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Two wires holding a weight

  1. Nov 7, 2004 #1
    If two wires are holding a 500N weight up, will the tension force in each rope be greater if they are both held taught to the side than if they are both simply holding it straight up? For holding it straight up, the tension force is simply 250N in each, derived from 2Ftsin90 - 500 = 0. To find the tension force when they are both held to the side, I wrote 2Ftcos0 - 500 = 0. However, this gets the same result for tension force as the first example. It seems to me that the tension force would be greater when the two wires are held to the side. Could someone explain whether I'm right, and if so, what I'm doing wrong? Interesting sentence there...

    [edit]
    I think that the second equation is wrong, and the equation should always be the first one. On solving for an angle of 120 between the wires, the equation 2Ftsin30 - 500 = 0 got me 500N for the tension force on each cord. On upping the angle between them to 170, the equation 2Ftsin5 - 500 = 0 gets 2868N! The force seem to approach infinity as the angle between the cords approaches 180. Does this mean that it's impossible to have two cords pulling perfectly horizontally on an object and keeping it at equilibrium?
     
    Last edited: Nov 7, 2004
  2. jcsd
  3. Nov 7, 2004 #2

    arildno

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    "Does this mean that it's impossible to have two cords pulling perfectly horizontally on an object and keeping it at equilibrium?"

    That is correct, when the tensile force is regarded as PARALLELL to the orientation of the cord. You need some vertical component of the tensile force in order to achieve equilibrium (i.e, balancing gravity in this case).

    Since a horizontal cord doesn't have a non-zero component of tensile force, the ONLY vertical force acting on the object is gravity, i.e, the object is not in equilibrium.
     
  4. Nov 7, 2004 #3
    Thanks. I didn't expect such a tricky question from such a simple-looking problem. Our teacher's a trickster!
     
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