Block $A$ of mass $3 \mathrm{kg}$ rests on another block $B$ of mass $7 \mathrm{kg}.$ The coefficient of friction between $A$ and $B$ is $0.4$ while the coefficient of friction between $B$ and the horizontal floor on which $B$ rests is $0.55.$ Find the force of friction between $A$ and $B,$ when a horizontal force of $50 \mathrm{N}$ is applied on the block $B.$
(Use $\mathrm{g}=10 \mathrm{m}/{\mathrm{s}}^2$)

In the above question, all the forces are acted in the lower block and there is no force acting on the upper block, now the value friction force will be, 

$f_l=\left(m_1+m_2\right)g{\mu }_l$, here ${\mu }_l=0.55, m_1=3 \mathrm{kg}$ and $m_2=7 \mathrm{kg}$, now the friction force between lower block and surface, 
$f_l=\left(3+7\right)10\times 0.55=55 \mathrm{N}$,

Which is greater than, applied force so no blocks will move and no force will act in the upper block so the net friction force between the lower block and the upper block will be zero.