The physics of a Zipline

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Main Question or Discussion Point

Hey! im new on the forum and I joined because I really enjoy physics but I have a horrible teacher. I was wondering if anyone could help me on a question
how do I find the velocity at the end of the zipline?
how do I find the momentum of the object going down the zipline after it is deployed and before it lands?
and does the conservation of momentum work for a zipline.
thanks!
delta

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phinds
Gold Member
2019 Award
Draw a diagram, labeling all the distances and any forces you can. That's where you always start with this kind of problem.

Draw a diagram, labeling all the distances and any forces you can. That's where you always start with this kind of problem.
I drew a FBD that showed the forces along with the overall structure of the system but where do I go from there? Is there an equation i should use? also thank you so much for responding it means alot to me!
delta

phinds
Gold Member
2019 Award
I drew a FBD that showed the forces along with the overall structure of the system but where do I go from there? Is there an equation i should use? also thank you so much for responding it means alot to me!
delta
Pics. We must have pics. No pics, it didn't happen. (better still, instead of a pic of a free-hand drawing, use a graphics tool to create a nice neat diagram)

Drakkith
Staff Emeritus
Is this a specific homework question? If so it would help to post the exact question. Right now there's not enough information to start. What kind of zipline is it? How is the object hanging from the zipline? Is it just a simple loop of rope/cable, or is there a wheel? Do you need to account for any forces other than gravity and friction (such as air resistance)? Is this a frictionless zipline?

tech99
Gold Member
I drew a FBD that showed the forces along with the overall structure of the system but where do I go from there? Is there an equation i should use? also thank you so much for responding it means alot to me!
delta
You will notice that if the zip line does not sag, you get infinite force in the wire. So it must have some sag.
In other respects it is like an inclined plane; it is just calculating the wire tension that is more complicated.

sophiecentaur
Gold Member
Forget Forces. Use Energy in your calculation of a good ballpark figure. If you know the biggest drop in height - from start position to lowest point on the trajectory. Then you can equate the Gravitational Potential Energy (mgΔh) with the Kinetic Energy (mv2/2). m cancels out and you can then work out v.
It assumes 100% efficiency, of course but using Forces involves all sorts of assumptions, like direction of the wire as you go along and the time throughout the journey is much more difficult.
Momentum at any time is just mv (you found v already). To take account of the dip in the wire, the Energy calculation again can give you the resulting lower speed when you reach the end.
Conservation of Momentum hardly applies here because the passenger is under an external force (gravity) so you would need to include the Earth in a calculation.

Forget Forces. Use Energy in your calculation of a good ballpark figure. If you know the biggest drop in height - from start position to lowest point on the trajectory. Then you can equate the Gravitational Potential Energy (mgΔh) with the Kinetic Energy (mv2/2). m cancels out and you can then work out v.
It assumes 100% efficiency, of course but using Forces involves all sorts of assumptions, like direction of the wire as you go along and the time throughout the journey is much more difficult.
Momentum at any time is just mv (you found v already). To take account of the dip in the wire, the Energy calculation again can give you the resulting lower speed when you reach the end.
Conservation of Momentum hardly applies here because the passenger is under an external force (gravity) so you would need to include the Earth in a calculation.
Hey! thanks for your help, I actually was able to figure it out on my own but that you so much!!

You will notice that if the zip line does not sag, you get infinite force in the wire. So it must have some sag.
In other respects it is like an inclined plane; it is just calculating the wire tension that is more complicated.
Thanks so much!!! I acc ended up figuring it out
You will notice that if the zip line does not sag, you get infinite force in the wire. So it must have some sag.
In other respects it is like an inclined plane; it is just calculating the wire tension that is more complicated.
thanks so much for your help, I acc figured it out but thank you so much!!

Pics. We must have pics. No pics, it didn't happen. (better still, instead of a pic of a free-hand drawing, use a graphics tool to create a nice neat diagram)