Question

Put a uniform meter scale horizontally on your extended index fingers with the left one at $0.00cm$ and the right one at $90.00cm$. When you attempt to move both the fingers slowly towards the center, initially only the left finger slips with respect to the scale and the right finger does not. After some distance, the left finger stops and the right one starts slipping. Then the right finger stops at a distance $x_R$ from the center $(50.00cm)$ of the scale and the left one starts slipping again. This happens because of the difference in the frictional forces on the two fingers. If the coefficients of static and dynamic friction between the fingers and the scale are $0.40$ and $0.32$, respectively, the value of $x_R$ (in $cm$ ) is

Answer 1

Answer: 25.6

Left finger will keep slipping till the kinetic friction is less than the static friction on right finger.
After that right finger will start slipping.
It will slip till the kinetic friction on it is less than static friction on the left one.
$N_{L}x=N_R\times 0.4$
$N_L+N_R=Mg$
$\Rightarrow N_L=\frac{Mg\times 0.4}{x+0.4}$ and $N_R=\frac{Mgx}{x+0.4}$
${\mu }_k{N}_L\ge {\mu }_5{N}_R$
$\frac{0.32\times Mg\times 0.4}{x+0.4}\ge \frac{0.4\times Mgx}{x+0.4}$
$x=\frac{0.4\times 0.32}{0.4}=0.32m$
Now right finger will start moving and it will stop, when kinetic friction on it will be more than static friction on left finger.
$N_L\times 0.32=N_R{X}_R$
$N_L+N_R=Mg$
and ${\mu }_k{N}_R\ge {\mu }_g{N}_L$
$0.32\times \frac{Mg\times 0.32}{X_R+0.32}\ge 0.4\frac{Mg\times X_R}{X_R+0.32}$
$X_R=\frac{0.32\times 0.32}{0.4}=0.256m$
$=25.6cm$