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

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

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