How can I calculate the force of a falling object?

  • #1
LugubriuousLamia
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Today I was thinking how would I calculate the amount of force I would experience if I were to jump out of a second floor building in the event of a fire and if I had to jump how much force would be applied to me? I was able to find the height of the building by dropping an object from rest out of the window and found that the object hit the ground 1.63 seconds after dropping it.

This allowed me to use the equation
d=vi*t+.5at^2 and I was able to calculate the height of the building to be 13.03 m which seems reasonable.

I then went to calculate force and I only have the formula F=m*a which I know my mass is around 67 kg and my acceleration would be gravity. But this force is 657.27 N which is the same as the force of me standing on the ground.

I then decided to try and calculate the work done on me by the ground and my kinetic energy which then produced values of 8564 Joules and 8576 joules respectively. However this does not tell me the force at which I would hit the ground. It only tells me the energy I would apply to the ground supposing I remain inert when hitting the ground and conversely the energy that would be applied to me when I hit the ground.

I just want to know how much force I would hit the ground with and I can't seem to figure out how much force that would be. I know I would hit the ground with a final velocity of ~16 m/s and I know the acceleration is due to gravity and I know the time I would fall is 1.63 seconds but I can't find a way to calculate the amount of Newtons that would be applied to me. And I know my acceleration is not going to change when I fall. But I also know from my lived experience that when I hit the ground falling from 13 meters I am going to be in much more pain than when I hit the ground falling from say 1 meter.

In short how can i calculate the amount of force I would hit the ground with because it seems I can find the velocity I hit the ground with, the amount of energy I would apply to the ground and thus the amount of energy the ground would apply back on me but I can't find the force that I would smack the ground with.

Thank you in advance!
 

Answers and Replies

  • #2
berkeman
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The better approach is to calculate the "impulse" when you hit different kinds of things on the ground. The impulse is much higher when you land on concrete compared to a pile of foam or a stunt airbag...

https://en.wikipedia.org/wiki/Impulse_(physics)

http://www.stunt.eu.com/equipments/air-bag.png

air-bag.png
 

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  • #3
FactChecker
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Suppose you accumulate a velocity, V, as you fall. Also suppose you land on your feet and your center of gravity is 3 feet above the ground. If you smash flat, you would decelerate from V to 0 within 3 feet. You can work with that to calculate an acceleration (a deceleration is a negative acceleration) if the acceleration is constant over that 3 feet. That can be turned into a force.

You have not given a mass to calculate the force. It is probably better to express the acceleration in Gs and leave it at that anyway.
 
  • #4
LugubriuousLamia
10
3
Ok, to interpret this formula should I interpret the impulse as how long the force is applied to my body or how long my body would be in free fall motion and then stopping when I would hit the ground. I imagine the impulse would be the duration of how long the force would be applied to my body and if this is the case how could I possible interpret that? It would seem that if I were to hit the ground the time would be from when I hit the ground to when I stop moving toward the ground and the reaction force would be done being applied to me, which still can't be reconciled in my brain because the reaction force will be applied no matter what. I am at a mental pretzel.
 
  • #5
LugubriuousLamia
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In response to fact checker, that seems like a good idea. Only how long would this deceleration take do you think? I am assuming .1 seconds and that would then result in a 10,720 N force being applied but I have no idea how I could interpret this. I guess my reverse ask is how do I know how long that force would be applied assuming I act as an inert object. I am not looking to complicate this by assuming I bend my knees and roll out of this motion because that would increase the duration of the force being applied.
 
  • #6
Dale
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You are interested in the force from the ground. So the time is from the moment when you first touch the ground until you (including your internal organs) come to a complete stop.
 
  • #7
LugubriuousLamia
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Thank you Dale! That is how I imagined it would happen. So what do you think would be a good estimate for this hypothetical situation?
 
  • #8
Dale
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Thank you Dale! That is how I imagined it would happen. So what do you think would be a good estimate for this hypothetical situation?
That can vary a lot depending on if you are landing head-first, feet-first, flat, and on concrete, air cushion, net, etc.
 
  • #9
Stephen Tashi
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You seem to be asking for a value of "the" force needed to bring you to a halt as you hit the ground. The force-on-your-body vs time values change as you come to a halt and they depend on the type of ground you hit A simple model is that both you and the ground are like springs. Hard ground is like a stiff spring. Loose sand is like a weaker spring. The "impulse" model is that there is some constant average force acting on you for some given time interval. So if you are interested is an average value, use ## F \triangle T = M \triangle V ##, (Force)(time interval) = (mass) (change in velocity)
 
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  • #10
CWatters
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+1

It's possible to get extremely high rates of deceleration (thousands of g) if you drop steel onto steel because both are hard so the stopping distance is very short.

If you landed upright with rigid legs and a straight back how much would they allow your body to compress before stopping? Perhaps a few inches? If you land on your head that might stop in a very short distance/time while the rest keeps moving. You will have to make some assumptions.
 
  • #11
sophiecentaur
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The better approach is to calculate the "impulse" when you hit different kinds of things on the ground. The impulse is much higher when you land on concrete compared to a pile of foam or a stunt airbag...

https://en.wikipedia.org/wiki/Impulse_(physics)

http://www.stunt.eu.com/equipments/air-bag.png

View attachment 232851
The Impulse (force times time) is the change in Momentum and the momentum on impact is the same in both cases. The point about the impulse is that, if the time is longer, the force is less. A soft landing provides a longer contact time so the force can be (acceptably) lower.

You can treat an impact in two ways - based on eliminating Momentum or eliminating the Kinetic Energy. Using Impulse and Momentum involves knowing the time to slow the body down but you would need to know the duration of the impact. This could be difficult to measure. OTOH, if you work in terms of Kinetic Energy, you can deal with the Work Done to stop the body. Work is Force times Distance and it is possible to estimate the distance traveled - say when an air bag is used:

Take a body of 90kg mass (average to large man) and drop him 30m. Over that distance, there is a force of 90g (g doesn't stand for grammes) (=900N) on him (g=10N/kg - near enough). You don't need to calculate the actual Kinetic Energy the body has, on impact - just equate the work done by gravity with the work done on the air bag. If the air bag stops his fall in 1m then the ratio of the forces will be the inverse of the ratio of the distances.
900 X 30 = Fair bag X 1
so Fair bag =27000N
or, alternatively, the acceleration (deceleration) force from the air bag would be 30g. Perhaps OK for a single impact but that off the head value of 1m was perhaps a bit short and 30m is a high fall. Increase the braking distance to 3m and the force goes down to 9000N and the gForce goes down to 10g. This all assumes constant force on the body. The peak forces involved could be higher but the method is good for ball-park figures.
 
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  • #12
Victor_W
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Today I was thinking how would I calculate the amount of force I would experience if I were to jump out of a second floor building in the event of a fire and if I had to jump how much force would be applied to me? I was able to find the height of the building by dropping an object from rest out of the window and found that the object hit the ground 1.63 seconds after dropping it.

This allowed me to use the equation
d=vi*t+.5at^2 and I was able to calculate the height of the building to be 13.03 m which seems reasonable.

I then went to calculate force and I only have the formula F=m*a which I know my mass is around 67 kg and my acceleration would be gravity. But this force is 657.27 N which is the same as the force of me standing on the ground.

I then decided to try and calculate the work done on me by the ground and my kinetic energy which then produced values of 8564 Joules and 8576 joules respectively. However this does not tell me the force at which I would hit the ground. It only tells me the energy I would apply to the ground supposing I remain inert when hitting the ground and conversely the energy that would be applied to me when I hit the ground.

I just want to know how much force I would hit the ground with and I can't seem to figure out how much force that would be. I know I would hit the ground with a final velocity of ~16 m/s and I know the acceleration is due to gravity and I know the time I would fall is 1.63 seconds but I can't find a way to calculate the amount of Newtons that would be applied to me. And I know my acceleration is not going to change when I fall. But I also know from my lived experience that when I hit the ground falling from 13 meters I am going to be in much more pain than when I hit the ground falling from say 1 meter.

In short how can i calculate the amount of force I would hit the ground with because it seems I can find the velocity I hit the ground with, the amount of energy I would apply to the ground and thus the amount of energy the ground would apply back on me but I can't find the force that I would smack the ground with.

Thank you in advance!
I think you're actually looking for the amount of energy that your mass would have after free falling 13 meters. I weigh 225 lbs and after falling 10 meters I would have 10,000.7 joules (or 10.007 kN) of energy that I now need to bring to a stop. How I dissipate that energy depends on whether I land flat, standing, on concrete, soft garden soil, a trampoline, etc.
 
  • #13
sophiecentaur
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I think you're actually looking for the amount of energy that your mass would have after free falling 13 meters.
Definitely! But alternatively, the Momentum when you arrive at the ground can also be used. The two approaches are:
1. Kinetic Energy (mv2/2) can be dissipated over a distance d, where the force F is KE/d
2. The Momentum (mv) can be shed over a time t where the Force is mv/t.

It comes down to the same think for the body hitting the ground but, in many ways, it's easier to control the 'depth of cushioning', rather than a 'time of braking'. Using a retro rocket could be easier to plan by considering Momentum.

PS Historically, there was a big debate as to whether Momentum or KE was more important descriptions of Motion.
 
  • #14
erobz
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I think its ok to assume impacting some type of spring ( what type of spring can be altered to model the landing conditions)

Lets assume for simplicity:
  1. That the force of your weight will be negligible in comparison to the spring force
  2. The spring is a linear spring
  3. Ignore drag
You get the following equation of motion, where ##x## will be measured from the top of the unstretched spring (where you come in contact with whatever it is you are hitting).

## x \downarrow^+##:

$$ m \ddot x = -kx $$

Which for initial conditions:

## x(0) = 0 ##

## \dot x (0) = \sqrt{2gh} ##

Has the solution:

$$ x(t) = \sqrt{ \frac{2mgh}{k} } \sin \left( \sqrt{ \frac{k}{m} } t \right)$$

The force you would feel against you from the impact would be:

$$F(t) = - \sqrt{ 2mghk } \sin \left( \sqrt{ \frac{k}{m} } t \right)$$

Where:

##m## is your mass
##g## is the acceleration due to gravity
## h## is the height you jumped from
##k## is the spring constant
 
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  • #15
erobz
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You could look at this like the person is the “spring” too, and what you are hitting is rigid. Like compacted ground. I don’t think it’s going to be terribly realistic either way. At some point the jumpers legs are just going to shatter to pieces, probably somewhere in between a fist and second story jump?
 
  • #16
malawi_glenn
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Damped spring is better model
 
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  • #17
sophiecentaur
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Lets assume for simplicity:
  1. That the force of your weight will be negligible in comparison to the spring force
  2. The spring is a linear spring
  3. Ignore drag
I don't like 2. or 3. There are not many linear springs in practice; you need a very long spring and operate over a limited range to approximate to linear

Damped spring is better model
Yep. You wouldn't want to be thrown back up where you started. The problem with a simple spring could be that the acceleration would not be uniform. A friction or plastic deforming medium would probably give a better 'ride'; a pile of cardboard boxes is commonly used in film stunts. Difficult to characterise though.
 
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  • #18
malawi_glenn
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A pile of cardboard boxes is commonly used in film stunts. Difficult to characterise though.
Now I got an idea for a lab for next semester :) thanks
 
  • #19
sophiecentaur
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Now I got an idea for a lab for next semester :) thanks
As I commented, characterising the system could be hard. You would need to measure and record the force on the bottom of the pile over the time of the impact. It could be done with a Force transducer and that would reveal the progress of the retarding force all the way. You'd have to find what equipment is available for you.
Good luck - you should enjoy it.
PS there are other methods for buffer design. They use water pushed out of a hole in a cylinder or even a rubber ball. Was it Volvo that had water filled 'bumpers' to reduce collision damage?
 
  • #20
erobz
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I don't like 2. or 3. There are not many linear springs in practice; you need a very long spring and operate over a limited range to approximate to linear


Yep. You wouldn't want to be thrown back up where you started. The problem with a simple spring could be that the acceleration would not be uniform. A friction or plastic deforming medium would probably give a better 'ride'; a pile of cardboard boxes is commonly used in film stunts. Difficult to characterise though.

Yeah, I feel I said as much in a later post ( although I don’t see why 3 is an issue). My thoughts were to give them something foundational to calculate.

Since @berkeman mentioned the inflatable landing bag we could just replace it with a non-linear gas spring, and since @malawi_glenn mentioned about the damping:

$$ m \ddot x + \beta \dot x + \frac{\kappa}{l_o - x} = 0 $$

If anyone wants to give them this to calculate?
 
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  • #21
sophiecentaur
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although I don’t see why 3 is an issue
But the motion is eliminated by the presence of drag. Without it, you go back up! However, I imagine that some sort of arrestor hook could grab the compressed spring and allow the person to step off safely.
 
  • #22
hutchphd
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I find the most useful absolutte lower bound on the estimate of required acceleration after a fall is given by assuming uniform deceleration a over a characteristic distance l. Then by energy conservation $$mgh=mal$$ where l might be the length of your leg extension. This seems as good an estimate as any and it is not difficult to calculate. So a ten foot fall will produce at least 5g assuming l=two feet. If you land on your butt then l=3 inches and consequentially you experience at least 40g.
 
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  • #23
erobz
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But the motion is eliminated by the presence of drag. Without it, you go back up! However, I imagine that some sort of arrestor hook could grab the compressed spring and allow the person to step off safely.
I think the motion is arrested by the damping. When I said ignore drag, I meant during the fall.

However, when I was age 15 I was involved in a dirt bike crash. I was T-boned by a trike ( carrying two occupants ) moving around 30-40 mph. Eyewitness's at the scene said I was launched about 15 ft into the air, and when I came down, I bounced. All of us lived.

There is a video montage of people bouncing after impact on reddit. Obviously, none of them show a human bouncing like a super ball. Most show some level of restitution, so the heavily damped bounce seems reasonable.
 
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  • #24
sophiecentaur
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I find the most useful absolutte lower bound on the estimate of required acceleration after a fall is given by assuming uniform deceleration a over a characteristic distance l. Then by energy conservation $$mgh=mal$$ where l might be the length of your leg extension. This seems as good an estimate as any and it is not difficult to calculate. So a ten foot fall will produce at least 5g assuming l=two feet. If you land on your butt then l=3 inches and consequentially you experience at least 40g.
A nice bit of Physics but I can’t help feeling a bit creepy. It’s like going to a party as a crash test dummy. 😱
 
  • #25
256bits
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Obviously, none of them show a human bouncing like a super ball. Most show some level of restitution,
You must mean some sort of cartwheeling of the body.
I wouldn't expect a drop straight down would produce much of a bounce.
 
  • #26
erobz
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You must mean some sort of cartwheeling of the body.
I wouldn't expect a drop straight down would produce much of a bounce.
No, I mean there is some small bounce. It’s clearly visible in videos (maybe not in second story falls), from falls of lower height. Obviously, it is going to depend on a lot of factors, but I think that is what we are trying to figure out.
 
  • #27
jbriggs444
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$$mgh=mal$$ where l might be the length of your leg extension. This seems as good an estimate as any and it is not difficult to calculate. So a ten foot fall will produce at least 5g assuming l=two feet. If you land on your butt then l=3 inches and consequentially you experience at least 40g.
There is a potential off-by-one-g error there depending on whether you want to count the one gee of acceleration that you "experience" simply from sitting stationary on your butt -- for a total of 41g worth of experienced acceleration measured against the free fall frame.
 

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