I'm trying to determine the impact force of an object when it comes to a sudden stop after a fall. The object weighs approximately 500lbs and has a safety cable designed to catch it in the event of a fall. I have looked at several sources on the web and have seen 2 schools of thought. One group seems to think you must have info on the time it takes to stop or the distance it travels or stretches the cable in order to solve the problem. Another group suggests an analytical solution, assuming K=EA/L, F= PE+KE, F=KD, and solving the quadratic. Can anyone offer insight into whether they feel the analytical method is legitimate. I understand it and can't see why it's wrong, but I also don't want to dismiss the group saying there must be some info about what happens after the impact. Thanks in advance.
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Falling Object Calculation
- Thread starter cjholl0
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Compositepro
Chemical
You are neglecting energy dissipation. If there were none, the object would bounce back to its original height, forever. Obviously that does not happen.
"K=EA/L, F= PE+KE, F=KD"
Not sure what all your symbols mean, but the second expression is incorrect, since PE and KE are energies, not forces. The last one is partially correct, if you know KE, then KE should equal 1/2K*x^2, solve that for x, and F is then equal to K*x. That's assuming the cord has a "constant" spring constant.
TTFN
FAQ731-376
Not sure what all your symbols mean, but the second expression is incorrect, since PE and KE are energies, not forces. The last one is partially correct, if you know KE, then KE should equal 1/2K*x^2, solve that for x, and F is then equal to K*x. That's assuming the cord has a "constant" spring constant.
TTFN
FAQ731-376
- Thread starter
- #4
cjholl0,
Look up strain energy in your mechanics of materials textbook. All of your kinetic energy must be converted to strain energy in your cable and mechanism, if your 500lb object is not to hit the ground and whatever is on the ground.
I assume you know how to work out the kinetic energy of a dropping object?
JHG
Look up strain energy in your mechanics of materials textbook. All of your kinetic energy must be converted to strain energy in your cable and mechanism, if your 500lb object is not to hit the ground and whatever is on the ground.
I assume you know how to work out the kinetic energy of a dropping object?
One group seems to think you must have info on the time it takes to stop or the distance it travels or stretches the cable in order to solve the problem. Another group suggests an analytical solution, assuming K=EA/L, F= PE+KE, F=KD, and solving the quadratic.
To paraphrase, one group thinks you must have info on the time it takes to stop or the distance it travels or stretches the cable.
Another group thinks you must calculate the distance it travels or stretches the cable.
To paraphrase, one group thinks you must have info on the time it takes to stop or the distance it travels or stretches the cable.
Another group thinks you must calculate the distance it travels or stretches the cable.
cjholl0,
This could be quite simple.
I lift a weight of 500lb up 10ft. I have exerted 500lb[×]10ft=5000ft.lb of work, and this now is potential energy.
I release the weight and let it drop 10ft. My work, and the weight's potential energy now are kinetic energy, which equals 5000ft.lb.
My retaining line reaches its limit, the weight decelerates to a halt in 2ft. This stores 5000ft.lb in my retaining line as strain energy.
Strain Energy = Force [×] displacement / 2
Force = 2 [×] 5000 ft.lb / 2ft = 5000lb.
Spring Rate = Force / Displacement = 5000lb/2ft = 2500lb/ft.
Can you do this?
JHG
This could be quite simple.
I lift a weight of 500lb up 10ft. I have exerted 500lb[×]10ft=5000ft.lb of work, and this now is potential energy.
I release the weight and let it drop 10ft. My work, and the weight's potential energy now are kinetic energy, which equals 5000ft.lb.
My retaining line reaches its limit, the weight decelerates to a halt in 2ft. This stores 5000ft.lb in my retaining line as strain energy.
Strain Energy = Force [×] displacement / 2
Force = 2 [×] 5000 ft.lb / 2ft = 5000lb.
Spring Rate = Force / Displacement = 5000lb/2ft = 2500lb/ft.
Can you do this?
GregTirevold
Mechanical
Hmmm, normally safety chain/cable doesn't let the object fall very far to prevent the sudden jerk snap as it goes taught.
I suppose if you want to estimate, I'd do this:
Find the Tensile Modulus of Resilience of the cable Ur=(0.5*Sy^2)/E
Sy = Yield Stress of material
E = Young's Modulus of material
Ur = Modulus of Resilience, capacity to store energy in the elastic range of the material and still return to its original shape(i.e. the trianglar area under the stress-strain curve from the origin to the yield point)
Find the "volume" of the cable Vol=A*L
A = cross-sectional area
L = length of cable
Vol = volume of material available to store the kinetic energy
Find the maximum velocity your object can reach before exceeding this limit by setting up the equation (0.5*m*v^2)=Ur*Vol and solving for v.
This would give you a safety cable that would stretch but not yield. I suppose you could find the area under the remainder of the stress-strain curve if you wanted to find out how much before it broke but I think I'd stick with keeping it under yield point.
I suppose if you want to estimate, I'd do this:
Find the Tensile Modulus of Resilience of the cable Ur=(0.5*Sy^2)/E
Sy = Yield Stress of material
E = Young's Modulus of material
Ur = Modulus of Resilience, capacity to store energy in the elastic range of the material and still return to its original shape(i.e. the trianglar area under the stress-strain curve from the origin to the yield point)
Find the "volume" of the cable Vol=A*L
A = cross-sectional area
L = length of cable
Vol = volume of material available to store the kinetic energy
Find the maximum velocity your object can reach before exceeding this limit by setting up the equation (0.5*m*v^2)=Ur*Vol and solving for v.
This would give you a safety cable that would stretch but not yield. I suppose you could find the area under the remainder of the stress-strain curve if you wanted to find out how much before it broke but I think I'd stick with keeping it under yield point.
Just think of a falling object mathematically as the phenenoma of motion in a uniform gravitational field. What is it doing?
So you are correct, potential energy is being converted to kinetic, the object has velocity at a neglegable time prior to impact. At impact the velocity goes to zero as energy is transferred to the mating object, so the collision may be elastic or inelastic. Then the velocity is reversed and the object bounces backwards.
The trick is at the precise moment of impact, the energy transfer part. Typically this is handled as an impulse calculation, but this requires some knowledge of the dent size or at least the contact time, which may be problematic in itself.
Most dynamic textbooks handle the problem as an elastic or inelastic collision. This would be my preferred approach, although the impulse thing works really good if you have a feel for arrestation of the falling body.
Kenneth J Hueston, PEng
Principal
Sturni-Hueston Engineering Inc
Edmonton, Alberta Canada
So you are correct, potential energy is being converted to kinetic, the object has velocity at a neglegable time prior to impact. At impact the velocity goes to zero as energy is transferred to the mating object, so the collision may be elastic or inelastic. Then the velocity is reversed and the object bounces backwards.
The trick is at the precise moment of impact, the energy transfer part. Typically this is handled as an impulse calculation, but this requires some knowledge of the dent size or at least the contact time, which may be problematic in itself.
Most dynamic textbooks handle the problem as an elastic or inelastic collision. This would be my preferred approach, although the impulse thing works really good if you have a feel for arrestation of the falling body.
Kenneth J Hueston, PEng
Principal
Sturni-Hueston Engineering Inc
Edmonton, Alberta Canada
btrueblood
Mechanical
The effective modulus for a cable is far less than the apparent (E) modulus. I.e. twisted wire is springier than straight bars.
Cockroach,
As you probably observed, my calculation above assumes an elastic safety cable. Hopefully, the OP will note what happens immediately after his weight comes to a halt.
Would one design for an inelastic safety cable? It does not sound like a good idea to me.
JHG
As you probably observed, my calculation above assumes an elastic safety cable. Hopefully, the OP will note what happens immediately after his weight comes to a halt.
Would one design for an inelastic safety cable? It does not sound like a good idea to me.
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