Q21

​Q21. The friction coefficient between the board and the floor shown in figure (6-E7 ) is $\mathrm{\mu }$. Find the maximum force that the man can exert on the rope so that the board does not slip on the floor.

                    

 

Let T be the maximum force exerted by the man on the rope.

From the free body diagram,
R + T = Mg
⇒ R = Mg − T                     (1)
Again,
R₁ − R − mg = 0
⇒ R₁ = R + mg                   (2)
and
T − μR₁ = 0

From Equation (2),
T − μ(R + mg) = 0
⇒ T − μR − μ mg = 0
⇒ T − μ(Mg − T) − μmg = 0
T − μMg + μt − μmg = 0



⇒ T (1 + μ) = μMg + μmg
$\Rightarrow T=\frac{\mathrm{\mu } \left(M+m\right)g}{1+\mathrm{\mu }}$
Therefore, the maximum force exerted by the man is $\frac{\mathrm{\mu }\left(M+m\right)g}{1+\mathrm{\mu }}$.
Regards