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I got caught up with a conceptual question dealing with a practice problem with connected mass mechanics.

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Looking at the solution lecture notes is the following image for the Free Body diagrams for each mass, and the pulley, respectively:

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Obviously, the Free Body diagrams on mass 1 and 2 are pretty straight-forward, but the situation with the pulley confuses me. $T_2$ going down is fine. The mass is falling, the system is overall causing $M_2$ to fall down, the string applies a force that way, sure. But $T_1$ confuses me. The mass is moving towards the pulley once we let the system act, so why is there a tension being applied that way? The directions for $T_1$ for mass 1 and the pulley are opposite for direction, and I'm not really sure why. Overall, the pulley is rotating clockwise, so this has to be the case -- $M_1$ will move towards the pulley, $M_2$ will drop. So why is $T_1$ facing that way? Intuitively speaking, that is.

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    $\begingroup$ What is the problem with T1? Based on Newton's third law it should be leftwards. $\endgroup$
    – Soroush khoubyarian
    Commented May 5, 2017 at 1:05
  • $\begingroup$ Imagine two people playing tug-of-war where the person on the right is winning. They're pulling rightward on the rope, which means the rope is pulling leftward on them. Also, whether they're winning or losing is completely irrelevant. $\endgroup$
    – knzhou
    Commented May 5, 2017 at 1:18
  • $\begingroup$ If you replace the people with a block and pulley, you get this problem. $\endgroup$
    – knzhou
    Commented May 5, 2017 at 1:18
  • $\begingroup$ I was going to answer this then I saw this comment. Maybe you can expand this into an answer. $\endgroup$
    – Kalpak Gupta
    Commented May 5, 2017 at 3:08
  • $\begingroup$ In a "frictionless pulley" does the wheel turn freely on the axle, or does the string slide without friction over the wheel, or both? $\endgroup$
    – DJohnM
    Commented May 5, 2017 at 4:55

3 Answers 3

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The force $T_1$ labelled $D$ in the diagram is the force on the pulley due to the string and it is trying to accelerate the pulley to the left, which is to be expected as the pulley is feeling a pull to the left from the mass $m_2$ communicated via the string.

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The force $T_1$ labelled $A$ in the diagram is the force on the mass $m_1$ due to the string and it is trying to accelerate the mass to the right, exactly as you expected.

Perhaps what has confused you is the missing (trivial) FBD for the string with forces $T_1$, labelled $B$ and $C$ in the diagram, acting on the string.
In the context of the question those to forces are shown to be equal in magnitude because it is assumed that the mass of the light string is much less than the masses of $m_1$ and $m_2$.

So the directions of all the forces labelled $T_1$ are consistent with the forces labelled $A$ and $B$ and $C$ and $D$ being Newton's third law pairs.

What is also missing from your FBDs is a force on the pulley pointing in a north-easterly direction.

Update as the result of some comments and questions.

Why does the mass of the light spring being much less than that of m1 and m2 show that the two forces are equal in magnitude? Also, is B and C both T1?

Perhaps I phrased that badly.
Assume that the magnitude of $T_{1\rm A} = 100\, \rm N$ and this force is the force accelerating mass $m_1$ to the right.
The magnitude of $T_{1\rm B}$ will be the same as the magnitude of $T_{1\rm A}$.
Suppose that the string had some mass but much less than $m_1$, say $\frac {m_1}{1000}$, then the magnitude of force $T_{1\rm C} = 100.1 \,\rm N$ to ensure that the string accelerates at the same rate as mass $m_1$.
So what I am saying is that the difference between $T_{1\rm C}$ and $T_{1\rm B}$ which is equal to $0.1 \,\rm N$ is much less that $100 \,\rm N$ and can be neglected.

Additionally, what is causing this north-easterly directed force?

The north-easterly force is the force that the support for the pulley exerts on the axle of the pulley.
If you did not have that force there then the resultant of forces $T_{1\rm D}$ and $T_1$ is a south-westerly force which would cause the pulley to accelerate in that direction.

Also, if my third law force pairs thinking is correct, then why is that not applied to T2 on the pulley? It pulls down on the pulley, just as the mass goes down.

The downward force $T_2$ acting on the pulley is due to the vertical part of the string.
The upward force $T_2$ acting on mass $m_2$ is due to the vertical part of the string.
As drawn these are no a Newton third law pair.
The FBD is the vertical string is missing.

Very often when such FBDs are drawn the FBDs for the strings are omitted because they do not add much to the actual calculation in hand.

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  • $\begingroup$ Why does the mass of the light spring being much less than that of $m1$ and $m2$ show that the two forces are equal in magnitude? Also, is B and C both $T1$? I suppose a string can only really be taught if its feeling a force on it from both sides, anyway. I usually don't worry about third law pairs in mechanics because we only talk about one object in a lot my examples but here, if we have more than one object, they're necessary. Hopefully I'm right by thinking that? Additionally, what is causing this north-easterly directed force? $\endgroup$
    – sangstar
    Commented May 5, 2017 at 14:02
  • $\begingroup$ Also, if my third law force pairs thinking is correct, then why is that not applied to $T2$ on the pulley? It pulls down on the pulley, just as the mass goes down. $\endgroup$
    – sangstar
    Commented May 5, 2017 at 14:10
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Intuitively speaking, the mass $m_1$ is decreasing the speed. So it must reduce the speed of the system by exerting a force on the pulley that opposes that direction of movement. So $T1$ must be leftwards.

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  • $\begingroup$ Why is m1 decreasing the speed, though? The surface is frictionless? $\endgroup$
    – sangstar
    Commented May 5, 2017 at 1:16
  • $\begingroup$ Even if it's friction less, it still has mass. If you want to calculate the acceleration of the system(provided that the pulley's mass is negligible) we must write: $a = \dfrac{m_2 \times g}{m_1 + m_2}$ $\endgroup$
    – Soroush khoubyarian
    Commented May 5, 2017 at 1:20
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Strings cannot push unlike a coiled spring.Hence if we show the forces acting on a point due to a string the tension has to be pulling that point not pushing it.

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