Question

Two block $A$ and $B$ of mass $10\text{kg}$ and $20\text{kg}$ respectively, are arranged as shown in figure on a smooth horizontal surface. A constant force $F=60\text{N}$ acts on block $A$. Find the minimum coefficient of static friction (${\mu }_{\text{min}}$) so that both the block moves together without slipping? Take $g=10{\text{m/s}}^2$.

• A. $0.2$
• B. $0.4$
• C. $0.3$
• D. $0.1$

Answer 1

The correct option is B $0.4$

Friction force $f$ will support the motion of block $B$ and simultaneously opposes the motion of block $A$. Hence both blocks will move with same acceleration $a$.

$\Rightarrow$ To find the ${\mu }_{\text{min}}$, friction will have to act at it's maximum value $\therefore f={\mu }_{\text{min}}N$



From FBD of blocks applying equation of dynamics in direction of acceleration:
For block $A$
$F-f=m_{A}a...\left(i\right)$
For satisfying the equilibrium condition in vertical direction:
($N=m_{A}g$)
$\therefore f={\mu }_{\text{min}}{m}_{A}g.......\left(ii\right)$
For block $B$:
$f=m_{B}a...\left(iii\right)$
From equation $\left(i\right)$ and $\left(ii\right)$, $\left(iii\right)$:
$60-m_{B}a=m_{A}a$
$\Rightarrow 60=(10+20)a$
$\therefore a=2{\text{m/s}}^2$
Hence from Eq. $\left(iii\right)$
$f=m_{B}a=20\times 2=40\text{N}$
$\Rightarrow {\mu }_{\text{min}}{m}_{A}g=40$
${\mu }_{\text{min}}\times 10\times 10=40$
$\therefore {\mu }_{\text{min}}=0.4$