Question

In the figure, $m_A=2\text{kg}$ and $m_B=4\text{kg}$. The minimum value of $F$ for which $A$ starts slipping over $B$ is $(g=10{\text{ms}}^{-2})$

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

Answer 1

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

Maximum frictional force between $A$ and $B$ is
$f_1={\mu }_1{m}_{A}g=\left(0.2\right)\left(2\right)\left(10\right)\text{N}$
$\Rightarrow f_1=4\text{N}$
Limiting friction between $B$ and ground is
$f_2=\left(0.4\right)\left(6\right)\left(10\right)=24\text{N}$

For $A,4=2a_A=2a$
$\Rightarrow a=2{\text{ms}}^{-2}$
For $B,F-28=4a$
$\Rightarrow F-28=8$
$\Rightarrow F=36\text{N}$
Hence, the correct answer is (b).