Question

In the given figure below, a force of 120 N is applied horizontally on the upper block. Find the magnitude and direction of acceleration of the blocks A ($a_A$) and B ($a_B$).

• A. $a_A=5{\text{m/s}}^2,\text{right}$
  $a_B=9{\text{m/s}}^2,\text{right}$
• B. $a_A=9{\text{m/s}}^2,\text{right}$
  $a_B=5{\text{m/s}}^2,\text{right}$
• C. $a_A=9{\text{m/s}}^2,\text{right}$ $a_B=1.5{\text{m/s}}^2,\text{right}$
• D. $a_A=1.5{\text{m/s}}^2,\text{right}$ $a_B=9{\text{m/s}}^2,\text{right}$

Answer 1

The correct option is C $a_A=9{\text{m/s}}^2,\text{right}$ $a_B=1.5{\text{m/s}}^2,\text{right}$

The FBDs of the blocks are as shown


We have from the FBDs as
$N_1=m_{A}g=10g=100N$
$N_2=N_1+20g=100+200=300N$
In the above situation, we see that the maximum possible value of friction between the blocks is $f_{\text{max}}={\mu }_s{N}_1=0.3\times 100=30\text{N}$

Considering that there is no slipping between the blocks, the acceleration of system will be
$a=\frac{120}{20+10}=4{\text{m/s}}^2$

($\because$ only friction force between block $A$ and block $B$ is responsible for producing acceleration in block $B$)

$\Rightarrow f=m_{B}a=20\times 4=80\text{N}$
$\because$ $f>(f_s{)}_{max}$, it means our assumption is wrong and we can conclude that the blocks do not move together.

Now drawing the FBD of each block including horizonatal forces only, we have



$a_A=\frac{120-f_{\text{max}}}{m_A}=\frac{120-30}{10}=9{\text{m/s}}^2$ towards right
and, $a_B=\frac{f_{\text{max}}}{m_B}=\frac{30}{20}=1.5{\text{m/s}}^2$ towards right.