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Mass of A Hanging Spotlight?

  1. Jun 13, 2009 #1
    1. The problem statement, all variables and given/known data

    A spotlight hanging from two cables, given angles and other information, (see picture below)

    2. Relevant equations

    F=mg

    3. The attempt at a solution

    I re-wrote my initial equations on a clean sheet of paper.

    Basically I just tried to isolate T2 from the X forces equation, and then substitute that into the Y forces equation. The questions asks for the tension in the other cable and the mass of the light.

    The only given values are T1 being 140N and the angles of the cables.

    The actual answers are supposed to be 18 kg and 91.4N. I'm lost on how to properly solve this.

    attachment.php?attachmentid=19299&stc=1&d=1244869472.jpg
    1. The problem statement, all variables and given/known data



    2. Relevant equations



    3. The attempt at a solution
     

    Attached Files:

  2. jcsd
  3. Jun 13, 2009 #2

    rock.freak667

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    Homework Helper

    Since I can't see the picture as yet, I will try to explain this best as I can.


    You are given tensions at specific angles. The weight acts downwards. There is no net movement in the x-direction.


    resolve the two tensions into x and y components. All the forces in the x and y directions are in equilibrium. If you are given one tension and the two angles, then when you resolve in the x-direction, you should be able to get the value of the second tension.

    Then just form the equilibrium equation in the y-direction and substitute the the value of the second tension to find the unknown.
     
  4. Jun 13, 2009 #3

    ideasrule

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    Despite not being able to see the picture, he got it exactly. To the original poster: this is exactly how you do the question. :approve:
     
  5. Jun 13, 2009 #4
    I added the image as an attached file:

    https://www.physicsforums.com/attachment.php?attachmentid=19299&d=1244869468

    Thanks, I'll try that solution and come back.


    EDIT:

    It failed again. If my physics are right though then I don't need to worry about it.
    The image is rather large so I just linked to it.

    http://i35.photobucket.com/albums/d194/Lancelot59/Hosted%20Images/scanned_0006.jpg [Broken]
     
    Last edited by a moderator: May 4, 2017
  6. Jun 13, 2009 #5

    LowlyPion

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    Looks OK.
     
  7. Jun 13, 2009 #6
    Alright, then the answers on the key must be wrong.

    Thanks for your help.
     
  8. Jun 13, 2009 #7

    LowlyPion

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    I did just double check the answers given and they are correct, if the angles of 30 and 50 are with the vertical and not the horizontal as shown in your drawing.
     
  9. Jun 13, 2009 #8
    The initial problem gave the angles of 40 and 60 on the left and right respectively, and those angles were made by the cables with the roof.

    I just solved for the angles at the horizontal for the free-body diagram...which I did incorrectly. That's what the problem was.

    Late night physics apparently doesn't go too well. Thanks for helping me with that.
     
  10. Jun 13, 2009 #9
    There's a really elegant way to solve this, and it's the same for any angles.
    The spotlight is at equilibrium. That means that the net force acting on it is 0.
    The net force, is a vectorial sum of the force vectors acting on the spotlight.

    If a vector sum is 0, that means that if you add up the vectors, head to tail, you'll get an enclosed shape (The total translation is 0).
    That enclosed shape is a triangle whose sides are the vectors T1, T2 and mg. Find the angles of the triangle, and apply the law of sines to them:
    Where the sides of the triangle are a, b and c and the angles opposite those sides are A, B and C, then the law of sines states that:
    a/sinA = b/sinB = c/sinC

    From there, you can find the tension in the second rope and the weight of the spotlight.
     
  11. Jun 13, 2009 #10
    We were taught how to solve these problems geometrically, but I'm personally not big on it. I like my vector components :D.

    The method works, but it's more of a personal preference type thing. This was just a result of me making a calculation error.
     
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