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A linear differential equation problem

  1. Oct 18, 2015 #1
    1. The problem statement, all variables and given/known data

    A uniform 10-foot-long heavy rope is coiled loosely on the ground. One end of the rope is pulled vertically upward by means of a constant force of 5lb. The rope weighs 1lb/ft. Use newton's second law to determine a differential equation for the height x(t) of the end above ground level at time t. Assume that a positive direction is upward.
    2. Relevant equations
    The answer says it's [itex] x \frac{d^{2}x}{dt^2} + \left ( \frac{dx}{dt} \right )^{2}+32x=160 [/itex]

    3. The attempt at a solution
    Since newton's second law is F=ma, I tried this:
    a(Acceleration) is position x differentiated twice, so [itex]a=\frac{d^{2}x}{dt^2}[/itex]
    m=x, and force is 5-x. so, the equation becomes
    [itex]5-x=x \frac{d^{2}x}{dt^2}[/itex]
    is anything wrong?
     
    Last edited: Oct 18, 2015
  2. jcsd
  3. Oct 18, 2015 #2

    Orodruin

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    Yes, you cannot use F = ma, which is only true when the mass in question is constant. You need to use the second law on the more general form F = dp/dt, where p is the momentum.
     
  4. Oct 18, 2015 #3
    may I ask you how to apply it?
     
  5. Oct 18, 2015 #4

    Orodruin

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    Try expressing the momentum of the rope in terms of x and its derivatives.
     
  6. Oct 18, 2015 #5
    you mean this equation? this is from Wikipedia, and it's based on momentum equation p=mv, thus F=dp/dt.
    [itex] F=m(t)\frac{dv}{dt} - u\frac{dm}{dt} [/itex]
    When I put this I that question, I found something like this.
    [itex] 5-x= x\frac{d^{2}x}{dt^{2}} - \left ( \frac{dx}{dt} \right )^2 [/itex]
    Looks similar to the 'answer' in my book, but still not getting it. I suspect that the left side of the equation have been multiplied by 32(Gravitational acceleration in English unit, I think) but hard to find where to put it.

    Type your equation in this box
     
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