# I Is this proof regarding massless strings correct?

#### walking

I am reading the Tipler and Mosca textbook and am on the part about massless strings. I understand that in real life a string has more tension at the top than the bottom because the top part has to support a greater mass of rope. However, in other examples such as pulling a sled with a rope I can't seem to see how the tension varies intuitively. I tried proving that if the string is considered as being massless, the tension would be negligible, and I am posting this to ask if my proof is correct.

Proof: Consider a rope being pulled with tension T at an angle $\theta$ to the horizontal. To simplify the situation, I split the rope up into a finite number of segments. I pull on segment 1, which then pulls on segment 2, etc all the way down to the last segment which pulls on the sled. When I exert a force T on segment 1, let the force that segment 1 exerts on segment 2 be T'. Then segment 2 exerts an equal and opposite force T' on segment 1. Hence segment 1 is being pulling simultaneously to the left and right. Also, segment 1 has a weight mg acting downwards. The component of this weight in the direction parallel to the rope (ie "down the rope") is $mg\sin \theta$. The tensions and weight are the only forces acting on segment 1, hence $mg\sin \theta+T'=T$, giving $T-T'=mg\sin \theta$. Now because m is negligible this is roughly $T-T'\approx 0$ ie $T\approx T'$. Since this proof can be applied to any segment of rope, it follows that the tension is the same throughout the rope. $\square$

If the proof is correct, I am still wondering if a special proof is needed for each situation in which tension is applied to ropes. Is there a general proof that deals with all cases? For example, one can give a separate proof dealing with the case of a rope hanging from a ceiling in order to show that the tension is the same throughout if the rope is considered massless, but this doesn't prove the rule in general: if a string is massless then the tension is the same throughout.

Thanks

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#### Nugatory

Mentor
I am still wondering if a special proof is needed for each situation in which tension is applied to ropes
There's a very convincing general proof by contradiction: assume that the tension is not the same everywhere; derive from that assumption and $F=ma$ an impossible result; therefore the assumption is false.

#### walking

There's a very convincing general proof by contradiction: assume that the tension is not the same everywhere; derive from that assumption and $F=ma$ an impossible result; therefore the assumption is false.
Do you know where I can find this proof?

#### Let'sthink

Newton's law tells us that the tension in the string that pulls some object with with constant acceleration, say a, toward right cannot be uniform as some force has to be there which generates the acceleration of a tiny piece of string which is pulled by the right part, say with "Tr" right ward and left part, say "Tl" left word. So Tr > Tl. But if mass of the string is assumed to be zero then zero force can produce finite acceleration in a zero mass in the limiting case mathematically and the limit can be anything different for different cases.

#### jbriggs444

Do you know where I can find this proof?
Recall that the "F" in F=ma is the vector sum of all external forces acting on an object.

Suppose that m=0. Then it follows that F=0. So the sum of all external forces on a length of string is zero. There are only two such forces: the force F1 applied at the one end and the force F2 applied at the other end. Both are equal in magnitude to the tensions at the respective ends of the string, T1 and T2.

$$F_1+F_2=0$$ $$F_1=-F_2$$ $$|F_1| = |F_2|$$ $$T_1=T_2$$

#### sophiecentaur

Gold Member
Do you know where I can find this proof?
It strikes me that, if any actual proof were needed, it would be required to show that there would actually be a difference in tension. Not every step in the explanation of a Physics process needs proof. I'd say this is more axiomatic than 'testable'.
If the tensions were not the same, Newtons Laws tell us that the string would accelerate off in some direction and that is hardly likely.

#### walking

It strikes me that, if any actual proof were needed, it would be required to show that there would actually be a difference in tension. Not every step in the explanation of a Physics process needs proof. I'd say this is more axiomatic than 'testable'.
If the tensions were not the same, Newtons Laws tell us that the string would accelerate off in some direction and that is hardly likely.
But in this stackexchange post, https://physics.stackexchange.com/questions/308011/rigorously-proving-equal-tension-on-both-ends , dmkee says that "Once the pulleys have non-trivial mass and the rope is accelerating than the segments must be assumed to have different tensions even if the rope is light." Does this mean that massless strings do not always have the same tension throughout? I'm a bit confused considering the proof given by jbriggs.

#### sophiecentaur

Gold Member
But in this stackexchange post, https://physics.stackexchange.com/questions/308011/rigorously-proving-equal-tension-on-both-ends , dmkee says that "Once the pulleys have non-trivial mass and the rope is accelerating than the segments must be assumed to have different tensions even if the rope is light." Does this mean that massless strings do not always have the same tension throughout? I'm a bit confused considering the proof given by jbriggs.
You should perhaps try to see reasons why this should be OK, rather than scanning the detailed wording of that article and seeking some paradox. @jbriggs444 is just quoting the very basic Newtonian ideas; can you really be questioning them is such a simple situation? f=ma is something we can all agree on, surely. If m is zero then how can f (the net force on the string) be non-zero? If the net force is zero then that means the two tensions are the same. If, instead of having a string over the pulleys, you hold the pulleys together or use ideal gears (eliminating the string) then the tangential forces would have to be the same (N3 etc.).
The word "light" is used in the article and light doesn't mean massless so some tension difference could apply in this (totally different) scenario.

#### walking

You should perhaps try to see reasons why this should be OK, rather than scanning the detailed wording of that article and seeking some paradox. @jbriggs444 is just quoting the very basic Newtonian ideas; can you really be questioning them is such a simple situation? f=ma is something we can all agree on, surely. If m is zero then how can f (the net force on the string) be non-zero? If the net force is zero then that means the two tensions are the same. If, instead of having a string over the pulleys, you hold the pulleys together or use ideal gears (eliminating the string) then the tangential forces would have to be the same (N3 etc.).
The word "light" is used in the article and light doesn't mean massless so some tension difference could apply in this (totally different) scenario.
Excuse my ignorance (I am a beginner), I didn't know light and massless were different things. I thought massless itself only meant "negligibly small", since this is what the Tipler and Mosca textbook says in its discussion of tension.

#### Nugatory

Mentor
But in this stackexchange post, https://physics.stackexchange.com/questions/308011/rigorously-proving-equal-tension-on-both-ends , dmkee says that "Once the pulleys have non-trivial mass and the rope is accelerating than the segments must be assumed to have different tensions even if the rope is light." Does this mean that massless strings do not always have the same tension throughout? I'm a bit confused considering the proof given by jbriggs.
It depends on whether you interpret "throughout" as meaning "throughout each segment", which is what everyone means when they talk about uniform tension in a massless rope, or misinterpret it as meaning "everywhere, even in the different segments on different sides of the pulley". When the pulley has non-trivial mass we need a net force to accelerate it; this force comes from the segment of rope on one side pulling harder than the other.

It's easiest to think of this problem as if we had two ropes, both fastened to the same point on the rim of the pulley and pulling in opposite directions. What does the pulley do if the tensions are the same? What does it do if they are different?

#### sophiecentaur

Gold Member
Excuse my ignorance (I am a beginner), I didn't know light and massless were different things. I thought massless itself only meant "negligibly small", since this is what the Tipler and Mosca textbook says in its discussion of tension.
There is a massive difference in principle but not a lot in practice. However in a question like yours, in which you seems to be looking for loopholes in the basic theory, the possibility of a finite difference would depend on a finite (i.e. non-zero) mass for the string. You need to decide if you are going for ideally zero mass or a relatively small mass before you can get a real answer.
Iff the mass is "negligibly small" then the difference in Tension will also be negligibly small.

#### haruspex

Homework Helper
Gold Member
2018 Award
But in this stackexchange post, https://physics.stackexchange.com/questions/308011/rigorously-proving-equal-tension-on-both-ends , dmkee says that "Once the pulleys have non-trivial mass and the rope is accelerating than the segments must be assumed to have different tensions even if the rope is light." Does this mean that massless strings do not always have the same tension throughout? I'm a bit confused considering the proof given by jbriggs.
The proof given by @jbriggs444 is in regards to a segment of string on which the only external forces are applied at the ends. If the string is of negligible mass then any non-negligible net applied force would result in a huge acceleration.
In the case of a massive accelerating pulley, with the string not slipping on the pulley, the inertia of the pulley applies a frictional force to the string. So now you must consider the string in two segments, one each side of the pulley. Within each segment the tension is constant, but there will be a difference between the two tensions.

#### Let'sthink

Recall that the "F" in F=ma is the vector sum of all external forces acting on an object.

Suppose that m=0. Then it follows that F=0. So the sum of all external forces on a length of string is zero. There are only two such forces: the force F1 applied at the one end and the force F2 applied at the other end. Both are equal in magnitude to the tensions at the respective ends of the string, T1 and T2.

$$F_1+F_2=0$$ $$F_1=-F_2$$ $$|F_1| = |F_2|$$ $$T_1=T_2$$
I am talking of the case when common acceleration of the two masses and the string is non zero, Sir.

#### Nugatory

Mentor
I am talking of the case when common acceleration of the two masses and the string is non zero, Sir.
The argument still stands: $F=ma$ with $m=0$ implies that $F=0$; the net force on any segment of string must be zero, which implies uniform tension within that segment.

#### Mister T

Gold Member
Does this mean that massless strings do not always have the same tension throughout?
Correct. The pulley makes contact with the string and exerts forces on it.

#### sophiecentaur

Gold Member
I am talking of the case when common acceleration of the two masses and the string is non zero, Sir.
I can see that you feel a need to find something wrong with what has been said here. The core of the argument is F=ma. If m is zero then F will always be zero (under any value of acceleration). Can you possibly disagree with that?
The high number of posts in this thread are about of all proportion with the explanation in terms of Newton's Laws. Just sit and contemplate F=ma for a while.

#### Let'sthink

I can see that you feel a need to find something wrong with what has been said here. The core of the argument is F=ma. If m is zero then F will always be zero (under any value of acceleration). Can you possibly disagree with that?
The high number of posts in this thread are about of all proportion with the explanation in terms of Newton's Laws. Just sit and contemplate F=ma for a while.
What do you say about T in the string moving with common acceleration, when string is not mass less? Will it be uniform? Contemplate on F = ma and tell me Sir.

#### haruspex

Homework Helper
Gold Member
2018 Award
I am talking of the case when common acceleration of the two masses and the string is non zero, Sir.
@jbriggs444 was responding to post #3, not to your post #4.

#### Nugatory

Mentor
What do you say about T in the string moving with common acceleration, when string is not mass less? Will it be uniform?
Of course not. Consider any two points on one segment of the string. The string is not massless, so the bit of string between those two points has non-zero mass. It is also accelerating, so $F=ma$ tells us that $F$ is non-zero. This non-zero net force is the difference between the tension at those two points; if that difference is non-zero the tensions are different and therefore the tension is not uniform.

It would be a good exercise to derive the formula for the tension at any point; if the density of the string is constant (that is, the mass of any piece of lengthn$L$ is equal to $\rho{L}$ where $\rho$ is a constant) you can do this with just algebra, no calculus needed.

#### sophiecentaur

Gold Member
This is now a different problem. The new constraints assume no slippage over the pulleys. Tension will be the same at each end of a section between pulleys MINUS any ma and mg (weight) factor.
This is not a ‘proof’ we are after. It’s a calculation.

"Is this proof regarding massless strings correct?"

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