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Homework Help: Tension problem! help!

  1. Nov 12, 2003 #1
    Can anyone help me solve this problem that i am a little confused about? i think the answer is tension=mgsin"theta", am i close at all?!


    According to the picture, from point A to point B is a string that is attached to wall. The string is twice the distance from Point A to Point O, and Point A to Point O's distance is the same as the distance between Point B Point O. If i put k upto the string, which has a box M that is attached to it. when it stops rolling on the string. what is the Tension on the string? (no friction applied)

    Attached Files:

  2. jcsd
  3. Nov 13, 2003 #2
    just click the webpage below to check the answer,but I only provide answer for tension,since I am sure you will be able to find where the box stops sliding ,after you can find tension.hope it will benefit you

    Last edited by a moderator: Nov 13, 2003
  4. Nov 13, 2003 #3
    thank you!

    thank you so much for your help!!!
  5. Nov 14, 2003 #4
    hi, what i dont understand is how do you know that theta equals to beta? the 2 angels i mean.

    btw, i like your icon, vic is =)
  6. Nov 14, 2003 #5


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    α and β are the same because:
    The tension from K to A must be equal in magnitude to the tension from K to B because they are the same piece of rope and the horizontal components of the tension must cancel.
  7. Nov 14, 2003 #6
    yes, that is what i thought. as long as it is the same rope, then tension anywhere should be the same on the rope. but is there anyway that i can proof this theory.
  8. Nov 15, 2003 #7
    it is proved by observation.
    the system more or less is like pulleys

    (note:eek:bservation only focus on string and the ring H)

    in the system,the moving part is actually the ring,but relative to ring ,the string is moving.

    relative to the ring the string will keep moving until the tension in the string balance.because if the tension is not the same,then there will be net force produced ,which cause the string moves.

    (now we look from particle view)
    (note: open the graph attachment now)
    don't forget that string also made up of particles.the graph is a picture of a string which is broken down into 4 particle for simplicity.

    ***now look at the third particle
    the forth particle attract the third particle with F6 N and second particle attract thrird particle with F5 N
    since the string in equilibrium,it implies that the third particle does not move,and it happens if F5=F6

    ***now look at the second particle
    the third particle attract the second particle with F4 N and first particle attract second particle with F3 N.Also since the string in equilibrium,it implies that the second particle does not move,and it happens if F3=F4

    ***we know that attraction second particle to the third particle should be the same with the attraction of third particle to the second particle.That is to say F4=F5

    so F3=F4=F5=F6 and so on along the string.

    hence ,because the attractions between particles along the string are the same so we can conclude that the tensions along the string are the same(provided friction force neglected).

    I hope they are clear

    anyway how do you know vic chou?which country are you from? =)

    Attached Files:

    Last edited by a moderator: Nov 15, 2003
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