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Having trouble solving this problem due to its wording

  1. May 20, 2015 #1
    1. The problem statement, all variables and given/known data
    Two 300kg lonely obelisk (telephone poles) long to make a connection without the intermediary of copper cable of love (telephone wire). If the cable is stretched taut that connects them is 20m, then what is the force that will forever draw them together but may never be enough to enable them to become as one?

    2. Relevant equations
    (I believe its tension)
    T1=T2+F
    T1-T2=F


    3. The attempt at a solution
    SinΘ*300kg-SinΘ*300kg=F
    Since the masses are the same the net force is 0. But I am still not confident that I am doing right.
     
  2. jcsd
  3. May 20, 2015 #2

    CWatters

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    I don't think there is enough info to calculate an answer. Unless the answer is something simple like... It takes an infinite force to pull the wire taut/horizontal.
     
  4. May 20, 2015 #3

    DaveC426913

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    I am having trouble even picturing this scenario, let alone solving it.

    Are they vertical? Are they 20m apart? Is that why a cable between them is taut? What force is drawing them together? The tension on the cable?


    I think this statement should tell you something's wrong with your logic. The ratio of the masses should be irrelevant.
     
  5. May 20, 2015 #4

    gneill

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    This seems to be more of a puzzle problem than anything else. You aren't given any details about extant forces like the cable tension (and it would go slack anyways if the poles bent even a little bit towards each other). So tension can be crossed off the list.

    So consider the specific details that you are given.

    It is possible that you are only meant to identify the force involved and not necessarily calculate it...
     
  6. May 20, 2015 #5

    CWatters

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    Think I might have an idea....

    Both the mass and strength of the wire are proportional to it's diameter. So with luck the maximum allowable tension is independent of diameter!

    Edit: Humm I can't make that work.
     
    Last edited: May 20, 2015
  7. May 20, 2015 #6

    CWatters

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    Ignore my post above. Have figured it out. Quite a nice problem when you see the light.

    gneill - I've sent you a PM.
     
  8. May 20, 2015 #7
    Thank you guys! I think this is more of a conceptual problem, more than one that gives an actual answer. I'm just telling her the answer is tension. Its not actually for my it for my cousin. I have a degree in physics and have never come across a problem worded in this way. I'm blaming common core for this one.
     
  9. May 20, 2015 #8

    gneill

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    You may want to rethink that. What happens to the cable tension if the poles bend towards each other? Can the cable tension last "forever"?
     
  10. May 20, 2015 #9
    No, gneill the cable will eventually break. Or the poles will start to bend towards each other and the cable will start to sag, but that would require more information. I'm going to assume that it breaks because the cable cannot withstand the force of 2,940 N on each side of it.
     
  11. May 20, 2015 #10

    gneill

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    How do you come up with 2,940 N for the tension? Are the posts hanging from the cable somehow? What then is the cable attached to? Are the posts not driven into the ground like telephone poles? How do you decide what the failure tension is for the cable if no details are given?
     
  12. May 20, 2015 #11
    I got 2940 N by multiplying the 300, the mass of the obolus by g, acceleration due to gravity.

    I was staring to thing that this was a young modulus problem but we don't know the cross section of the string. And we don't know the original length of the string for it to be stretched 20m
     
  13. May 20, 2015 #12

    gneill

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    What direction does the force due to gravity act?
     
  14. May 20, 2015 #13
    Downward, but isn't tension equal to force on the body, which is its weight.
     
  15. May 20, 2015 #14

    gneill

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    Downward is correct. The tension acts horizontally though, so they are unrelated. Here the tension is provided by the poles' stiffness -- a sort of spring action. The poles are kept from tipping by being planted and held in the ground, but they will have some amount of flexibility along their length since no material is perfectly rigid.
     
  16. May 20, 2015 #15
    So are we to say that the cable will bring the poles slightly together. What does that have to do with the 20m?
     
  17. May 20, 2015 #16

    gneill

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    Yes. At least their top ends anyways; their bases are fixed.
    Nothing. The mass, too, is essentially irrelevant to the tension (other than helping to keep the poles seated in the ground).

    There is not enough information given to calculate the tension. All you can do is recognize that there is tension and identify its source.

    There is another force acting to pull the poles together but it will be entirely negligible. That force you do have enough information to at least estimate.
     
    Last edited: May 20, 2015
  18. May 20, 2015 #17
    And what force would that be? Gravity? Friction? Normal?
     
  19. May 20, 2015 #18

    gneill

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    What force do you have enough information about that you might calculate (or estimate) a value for it?
     
  20. May 20, 2015 #19
    Since we have the mass, I'm guessing gravity.

    But that 20m is still being neglected.
     
  21. May 20, 2015 #20

    gneill

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    Good guess :smile:
    Is it? What's the equation for gravitational force?
     
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