Question

In the figure shown, find the value of frictional force between $A$ & $B$ and $B$ and the surface respectively. Take $g=10{\text{m/s}}^2$.

• A. $4\text{N},6\text{N}$
• B. $6\text{N},4\text{N}$
• C. $2\text{N},4\text{N}$
• D. $2\text{N},6\text{N}$

Answer 1

The correct option is D $2\text{N},6\text{N}$

The FBDs of the block are as shown
Given applied external force $F=24N$


We have from the FBDs
$N_1=2g=20\text{N}$
$f_{\text{1 max}}={\mu }_1{N}_1=\left(0.2\right)20=4\text{N}$
$N_2=N_1+4g=60\text{N}$
$f_{\text{2 max}}=\left(0.1\right)60=6\text{N}$
Since, $F>f_{\text{2 max}}=6\text{N}$, $4\text{kg}$ block will slide relative to the surface.
Assuming, both the blocks move together, then common acceleration
$a_c=\frac{F-f_{\text{2 max}}}{m_1+m_2}=\frac{24-6}{6}=3{\text{m/s}}^2$

Now, for block $A$ we have
$f_1=m_A{a}_c=2\times 3=6N$
$\because f_1>f_{\text{1 max}}$, it means our assumption is wrong. Hence, the two blocks will slip on each other and they will move separately.
$\therefore$ we have
$f_1={\mu }_k{N}_1=0.1\times 20=2\text{N}$
and, $f_2={\mu }_k{N}_2=0.1\times 60=6\text{N}$