Wedge is fixed on horizontal surface. Triangular block A of mass M is pulled upward by applying a constant force F parallel to incline of the wedge as shown in the figure and there is no friction between the wedge and the block A, while coefficient of friction between A and block B of mass m is $μ .$ If there is no relative motion between A and B then frictional force developed between A and B is

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$[\frac{F + (m + M) g \sin θ}{(m + M)}] m \cos θ$

$μ m g$

$[\frac{F - (m + M) g \sin θ}{(m + M)}] m \cos θ$

$μ m g / 2$

answer is C.

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Detailed Solution

$a = [\frac{F - (m + M) g \sin θ}{(m + M)}] S o, f = m a \cos θ, a = [\frac{F - (m + M) g \sin θ}{(m + M)}] m \cos θ$

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