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Homework Help: Chain being pulled up

  1. Nov 20, 2014 #1
    Hey good people, im new here and i found that you help people, i hope you can help me with this
    ive been triyng to solve this for a while but with no luck

    1. The problem statement, all variables and given/known data

    A chain of mass m0 per unit length is loosely coiled on the floor. If one of the end is subjected to a constant force P, when y = 0 , determine followings when P = 10N and m0 = 0.2kg / m

    (a) Determine the maximum value of chain length ymax
    (b) Determine the velocity of the chain as a function of y while 0 ≤ y ≤ ymax .
    (c) Determine the acceleration of the chain as a function of y while 0 ≤ y ≤ ymax .
    (d) Plot velocity and acceleration of hook chain as a function as a function y in all the cases using MATLAB while 0 ≤ y ≤ ymax .

    Im only interested in Part (a) .. i can do the rest by myself.

    2. Relevant equations

    P=mv, p=momentum , m=mass , v= velocity
    u=m/y, u=linear density , m=mass , y= length

    3. The attempt at a solution

    i started with the momentum equation i derive it using the chain rule ... i worked it out i end up with deferential equation

    now im stuck at this point, im not sure if it correct or not ... is it possible to solve it without using deferential equation ?
    Last edited: Nov 20, 2014
  2. jcsd
  3. Nov 20, 2014 #2
    What is the velocity of the chain at ##y_\text{max}##? Why?

    What is the total energy of the chain? Can you relate that with ##P##?
  4. Nov 20, 2014 #3

    Vf=Vi+at since Vi=0 >>> Vf=at ? or v=ds/dt

    KEi + PEi + Wext = KEf + PEf
    initially it was at rest v=0 and hight was 0 that gives us
    Wext = KEf + PEf => W = 0.5mv^2+mgh

    i dont have enough knowns to use this ...
  5. Nov 20, 2014 #4
    What does ##y## mean in #1? And what is ##y_\text{max}##? Why does it have the "max" label?

    And what is this ##h## that you used in #3?
  6. Nov 20, 2014 #5
    in this case y is the length, the question stated that initially it was at elevation of 0 so length will be 0, y max is the maximum length the chain will reach up the ground ... here is where my confusion start, im not sure if the chain will stop going up at any point, so from this i concluded the answer will be an equation with respect to time

    h is the length
  7. Nov 20, 2014 #6
    So, is this "length" really the "height"?

    What is the velocity of the chain when it reaches the max height?

    So is it the same as ##y##?
  8. Nov 20, 2014 #7
    the height of a point at the tip of the chain off the ground is the length of the chain,
    the force is constant velocity will keep increasing

  9. Nov 20, 2014 #8
    If the velocity is increasing, how can there be a maximum height?
  10. Nov 20, 2014 #9
    i asked myself the same thing, but that's what the question asks .. it beats me really
  11. Nov 20, 2014 #10
    So perhaps the velocity is not always increasing? What makes you think it must be ever increasing in the first place?
  12. Nov 20, 2014 #11
    Since F=ma , if the force is constant and it is as stated in the problem, then there is constant acceleration and therefore changing velocity according to a=dv/dt
  13. Nov 20, 2014 #12
    If ##m## is constant. Is it constant?
  14. Nov 20, 2014 #13
    hmmm ... no, m increase as length increase
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