Force exerted on a rope by a falling mass

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Nicho1410
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Hi

I'm wondering, if I had a mass of 10kg and I dropped it 10 metres, how would I work out how much force is exerted on a rope? The rope is made of rubber bands
 
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You wouldn’t be trying to continue your bungee-jumping thread now, would you?

If the other end of the elastic line is attached to a fixed point and we assume that the line is an ideal spring with spring constant ##k## we can use the differential equation for a harmonic oscillator to calculate the forces and velocities.

However, this calculation will only be as good as the assumptions that go into it. You don’t have an ideal spring here. Is ##k## constant across the entire range of displacements? How much can the line stretch before it breaks?
 
By the way, does this have anything to do with
I have been asked to build a prototype structure for bungee jumping.
Although, I am struggling with the calculations for the maximum force that the whole structure will have to sustain.
For the bungee jumping rope, I am using elastic bands with constant k.
a thread that was closed by a moderator -- with clear argumentation ?

[edit] ah, mentor was here too ...
 
Nicho1410 said:
Hi

I'm wondering, if I had a mass of 10kg and I dropped it 10 metres, how would I work out how much force is exerted on a rope? The rope is made of rubber bands
The force on the rubber band is given by ##F=-kx ## where k is the stiffness of the rubber band measured in N/m and x is the distance that the rubber band is stretched.

The easiest way to calculate the distance the rubber band is stretched is to use conservation of energy. The change in the mass’ gravitational potential energy ##mgh## is equal and opposite to the change in the elastic potential energy ##\frac{1}{2} k x^2##.
 
Be advised that:
1) Rubber bands have nonlinear force vs displacement. The F = -kx equation does not apply. The peak force will be higher, possibly much higher, than that calculated assuming a linear spring. Anybody designing a real world system needs to know what that implies, how to measure it, and how to deal with it. If not, they are not qualified.
2) Polyurethane bands have a strain rate effect, as do many elastomeric materials. Rubber bands may also have a strain rate effect. Anybody designing a real world system needs to know what that is and how to deal with it. If not, they are not qualified.
3) Elastomeric materials are subject to creep. Same comment as #2.
 
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BvU said:
Hello Nicho, ##\quad## :welcome: ##\quad ## !

And is the rubber band rope attached to anything else than the falling mass ?
The rope is held by a fixed structure from where the mass is dropped. I wanted to find out the force applied to the rope and consequently, to the structure.
 
BvU said:
By the way, does this have anything to do with
a thread that was closed by a moderator -- with clear argumentation ?

[edit] ah, mentor was here too ...
Yes it does, but I was just asking for help on the calculations and not the structure itself.
 
BvU said:
I'd like to know about the context. Is this a homework exercise or a real world enterprise ? PF has rules for both.
It is a university group project.
 
Dale said:
The force on the rubber band is given by ##F=-kx ## where k is the stiffness of the rubber band measured in N/m and x is the distance that the rubber band is stretched.

The easiest way to calculate the distance the rubber band is stretched is to use conservation of energy. The change in the mass’ gravitational potential energy ##mgh## is equal and opposite to the change in the elastic potential energy ##\frac{1}{2} k x^2##.
Thank you, I understand now.
 
Nicho1410 said:
It is a university group project.
Do you have access to the equipment used in an introductory physics class? If so, use a force sensor to measure the force. Make a graph of the force versus time. You could also connect a motion sensor and make a graph of the force versus the position.

See if the values you measure match the values you calculate.
 
Nicho1410 said:
if I had a mass of 10kg
BvU said:
Sounds like fun. But what if someone breaks his/her neck ?
Don't stand underneath looking up and you will be pretty safe in that regard.
 
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