Why is the tension on both sides of an Atwood machine identical? The field forces $$F_{g1}$$ and $$F_{g2}$$ push down on Block 1 and Block 2, respectivley, where $$F_{g1}=m_1g$$$$F_{g2}=m_2g$$ Since the pully system reverses the direction of each force, wouldn't the following be true? $$T_1 = F_{g2} = m_2g$$$$T_2 = F_{g1} = m_1g$$ And since $$m_1 \neq m_2$$, wouldn't $$T_1 \neq T_2$$?
My textbook states that tension is the same throughout the whole string, but I can't wrap my head around why this is so. If $$m_1 \neq m_2$$, how could the same force $$T$$ accelerate both of them an equal amount? Wouldn't Block 2 require a greater force?

Tension in a rope is another word for the force the rope exerts.

Forces always occur in equal and opposite pairs. This is true for a rope. There are two ends. The ends pull two objects toward each other with equal force.

Instead of masses hanging from a pulley, suppose two massive people named 1 and 2, had a tug of war. Suppose 1 pulled harder than 2.

So if 1 is pulling harder on the rope than 2, why is the rope pulling equally hard on 1 and 2? Why doesn't the rope pull back with the same force as each person pulls?

That would happen if 1 and 2 were on opposite sides of a wall and pulled on separate ropes tied to the wall. The wall would pull back just hard enough to keep them both still. It would pull back just as hard as each pulled on the rope.

But in a tug of war, both 1 and 2 will move if 1 pulls harder. They will have the same velocity and acceleration in the direction 1 is pulling, and they will stay a constant distance apart.

Suppose 2 wasn't pulling. He was just standing on frictionless ice and letting 1 pull him with force $$F_1$$. What is the force in the rope? You can see the force is between $$F_1$$ and $$0$$.

It isn't $$0$$ on 2's end. The rope is pulling him, making him accelerate.

It isn't $$0$$ on 1's end. He is accelerating forward, but the rope is pulling him back. He would accelerate faster if not for the rope.

Likewise, you can see it isn't $$F_1$$ on 1's end. It isn't enough to hold 1 still.

It isn't $$F_1$$ on 2's end. 1 is accelerating a total mass of $$m_1 + m_2$$. The acceleration for both is $$a_1 = a_2 = F_1/(m_1+m_2)$$. This is smaller than if 1 stood still and pulled 2 toward him with force $$F_1$$. 2 would accelerate with $$a_2 = F_1/m_2$$.

So how big is $$T_2$$, the force from the rope that accelerates 2?

$$T_2 = m_2a_2 = F_1 \frac{m_2}{m_1+m_2}$$

How big is $$T_1$$, the force from the rope that holds 1 back?

$$F_1 - T_1 = m_1a_1 = F_1\frac{m_1}{m_1+m_2}$$

or

$$T_1 = F_1 - F_1\frac{m_1}{m_1+m_2} = F_1\frac{m_2}{m_1+m_2} = T_2$$

You can do similar calculations when both 1 and 2 pull.

Equality of the two tensions only holds in the special case of a massless wheel in the pulley. If you read forward in your textbook to the section which deals with massive pullies, you will find that indeed these tensions should not generally be the same and it's only in the special case where the wheel is massless that they work out to be the same.

So in the end, I think you're right to be confused about this...the textbook hasn't yet given you all the tools needed to actually justify the assumption that the tensions work out to be equal.

(By the way, the section with massive pullies is likely titled "torque" or something similar. If not, check for "torque" in the index at the back. The massive pullies will be somewhere around there if the text is at all standard).

• I guess I'm even more so confused with the concept of tension itself. Even if I disregard the pully and consider a horizontal string where one side is pulled to the left $m_1g$ Newtons and the other side to the right $m_2g$ Newtons, How is any point along the string "not moving" as I've heard in other explanations? The whole string is moving to the right, as that force is greater. Nov 17 '20 at 22:24
• Please see my answer regarding the assumption of a massless, perfectly rigid string. Nov 17 '20 at 22:44
• @JJohnHipp It is not that no point along the string is not moving, it's that no point on the string is moving relative to any other point. This is the rigidity assumption John mentions. Nov 17 '20 at 22:52

how could the same force T accelerate both of them an equal amount?

This part of your question holds the key - specifically the word acceleration.

Lets start with your second question first: Why are the accelerations the same? This is a result of the constraint that the string is a fixed length. As long as this is true then for each metre that the larger mass moves the smaller mass moves one metre in the same time. The distances travelled are the same in the same time so the speeds are the same and likewise the (magnitudes of the) accelerations.

What your question assumes, without explicitly stating, is that $$T_1=m_1g$$ and $$T_2=m_2g$$. There is no reason that this has to be true and if it were true then the forces on each mass would be balanced and they would not accelerate.

The tension will have a magnitude somewhere between the weights of the two masses so for the heavier mass the net force is down and for the lighter mass the net force is up.

• Why is $T_1$ not equal to $m_2g$? Isn't the first mass pulled up by the force due to gravity of the second mass, since the pully reverses its direction? Nov 18 '20 at 0:15
• The size of the tension force on either side of the pulley is the same (ideally) but the size of the tension force on pulling up on m1 is not the equal to the force of gravity on m1 - if it were m1 wouldn't accelerate. So if the amount of tension on the left side isn't equal to the weight of m1 and the tension is the same on both sides of the pulley then its not going to be equal to m1g on the side pulling up m2 Nov 18 '20 at 0:18
• Is the force of the weight of the second mass pulling up on the first mass via the pulley not tension at all? Nov 18 '20 at 0:28

In addition to the assumption of a massless wheel as Myers points out in his earlier answer, the analysis you refer to also assumes the string is massless and perfectly rigid (no stretching/compression). With this assumption, the tension at both ends of the string is the same. Specifically, T1 - T2 = ma where T1 and T2 are the tensions at each end of the string, m is the mass of the string, and a is the acceleration of the string. As m approaches zero, T2 approaches T1 even though the string is accelerating (a is not zero).

Also, when you later consider the mass of the pully- in which case the tensions at the ends of the string are not the same- you will see the implicit assumption that the tensions in the $$\textit{string}$$ are the same as the forces of the string on the $$\textit{pulley}$$. This is true, but not at all obvious as discussed in detail in the following reference. https://www.researchgate.net/publication/318107848_Force_and_torque_of_a_string_on_a_pulley.

You can search this site for detailed discussions of the case considering the mass of the pulley; e.g., search for "Atwood friction".

• So in this example, the forces pulling on both ends of the string are $m_1g$ ($T_2$) and $m_2g$ ($T_1$) in the opposite direction? Nov 18 '20 at 0:11
• The forces on the ends of the string are T1 and T2 in opposite directions; for a massless pully these forces are equal, for a pully with mass these forces are not equal. You can search this site and/or a good physics book to help you see how to calculate T1 and T2 in terms of $m_1g$, $m_2g$,and the mass of the pully and its inertia. Nov 18 '20 at 0:40
• If the string is perfectly rigid every point has both $T_1$ and $T_2$ applied to it, but since it is also massless, wouldn't $T_1$ and $T_2$ cancel out, giving a net force of 0 for every point in the string? I don't understand how it can be accelerating if both the tensions are equal and opposite Nov 18 '20 at 0:50
• Furthermore, wouldn't the acceleration of the string be $\frac{T_1 - T_2}{0}$? Nov 18 '20 at 1:11
• Consider this simple case; string laid out horizontally, no pulley, no gravity. Let T1 be the tension on the left end and T2 the tension on the right end. From a force balance, the acceleration of the string to the right is ma = (T2 - T1) (a is negative if T1 exceeds T2) where m is the mass of the string. Since m is very small, T2 -T1 is very small no matter what a is. That is, $a = \lim_{m \to 0}[(T_2 - T_1)/m] = \lim_{m \to 0}[ma/m] = a$ Nov 18 '20 at 3:47

Since the pully system reverses the direction of each force, wouldn't the following be true?

T1=Fg2=m2g

T2=Fg1=m1g

Technically you are correct. But a few changes have to be made to your statement.

The blocks are not at rest and both travel with acceleration $$a$$. We can assume that $$M_1$$ goes up and $$M_2$$ goes down. Since $$M_1$$ goes up, it's weight increase ($$W_{M_1}=M_1(g+a)$$). While $$M_2$$ goes down, its weight decreases ($$W_{M_2}=M_2(g-a)$$)

$$\therefore T_1=W_{M_2}=M_2(g-a)$$ and $$T_2=W_{M_1}=M_1(g+a)$$(modified version of your quoted equations).

Now, $$T_1$$ supports $$W_{M_{1}}$$,

$$\therefore T_1=W_{M_{1}}=M_1(g+a)=T_2$$

Its because the pulley in an atwood's machine is an Ideal pulley. It has no mass and its frictionless. This means that the rope is only going to slip over the pulley freely without rotating it at all. In that case the rope is completely isolated from the pulley and tension should be uniform throughout.
Note that frictionless means that no friction between rope and pulley

Had it been a rolling pulley (one with friction along rim so that rope does not slip), the tension on both ends would be different due to friction. and it would be that difference that help it turn.

Everywhere in mechanics a frictionless pulley implies that its a "slipping" pulley and not a rolling pulley (as we see in our daily life). These frictionless pulleys are kept only to change the direction of pull keeping the tension in the string same.