Question

On a smooth horizontal surface two blocks of mass 2 kg and 8 kg are placed as shown. Two horizontal forces of 10 N and 50 N are being applied as shown. What is the normal reaction between the blocks ?

• A. 16 N
• B. 20 N
• C. 18 N
• D. None

Answer 1

The correct option is C 18 N

$\text{Step 1: Identify all the forces and predict the direction of motion}$

The surface is smooth and the Force towards left is greater than the force towards right, So both blocks will move towards left with same acceleration, say $a$.

$\text{Step 2: Draw F.B.D. and apply Newton's second law}$

$\text{For System(Block 1+2):}$ As both blocks will move together with same acceleration, therefore we can assume them as System.

Therefore, applying Newton's second law on the system (Taking left direction as positive)

$\Sigma F=ma$

$\Rightarrow 50N-10N=10kg\times a$

$\Rightarrow a=4m/s^2$

Therefore, both the blocks will move with an acceleration of $4m/s^2$

$\text{For block 1:}$$N_{12}$ is normal on block $1$ due to block $2$ and $N_{21}$ is normal on block $2$ due to block $1$

Both are action reaction pair so $N_{12}=N_{21}$

Apply Newton's second law on block $1$ (Taking left direction as positive)

$\Sigma F=m_{1}a$

$\Rightarrow N_{12}-10N=m_{1}a$ $....\left(1\right)$

$\text{Step 3: Solving above equations to get the required value}$

Put $m_1=2kg$ and $a=4m/s^2$ in equation $\left(1\right)$

$N_{12}-10N=2kg\times 4m/s^2$

$\Rightarrow N_{12}=18N$

$\therefore N_{12}=N_{21}=18N$

Hence, Normal force between the blocks is $18N$

Hence option C is correct