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

[Steve4Physics]

No i am telling @haruspex to give diagram like you did.

I would advise 'asking' rather than 'telling'!

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 $F_{MR}$

There is no difference, 'is equal to' means having the same value.

But remember, different vectors can have the same magnitude but the vectors will be unequal if their directions are different.

$\vec T$ is the tension acting upwards on the rope (where it is hed by the monkey).
$\vec F_{MR}$ is the frictional force acting downward on the rope.
The rope is in equilibrium. To be rigorous we should write:
$\vec T+ \vec {F_{MR}}$ = 0
which gives
$\vec T = -\vec {F_{MR}}$
Less rigorously we could write this as:
$T = -F_{MR}$

In terms of magnitudes we can write:
$|T| = |F_{MR}|$

If we are using '$T$' and '$F_{MR}$' to represent magnitudes, we should state this to avoid confusion/ambiguity. Then we can write:
$T - F_{MR}=0$ which gives
$T = F_{MR}$.

We could also use the unit vector ($\hat k$) but this might be a bit over-the-top for this type of problem.

 

[rudransh verma]

I would advise 'asking' rather than 'telling'!There is no difference, 'is equal to' means having the same value.

But remember, different vectors can have the same magnitude but the vectors will be unequal if their directions are different.

$\vec T$ is the tension acting upwards on the rope (where it is hed by the monkey).
$\vec F_{MR}$ is the frictional force acting downward on the rope.
The rope is in equilibrium. To be rigorous we should write:
$\vec T+ \vec {F_{MR}}$ = 0
which gives
$\vec T = -\vec {F_{MR}}$
Less rigorously we could write this as:
$T = -F_{MR}$

In terms of magnitudes we can write:
$|T| = |F_{MR}|$

If we are using '$T$' and '$F_{MR}$' to represent magnitudes, we should state this to avoid confusion/ambiguity. Then we can write:
$T - F_{MR}=0$ which gives
$T = F_{MR}$.

We could also use the unit vector ($\hat k$) but this might be a bit over-the-top for this type of problem.

Yes! Thank you! @haruspex should understand this. People like us are in their elementary stage. Sticking to one rule will help in long run like something like up is +ve sign convention and not changing signs every time. Writing casually in my views should be avoided because it hinders in what one is trying to say. It should be rigorous.

 

[kuruman]

$\vec T+ \vec {F_{MR}}$ = 0
which gives
$\vec T = -\vec {F_{MR}}$
Less rigorously we could write this as:
$T = -F_{MR}$

I agree with first two equations, but the one following "less rigorously" can be confusing to a novice. That's because a symbol with an arrow over it denotes a vector whilst the same symbol without the arrow denotes the magnitude of the vector. In one dimension, the "less rigorous" approach that you show confuses the magnitude of the vector, which is always positive, with the component of the vector which could be positive or negative. All this can easily be sorted out by using, rigorously, unit vectors and subscripts for components, both which are routinely dropped in the less rigorous approach.

Starting with
$\vec T+ \vec {F}_{MR}=0$, we rewrite the one-dimensional vectors in terms of their components as
$T_y~\hat y+ F_{MR,y}~\hat y=0$
Here is where we drop the unit vectors and write
$T_y+ F_{MR,y}=0~\implies~T_y=-F_{MR,y}$
which shows that when two one-dimensional vectors add to zero, their components have the same magnitude and opposite signs.

The interpretation of vectors in FBDs is that the label of the arrow denotes the magnitude while the direction of the arrow denotes the direction of the vector. So if we have the FBD of this, we would see two vectors labeled $T$ and $F_{MR}$ pointing in opposite directions.
Starting again with
$\vec T+ \vec {F}_{MR}=0$, we rewrite the one-dimensional vectors in terms of their magnitudes as
$\vec T=T~(+\hat y)~;~~ \vec {F}_{MR}=F_{MR}~(-\hat y)$
Then
$T~(+\hat y)+F_{MR}~(-\hat y)=0$
$T~(+\hat y)=-F_{MR}~(-\hat y)=F_{MR}(+\hat y)$
Again we drop the unit vectors and write
$T=F_{MR}$
which shows that when two vectors point in opposite directions and add to zero, they have the same magnitude.

In short, when we write an equation relating two or more one-dimensional vectors, we have to be clear whether the equation relates components or magnitudes.

 

[haruspex]

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.

View attachment 296996

That is still answered by my last option, consider the piece of rope the monkey clutches as part of a monkey+rope piece system.

 

[haruspex]

But remember, different vectors can have the same magnitude but the vectors will be unequal if their directions are different.

Yes, but for tension it is a little different. Tension and force have the same dimension, but the notion of direction is quite different. Tension is not a vector.
Many students start off thinking that if there is a force magnitude F pulling on each end of a rope then the tension should be 2F. Strictly speaking, tension could have been defined that way, but it is not. So saying they have the same magnitude is just (the universally agreed) convention.

Sticking to one rule will help in long run like something like up is +ve sign convention and not changing signs every time.

More easily wished than done. In 3D problems, what could the fixed rule be for horizontal forces? Or maybe the context is gravity-free, so there is no vertical. And advisors on PF don't know what the students have been taught to do, and don't all have the same ways of working.

 

[rudransh verma]

Tension is not a vector.

Oh yes! That’s what I was missing. Tension is scalar. In whatever direction(angle we change) we pull the rope that is tied to a weight through a pulley, tension doesn’t change.
Can you elaborate?

 

[rudransh verma]

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...”

So you mean tension is pulling the hand of the holder and the weight inwards. That is what tension is . It is a pulling force.

 

[Lnewqban]

Sorry, rudransh, I can't understand your question.
That statement that you have just posted is only a quotation from this link:

https://courses.lumenlearning.com/physics/chapter/4-5-normal-tension-a

What is about the concept of tension that is still troubling you after so many posts of this thread?

The way I understand the tension inside the rope (perhaps incorrectly):
The cohesion of the fibers, which is the natural attraction of similar molecules for each other, resists the action of the forces trying to separate those so many molecules.

The molecules of that rope do not allow much distance to grow between each other, reason for which the rope does not stretch as much as a rubber band or metal spring would do under stress.

If the magnitude of the external forces acting on the rope becomes greater than the cohesion of all the molecules contained in the plane of any cross-section of the rope, those molecules get separated and the rope breaks.

The rope of your example is no more no less than a solid link between the two forces in question: the static weight of the monkey and the reaction force at the anchor located at the top.
It could be long, short, that does not change the magnitude and direction of those external forces.

Or it could be a chain, a wire-rope, an ivy, a bungee cord, a spring, a long piece of wood, or any other thing able to transfer forces, creating internal tensions in the process.

Even if there is no rope and the monkey directly grabs the anchor or branch, its weight and the reaction at the point of anchorage would still be the same.

 

[rudransh verma]

That statement that you have just posted is only a quotation from this link:

I just want to be verified that I understand it right. All rope does when you say “The tension force pulls outward along the two ends of a rope” is that it is pulling inwards what is attached to it on both ends when the rope is under tension?

 

[jbriggs444]

I just want to be verified that I understand it right. All rope does when you say “The tension force pulls outward along the two ends of a rope” is that it is pulling inwards what is attached to it on both ends when the rope is under tension?

The attachment points pull outward on the rope. The rope pulls inward on the attachment points. For an ideal (massless) rope, the magnitudes of all four of those forces are identical.

One can see that they are identical by applying Newton's third law at the two attachment points ends and Newton's second and third laws in a daisy-chain all the way along the rope.

 

[Lnewqban]

I just want to be verified that I understand it right. All rope does when you say “The tension force pulls outward along the two ends of a rope” is that it is pulling inwards what is attached to it on both ends when the rope is under tension?

I believe that I did not say that, it is a statement in that linked website.

The internal forces in the rope ("molecular tension") prevent the rope from being ripped apart by the external forces acting at both ends.
The internal forces are resistive, are a natural reaction to external energy.
If the rope breaks, the energy of the external forces did the destructive work.

 

[rudransh verma]

I believe that I did not say that.

I mean the link says that.

 

[Lnewqban]

I mean the link says that.

As the magnitudes of external forces and internal forces are similar, they use tension and force alternately.
Again, I may be wrong.

 

[kuruman]

To @rudransh verma :
The important thing to remember about tension in a rope is that a force is associated with it. To find its direction, first you need a system on which this force acts. Once you have defined the system, the force due to tension acts along the rope or string in a direction away from the system. In short, you cannot push with a rope.

For example, in the climbing monkey problem we can choose to define the monkey as the system (see post #48.) The force due to the rope tension is up and away from the monkey. We can also choose as our system a 10 cm piece of rope above the point of contact with the monkey's hand. In that case, there is a down force on the 10 cm piece of rope due to the monkey's hand (equal and opposite to the one exerted by the rope on the monkey and away from the system), and an "up" force, also away from the system exerted by the rope that is above the 10 cm system. We can also choose a piece of rope of any length as the system. There will be two forces at each end of the system, both pointing away from the system.

At this stage, I think that's all you need to know to figure out the direction of the force due to tension in order to use it in free body diagrams. Knowing the details of the molecular forces that give rise to this force is not necessary for drawing and interpreting FBDs just like it is not necessary to know exactly how a car engine works in order to drive one to a friend's house. Defining the system before you attempt to draw the force due to tension is as important as knowing what a car looks like before you get in it.

 

[rudransh verma]

The attachment points pull outward on the rope. The rope pulls inward on the attachment points. For an ideal (massless) rope, the magnitudes of all four of those forces are identical.

One can see that they are identical by applying Newton's third law at the two attachment points ends and Newton's second and third laws in a daisy-chain all the way along the rope.

Can you clarify my post#50

 

[jbriggs444]

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 $T=F_{MR}=-mg$

You said you do not want casual. Ask and you shall receive.

For Tension as a stress in a rope, it has neither direction "up" nor direction "down". It has only direction "vertical". In three dimensions, tension has a coordinate representation as a 3 by 3 matrix. If you choose a coordinate system in which the first coordinate is vertically aligned with the vertical rope, the stress is uniform and the cross-sectional area of the rope is $a$ then this tensor will be $$
\begin{bmatrix}
-T/a & 0 & 0\\
0 & 0 & 0\\
0 & 0 & 0
\end{bmatrix}$$More generally, if one chooses a poorly aligned coordinate system, the tension will manifest as two or three negative terms on the main diagonal that sum (via the Pythagorean theorem) to the magnitude of the tension (per unit area) in the rope.

If one wants to recover the force transmitted by a uniform tension through an imaginary dividing line in the rope, one multiplies the stress tensor by the directed area of a cross-section through the rope.

A "directed area" is a planar area represented as a vector. The area is represented by the magnitude of this vector. The direction is represented by the direction of the vector (normal to the plane). If one multiplies a 3x3 stress tensor by a 1x3 directed area, one recovers a 3x1 force vector. That vector is the force resulting from the tension acting through the area.

Even less casually, tension need not be uniform across the rope. One could instead take a surface integral, adding up local stress times incremental directed area across a surface that bisects the rope.

What this means for a rope attached to the monkey is that you multiply the stress tensor in the body of the rope by a directed area pointing outward at the monkey and recover a vector force that points inward toward the rope. This is the force of rope on monkey.

More casually, we have a useful invariant. No matter where we choose to slice an ideal massless rope with a directed surface between the potion exerting a force and the portion on which a force is being exerted, the resulting vector force will be aligned with the rope and will have a direction opposite to the direction of the slice. [It does not matter whether you slice neatly at right angles, diagonally or along a jagged curve. The result is always the same -- aligned with the rope, not with the cut].

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.

Casually speaking, yes.

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?

Yes, there is one force from rope on attached object. Though, of course, Newton's third law still applies. There is an equal and opposite force from attached object on rope.

 

[rudransh verma]

You said you do not want casual. Ask and you shall receive.

Ok! Very clever. Thanks man!