Question

The friction coefficient between the board and the floor shown in diagram is $\mu$ . Find the maximum force that the man can exert on the rope, so that the board does not slip on the floor.

• A. $\frac{\text{μ}\left(\text{m}+\text{M}\right)\text{g}}{\left(2+\text{μ}\right)}$
• B. $\frac{\text{μ}\left(\text{m}+\text{M}\right)\text{g}}{\left(1+\text{μ}\right)}$
• C. $\frac{\text{μ}\left(\text{m}+\text{M}\right)\text{g}}{\left(2-\text{μ}\right)}$
• D. $\frac{\text{μ}\left(\text{m}+\text{M}\right)\text{g}}{\left(1-\text{μ}\right)}$

Answer 1

Answer: (B)

The forces acting on the system are shown in the diagram
For vertical equilibrium of the point $P$
$T=F$ ....(i)

And for the vertical equilibrium of the system

$R+T=(m+M)g$ , i.e., $R=(m+M)g-T$ ....(ii)
Now the system will not move horizontally till
$T<f_L$
i.e., $T<\mu [\left(m+M\right)g-T]$ $\left(f_L=\mu R\right)$
which on simplification gives
$T<\frac{\mu \left(\text{m}+\text{M}\right)\text{g}}{\left(1+\mu \right)}$
So in the light of equation (i), we get
$F_{\text{max}}=\frac{\mu \left(\text{m}+\text{M}\right)\text{g}}{\left(1+\mu \right)}$