Question

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

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

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
$\mathrm{f}_2=\mu_2\left(\mathrm{~m}_{\mathrm{A}}+\mathrm{m}_{\mathrm{B}}\right) \mathrm{g}$
$f_{2}=0.4\left(2 + 4\right)\left(10\right)$
$f_{2}=24 \, N$
Therefore for the entire system,
$F-f_{2}=\left(m_{A} + m_{B}\right)a$
$F-24=\left(2 + 4\right)\left(2\right)$
$F=36 \, N$