In the given arrangement,find the maximum value of $F$ for which there is no relative motion between the blocks.

In the shown arrangement,the mass of $A = 1\,kg$ and the mass of $B = 2\,kg$. The coefficient of friction between $A$ and $B$ is $\mu = 0.2$. There is no friction between $B$ and the ground. $A$ horizontal force $F = 5\,N$ is applied to block $B$. The frictional force exerted by $A$ on $B$ is equal to:

Two blocks $A$ and $B$ of masses $5 \,kg$ and $3 \,kg$ respectively rest on a smooth horizontal surface with $B$ over $A$. The coefficient of friction between $A$ and $B$ is $0.5$. The maximum horizontal force (in $kg \,wt.$) that can be applied to $A$,so that there will be motion of $A$ and $B$ without relative slipping,is

Two blocks $A$ and $B$ of masses $4 \ kg$ and $6 \ kg$ are as shown in the figure. $A$ horizontal force of $12 \ N$ is required to make $A$ slip over $B$. Find the maximum horizontal force $F_B$ that can be applied on $B$ so that both $A$ and $B$ move together. (Take $g=10 \ m \ s^{-2}$) (in $N$)

$A$ body $B$ lies on a smooth horizontal table and another body $A$ is placed on $B$. The coefficient of friction between $A$ and $B$ is $\mu$. What acceleration given to $B$ will cause slipping to occur between $A$ and $B$?

When $F = 2\,N$,the frictional force between the $10\,kg$ block and the $5\,kg$ block is $..........\,N$. (Given: $\mu_s = 0.1$ between blocks,$\mu_s = 0.3$ between $5\,kg$ block and ground).