Question

Find the accelerations $a_1,a_2$ and $a_3$ of the three blocks as shown in figure, if a horizontal force of $10\text{N}$ is applied on the $3\text{kg}$ block. The coefficient of static friction ($\mu$) for all contact surfaces are shown in the figure. Take $g=10{\text{m/s}}^2$.

• A. $a_1=a_2=a_3=\frac{5}{6}{\text{m/s}}^2$
• B. $a_1=\frac{10}{3}{\text{m/s}}^2,a_2=\frac{5}{6}{\text{m/s}}^2,a_3=\frac{10}{3}{\text{m/s}}^2$
• C. $a_1=\frac{10}{3}{\text{m/s}}^2,a_2=\frac{10}{3}{\text{m/s}}^2,a_3=\frac{5}{6}{\text{m/s}}^2$
• D. $a_1=\frac{5}{6}{\text{m/s}}^2,a_2=\frac{5}{6}{\text{m/s}}^2,a_3=\frac{10}{3}{\text{m/s}}^2$

Answer 1

The correct option is A $a_1=a_2=a_3=\frac{5}{6}{\text{m/s}}^2$

When force of $10N$ is applied on $3kg$ block, let us assume that all three blocks moves together with common acceleration $a$, then
$a=\frac{F}{m}=\frac{10}{2+3+7}=\frac{10}{12}=\frac{5}{6}{\text{m/s}}^2$

Friction between $2\text{kg}$ block and $3\text{kg}$ block:


Required frictional force on $2\text{kg}$ block
$f_1=m_{1}a=2\times a=2\times \frac{5}{6}=\frac{5}{3}N$
Maximum frictional force on $2\text{kg}$ block
$f_{1max}={\mu }_1{N}_1={\mu }_1{m}_{1}g=0.2\times 2\times 10=4\text{N}$
$\because (f_{1max}>f_1$, hence block will move with that acceleration.

Friction $f_2$ between $3\text{kg}$ and $7\text{kg}$ block


Required frictional force on $7\text{kg}$ block
$f_2=7a=7\times \frac{5}{6}=\frac{35}{6}\text{N}$
Maximum frictional force between $3\text{kg}$ and $7\text{kg}$ block
$f_{2max}={\mu }_2{N}_2={\mu }_2(m_1+m_2)g$
$=0.3\times (2+3)\times 10=15\text{N}$
$\because (f_2{)}_{max}>f_2$, hence both blocks will move with that acceleration.

Therefore, no sliping occurs anywhere, and all the $3$ blocks will move with common acceleration of $\frac{5}{6}{\text{m/s}}^2$
$\Rightarrow$Option $A$ is correct