Question

Three block, $A,B$ and $C$ of masses $1kg,2kg$ and $3kg$ respectively are arranged as shown in the figure. The coefficient of friction between different surface are shown in figure. A force of magnitude $20N$ acts on block $B$ in horizontal direction. The acceleration of block $A$ is

• A. $5m/s^2$
• B. Zero
• C. None of these
• D. $\frac{8}{3}m/s^2$

Answer 1

The correct option is D

$\frac{8}{3}m/s^2$

Draw FBD of blocks.
Given:
$m_A=1kg,m_B=2kg,m_C=3kg,{\mu }_A=0.5,{\mu }_B=0.4,{\mu }_C=0.3$
Let,${\mu }_A$ be coefficient of frition between $A$ and $B$,
${\mu }_B$ be coefficient of of frition between $B$ and $C$,
${\mu }_C$ be coefficient of of frition between $C$ and ground,
Let$f_A$ be friction between block $A$ and b;lock $B$,
$f_B$ be friction block $B$ and block $C$
$f_C$ be friction block $C$ and ground

Find limiting values of friction and find if there is relative motion or not.
$(f_A{)}_{max}={\mu }_A{N}_A=0.5\times 1\times 10=5N(f_B{)}_{max}={\mu }_B{N}_B=0.4\times (1+2)\times 10=12N(f_C{)}_{max}={\mu }_C{N}_C=0.3\times (1+2+3)\times 10=18N$


Diagrame

$a_{together}=\frac{20-12}{1+2}=\frac{8}{3}m/s^2$
Maxximum acceleration that the topmost block can have due to friction only$={\mu }_{A}g=5m/s^2$
$\therefore$ we can conclude that our assumption is correct and both blocks will move together , as required friction is less than maximum friction.
So, both blocks will have same acceleration ahich is equal to $\frac{8}{3}m/s^2$.