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(Easy) Gravitational Force at an angle?

  1. Sep 4, 2014 #1
    Hi Guys,

    I'm working on the theory for a piece of gym equipment, but im afraid my physics-capabilities are somewhat limited.

    In order to illustrate my question, I have made a drawing, please look at the attachment.

    Excuse my english as well, it is not my first language.

    So on the image you see a long cable pulled around 3 wheels. For the sake of theory let's assume there is no friction caused by the wheels.

    How much weight do you need to put on the cable (Side C) in order to even out the weight distribution?

    At first I thought 10 KGs, but I guess the angles make a lower requirement on side C than side A?

    Is there someone who can help me confirm this? Preferrably with some kind of math behind it.

    Thanks in advance!

    - Peter
     

    Attached Files:

  2. jcsd
  3. Sep 4, 2014 #2

    NascentOxygen

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    Staff: Mentor

    Hi jinglepeter. http://img96.imageshack.us/img96/5725/red5e5etimes5e5e45e5e25.gif [Broken]

    To keep that cable from moving, you must oppose the weight acting in line with the cable, viz., 10. sin 30°, and this equals 5kg

    So you could hang a 5kg weight from the vertical part.

    Goog luck with your creation!
     
    Last edited by a moderator: May 6, 2017
  4. Sep 4, 2014 #3

    Nugatory

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    Staff: Mentor

    If you actually build this device, you will find that the cable sags under the load so that the angle the cable makes with the top pulley is not the same as the angle between the pulleys, and changes with the position of the 10kg weight. Unless the cable is unrealistically strong and maintained under an unrealistic tension, I don't think you'll be able to find any single weight value that works.
     
  5. Sep 4, 2014 #4
    Hi,

    Thank you so much for your answer - Would that also mean that if the angle was 15%, then I would only need 2,5 kg?
     
  6. Sep 4, 2014 #5
    The cable would sag under a bit yes. There should only be strong tension on line A and C.

    While before it was calculated with 30 degrees - arent we supposed to calculate with 60 degrees, as that's the actual angle from the top piece?
     
  7. Sep 4, 2014 #6

    NascentOxygen

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    Staff: Mentor

    10 sin 15° = 2.59
     
  8. Sep 4, 2014 #7
    Thanks again,

    Can you look over the calculation for the first one again? Aren't we supposed to use 60 degrees - as that's the angle for the top pulley?
     
  9. Sep 4, 2014 #8

    Nugatory

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    Staff: Mentor

    You can use the sine of the side angle or the cosine of the top angle - you'll end up with the same value either way.
     
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