Question

Three-blocks $A$,$B$ and $C$ of mass $4\text{kg}$, $2\text{kg}$ and $1\text{kg}$ respectively are in contact of a frictionless surface as shown. If a force of $14\text{N}$ is applied on the $4\text{kg}$ block, then the contact force between $A$ and $B$ will be

• A. $6\text{N}$
• B. $8\text{N}$
• C. $18\text{N}$
• D. $2\text{N}$

Answer 1

The correct option is A $6\text{N}$


Writing force equations (assuming common acceleration to be $a$),
$14-N=4a$
$N-N_1=2a$
$N_1=1\times a$

Solving, we get $N=6N$

Alternate solution:
Since all the three blocks are going to move with common acceleration, we can consider them as one system.
$a=\frac{\text{Net external force}}{\text{Total mass}}=\frac{14}{7}=2m/s^2$

Now if you consider blocks B and C together as a system, the net external force acting on it is contact force $N$.


Therefore $N=a(2+1)=6N$