Tension in a string being pulled from both ends

[PeroK]

View attachment 109644

I understand that the tension is equal to W, but, I do not understand why it equals W as oppose to it equaling 2W. I need to have the explanation of why tension would equal W to get points for corrections on my test.

If it was $2W$ each mass would have an unbalanced upward force of $2W - 1W$ acting on it and each would acclerate upwards!

 

[haruspex]

I understand that the tension is equal to W, but, I do not understand why it equals W as oppose to it equaling 2W. I need to have the explanation of why tension would equal W to get points for corrections on my test.

See also section 2 of https://www.physicsforums.com/insights/frequently-made-errors-mechanics-forces/.

 

[Imtiaz Ahmad]

See also section 2 of https://www.physicsforums.com/insights/frequently-made-errors-mechanics-forces/.

If we do not apply any force on a string than what's the tension?

 

[haruspex]

If we do not apply any force on a string than what's the tension?

Why would there be any tension? Perhaps you need to describe the complete set-up you have in mind.

 

[Imtiaz Ahmad]

Why would there be any tension? Perhaps you need to describe the complete set-up you have in mind.

Yes, there is no tension
Tension produced in string due to our applied force so if we hang any things with an inextensible string the tension in the string is equal to weight of that thing but both have different direction

 

[haruspex]

both have different direction

What have different direction? Tension has an inward direction, as opposed to the outward direction of a compression. The forces exerted at the two ends of the (straight) string by the tension are in opposite directions. But it doesn't really mean anything to say that a tension is in the opposite direction to a force.

 

[Imtiaz Ahmad]

What have different direction? Tension has an inward direction, as opposed to the outward direction of a compression. The forces exerted at the two ends of the (straight) string by the tension are in opposite directions. But it doesn't really mean anything to say that a tension is in the opposite direction to a force.

Yes, that's my mean

 

[Quanta Ali Ahmed Jan]

http://www.Newtonsapple.org.uk/the-force-of-tension/

 

[Nutan]

But I still don't get this
If the string is being pulled by forces 60 and 70N from either ends then what's the tension
1.On first thought I think tension should be greater than 60N..because earlier in the discussion I saw when the string was being pulled with 60N from either sides tension was 60..now since I increased one of the forces then shouldn't there be greater stretched tendency sort of? I.e greater tension than 60N?
On second thought
I imagine a situation let there be a mass m , attached to it is an ideal string now apply force F on th side opposite to where the string is stuck directly on the block tension must be 0..you can do the math now when the string is being pulled in opposite direction of F with force f...then tension is f ... again you can do the math I've done it...connecting this to this question I asked let the original force "F" be 70 and "f" be 60 therefore I arrive at tension=60...as you can see I'm utterly confused with my own thought processes . Any help and explanation is greatly appreciated

 

[haruspex]

If the string is being pulled by forces 60 and 70N from either ends then what's the tension

What would be the net force on the string? What will happen in consequence?

 

[CWatters]

Consider what happens if you subject a string to 1,000,000N in one direction and 0.000,000,1N in the other. What do you think the tension in the string will be?

 

[NicholasJ]

Ok so I have a similar homework problem and I've read the thread but I'm still failing to understand. Sorry if it's dumb and repetitive, but if you had a string on a pulley with two weights on each side, each exerting a force of 80N would the tension on the string be the same throughout? Would the centre of the string have a tension of 0?

 

[haruspex]

would the tension on the string be the same throughout?

Yes.
Think of the string as made up of a lot of short segments. Each part pulls on each of its neighbours with the same force.

 

[CWatters]

+1

Break the problem into parts and think about the forces that adjacent parts apply on it.

 

[vela]

Ok so I have a similar homework problem and I've read the thread but I'm still failing to understand. Sorry if it's dumb and repetitive, but if you had a string on a pulley with two weights on each side, each exerting a force of 80N would the tension on the string be the same throughout? Would the centre of the string have a tension of 0?

I think students often get confused because they have only a nebulous idea of exactly what tension is. When you say the tension in the string is, say, 80 N, it means that the string exerts a force of 80 N on whatever it's attached to.

 

[NicholasJ]

I think students often get confused because they have only a nebulous idea of exactly what tension is. When you say the tension in the string is, say, 80 N, it means that the string exerts a force of 80 N on whatever it's attached to.

NOW I understand! Thank you!

 

[cryingoverphysics]

Homework Statement

If you have a string being pulled on each end by 60N, what is the total tension in the string?

Homework Equations

N/A[/B]

The Attempt at a Solution

The answer in my textbook says 60N, but it doesn't make a whole lot of sense to me (even though it's obviously right). If you're pulling both ends with 60 Newtons of force, why isn't it 120N of force, total, in the string? Why is it that the tension is only 60N? If one end were tied to a fixed object, and the string was pulled by 60N, then the total tension would also be 60, no? So then do you get the same result from pulling the other end rather than keeping it stationary?

If one end were to pulled by 70N, and the other by 60N, would the total tension be 70?

I find tension fairly confusing, and any help would very much appreciated.

The tension in a rope is dependent on the net force and the acceleartion of the system when the two opposing forces are different. What you have to do first is calculate the acceleration of the system.

So, imagine we have masses m1=3 pulling a force of F1=70N left and m2=4 pulling a force of F2=60N right. The net force of the system should be (F1-F2)/(m1+m2). That's an application of netforce=ma (F1+F2 but the two forces are opposing directions). Then, you can use the acceleartion for a net force equation on a system consisting of only mass 1.

So, there is a force left of F1, but the mass 1 is accelerating at a= (F1-F2)/(m1+m2). Therefore, there must be another tension force acting against the 70 newton force (F1) to cause it to accelerate. So, we have the equation F1-Ft=m1(F1-F2)/(m1+m2). Then, Ft=F1-m1(F1-F2)/(m1+m2).

Plugging everything in, we have Ft=70-3(70-60)/(3+4)=65.7N.

 

[PeroK]

The tension in a rope is dependent on the net force and the acceleartion of the system when the two opposing forces are different. What you have to do first is calculate the acceleration of the system. So, imagine we have masses m1=3 pulling a force of F1=70N left and m2=4 pulling a force of F2=60N right. The net force of the system should be (F1-F2)/(m1+m2). That's an application of netforce=ma (F1+F2 but the two forces are opposing directions). Then, you can use the acceleartion for a net force equation on a system consisting of only mass 1.
So, there is a force left of F1, but the mass 1 is accelerating at a= (F1-F2)/(m1+m2). Therefore, there must be another tension force acting against the 70 newton force (F1) to cause it to accelerate. So, we have the equation F1-Ft=m1(F1-F2)/(m1+m2). Then, Ft=F1-m1(F1-F2)/(m1+m2). Plugging everything in, we have Ft=70-3(70-60)/(3+4)=65.7N.

Note that this thread is five years old.

 

[berkeman]

Plugging everything in, we have Ft=70-3(70-60)/(3+4)=65.7N.

In addition to being a very old thread, we generally do not allow posting of solutions to schoolwork questions -- the student must do the bulk of the work on their homework questions.

But since this thread is so old that the OP already has solved the problem, your post is okay. Welcome to PF.