Question

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.

Short answer

a) The magnitude of the force between two blocks when force is applied to the larger block is 1.1N .

b) The magnitude of the force between two blocks when force is applied to the smaller block is 2.1 N .

c) Since the mass$m_1$ is greater than the mass$m_2$ , the magnitude of the force between two blocks when force is applied to the smaller block is greater than the magnitude of the force between two blocks when force is applied to the larger block.

Step-by-step solution

Given

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$f=3.2N$.

Understanding the concept of Newton’s laws

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.

(a) Find the magnitude of the force between the 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$1.1N$.

(b) Show that if a force of the same magnitude F is applied to the smaller block but in the opposite direction, the magnitude of the force between the blocks is 2.1 N

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

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

Explain the difference between the two answers

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.