Two blocks are in contact on a frictionless table. A horizontal force is applied to the larger block, as shown in Figure.

(a) If${\mathit{m}}_{\mathbf{1}}\mathbf{=}\mathbf{2}\mathbf{.}\mathbf{3}\mathit{k}\mathit{g}\mathbf{,}{\mathit{m}}_{\mathbf{2}}\mathbf{=}\mathbf{1}\mathbf{.}\mathbf{2}\mathit{k}\mathit{g}$ ,$\mathit{F}\mathbf{=}\mathbf{3}\mathbf{.}\mathbf{2}\mathit{N}$ find the magnitude of the force between the two blocks. (b) Show that if a force of the same magnitude Fis applied to the smaller block but in the opposite direction, the magnitude of the force between the blocks is , which is not the same value calculated in (a). (c) Explain the difference.

1. The mass of a larger block is$m_1=2.3kg$
2. The mass of a smaller block is$m_2=1.2kg$
3. The force is$f=3.2N$.

Newton’s second law states that the net force acting on the object is equal to the product of mass and net acceleration of the object. According to Newton’s third law, every action has an equal and opposite reaction.

Apply Newton’s second law to the applied force and Newton’s third law to the forces between two blocks.

The free body diagrams for both blocks are shown below. The force ${\overrightarrow{F}}_{12}$is the force applied by block 1 on block 2. So, block 2 also will apply equal and opposite force on block 1.

Hence, ${\overrightarrow{F}}_{12}=-{\overrightarrow{F}}_{\backslash 21}$. Refer to the free body diagram for both masses as below:

The force equation for block 1 is,

$F-F_{21}=m_{1}a$

And the force equation for block 2 is

$F_{12}=m_{2}a$

Rearranging this equation for acceleration, a we get

$a=\frac{F_{12}}{m_2}$

By using this in the equation of block 1, we get

$F-F_{21}=m_1\times \frac{F_{12}}{m_2}$

Now, if we consider only magnitude, we can write$F_{12}=F_{21}$

So,theabove equation will become

$$\begin{gathered} F_{12}=\frac{m_2}{\left(m_1+m_2\right)}F \\ =\frac{1.2kg}{\left(1.2kg+2.3kg\right)}3.2N \\ =1.1N \end{gathered}$$

Therefore, the magnitude of force between two blocks when force is applied to the larger block is$1.1N$.

When we apply force to the smaller block, the free body diagrams will be as follows:

The corresponding equation for force between block 1 and block 2 in this case will be

$$\begin{gathered} F_{21}=\frac{m_1}{\left(m_1+m_2\right)}F \\ =\frac{2.3kg}{\left(1.2kg+2.3kg\right)}\times 3.2N \\ =2.1N \end{gathered}$$

In part a, one of the equations of motion is, $F_{12}=m_{2}a$while in part b, one of the equations, will be,$F_{12}=m_{1}a$ .

So, as$m_1>m_2$, the force $F_{21}$in part b must be greater than the force $F_{12}$of part a.