Question

Three large blocks each having flat surfaces are kept stationary over one another as shown in the figure. A horizontal force of $20\text{N}$ acts on the middle block $B$, then choose the incorrect statement: (Take $g=10m/s^2$)

• A. Acceleration of the lowest block is $7{\text{m/s}}^2$
• B. Friction between middle and lower blocks is $12\text{N}$
• C. Acceleration of the upper most block is $2{\text{m/s}}^2$
• D. Friction between middle and upper blocks is $2\text{N}$

Answer 1

The correct option is B Friction between middle and lower blocks is $12\text{N}$

FBD for block $A$:

$N=1g=10\text{N}$
$f_{1max}=\mu N=0.2\times 10=2\text{N}$

FBD for Block $B$:

$N=2g=20\text{N}$
$f_{2max}=\mu N=0.8\times 20=16\text{N}$

FBD for block $C$:

$N=3g=30\text{N}$
$f_{3max}=\mu N=0.3\times 30=9\text{N}$

Since, the applied external force i.e $20\text{N}$ is greater than $f_{1max}$ and $f_{2max}$, the blocks will slide on each other with different acceleration.

Hence, acceleration of the lowest block $C$ is
$a_C=\frac{f_{2max}-f_{3max}}{m_C}=\frac{16-9}{1}=7{\text{m/s}}^2$
acceleration of upper most block $A$ is
$a_A=\frac{f_{1max}}{m_A}=\frac{2}{1}=2{\text{m/s}}^2$
and, acceleration of middle block $B$ is
$a_B=\frac{20-f_{1max}-f_{2max}}{m_B}=\frac{20-2-16}{1}=2{\text{m/s}}^2$