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In figure, the coefficient of friction between the floor and the block $B$ is $0.1$. The coefficient of friction between the blocks $B$ and $A$ is $0.2$. The mass of $A$ is $\frac{m}{2}$ and of $B$ is $m$. ........ $mg$ is the maximum horizontal force $F$ can be applied to the block $B$ so that two blocks move together ?
$0.15$
$0.05$
$0.1$
$0.45$
Solution
Here, $\mathrm{m}_{\mathrm{A}}=\frac{\mathrm{m}}{2}, \mathrm{m}_{\mathrm{B}}=\mathrm{m}$
$\mu_{\mathrm{A}}=0.2, \mu_{\mathrm{B}}=0.1$
Let both the blocks are moving with common
acceleration a. Then,
$a=\frac{\mu_{A} m_{A} g}{m_{A}}=\mu_{A} g=0.2 \mathrm{g}$
and $F-\mu_{\mathrm{B}}\left(\mathrm{m}_{\mathrm{B}}+\mathrm{m}_{\mathrm{A}}\right) \mathrm{g}=\left(\mathrm{m}_{\mathrm{B}}+\mathrm{m}_{\mathrm{A}}\right) \mathrm{a}$
$F=\left(\mathrm{m}_{\mathrm{B}}+\mathrm{m}_{\mathrm{A}}\right) \mathrm{a}+\mu_{\mathrm{B}}\left(\mathrm{m}_{\mathrm{B}}+\mathrm{m}_{\mathrm{A}}\right) \mathrm{g}$
$=\left(\mathrm{m}+\frac{\mathrm{m}}{2}\right)(0.2 \mathrm{g})+(0.1)\left(\mathrm{m}+\frac{\mathrm{m}}{2}\right) \mathrm{g}$
$=\left(\frac{3}{2} \mathrm{m}\right)(0.2 \mathrm{g})+\left(\frac{3}{2} \mathrm{m}\right)(0.1 \mathrm{g})=\frac{0.9}{2} \mathrm{mg}$
$=0.45 \mathrm{mg}$
