Tension in a rope hanging between 2 trees

[Manish_529]

Homework Statement

A uniform rope of weight
hangs between two trees. The ends of the rope are same height, and they each make an angle
with the trees. Find :

a): The tension at the either end of the rope.

b): The tension in the middle of the rope.

Relevant Equations

F=ma

I know that I can draw an FBD and apply Newton's 2nd law to find the relevant equations. But my question here is why is the mg vector or the weight of the entire rope same at every point on it I mean to say that if the mass of the entire rope is say M then how can a small point on the rope also have the same mass thus giving the same value for weight Mg(acting downwards) at every point on the rope?

 

[haruspex]

I know that I can draw an FBD and apply Newton's 2nd law to find the relevant equations. But my question here is why is the mg vector or the weight of the entire rope same at every point on it I mean to say that if the mass of the entire rope is say M then how can a small point on the rope also have the same mass thus giving the same value for weight Mg(acting downwards) at every point on the rope?

The short answer is that it doesn’t.
If the rope has density $\rho$ then a section of it length $ds$ has weight $\rho g\cdot ds$.
But to solve the question you do not need to consider small elements like that. You can treat the halves of the rope as the two rigid bodies.

 

[Orodruin]

Adding to the above: The infinitesimal FBD only needs to be considered if you need to know the shape of the rope. It will lead to a differential equation for the rope’s shape that is significantly harder to solve* than the algebra required to solve this problem.

* The solutions being catenary curves.

 

[Manish_529]

okay thank you. But I had an

The short answer is that it doesn’t.
If the rope has density $\rho$ then a section of it length $ds$ has weight $\rho g\cdot ds$.
But to solve the question you do not need to consider small elements like that. You can treat the halves of the rope as the two rigid bodies.

But what's the problem in choosing small elements?

 

[anuttarasammyak]

Balance of forces in horizontal and perpendicular directions.

 

[Orodruin]

okay thank you. But I had an

But what's the problem in choosing small elements?

There is no problem per se. It just leads you to a differential equation that you will need to solve. It is completely unnecessary to solve the problem at hand. It is not wrong, just not very useful.

 

[Manish_529]

okay thank you. But I had a different perspective (please correct me if I am wrong here) so, I was thinking that the entire mass is just concentrated at it's center of mass and the weight at the com is therefore mg can we are just parallelly shifting the mg vector from the com to the other points in the rope hence it remains the same in both the magnitude and direction.

 

[Orodruin]

okay thank you. But I had a different perspective (please correct me if I am wrong here) so, I was thinking that the entire mass is just concentrated at it's center of mass and the weight at the com is therefore mg can we are just parallelly shifting the mg vector from the com to the other points in the rope hence it remains the same in both the magnitude and direction.

Parallelly shifting a force acting on an object does not affect the force balance. It does affect the torque balance, but that is not necessary to consider in this problem.

 

[haruspex]

I was thinking that the entire mass is just concentrated at it's center of mass and the weight at the com is therefore mg can we are just parallelly shifting the mg vector from the com to the other points in the rope hence it remains the same in both the magnitude and direction.

That is rather different from what you wrote in post #1. There, you implied that mg was acting at every point of the rope (which would add up to infinite weight).
The entire mass is not concentrated at the mass centre.
For the purpose of force balance, you can treat the weight as concentrated anywhere.
For the purpose of torque balance, you can treat the weight as concentrated at the mass centre (indeed, that is the definition of mass centre).
In both cases, it just a matter of algebraic equivalence: the mass (and therefore weight) is not really concentrated at those places.

 

[Manish_529]

That is rather different from what you wrote in post #1. There, you implied that mg was acting at every point of the rope (which would add up to infinite weight).
The entire mass is not concentrated at the mass centre.
For the purpose of force balance, you can treat the weight as concentrated anywhere.
For the purpose of torque balance, you can treat the weight as concentrated at the mass centre (indeed, that is the definition of mass centre).
In both cases, it just a matter of algebraic equivalence: the mass (and therefore weight) is not really concentrated at those places.

So, for force balancing the weight can be taken on any point on the body, but for torque balancing it needs to be taken at it's Center of mass. Did I get it right?

 

[Manish_529]

Balance of forces in horizontal and perpendicular directions.
View attachment 350705

Can you please solve it this way and send the solution I tried it this way but failed

 

[jbriggs444]

Can you please solve it this way and send the solution I tried it this way but failed

That is not the way these forums work. You must show effort. We can assist. But we do not spoon feed answers.

You say you "tried it this way". Show us that attempt.

 

[haruspex]

So, for force balancing the weight can be taken on any point on the body, but for torque balancing it needs to be taken at it's Center of mass. Did I get it right?

Yes.

 

[Manish_529]

Yes.

But that's exactly what my question was that why does the above happen?

 

[haruspex]

So, for force balancing the weight can be taken on any point on the body, but for torque balancing it needs to be taken at it's Center of mass. Did I get it right?

Yes

But that's exactly what my question was that why does the above happen?

Because gravitational force is proportional to the mass, and forces in the same direction add by adding their scalar values.
If a body of mass M is composed of particles of mass $\{m_i\}$ then the total gravitational force on it is $\Sigma gm_i=g\Sigma m_i=gM$.

 

[Manish_529]

The total gravitational force on the entire body surely is Mg but what about a small part in it that's not Mg rather it's dm(g) where dm is the mass of the small element and that's what I am asking why is the gravitational force even on a small part of the rope taken as Mg

 

[haruspex]

why is the gravitational force even on a small part of the rope taken as Mg

I answered that in post #2: it isn't. Why do you think it is?

 

[Orodruin]

The total gravitational force on the entire body surely is Mg but what about a small part in it that's not Mg rather it's dm(g) where dm is the mass of the small element and that's what I am asking why is the gravitational force even on a small part of the rope taken as Mg

It isn't.

 

[Lnewqban]

... and that's what I am asking why is the gravitational force even on a small part of the rope taken as Mg

Are you referring to any diagram that shows that?
If so, could you post it here?

A chain is only as strong as its weakest link.
The internal tension along the center line of the rope, which is caused by gravity acting on all the little sections, is the same for any particular section.

 

[Orodruin]

The internal tension along the center line of the rope, which is caused by gravity acting on all the little sections, is the same for any particular section.

The tension is higher at the endpoints than it is at the middle of the rope.

 

[jbriggs444]

The tension is higher at the endpoints than it is at the middle of the rope.

One can see this almost immediately if one contemplates a horizontal force balance. The horizontal component of the tension force must be constant across the entire span. Else a segment somewhere would be under an unbalanced horizontal net force.

Since the angle of the dangle changes from one segment to another, the local tension must also change in inverse proportion to the cosine (or sine depending on how you measure) of the angle. That way the horizontal component can remain constant.

 

[Orodruin]

One can see this almost immediately if one contemplates a horizontal force balance. The horizontal component of the tension force must be constant across the entire span. Else a segment somewhere would be under an unbalanced horizontal net force.

Indeed, and it is the difference in the angle on opposite sides of any segment of rope that provides the vertical force required to keep it from falling.

 

[Manish_529]

I answered that in post #2: it isn't. Why do you think it is?

I was looking up at the solution to this problem over there this problem was solved by taking Mg as the weight at the end points or the middle (depending on which point we are finding tension at) and then breaking down the tension force at those points to balance them and then find the value of tension.

 

[haruspex]

I was looking up at the solution to this problem over there this problem was solved by taking Mg as the weight at the end points or the middle (depending on which point we are finding tension at) and then breaking down the tension force at those points to balance them and then find the value of tension.

Ok, but that is not taking the weight as acting at every point simultaneously. As I wrote, you can choose where it acts (for the purpose of force balance) but if taking it as acting at multiple points concurrently then you must apportion the weight between them.

 

[Orodruin]

I was looking up at the solution to this problem over there this problem was solved by taking Mg as the weight at the end points or the middle (depending on which point we are finding tension at) and then breaking down the tension force at those points to balance them and then find the value of tension.

As has been pointed out repeatedly in this thread, it is irrelevant where a force acts for the purposes of force balance. The only thing that matters is picking a relevant subsystem, drawing the free body diagram of said system, and make sure the forces balance out to zero.

We do not have access to the solutions you refer to, which makes it impossible for us to tell you exactly where your misinterpretation arises. But one thing is certain, you do not have mg acting along every point of the rope.