Question

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

Answer 1

Let the maximum force is $F$.

When the force is applied then effective weight on the board will be $Mg-F$.

Total force applied on the floor is given as,

$F_t=\left(Mg-F\right)+mg$

The maximum friction force is given as,

$F_f=\mu \left(Mg-F+mg\right)$

So, the maximum friction force will be equal to the horizontal pull by the string that is equal to $F$.

The expression can be written as,

$\mu \left(Mg-F+mg\right)=F$

$F=\frac{\mu \left(M+m\right)g}{\left(1+\mu \right)}$

Thus, the maximum force is $\frac{\mu \left(M+m\right)g}{\left(1+\mu \right)}$.