Question

Two blocks A and B each of mass m are placed on a smooth horizontal surface. Two horizontal forces $F$ and $2F$ are applied on the blocks A and B respectively as shown in figure. The block A does not slide on block B. Then the normal reaction acting between the two blocks is:

• A. $F$
• B. $\frac{F}{2}$
• C. $\frac{F}{\sqrt{3}}$
• D. $3F$

Answer 1

The correct option is D $3F$

As block A does not slide on block B this means both blocks have the same acceleration along horizontal.

Net force on the system is $2F-F=F$

Therefore, Acceleration is $F/(m+m)=F/2m$.

Now taking of individual blocks. Let Normal between block be $N$.

Block on right:

Along Horizontal: $2F-Ncos60=ma\Rightarrow 2F-m\left(F/2m\right)=N\left(1/2\right)\Rightarrow \frac{3}{2}F=\frac{1}{2}N\Rightarrow N=3F$

Just to check we can calculate $N$ at block on right also;

$Ncos60-F=ma\Rightarrow F+m\left(F/2m\right)=N\left(1/2\right)\Rightarrow \frac{3}{2}F=\frac{1}{2}N\Rightarrow N=3F$