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

Explanation

The force acting on the system are shown in Fig. 7.47

For vertical equilibrium of the point P
...... (i)
And for 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 lt f_(L)`
i.e., `" " T lt mu[(m + M)g - T] " " (f_(L) = muR)`
which on simplification gives
`T lt (mu (m + M)g)/(( 1 + mu))`
So , in the light of eqn. (i), we get
`F_("max") = (mu (m + M)g)/((1 + mu))`