Question

In the figure shown below, $m_A=2\text{kg}$ and $m_B=4\text{kg}$. For what minimum value of $F$, $A$ starts slipping over $B$. (Take $g=10{\text{m/s}}^2$)

• A. $24\text{N}$
• B. $36\text{N}$
• C. $12\text{N}$
• D. $20\text{N}$

Answer 1

The correct option is B $36\text{N}$

The FBDs of the blocks are as shown



We have from the FBDs
$N_1=m_{A}g=2g=20N$
$N_2=N_1+m_{B}g=20+4g=60N$
Maximum frictional force between $A$ and $B$ could be
$f_{\text{1 max}}={\mu }_1{N}_1=\left(0.2\right)\left(20\right)N$
$\Rightarrow f_{\text{1 max}}=4\text{N}$

As the block $A$ will just start slipping on block $B$, its acceleration would be
$a_A=\frac{f_{\text{1 max}}}{m_A}=\frac{4}{2}=2{\text{m/s}}^2$

Since, $A$ has just started slipping on $B$, we can consider $(A+B)$ as the system whose acceleration would be $a_A$.
$\because (f_2{)}_{\text{max}}={\mu }_2{N}_2=0.4\times 60=24\text{N}$
Thus, we can write the force equation over the system as
$F-(f_2{)}_{\text{max}}=(m_A+m_B)a_A$
$\Rightarrow F-24=6a_A=6\times 2=12$
$\therefore F=36\text{N}$