What is the tension in a rope when a monkey accelerates up?

[rudransh verma]

Homework Statement

A monkey of mass m is climbing up the rope with uniform speed, the tension in the rope will be

Relevant Equations

F=ma

I think the tension in the rope will be equal to its weight , mg.
I want to ask what if the monkey accelerates up with acceleration a, then what will be the tension in the rope?

 

[phinds]

What do you think it will be and why?

 

[rudransh verma]

What do you think it will be and why?

I think whatever force monkey applies to accelerate will just add to the mg. So increased tension, mg+ma.
Monkey applies force ma in downward direction. Rope applies same opposite force on monkey. Newton’s third law. So tension is mg+ma.

 

[phinds]

Monkey applies force ma in downward direction. Rope applies same opposite force on monkey. Newton’s third law. So tension is mg+ma.

Looks to me as though you have two statements here
1) F = ma (your first two sentences)
2) F = ma +mg (your last sentence)

Which is it?

 

[rudransh verma]

2) F = ma +mg (your last sentence)

Tension in rope.

 

[phinds]

Can you draw a force diagram showing all the forces acting?

 

[rudransh verma]

Can you draw a force diagram showing all the forces acting?

 

[haruspex]

As you have been told several times, I believe, a FBD should show one system (often one rigid body) and all the forces acting on it; no internal forces. That diagram shows the gravity acting on the monkey and the monkey's force on the rope.

 

[rudransh verma]

the monkey's force on the rope.

And I think this will increase the tension.
Are you saying monkeys force is internal?
Edit: Free body diagram alone doesn't give the answer. Force applied by monkey is to be considered.
Look at the beauty of my FBD. So pretty.

 

[Lnewqban]

And I think this will increase the tension.
Are you saying monkeys force is internal?
Edit: Free body diagram alone doesn't give the answer. Force applied by monkey is to be considered.
Look at the beauty of my FBD. So pretty.

Your free body diagram should be essentially similar to the one in your other thread, because both situations are very similar.
https://www.physicsforums.com/threads/forces-within-an-elevator-cab.1009507/

The rope “feels” a heavier monkey only when the creature is accelerating upwards (monkey mass x [a+g]).

When the monkey is either hanging in a static position, or climbing at a steady pace, the rope “feels” only the natural weight (monkey mass x g).

 

[Doc Al]

And I think this will increase the tension.
Are you saying monkeys force is internal?
Edit: Free body diagram alone doesn't give the answer. Force applied by monkey is to be considered.
Look at the beauty of my FBD. So pretty.

You really need to start paying attention to @haruspex (and others).

A free-body diagram of the monkey should show only forces acting on the monkey. (Only two forces here.) Then you can apply Newton's 2nd law to those forces.

 

[phinds]

Look at the beauty of my FBD. So pretty.

Yep. Too bad beauty doesn't count and it is functionally wrong. See post #11

 

[haruspex]

And I think this will increase the tension.
Are you saying monkeys force is internal?
Edit: Free body diagram alone doesn't give the answer. Force applied by monkey is to be considered.
Look at the beauty of my FBD. So pretty.

Let me explain in more detail.

If the FBD is for the system "monkey", it should show all and only forces acting on the monkey, not forces the monkey exerts on anything else. You have shown the force the monkey exerts on the rope, when you should show the opposite force.

If the FBD is for the system "monkey+rope", it should show all and only forces acting on that system, not any forces internal to that system, so not forces between monkey and rope.

 

[rudransh verma]

You have shown the force the monkey exerts on the rope, when you should show the opposite force.

Oh yes you are absolutely right.

 

[haruspex]

Oh yes you are absolutely right.

Much better, but now you have shown it twice, once as "force from rope" and once as T.
So what equation can you write from that?

 

[rudransh verma]

Much better, but now you have shown it twice, once as "force from rope" and once as T.
So what equation can you write from that?

$T=F_{RM}+mg$ where $F_{MR}$ is force applied on monkey by rope.
Edit: @haruspex Is it right?

 

[Doc Al]

Realize that the force the rope exerts on the monkey is the tension in the rope. Don't count it twice. Only two forces act on the monkey.

 

[Steve4Physics]

No one has mentioned friction. I'd look at the system is like this:

- for the monkey climbing (or hanging onto) the rope, the only upwards force on the monkey is the upwards friction of the rope on the monkey, $F_{rm}$;

- for the rope (assumed of negligible weight), the only downwards force on the rope is the downwards friction of the monkey on the rope $F_{mr}$.

$F_{rm}$ and $F_{mr}$ are a Newton’s 3rd law pair so have equal magnitudes.

With this view, the tension doesn’t directly act on the monkey, so it should not be shown on the monkey's FBD.

Similarly, the monkey’s weight doesn’t directly act on the rope, so it should not be shown on the rope's FBD.

 

[haruspex]

Wh

$T=F_{RM}+mg$ where $F_{MR}$ is force applied on monkey by rope.
Edit: @haruspex Is it right?

What part of "but now you have shown it twice, once as 'force from rope' and once as T" did you not understand?

 

[rudransh verma]

What part of "but now you have shown it twice, once as 'force from rope' and once as T" did you not understand?

Realize that the force the rope exerts on the monkey is the tension in the rope. Don't count it twice. Only two forces act on the monkey.

I am thinking like tension in string will be how much it’s being stretched by Monkey. So in non accelerating case it’s just the weight of the monkey but otherwise it’s applying a force on the rope to pull himself up. So tension will be simply $T=F_{RM}+mg$. $F_{RM}$ is just what monkey is applying on rope.
But according to you I’ll be wrong and what you want is $T=mg+ma$

 

[haruspex]

But according to you I’ll be wrong and what you want is $T=mg+ma$

Quite so.

The flaw in your thinking is that once again you are mixing forces on different components.
Strictly speaking, tension is not a force. It is better thought of as a pair of equal and opposite forces. From the rope's perspective, those are forces tending to stretch the rope, so the force at the monkey end points down, i.e. it is the force the monkey exerts on the rope. $T=F_{MR}$.

The force the rope exerts on the monkey is equal and opposite to that. $F_{MR}=-F_{RM}$.

The only other force on the monkey is gravity, therefore in equilibrium $mg=-F_{RM}$.

 

[Doc Al]

But according to you I’ll be wrong and what you want is $T=mg+ma$

You need to stop trying to "guess" the answer. Instead, use simple physics to solve for the answer.

Here there are only two forces acting on the monkey (as @haruspex has explained many times): The rope pulling up (which can be described as the "tension" in the rope) and gravity pulling down. Now apply Newton's 2nd law to solve for that upward tension: $$\Sigma F = ma$$ $$T - mg = ma$$ thus $$T = mg + ma$$.

 

[rudransh verma]

is better thought of as a pair of equal and opposite forces.

You say pair. Are you talking about T and $F_MR$ as pair?

the monkey end points down, i.e. it is the force the monkey exerts on the rope. T=FMR.

The force the rope exerts on the monkey is equal and opposite to that. FMR=−FRM.

The only other force on the monkey is gravity, therefore in equilibrium mg=−FRM.

You are using different definition for $F_{MR} and F_{RM}$. You seem right! $F_{MR}$ is what creates tension. $T=F_{MR}=mg$ in equilibrium condition. But in accelerating case, $T=F_{MR}= mg+ma$. It’s just that in acceleration monkey applies more force which is true if we see. Tension increases!

My eqn is wrong probably because in equilibrium condition we cannot apply more force than our weight.

 

[Lnewqban]

You say pair. Are you talking about T and $F_MR$ as pair?

Copied from
https://courses.lumenlearning.com/physics/chapter/4-5-normal-tension-and-other-examples-of-forces/

“A tension is a force along the length of a medium, especially a force carried by a flexible medium, such as a rope or cable. The word “tension” comes from a Latin word meaning “to stretch.” Not coincidentally, the flexible cords that carry muscle forces to other parts of the body are called tendons. Any flexible connector, such as a string, rope, chain, wire, or cable, can exert pulls only parallel to its length; thus, a force carried by a flexible connector is a tension with direction parallel to the connector. It is important to understand that tension is a pull in a connector. In contrast, consider the phrase: “You can’t push a rope.” The tension force pulls outward along the two ends of a rope. Consider a person holding a mass on a rope as shown...”

 

[Doc Al]

“You can’t push a rope.”

One of my physics 101 professor's favorite sayings.

 

[Lnewqban]

Also, please see this old thread:
https://www.physicsforums.com/threads/trouble-with-the-concept-of-tension.985741/

 

[haruspex]

You say pair. Are you talking about T and F_MR as pair?

No, that's not what I meant.
If a rope is under uniform tension T then you could consider any short section of it as a body in its own right and find that each end is subject to a force T acting away from it. So uniform tension is like equal and opposite pairs of forces acting all the way along its length.
If the tension is not uniform (e.g. massive rope hanging vertically) then the pairs are not quite equal.

 

[rudransh verma]

end points down, i.e. it is the force the monkey exerts on the rope. T=FMR.

The force the rope exerts on the monkey is equal and opposite to that. FMR=−FRM.

Can I say $T=F_{RM}$ since T is acting everywhere.

So tension will be simply T=FRM+mg. FRM is just what monkey is applying on rope.
But according to you I’ll be wrong an

$T=F_{MR}$. So in static case $F_{MR}=mg$ due to this, $F_{RM}=mg$. Monkey moves up with constant speed.
In accelerating case, $F_{MR}=mg+ma$ due to this $F_{RM}=mg+ma$. Monkey accelerates.
Edit:
$T=F_{MR}$. So in static case $F_{MR}=-mg$ due to this, $F_{RM}=mg$. Monkey moves up with constant speed.
In accelerating case, $F_{MR}=-mg-ma$ due to this $F_{RM}=mg+ma$. Monkey accelerates.

 

[haruspex]

Can I say $T=F_{RM}$ since T is acting everywhere.

$T=F_{MR}$. So in static case $F_{MR}=mg$ due to this, $F_{RM}=mg$. Monkey moves up with constant speed.
In accelerating case, $F_{MR}=mg+ma$ due to this $F_{RM}=mg+ma$. Monkey accelerates.

That's all correct if in each case you are taking the positive direction as the direction in which you expect the force or acceleration to act. I.e. taking up as positive for $T, F_{RM}, a$, but down as positive for $F_{MR}, g$.
In some problems it might not be obvious which way some forces or accelerations (or displacements or velocities) act, so a useful practice is to fix on one direction as positive for everything, usually up. In that convention (and taking T as a force acting on the rope, so actually down) you would write:
$T=F_{MR}=-F_{RM}$ and $ma=-mg+F_{RM}$, leading to $T=-(mg+ma)$.

 

[rudransh verma]

That's all correct if in each case you are taking the positive direction as the direction in which you expect the force or acceleration to act. I.e. taking up as positive for T,FRM,a, but down as positive for FMR,g.

I took up as +ve every time.

actually down) you would write:
T=FMR=−FRM and ma=−mg+FRM, leading to T=−(mg+ma).

In post#24 in diagram we can see T is upwards , downwards, acting on the rope , is the force that stretch the rope as you said. So all these are tension.
And in your post #21 you write $T=F_{MR}$. Then in above post you write $T=F_{MR}=-F_{RM}$.
Isn't this should be $-T=-F_{MR}=F_{RM}$ taking T as the force applied on rope(taking up as +ve)?
And we can also write $T=-F_{MR}=F_{RM}$ taking T as the force the rope applies.

 

[haruspex]

In post#24 in diagram we can see T is upwards , downwards, acting on the rope , is the force that stretch the rope as you said. So all these are tension.
And in your post #21 you write $T=F_{MR}$. Then you write $T=F_{MR}=-F_{RM}$. Is this because you took Tension to be the force the monkey applies. And we can also write $T=-F_{MR}=F_{RM}$ taking T as the force the rope applies.

Yes.

I took up as +ve every time.

No you didn't. In post #28 you wrote $F_{MR}=mg$. By convention, g is always positive, so with up positive, weight is -mg. So it becomes $F_{MR}=-mg$.

 

[rudransh verma]

ctually down) you would write:
T=FMR=−FRM and

Isn't this should be $T=−F_{MR}=F_{RM}$ taking T as the force applied on rope(taking up as +ve)?

 

[rudransh verma]

No you didn't. In post #28 you wrote FMR=mg. By convention, g is always positive, so with up positive, weight is -mg. So it becomes FMR=−mg.

I used Newtons second law wrongly. I had taken the forces from different bodies when it has to be one body. We cannot apply this law here. I got you. You are merely stating the value of $F_{MR}$ as $-mg$ only. Right?
I edited the post.

 

[haruspex]

Isn't this should be $T=−F_{MR}=F_{RM}$ taking T as the force applied on rope(taking up as +ve)?

If I understand your notation, $F_{MR}$ means the force the monkey exerts on the rope. Taking T as the force applied on rope, necessarily, $T=F_{MR}$, regardless of sign convention.

 

[rudransh verma]

If I understand your notation, $F_{MR}$ means the force the monkey exerts on the rope. Taking T as the force applied on rope, necessarily, $T=F_{MR}$, regardless of sign convention.

Yeah! but T as vector is $-F_{MR}$. Right?

 

[vcsharp2003]

If I understand your notation, $F_{MR}$ means the force the monkey exerts on the rope. Taking T as the force applied on rope, necessarily, $T=F_{MR}$, regardless of sign convention.

How about the static friction force acting upwards on the monkey's hands/legs from the rope's surface as it climbs up the rope? The rope and monkey's hands interface cannot be smooth else monkey would slip down as it tried to climb up.

Or we are already considering this friction force when we say force exerted by the rope on the monkey? How is tension in rope related to this frictional force? Probably equal since rope is stationary as monkey climbs up.

 

[Steve4Physics]

How about the static friction force acting upwards on the monkey's hands/legs from the rope's surface as it climbs up the rope? The rope and monkey's hands interface cannot be smooth else monkey would slip down as it tried to climb up.

Or we are already considering this friction force when we say force exerted by the rope on the monkey? How is tension in rope related to this frictional force? Probably equal since rope is stationary as monkey climbs up.

I previously (Post #18) pointed out the role of friction, but no one seems to want to refer to friction!

 

[rudransh verma]

@Steve4Physics (taking up as +ve)
1. If tension is the pair of forces which stretch the rope then why do we show in diagrams the tension inwards?

2. Tension is the force applied by monkey on rope. It is pair of outward forces. Magnitude wise $T=F_{MR}$. But vectorially $-T=-F_{MR}$.
Can I say $T=F_{RM}$?

3. What is the meaning of each?
$-T=-F_{MR}$.
$T=-F_{MR}$.
$-T=F_{MR}$.
$T=F_{MR}$.

 

[vcsharp2003]

With this view, the tension doesn’t directly act on the monkey, so it should not be shown on the monkey's FBD.

Very true.

If we look at the monkey as the isolated system in our free body diagram, then surely it's the friction force from rope that's acting on the monkey and the tension force in the rope is not acting directly on the monkey. We shouldn't be caring about the tension force in our free body diagram.

 

[haruspex]

How about the static friction force acting upwards on the monkey's hands/legs from the rope's surface as it climbs up the rope? The rope and monkey's hands interface cannot be smooth else monkey would slip down as it tried to climb up.

Or we are already considering this friction force when we say force exerted by the rope on the monkey? How is tension in rope related to this frictional force? Probably equal since rope is stationary as monkey climbs up.

I don't see that how the monkey is attached to the rope is of any interest. Could be friction, could be the rope is tied around the monkey (do we need to get into how the friction stops the knot coming undone?), or maybe the monkey sits in a light bucket.
Or you could choose to consider the piece of rope the monkey clutches as part of a monkey+rope piece system.

 

[haruspex]

@Steve4Physics (taking up as +ve)
1. If tension is the pair of forces which stretch the rope then why do we show in diagrams the tension inwards?

2. Tension is the force applied by monkey on rope. It is pair of outward forces. Magnitude wise $T=F_{MR}$. But vectorially $-T=-F_{MR}$.
Can I say $T=F_{RM}$?

3. What is the meaning of each?
$-T=-F_{MR}$.
$T=-F_{MR}$.
$-T=F_{MR}$.
$T=F_{MR}$.

1.
It depends what body is being considered in the FBD. Whether it is something pulled by an end of the rope or the rope itself, it will point away from that body.

2., 3.
'T' is being used to mean two things here.

On the one hand it can mean the tension, which exists everywhere along the rope acting on each short element as a pair of opposing forces rather than as a single force. In this meaning, direction is either outward from each rope element (tension) or inward on it (compression).

On the other hand, it can mean one of that pair at an end of the rope, and it is up to you to decide which of the pair it stands for. When you write a force equation involving tension you are necessarily taking this second view.
I chose above to consider it as the force the monkey exerts on the rope.

 

[Steve4Physics]

@Steve4Physics (taking up as +ve)
1. If tension is the pair of forces which stretch the rope then why do we show in diagrams the tension inwards?

Did you read the link in @Lnewqban;s Post #26?

Tension is referred to as a ‘force’ but is a bit more complicated than a simple force.

Say we have rope, AB represented as A------B .

Pull A left with a force of 10N and pull B right with another force of 10N.

10N←A------B→10N

The system is in equlibrium and we have an ‘inwards force’ inside the rope, which stops A and B moving apart. This ‘inwards force’ is called the tension.

At the ends, A is being pulled right by tension and B is being pulled left. We say the tension (inside the rope) is T=10N and we represent it by a pair of inwards arrows:

10N←A→--←-B→10N

The forces at point A are: 10N←A→T

The forces at point B are: T←B→10N

(Note, the external applied 10N forces are not called tension, though they are numerically equal to the tension here.)

2. Tension is the force applied by monkey on rope. It is pair of outward forces.

I don't know what you mean by 'It is a pair of outwards forces.'.

Magnitude wise $T=F_{MR}$. But vectorially $-T=-F_{MR}$.
Can I say $T=F_{RM}$?

3. What is the meaning of each?
$-T=-F_{MR}$.
$T=-F_{MR}$.
$-T=F_{MR}$.
$T=F_{MR}$.

Sorry, I don't understand the questions.

 

[Steve4Physics]

I don't see that how the monkey is attached to the rope is of any interest. Could be friction, could be the rope is tied around the monkey (do we need to get into how the friction stops the knot coming undone?), or maybe the monkey sits in a light bucket.
Or you could choose to consider the piece of rope the monkey clutches as part of a monkey+rope piece system.

It’s simply the desire to correctly name the actual physical forces which operate.

Also, recognising the role of friction might be important in some (slightly harder) problems, e.g. finding the monkey’s maximum upwards acceleration.

 

[vcsharp2003]

I don't see that how the monkey is attached to the rope is of any interest. Could be friction, could be the rope is tied around the monkey (do we need to get into how the friction stops the knot coming undone?), or maybe the monkey sits in a light bucket.
Or you could choose to consider the piece of rope the monkey clutches as part of a monkey+rope piece system.

In the absence of a diagram, it can get confusing to understand the question. The scenario that I had in mind while looking at the question is as below, where frictional force makes sense.

I'm not sure if the scenario of question is different from what I thought.

 

[rudransh verma]

You mean you used T tension as a condition in rope and wrote Tension as a condition is $T=F_{MR}$ whereas I used it as a single force F

In this meaning, direction is either outward from each rope element (tension) or inward on it (compression).

Can you show a diagram?

The system is in equlibrium and we have an ‘inwards force’ inside the rope, which stops A and B moving apart.

This is understandable but what you are telling is a bit complicated without diagram

On the other hand, it can mean one of that pair at an end of the rope, and it is up to you to decide which of the pair it stands for.

Again Please show a diagram

 

[Steve4Physics]

This is understandable but what you are telling is a bit complicated without diagram

I thought I already had (in Post #42) included diagrams. Here they are again:

10N←A------B→10N (external forces on rope being pulled outwards)

10N←A→--←-B→10N (internal force (tension, T) inside the rope shown)

10N←A→T (forces acting at point A)

T←B→10N (forces acting at point B)

You might like to watch a video I did about tension (the sound is a bit crackly):

 

[rudransh verma]

I thought I already had (in Post #42) included diagrams. Here they are again:

No i am telling @haruspex to give diagram like you did. But thanks for additional information.
By the way can you tell me when we write something like $T=F_{MR}$ then are we saying $T$ is equal to $F_{MR}$ or the value of $T$ is $F_{MR}$

 

[kuruman]

No i am telling @haruspex to give diagram like you did. But thanks for additional information.
By the way can you tell me when we write something like $T=F_{MR}$ then are we saying $T$ is equal to $F_{MR}$ or the value of $T$ is $F_{MR}$

Shown below is my diagram adapted from @vcsharp2003's. We know that there are two items that exert a force on the monkey, the rope and the Earth. The forces are shown as arrows in the diagram. You can label them any way you wish. We also know that the monkey is moving up at constant speed. That's all you need to answer the question.

 

[jbriggs444]

No i am telling @haruspex to give diagram like you did. But thanks for additional information.
By the way can you tell me when we write something like $T=F_{MR}$ then are we saying $T$ is equal to $F_{MR}$ or the value of $T$ is $F_{MR}$

When we write equalities in physics, we normally mean that the two quantities are identical. Not that they are literally one and the same thing.

We do hold than when two entities are literally one and the same thing, they have the same quantititative measure. So $F_{\text{MR}}=F_{\text{MR}}$ in both senses.

In first year physics, we often play a little bit fast and loose with the distinction between a vector value, a scalar value and a tensor value. We do not even explain to students what a "tensor" is and how it might be considered to have a single numeric value.

So we can casually write that $T=F_\text{MR}$ when we may mean that the downward vertical force ($T$) exerted to anchor the bottom end of the rope is literally the same force that the monkey applies on the rope.

The tension in an ideal rope is the same as the scalar force applied at either end away from the middle.

 

[rudransh verma]

So we can casually write that T=FMR when we may mean that the downward vertical force (T) exerted to anchor the bottom end of the rope is literally the same force that the monkey applies on the rope.

The tension in an ideal rope is the same as the scalar force applied at either end away from the middle.

This casual explanation is making my life very difficult from morning (and now its night time) because what I knew is Tension T's direction when the rope is under stress is upwards and the force applied by monkey is downwards. So $T=F_{MR}=-mg$
So tension is basically a force that the rope applies back when it is under stress. It is an inward force. Tension T's direction at end points of rope where its attached to the body and ceiling is inwards. Tension is what we pull something with not push.

This video explains nicely how force applied is transmitted from one end of rope to other under tension.
Someone said that tension is pair of opposite forces acting on small rope elements all along the rope cancelling each other. At the ends there is just one force acting inwards. Am I right?