Is this still an equivalent force?

[Bengineer]

Say I want to determine if a rope/ attachment equipment is strong enough for a person falling at some height. I am unable to have a mass fall at that height so I decide to up the mass and reduce the height.

Should I use kinetic energy or conservation of momentum?
Because the impulsive force should be F= change in momentum/change in time
But I also should be able to model the system with energy conservation however the force would be different because of different mass right? As F=MA

I am not actually going to test anything, I am just curious.

 

[jbriggs444]

Say I want to determine if a rope/ attachment equipment is strong enough for a person falling at some height. I am unable to have a mass fall at that height so I decide to up the mass and reduce the height.

Should I use kinetic energy or conservation of momentum?
Because the impulsive force should be F= change in momentum/change in time
But I also should be able to model the system with energy conservation however the force would be different because of different mass right? As F=MA

I am not actually going to test anything, I am just curious.

I would think that an energy metric would best reflect on the relevant capacity of the rope. You want to know how much energy it can absorb when it is jerked taut.

 

[jrmichler]

The rope is a spring. Whatever the rope is attached to is also a spring. And the falling mass on the rope is a spring. All of these springs are in series, so the total of them has a spring constant.

The falling mass has a kinetic energy at the point where the rope goes tight. Knowing the spring constant, there is an equation for energy vs amount of spring compression (or extension). If you equate the falling mass kinetic energy to the energy stored in the spring, you can calculate the maximum compression of the spring. The peak force is calculated from the maximum compression and the spring constant.

Or you can study mountain climbing equipment, looking at rope specifications. That will give a real world estimate of the necessary strength, an estimate that includes appropriate safety factors.

 

[Bengineer]

Since I a test at a shorter height would have less rope, therefore, less k would an accelerometer be useful in determining max force by the number of recorded G's times the mass of the object. Also, I could make an equivalent k It is just difficult to find k values of climbing rope per meter.

 

[PhanthomJay]

if The rope is short, it has greater stiffness and a higher k value than a long rope. The k value of the rope is roughly AE/L , where A is it’s cross section area, L is it’s unstretched length, and E is the equivalent Young Modulus of Elasticity.

 

[tech99]

When the mass is dropped, we have an oscillatory condition, where the mass will bounce up and down. The energy losses in the system are small and cause the oscillation to slowly damp out. So it seems reasonable to me to use energy conservation to estimate the initial force. If you can find the spring constant, then you can solve everything using PE=KE. In a practical situation with a steel wire rope and a person falling we may find a force of maybe ten times the weight.

 

[Lnewqban]

Say I want to determine if a rope/ attachment equipment is strong enough for a person falling at some height. I am unable to have a mass fall at that height so I decide to up the mass and reduce the height.
...
I am not actually going to test anything, I am just curious.

Do I understand the problem you present properly?
Mass A free-falling from height H1 produces a maximum tension force Tmax on restraining rope of length L1.
Not having height H1 available, you mimic the situation by using
Mass B free-falling from height H2 (H2<H1) produces a maximum tension force Tmax on restraining rope of length L2 (L2<L1).

 

[Bengineer]

Do I understand the problem you present properly?
Mass A free-falling from height H1 produces a maximum tension force Tmax on restraining rope of length L1.
Not having height H1 available, you mimic the situation by using
Mass B free-falling from height H2 (H2<H1) produces a maximum tension force Tmax on restraining rope of length L2 (L2<L1).

That is correct.

 

[zoki85]

If only all the ropes were like bungee jumping ropes that would make problem trivial. In reality many materials show nonlinear response long before they actually break . IOW they are far from ideal spring as they approach critical tension limit.

 

[Bengineer]

If only all the ropes were like bungee jumping ropes that would make problem trivial. In reality many materials show nonlinear response long before they actually break . IOW they are far from ideal spring as they approach critical tension limit.

Ok to eliminate the complexity of the different K constants and how they are not an ideal spring what about using the same rope. The shorter length will provide less stretch and be a more volatile result, therefore, giving a sort of an inherent safety factor.

 

[JBA]

For a drop test, the basic formula is PE = mgh (where PE changes linearly with h) and assuming you lift the weight to the same height for every test then the h will automatically be shorter and the resulting KE will be less for the shorter rope by the ratio of the two rope lengths.

 

[zoki85]

Ok to eliminate the complexity of the different K constants and how they are not an ideal spring what about using the same rope. The shorter length will provide less stretch and be a more volatile result, therefore, giving a sort of an inherent safety factor.

As long as you're dealing with an elastic case and the same rope, K can be eliminated as pointed out in post #5 (K=E⋅A/L). The modulus of elasticity E is an inherent property of a material. From formulae of energy of falling mass and absorbed potential energy of the rope you calculate stretch (elongation) Δx