Most of the time when I was solving physics problems I assumed that the tensions between several objects cancel each other if we consider a system of the objects and strings. I thought that was because tension is an internal force. Is this a correct understanding?

I know that the work done by a string is always zero, but I'm not so sure about the internal force and cancellation.

enter image description here

If you look at the diagram above, the tension of the $m_1$ is in the positive $x$ direction, while that of the $m_2$ is in the negative $y$ direction. Since force is a vector quantity, even if you consider a system consisting of the $m_1$, $m_2$, and the string, the two tensions cannot cancel each other therefore not a internal force, right?

My intuition says that they can still cancel each other, but I can't understand why. Can anyone help me?

  • $\begingroup$ Don't forget that Tension will also act on the pulley in -Y and -X direction. $\endgroup$ Apr 15, 2021 at 11:44

2 Answers 2


Tension, like any force, can be either internal or external. What determines if a force is internal or external is the definition of the system boundaries. The type of force is irrelevant.

For example, in this problem if each mass is considered its own system then the tension is external, but if the system is the pair of masses and the string all together then tension is internal.

You are free to choose the system boundaries as you like, and you can even change the system boundaries during your analysis as needed. So you can always make any force either internal or external as needed.

Now, you may be confused that the tension appears to not be an internal force even if the system includes both masses and the string. Internal forces always have an internal Newton's 3rd law pair associated with them, and the tension as drawn is not equal and opposite here. So even if the string+masses system is used, how can tension be internal? In this case the upward tension on $m_2$ forms a Newton's 3rd law pair with the downward tension on the string, and the rightward tension on $m_1$ forms a Newton's 3rd law pair with the leftward tension on the string. So the tension is still a valid internal force, but the corresponding internal 3rd law pairs are simply not drawn in this diagram.

  • $\begingroup$ I know what you're saying. I didn't specify the system in the title to make it short.. What I'm asking is that why the tension is an internal force if you choose a system of the masses and string, given that the two tension vectors acting on A and B does not cancel out. $\endgroup$
    – DH K
    Apr 15, 2021 at 11:42
  • $\begingroup$ @DHK no problem, I have updated my answer with a paragraph discussing that issue. $\endgroup$
    – Dale
    Apr 15, 2021 at 12:17

The tension force is internal force, it is constraint force, if the string is not elastic , otherwise "spring force"

I) string is not elastic

the equations of motion

$$m_1\,\ddot x +T=0\\ m_2\,\ddot y -T+m_2\,g=0$$

and because the string is not elastic , the string length $~L~$ is constant

$$x-y=L~\Rightarrow~ \dot x-\dot y=0\qquad,\ddot x-\ddot y=0$$

you have now three equations for three unknowns $~\ddot x~,\ddot y~, T$

thus the string tension is:

$$T= \left| \frac{m_1\,m_2}{m_1+m_2}\,g\right|$$

the work is:

$$W=\int T\,dl=\int T\,\left(\dot x-\dot y\right)\,dt=0$$

II) string is elastic

if the string in elastic with the expansion-related coefficient of elasticity $~$k then your equations of motion are

$$m_1\,\ddot x +k(x-y)=0\\ m_2\,\ddot y -k(x-y) +m_2\,g=0$$

the tension force is $$T=~k(x-y)\\ W=\left|\frac{k}{2}\left[x(t)^2-y(t)^2\right]\right|> 0$$

  • $\begingroup$ The elasticity of the string has nothing to do with whether or not to treat the tension as internal. $\endgroup$
    – Dale
    Apr 15, 2021 at 15:53
  • 1
    $\begingroup$ @Dale i see what you mean, you have always tension force, constraint force if the string is not elastic or spring force if the string is elastic, both are internal. I put addition modifications $\endgroup$
    – Eli
    Apr 15, 2021 at 16:56
  • 1
    $\begingroup$ Good changes!!! $\endgroup$
    – Dale
    Apr 15, 2021 at 17:04

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