  Question

A board of mass '$$M$$' is placed on a rough inclined plane and a man of mass '$$m$$' walks down the board. If the coefficient of friction between the board and inclined plane $$\mu$$, the acceleration of the man, such that plank does not slip, is given by: A
a(Mmm)(cosθ+μsinθ)g  B
a(M+mm)(sinθ+μcosθ)g  C
a(M+mm)(sinθ+μcosθ)g  D
a=(mM+m)(sinθ+μcosθ)g  Solution

The correct option is B $$a\le \left( \cfrac { M+m }{ m } \right) \left( \sin { \theta } +\mu \cos { \theta } \right) g$$Let $$F$$ be the interaction force in between the board and the man , $$f$$ is the friction force in between the board and inclined plane.Here the value of $$F$$ is   $$Mg\sin\theta-\mu(M+m)g\cos\theta\leq F \leq Mg\sin\theta+\mu(M+m)g\cos\theta ...(1)$$and      $$f=\mu N=\mu(M+m)g\cos\theta$$also, $$ma=F+mg\sin\theta \Rightarrow F=ma-mg\sin\theta$$now from (1), $$Mg\sin\theta-\mu(M+m)g\cos\theta\leq ma-mg\sin\theta \leq Mg\sin\theta+\mu(M+m)g\cos\theta$$or $$(M+m)g\sin\theta-\mu(M+m)g\cos\theta\leq ma \leq (M+m)g\sin\theta+\mu(M+m)g\cos\theta$$or$$(\frac{M+m}{m})(g\sin\theta-\mu g\cos\theta) \leq a \leq (\frac{M+m}{m})(g\sin\theta+\mu g\cos\theta)$$ Physics

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