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Homework Help: Mass Hanging in Center of Chain Problem

  1. Jun 14, 2012 #1
    1. The problem statement, all variables and given/known data
    A mass of 200 kg is hanging directly in the center of a chain; the chain makes a 20o angle from its horizontal. The chain will break if more than 2000 N of force are applied at any point on the chain. Will the chain break?

    2. Relevant equations

    3. The attempt at a solution
    2000N = 2Tf(sin20)
    Tf1 = 1000/sin20
    Tf1 = 1095 N
    Tf2 = 1095 N
    Tftotal = 1095N + 1095N
    Tftotal = 2190 N
    The chain will break.
  2. jcsd
  3. Jun 14, 2012 #2


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    Newton's second law is fine here. What it means in this case is that since nothing is accelerating, all the forces involved must sum up to 0 (the vector sum).
    The above equation is incorrect, as applied to this problem.
    I think you mean units meters per second squared.
    Okay, that's better! :approve: I assume the "2000 N" comes from mg = (200 [kg])(10 [m/s2]) = 2000 N. So far so good.
    Everything in your attempted solution is fine so far. :approve:
    Wait, something's not right here. Where did the 1095 N come from?

    And why are you breaking it into 2 parts? There's no need to do that because you've already accounted for the two sections of chain in your "2000N = 2Tf(sin20)" formula.
    So that's not quite right.

    If it helps things make more sense, take another look at your free body diagram (FBD).
  4. Jun 14, 2012 #3
    Hey thanks for the response!
    The 1095N comes from
    1000N/sin20 = Tf
    Tf = 1095N
    Is Tf the total tension force?
  5. Jun 14, 2012 #4


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    I think your calculator is set to radians. The problem statement said the angle was 20o, not 20 radians.
    You're the one that defined what Tf stands for. You tell me! :smile:

    Really though, make sure to draw all the forces out on your FBD. That way, if you forget what one of the variable names are, your FBD will remind you what it is. :smile:

    (But yes, I was assuming that when you set up your 2000 [N] = 2Tf(sin20) equation, that Tf is the magnitude of the tension on the chain. But you should check your FBD to be sure.)
  6. Jun 15, 2012 #5
    Attached is my FBD.

    You were spot on, I believe.

    I used a calculator where I could specify degrees, and I came out with 2923.8 N. In other words, the chain will break.

    Weirdest thing is that it was my cell phone's calculator. For what possible reason is that in radians?

    Attached Files:

  7. Jun 15, 2012 #6


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    That looks right to me. :approve:
    Working in radians make a lot of sense in many situations. But sometimes degrees is more convenient. Whichever cell phone you have (or whichever app you are using on your cellphone if that is the case), there should be a settings parameter that will allow you to specify radians or degrees, before using the trigonometric functions. Look in the cell phone calculator's settings menu or something like that and you should be able to change it back and forth as necessary.

    Or you can always convert back and forth manually using,
    [tex] \frac{\mathrm{AngDeg}}{180} = \frac{\mathrm{AngRad}}{\pi} [/tex]
  8. Jun 15, 2012 #7
    Thank you very much! I believe I understand tension force problems now!
  9. Jun 15, 2012 #8
    I do have one question, though. How is the magnitude of the total tension force greater than the magnitude of the weight? Where does this force come from? I read that tension force is the opposite of compression force. Is this related? Thanks again!
  10. Jun 15, 2012 #9
    If you pull directly an object up, the force must be equal to weight since total force is applied to lift the object.
    If you are not pulling directly, only part(component) doing the job, which opposite to the weight.
    The total force is the sum of its parts
  11. Jun 15, 2012 #10
    Only the vertical component of the chain has the equal half the magnitude of the weight, this vertical component is Tsin20, if you set that equal to (200*9.8)/2 and solve for tension, the tension will be greater.
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