A linear differential equation problem

[A330NEO]

Homework Statement

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.

Homework Equations

The answer says it's $x \frac{d^{2}x}{dt^2} + \left ( \frac{dx}{dt} \right )^{2}+32x=160$

The Attempt at a Solution

Since Newton's second law is F=ma, I tried this:
a(Acceleration) is position x differentiated twice, so $a=\frac{d^{2}x}{dt^2}$
m=x, and force is 5-x. so, the equation becomes
$5-x=x \frac{d^{2}x}{dt^2}$
is anything wrong?

 

[Orodruin]

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.

 

[A330NEO]

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.

may I ask you how to apply it?

 

[Orodruin]

Try expressing the momentum of the rope in terms of x and its derivatives.

 

[A330NEO]

Try expressing the momentum of the rope in terms of x and its derivatives.

you mean this equation? this is from Wikipedia, and it's based on momentum equation p=mv, thus F=dp/dt.
$F=m(t)\frac{dv}{dt} - u\frac{dm}{dt}$
When I put this I that question, I found something like this.
$5-x= x\frac{d^{2}x}{dt^{2}} - \left ( \frac{dx}{dt} \right )^2$
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.

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