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System: a rope and a spring

  1. Jan 21, 2015 #1
    1. The problem statement, all variables and given/known data

    You suspend a bowling ball of mass m using a spring and a rope. The spring is horizontal and has spring constant k; the rope makes an angle θ with the vertical.

    Answer all parts in terms of m, g, k, and θ.

    a) What is the tension T in the rope?

    b) Your arch nemesis cuts the rope. What is the acceleration of the ball immediately after the rope is cut?

    2. Relevant equations

    Tension: (m*g)/cos(theta)

    3. The attempt at a solution

    I answered a without a problem but I have no idea what to do with b. I tried to use the vertical force when a system is in equilibrium but it doesn't work. Combinations : g*sin(theta), g*cos(theta) don't work either. I guess it is probably simple but I am at my wits' end.
    The stretched distance is not given I am wondering what k is for. I could recover x by Hooke's law when the string is in a vertical position: x=-mg/k
     
  2. jcsd
  3. Jan 21, 2015 #2

    phinds

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    Draw a force vector diagram and show your work, then we can see how to guide you along.
     
  4. Jan 21, 2015 #3

    haruspex

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    Before the rope is cut, the system is in equilibrium. Cutting the rope removes that force. How does that affect the sum of forces? Do any other forces change immediately?

    (Out of interest, do you know whether you got part a) correct? If not, would you mind posting your answer?)
     
  5. Jan 22, 2015 #4
    Thank you very much. I will draw it and post it here. I guess I have a problem with directions.
    I thought I could use the formula from a somehow.
     
  6. Jan 22, 2015 #5
    Bowling ball on a rope.png

    I understand this is a spring and can be stretched. It is also a kind of a pendulum isn't it?

    By the way I have started to learn things like that in order to understand Newton's Principia and the eighteenth-century discussions about mechanics and metaphysics (e.g. the vis viva controversy.)
     
  7. Jan 22, 2015 #6

    haruspex

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    You can. Just try to answer my questions in post #3.
     
  8. Jan 22, 2015 #7
    From your free body diagram in post #5, what is the horizontal force exerted by the spring on the mass before the rope is cut? Does this force change instantly as soon as the rope is cut, or is initially still the same?

    Chet
     
  9. Jan 22, 2015 #8
    I thought this horizonal force had been initially the same. So I used the formula for the tension force.
    I got:

    sin(theta)*T=(sin(theta)*(m*g))/cos(theta)=tan(theta)*m*g

    Of course to solve it for acceleration I divided it by the mass

    a(x)=tan(theta)*g

    Well, it is still wrong. :(
     
  10. Jan 22, 2015 #9
    That's the correct horizontal component of the acceleration. What is the vertical component of the acceleration? From these findings, what is the resultant of the horizontal and vertical components of acceleration?

    Chet
     
  11. Jan 23, 2015 #10
    I got it. :) I messed up signs. :) Many thanks. :) :) :)
    Bowling ball on a rope2.png
     
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