# Force/Power pulling a rope (I get 2 solutions)

## Homework Statement

A uniform rope of mass λ per unit length is coiled on a smooth horizontal table. One end is pulled straight up with constant speed v0

Find the force exerted on the end of the rope as a function of height y and find the power delivered to the rope.

Fnet = ma
Fgravity = mg
ΔK = ∫Fnetdy
P = Fv

## The Attempt at a Solution

Solution 1:
"with constant speed v0" $\Rightarrow$ Fnet = 0 $\Rightarrow$
Fpull = Fgravity = m(y) * g = λyg
P = Fpullv = λgyv0
Solution 2:
ΔK = ∫Fnetdy
$\frac{1}{2}$m(y) * v02 = ∫Fnetdy
$\frac{1}{2}$m(y) * v02 = ∫(Fpull - Fgravity)dy
$\frac{1}{2}$λyv02 = ∫Fpulldy - ∫Fgravitydy
$\frac{1}{2}$λyv02 = ∫Fpulldy - m(y) * g
$\frac{1}{2}$λyv02 = ∫Fpulldy - $\frac{1}{2}$λy2g (where the half comes from the fact that gravity acts at the center of mass of the rope (y/2)).
$\frac{1}{2}$λyv02 + $\frac{1}{2}$λy2g = ∫Fpulldy
Differentiate both sides with respect to y:
$\frac{1}{2}$λv02 + λyg = Fpull
P = Fpullv = λv03 + λygv0

So which ones right? Please tell me why its right and also why the other one is wrong.

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Andrew Mason
Homework Helper
It may be easier to analyse it from an energy perspective rather than force. Would the power supplied not just be the rate of change of total energy: d/dt(KE + PE)?

AM

Ok, so if I do it from the power perspective the power is:
P = d($\frac{1}{2}$m(y) * v2 + m(y) * λg$\frac{y}{2}$) / dt
P = d($\frac{1}{2}$λyv2 + $\frac{1}{2}$λgy2) / dt
P = $\frac{1}{2}$λ(dy/dt)v2 + λgy(dy/dt)
dy/dt = v, by defintion, so:
P = $\frac{1}{2}$λv3 + λgyv

So does that mean my second answer is right? But then how is the rope being pulled at constant speed if the net force is nonzero (does the rope still on the table affect the already pulled rope)?

Andrew Mason
Homework Helper
Ok, so if I do it from the power perspective the power is:
P = d($\frac{1}{2}$m(y) * v2 + m(y) * λg$\frac{y}{2}$) / dt
Would it not be:

$$P = \frac{d}{dt}(\frac{1}{2}mv^2 + mgy) = \frac{d}{dt}\left(\frac{1}{2}(y\lambda/2) v^2 + (y\lambda/2) gy\right) = \left(\frac{1}{4}\lambda v^3 + \lambda gyv\right)$$

So does that mean my second answer is right? But then how is the rope being pulled at constant speed if the net force is nonzero (does the rope still on the table affect the already pulled rope)?
I get a factor of 1/4 not 1/2 for the first term. If the rope is being pulled at constant speed the force must be increasing with height.

AM

PhanthomJay
Homework Helper
Gold Member
I like solution 1 better. There is no KE change with time.

Andrew Mason
Homework Helper
I like solution 1 better. There is no KE change with time.
Sure there is. The rope on the table is not moving. The rope above the table is moving.

AM

Yeah solution 1 is silly because m is not constant, and F=ma assumes constant mass, so the more general form dp/dt = Fnet is necessary. The problem is when I do this method I get λv2 + λyg = Fpull (I showed the dp/dt = Fnet solution and the work-energy solution to my GSI and he still can't figure out the discrepancy). Just as a sanity check, without the gravitational field, the force still must be nonzero so that increases my confidence in the term with velocity.

The velocity term should have a $\frac{1}{2}$ not a $\frac{1}{4}$ because m(y) = λy (the mass of the entire lifted rope --not just the mass of the rope under the center of mass). However, the half in the gravity term comes from the fact that gravity acts on the center of mass of the lifted rope, not on the end. m(y) * g * h = m(y) * g * y/2 = (1/2)λy2.

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Andrew Mason
Homework Helper
Yeah solution 1 is silly because m is not constant, and F=ma assumes constant mass, so the more general form dp/dt = Fnet is necessary. The problem is when I do this method I get λv2 + λyg = Fpull (I showed the dp/dt = Fnet solution and the work-energy solution to my GSI and he still can't figure out the discrepancy).
In your second solution you state: ΔK = ∫Fnetdy

Would not the work done on the rope be equal to the increase in total energy of the rope? ie the increase in kinetic and potential energy.

The velocity term should have a $\frac{1}{2}$ not a $\frac{1}{4}$ because m(y) = λy (the mass of the entire lifted rope --not just the mass of the rope under the center of mass). However, the half in the gravity term comes from the fact that gravity acts on the center of mass of the lifted rope, not on the end. m(y) * g * h = m(y) * g * y/2 = (1/2)λy2.
Right you are. Replace the 1/4 with 1/2 in my post #4.

AM

PhanthomJay
Homework Helper
Gold Member
Sure there is. The rope on the table is not moving. The rope above the table is moving.

AM
Oh, yes, changing KE due to changing mass, sorry.

You right in the sense that the work done (∫Fpulldy) by the pulling force is equal to the net increase in total mechanical energy. But the total work done (∫Fnetdy) on the rope is equal to just the net increase in the kinetic energy. This can be seen if you replace ∫Fnetdy by: ∫Fpulldy - ∫Fgravitydy = ΔK. The second integral in that equation just becomes the negative of the change in potential energy so when you move it to the other side of the equals sign it becomes the change in the total mechanical energy.

I simply prefer doing things from a work-energy theorem perspective rather than a conservation of energy + non-conservative work perspective for some irrational reason.

EDIT: Just realized in the rope+earth system the ∫Fnetdy = Δ(Total Energy) is true, that is it would just reduce to the pulling force acting on the earth+rope system, while my equations just look at the rope system. I think this might be the error I have when I use dp/dt = Fnet.

BTW, Thanks for all the help guys!

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