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Q.

A boy of mass 4 kg is standing on a piece of wood having mass 5 kg. If the coefficient of friction between the wood and the floor is 0.5, the maximum force that the boy can exert on the rope so that the piece of wood does not move from its place is ______ N. (Round off to the Nearest integer)(Take  g=10ms−2)

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answer is 30.

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Detailed Solution

FBD for wood,Horizontal , T=fL=μNVertical, N=N1+mg⇒T=μN1+mg........(1)FBD for manVertical, N1=m1g−T .........(2)   put (2) in (1) ⇒T=μm1g−T +mg ⇒T=μm1g+mg1+μ=0.5×901.5=30N
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