# Question #d4070

Jan 11, 2017

The answer is very simple: the difference is in the object on which the forces act. An action-reaction pair is not canceled. On the other hand, balanced forces do.

#### Explanation:

Consider the following image:

Consider case 1:

On a body two forces of the same modulus and direction act but opposite directions, ${F}_{1}$ and ${F}_{2}$. Therefore, the resultant ${F}_{R}$ of the forces acting on said body is 0, because:

${F}_{2} = {F}_{1} \Rightarrow {F}_{R} = {F}_{1} - {F}_{2} = 0$

Instead, let's see what happens in case 2:

Here, we have a body $A$ that exerts a force ${F}_{A / B}$ on a body $B$. By virtue of Newton's Third Law, we can state that on the body $A$ will appear a force ${F}_{B / A}$ exerted by the body $B$.

That is, to the ${F}_{A / B}$ action responds an ${F}_{B / A}$ reaction.

But these forces, being of equal modulus and opposite senses, do not cancel out. Why? Because they act on different bodies . The resulting force on $A$ is ${F}_{R} = {F}_{B / A}$, while on $B$ there is a resulting force ${F}_{R} = {F}_{A / B}$. When acting on different bodies they are independent forces and each body will move according to the acceleration that each of these forces provoke on him, i.e.:

${a}_{A} = - \frac{{F}_{B / A}}{{m}_{A}} \textcolor{w h i t e}{\text{..................}} {a}_{B} = \frac{{F}_{A / B}}{{m}_{B}}$

Where we have assumed that the left-right direction is the positive.