Does the tension in the rope change as the transverse wave travels upward?

[Fluxxx]

Homework Statement

A rope of mass m is hanging down from the ceiling. Nothing is attached to the loose end of the rope. As a transverse wave travels upward on the rope, does the speed of the wave increase, decrease, or remain the same?

Homework Equations

$$v=\sqrt{\frac{F}{m/L}}$$
$$F=-ma$$

The Attempt at a Solution

As the wave travels upward, the second equation would imply that the force on the wire directed downwards (i.e. gravitational force) is larger, so the speed would be smaller than if you did the same thing when standing high and the rope was attached to the floor. But the answer given in the book is "Increases". Why?

 

[Orodruin]

What is the tension in the upper part of the rope? What is the tension at the hanging end of the rope?

 

[Fluxxx]

What is the tension in the upper part of the rope? What is the tension at the hanging end of the rope?

Tension is given by the second equation, and the rope has uniform mass/length, so why isn't it the same everywhere?

 

[Orodruin]

No, the tension is not constant. Which is the mass that should go into that equation? Consider the forces acting on the lower part of the rope (say of length x, where x < L).

 

[Fluxxx]

The only force acting on the rope is the gravitational force, isn't it? I.e. the second equation I wrote above. And that in this case is equal to the tension, right?

But the gravitational force only acts downwards, from top to bottom. Is there another force acting on the rope in the opposite direction? There must be, since the rope is stationary. So we have two tension forces, one up and one down?

Exactly which forces are involved here?

 

[Orodruin]

No, tension is not a force. It only relates to a force when you make a free body diagram of a particular part of the rope. What happens when you make a free body diagram on the lower part of the rope with length $x$? There are two forces acting on it, the gravitational force (on the lower part!) and the force from the upper part (which is equal to the tension at the point where the lower part meets the upper part). How must these forces be related for the rope to be in equilibrium?

 

[Fluxxx]

No, tension is not a force. It only relates to a force when you make a free body diagram of a particular part of the rope. What happens when you make a free body diagram on the lower part of the rope with length $x$? There are two forces acting on it, the gravitational force (on the lower part!) and the force from the upper part (which is equal to the tension at the point where the lower part meets the upper part). How must these forces be related for the rope to be in equilibrium?

Mustn't they be equal? So the sum of forces are zero. But in that case, won't we get F=0 in the equation above? For it to be correct only the upward force (what is that force called by the way?) is counted? But why isn't the downward force (gravitation) counted then?

 

[jbriggs444]

Mustn't they be equal? So the sum of forces are zero. But in that case, won't we get F=0 in the equation above?

The tension supporting a section of rope and gravity pulling down on that section of rope must sum to zero, yes. That's what F=ma is telling you. Because a is zero, net force F must be zero.

That's the F in the bottom equation in the original post -- net external force on an arbitrary section of rope. It allows you to compute the force from tension immediately above that section.

The F in the top equation in the original post is not the same F. It denotes the tension in the rope.

There is an important lesson here. Do not blindly manipulate symbols without knowing what they mean in context. This is the fallacy of "equivocation" -- using the same symbol with two distinct meanings as it were a single entity.

 

[Fluxxx]

The F in the top equation (...) denotes the tension in the rope.

No, tension is not a force.

A bit confusing! Is tension a force or not?

 

[jbriggs444]

A bit confusing! Is tension a force or not?

Strictly speaking, it is not a force because it is not a vector. It is a condition in the rope like a stress or a strain. The tension at a point in a rope is equal to the force with which the two halves of the rope that meet at that point are pulling on each other. But since those two forces are in opposite directions, there is no one defined direction for the tension to point in. That is one reason why it is not considered to be a "force".

However, it has the same units as a force and if you pull on a rope, the tension in the rope will be equal to the [magnitude of the] force of your pull.

In addition, the fact that you use the symbol "F" for a quantity does not make that quantity a force.

 

[Fluxxx]

Why does "F" in the following equation
$$v=\sqrt{\frac{F}{m/L}}$$
only denote tension, and not the sum of all forces? This equation is not my invention, it's how it's written in my textbook.
Would have been more correct like this?
$$v=\sqrt{\frac{F_{T}}{m/L}}$$
where $F_{T}$ = Tension force.

 

[Orodruin]

It does not matter what you call things if it is specified and clear what is being denoted. Your focus on this is distracting you from the actual problem at hand.

 

[Fluxxx]

It does not matter what you call things if it is specified and clear what is being denoted. Your focus on this is distracting you from the actual problem at hand.

Only it is not clear in this case, there is the problem of equivocation as jbriggs already pointed out.
By the way: The problem at hand is unimportant, only understanding is important! Understanding refers to the understanding of the one asking the question, and if that person wants to ask other follow-up questions other than the original question, in order to understand, then that's the way it should be.

 

[Orodruin]

So, back to the original question.

Mustn't they be equal? So the sum of forces are zero.

Yes. The net force on any part of the rope must be zero. What does this imply for the tension at a position a distance $x$ above the lower end of the rope?