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What is tension in a string/rope/wire/etc?

  1. Sep 29, 2011 #1
    This has always confused me, and I'd appreciate any help in clearing this up. Let's say you have 2 masses hanging from a rope, one below the other, like the attached diagram. What is the tension in T2? How would you go about finding it?

    I am under the impression it's the vector sum of the force of gravity and the force of the 10.0kg mass on the 15.0kg mass, but it doesn't seem right to me. Can someone please explain?

    Thank you very much!

    Attached Files:

  2. jcsd
  3. Sep 29, 2011 #2
    IF you were to draw a FBD on the lower weight there would be the force of gravity, since the object is not accelerating up or down the tension in the rope is equal in magnitude and opposite in direction to the force of gravity, so about 147N.
  4. Sep 29, 2011 #3
    Thanks for the answer!
    But what if it were accelerating (for example, let's say 1.0m/s^2 [up])?

    Also, I'm still not sure what tension in the rope actually MEANS, or what it IS.
  5. Sep 29, 2011 #4
    147.1 newtons [up], then...? So would that make the tension 0? But that doesn't make any sense...
    I'm confused. D:
  6. Sep 29, 2011 #5
    basically tension is how much "pulling force" a string has, ie if you were to tie a rope to a wall and pull just till there is no slack, the rope will be horrizontal but there is not much tension, if you keep pulling the tension increases which in turn causes a higher pulling force at the end of the rope on the wall.
    as for if the weight was accelerating at 1m/s^2
    Then to solve for tension Fnet=ma, so fnet would equal 15n, so the tension in the rope would be about 162N
  7. Sep 29, 2011 #6

    Ken G

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    If the mass is accelerating, you still follow Trevormbarker's instructions, but now the free-body diagram must generate a net force of 15 N upward, because of F = ma. So just add on 15 N more onto T2. It is important to recognize that the source of the tension is irrelevant here-- as is the presence of the other mass. If you have two forces on any object, and you know one of them, and you know the acceleration, then F1+F2 = ma so F2=ma-F1 is all you need to know. In a more complicated problem, you may need to know that the rope exerting T2 will often be regarded as massless, so cannot support any net force, so if there's T2 pulling at one end of the rope, there must be an equal and opposite T2 pulling at the other end. That isn't needed here because you didn't ask for T1.
    A rope does one thing-- it pulls along its own direction. It doesn't push, it doesn't pull sideways (if it's a straight rope-- a curved rope pulls in the direction it has at its end), ropes just can't do that. The tension is that one thing ropes do-- pull along their direction. This pull is exerted on any object that is attached to the rope, but its magnitude is not determined by the rope, it is determined by the larger requirements of the setup. In this sense, it is like the normal force-- these are forces of constraint, you don't know what they are until you analyze the full problem, but you know what direction they point. (Physically, tension appears because the rope is slightly stretched, like a rubber band, but that isn't necessary to know to figure out what the tension must be.)
  8. Sep 29, 2011 #7
    Thank you all for the informative answers. I think I understand it just a little better now.

    Would it be fair to say that tension can be considered the sum of the magnitude of the forces? Since in that example, gravity is -147N (let downward be negative) and net force is +15N, the total magnitude of force/total tension is 162N?

    Does tension have a direction, now that we mention it...? (Sorry for all the questions.)
  9. Sep 29, 2011 #8
    Yes, It is essentially 1-dimmensional, a good example is 2 cars with a rope attached to their bumpers driving away, if the rope is non elastic, then tension will increase immediatly and the force exerted by the tension in the rope will pull each car towards the other. If that makes sense.

    Edit, Sorry my multi Quote didnt work....
  10. Sep 29, 2011 #9
    So basically, the direction of tension is both directions along the plane? So if I were to write out an answer to the question in that example, do I say it is plus/minus 162N? Or just 162N?
  11. Sep 30, 2011 #10
    Well sorry I was explaining the possible directions of tension, but you know if your example the object was not accelerating so the forces must be balance, the direction of the tension (both t1 and t2) would be up
  12. Sep 30, 2011 #11

    Ken G

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    Because tension in a massless rope is always equal and opposite on the two ends, and always points along the direction of a straight rope, generally only the strength of the tension is reported, not its direction. Even if there is a bend in the rope, like if it goes over a pulley, the tension at each end is always a force exerted on something attached to the rope that pulls in the direction the rope extends at the point of the attachment, so it's never ambiguous the direction of tension. You can see this by breaking the rope up into tiny segments, and then the tension is the action/reaction pair each segment shares with its neighboring segments, always pointing in the local direction of the rope. It works out exactly as if the tension on the rope was only applied at its two ends, you simply have to notice that the source of the tension is whatever is attached to the ends of the rope, and the force of the rope on something attached at its ends is also an action/reaction pair with that source of tension in the rope.
  13. Sep 30, 2011 #12
    I think I get it now. Thank you all very much!
  14. Sep 30, 2011 #13
    Does T1 just = 98 N due to T2 cancelling the 147 N force exerted by the second mass or would you add the tension caused by the weight of the second mass to T1 making it 245 N?
  15. Sep 30, 2011 #14

    Ken G

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    The latter. The easiest way is to regard the whole system hanging down from T1 as a single object with mass 25 kg. But if you want to dot all the i's, write the free-body diagram for both masses separately, and note that T2 shows up in each diagram. The bottom diagram tells you T2, which you then plug into the top diagram. It's not unusual to have to solve for all the forces before you know the final answer.
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