Question

In the figure shown, the coefficient of static friction between $C$ and ground is $0.5$, coefficient of static friction between $A$ and $B$ is $0.25$, coefficient of static friction between $B$ and $C$ is zero. The minimum value of force $F$ (in Newton), to cause sliding between $A$ and $B$ is (Masses of $A,B$ and $C$ are respectively $2kg,4kg$ and $5kg$).

Answer 1

Since there is no friction between $B$ and $C$, there is no possibility of a force acting on $C$ and it will not move.

$A$ and $B$ will move together until $A$ reaches its maximum acceleration and friction between $A$ and $B$ reaches limiting value.
So,
$f_L^{AB}=m_A{a}_{max}$
$\Rightarrow \mu m_{A}g=0.25\left(20\right)=2a_{max}$
$\Rightarrow a_{max}=2.5m{s}^{-2}$
This acceleration is brought about by $F$ in (A + B) system.
$F=(m_A+m_B)a_{max}$
$F=(2+4)2.5$
$F=15N$

So, a minimum force of $15N$ has to be applied for the block $A$ to slip over block $B$.