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The concept of tension.

by mrproper
Tags: concept, tension
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mrproper
#1
Mar30-04, 01:11 AM
P: 5
Hi. I recently started learning physics on my own, and I'm having a bit of a problem with the concept of tension. That is, I'm not exactly sure what it is.

In the textbook I bought, tension is defined a bit hastility in a section entitled "Applications of Newton's Laws." I grasped the Three Laws very well, but the defintion of tension has me questioning a lot.

The formal defintion of tension is as follows:

When a rope attached to an object is pulling on the object, the rope exerts a force T on the object, and the magnitude T of that force is called the tension in the rope.
I'm not really sure what to make of this. I understand that if I pull the rope with a force of 100 N, the box will also have a pull of 100 N. However, because of Newton's Third Law, there is will be a 100 N force pulling on the right end of the rope because of me, and there will be a 100 N force pulling to the left of the rope because of the box. Wouldn't this mean the tension is 200 N? I generally think of tension as stretching, and viewing the tension as 100 N seems a bit like one-sided stretching, which doesn't make sense.

If someone has an explantion of tension that might shed some light on my current concept, I'd be really grateful if you shared it. As it is now, I'm stuck on an Atwood Machine problem because I'm unsure about the whole tension business.

NOTE: To give you an idea of my level, the questions I'm working on all neglect mass and say that the magnitude of the force exerted at any point along a rope is the same at all points on the rope.
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Michael D. Sewell
#2
Mar30-04, 02:33 AM
P: n/a
Quote Quote by mrproper
...a 100 N force pulling on the right end of the rope because of me, and there will be a 100 N force pulling to the left of the rope because of the box. Wouldn't this mean the tension is 200 N? I generally think of tension as stretching, and viewing the tension as 100 N seems a bit like one-sided stretching, which doesn't make sense.
If this problem involved compression would you get it? Think about how tension is different from compression.

If the box wasn't there, and you still pulled on the rope, would the 100N force
still be there? No.

If you had two ropes tied together between you and the box, would you now have 400 N of tension? No.

You need to review equilibrium. If the whole system is in equilibrium, then each part of the system must be in equilibrium.

You should also get some help from someone who has been through this before.

I hope this helps. Stick with it no matter how difficult it seems. You'll get it.
-Mike
Chi Meson
#3
Mar30-04, 07:36 AM
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Mike said it well. I can rephrase what he said slightly differently:

You pull on a rope with 100 N of force and your friend pulls on the other end so that there is no acceleration. How much force does your friend pull with? 100 N of course.

If you are pulling on your own rope end withh 100 N, how much must the rope be pulling on you? Again 100 N. How much does the rope pull on your friend? Once again, 100 N.

Now consider a spot in the middle of the rope. Cut the rope there and feel how much you get pulled in both directions. 100 N each way right? YOu would get the same result if you cut the rope at any spot, so you can think : "every point along the rope is being pulled equally in both directions (by the two adjacent points of the rope)meaning net force on each part of the rope is zero."

Ultimately: "Tension is in the rope, and the force of tension pulls equally at both ends of the rope."

jdavel
#4
Mar30-04, 10:09 AM
P: 618
The concept of tension.

mrproper, said: "I understand that if I pull the rope with a force of 100 N, the box will also have a pull of 100 N. However, because of Newton's Third Law, there is will be a 100 N force pulling on the right end of the rope because of me, and there will be a 100 N force pulling to the left of the rope because of the box. Wouldn't this mean the tension is 200 N?"

It seems that way doesn't it! But think about a slightly different situation. A rope is hanging from the ceiling, just dangling there. Assuming that you can neglect its mass, what 's the tension in the rope? Now attach a 100N weight to the end of the rope. What does it seem like the tension in the rope should be now? But what are the forces acting on the rope? Isn't this really the same as your problem, just turned sideways?
KSCphysics
#5
Mar30-04, 11:15 AM
P: 35
"every action has an equil and opposite reaction" meaning, you pull with 100N, its gonna pull back at 100N on you. and if you push with 100N, then the obj is gonna push back at 100N. this is only true in equilibrium.
Michael D. Sewell
#6
Mar30-04, 11:30 AM
P: n/a
Each one of you has approached this from a slightly different perspective. This should really help mrproper understand equilibrium and tension. This is physics forums at it's best. Good job. -Mike
mrproper
#7
Apr1-04, 03:49 AM
P: 5
Thanks everyone. You guys all gave good examples, and I understand it now.
Molydood
#8
Apr1-04, 05:46 AM
P: 117
I seems to have been explained very well now, but out of interest, this is how I see things:

as an example, imagine a 200N (20kg) pull on one end of a pulley and a 100N (10kg) pull on the other end of the pulley. Because this is out of balance, the 200N weight will pull the 100N weight, creating movement (the system is not in equilibrium).

Now try to imagine the maximum 'tension' in the cable; its only as strong as the weekest link, the 100N pull. If the weakest link cannot 'pull back' enough then the rope will not reach 200N tension and the system will start to move. The 100N surplus from the 200N weight then accelerates the 100N weight until the forces balance (air resistance etc.)

well, that's just the way i see it anyway, hope it helps,

regards,
Martin


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