Question

In the figure, $m_A=2kg$ and $m_B=4kg$ . For what minimum value of $F$ , $A$ starts slipping over $B\left(g=10m{s}^{-2}\right)$ ?

• A. $24N$
• B. $36N$
• C. $12N$
• D. $20N$

Answer 1

Answer: (B)

Maximum frictional force between $A$ and $B$ could be
${\mathrm{f}}_1={\mu }_1{\mathrm{m}}_{\mathrm{A}}\mathrm{g}=(0.2)(2)(10)\mathrm{N}$
Acceleration when frictional force is maximum,
$a=\frac{f_1}{m_A}$
$a=\frac{4}{2}=2m/s^2$
Fictional force between ground and B
${\mathrm{f}}_2={\mu }_2\left({\mathrm{m}}_{\mathrm{A}}+{\mathrm{m}}_{\mathrm{B}}\right)\mathrm{g}$
$f_2=0.4\left(2+4\right)\left(10\right)$
$f_2=24N$
Therefore for the entire system,
$F-f_2=\left(m_A+m_B\right)a$
$F-24=\left(2+4\right)\left(2\right)$
$F=36N$