masin45 .......(1)
& f+mgsin45=macos45 .......(2)
μN=macos45mgsin45
[for maximum acceleration, we have to take limiting case]
μ(mgcos45+masin45)=macos45mgsin45
[from (1)]
a=μgcos45+gsin45cos45μsin45
a=g(1+μ1μ)

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