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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:

136581_bbae90b0fc444a1189e36ba4ca6a3ecf.png


A
a(Mmm)(cosθ+μsinθ)g
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B
a(M+mm)(sinθ+μcosθ)g
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C
a(M+mm)(sinθ+μcosθ)g
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D
a=(mM+m)(sinθ+μcosθ)g
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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$$ (\frac{M+m}{m})(g\sin\theta-\mu g\cos\theta) \leq a \leq (\frac{M+m}{m})(g\sin\theta+\mu g\cos\theta)$$

178258_136581_ans_c73c970e63be4b26ac717a39f1bd7e2a.png

Physics

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