Question

In the figure shown, the coefficient of static friction between $B$ and the wall is $\frac{2}{3}$ and the coefficient of kinetic friction between $B$ and the wall is $\frac{1}{3}$. Other contacts are smooth. If the minimum force ${}^{\prime }{F}^{\prime }$ required to lift $B$ up is $\frac{xmg}{2}$, then find $x$. Mass of $A$ is $2m$ and the mass of $B$ is $m$. Take $\tan \theta =\frac{3}{4}$.

Answer 1

The F.B.D. of $A$ and $B$ are
For $A$ to be in equilibrium
$F=N\sin \theta$ ... (1)
For $B$ to just lift off
$N\cos \theta =mg+{\mu }_s{N}^{\prime }$
(here $\mu ={\mu }_s$) ......... (2)
For horizontal equilibrium of $B$
$N^{\prime }=N\sin \theta$ ... (3)
From (2) and (3)
$N(\cos \theta -{\mu }_{s}sin\theta )=mg$
or $N(\frac{4}{5}-\frac{2}{3}\times \frac{3}{5})=mg$
or $N=\frac{5}{2}mg$ ... (4)
From equation (1)
$F=N\times \frac{3}{5}$
$\Rightarrow$ $\therefore F=\frac{3}{2}mg$