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How large is the force that tightens the wire

  1. May 12, 2014 #1
    1. The problem statement, all variables and given/known data

    How large is the force that tightens the wire A (see picture), when the weight's mass is 670 kg? Guidance: Assume that the wires between the wheels are horizontal.

    2. Relevant equations
    G=mg?


    3. The attempt at a solution
    What seems weird to me is that none of the wheels look movable, wouldn't the tension force simply equal G in that case? I guess the weight is supported by three wires but the correct answer is 20 kN which seems very inexact if the answer is indeed G/3. Would appreciate a more thorough explanation :)
     

    Attached Files:

  2. jcsd
  3. May 12, 2014 #2

    Filip Larsen

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    Gold Member

    You may want to use the concept of virtual work (force times a small virtual displacement) and conservation of energy (work performed by virtual displacement of the weight equals work transmitted via A). You should arrive at the conclusion that the force is not mg/3 as you imply in your post but something else.
     
  4. May 12, 2014 #3

    NascentOxygen

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    Staff: Mentor

    Not my forte, but you can see that the steel weight sets the tension in the rope. This end of arm A is acted on by 3 equal forces, each the tension in the rope.

    You can sometimes see this sort of arrangement being used to maintain tension in high voltage power lines.
     
  5. May 12, 2014 #4
    This scenario is about an electric train so you're exactly right :) The a) part of the problem asks why this sort of arrangement would be appropriate.

    So is mg/3 the answer?
     
  6. May 12, 2014 #5

    Filip Larsen

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    No, that is not the correct answer.

    If you do not like to figure out how displacement relates to displacements at A (like I suggested before) you may also in this simple example look at the tension force in the string directly and make a free body diagram around the pulley at A so that the "cut" goes through the 3 strings in the middle. Since each time you "cut" a string in this system you must replace it with a force equal to the string tension you can directly count how many times more you need in order for a force at A to balance the forces from the string (this is a slightly different worded version of that NascentOxygen suggested).
     
  7. May 12, 2014 #6

    NascentOxygen

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    What value do you calculate to be the tension in the rope?
     
  8. May 12, 2014 #7
    I realize I missunserstood what the actual wire was, I understand now. The tension in the string is 670*9,81 and it affects wire A three times essentially. So 20kN. Thanks and sorry for the ignorance. :)
     
  9. May 12, 2014 #8

    NascentOxygen

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    No problem. https://imageshack.com/a/img848/9251/cool0038.gif [Broken]
     
    Last edited by a moderator: May 6, 2017
  10. May 12, 2014 #9
    But how does this work? How does one really draw the tension because if it's drawn like this, the tension in the lower part of the wire wouldn't be pulling on wire A? Tension is confusing me, I guess it tightens in both directions.
     

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  11. May 12, 2014 #10

    NascentOxygen

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    Under static equilibrium, the arrows all point to the right. Nothing moves, nothing turns.
     
  12. May 12, 2014 #11
    How can the last (lower) arrow point to the right? Wouldn't that force have the same direction as the force of the weight?
     
  13. May 12, 2014 #12

    NascentOxygen

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    As far as the pulley A is concerned, the load it supports pulls away on both ends of the rope looped over it. If the forces weren't balanced in opposition like that, the rope would unravel from the pulley.

    Besides, you can't push rope.

    Perhaps you should put together a couple of pulleys and string and observe how they work.
     
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