I want to know how high a 55kg object will need to fall from, such that when it hits the solid concrete floor, the force generated will be ca. 28g. The object has no wind resistance, and the floor won't move under impact.
Of course, it may be that at terminal velocity (10m/s), 28g can't be achieved.
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>> I want to know how high a 55kg object will need to fall from, such
>> that when it hits the solid concrete floor, the force generated will be ca. 28g.
>> The object has no wind resistance, and the floor won't move under impact.
>>
>> Of course, it may be that at terminal velocity (10m/s), 28g can't be achieved.
>>
Question can't be answered without knowing the nature of the object, ie. how much it deforms after impact.
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I guess we can assume that it has negligible deformation.
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Let me try
Potential energy at the start = m*g*h
Kinetic energy before impact = 0.5*m*v^2
Both must equate by conservation of energy law.
Assume g = 9.8 m/s^2 so calculate for h and v knowing the m.
Try this
www.livephysics.com/tools/mechanics-tools/solve-problem-related-impact-force-falling-object/
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FF is quite right (no surprise!), the question cannot be answered in its current form.
As an extreme example, if there is no deformation at all, and both object and floor are perfectly rigid, the object comes to rest instantaneously, and is subject to infinite force and infinite acceleration, but, for infinitesmal time.
To answer the question, the stiffness of the object / floor pair needs to be estimated. If the floor doesn't deform significantly compared to the object, then, only the properties of the object are required.
Dave, do you have any more context / information you can share?
A good estimate for impact force is F = v* sqrt(k*m)
where, in this case, the v is that given by equating potential and kinetic energy, i.e., v=sqrt(2*g*h)
k is the stiffness, and m, the mass of the impacting object.
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>> FF is quite right (no surprise!),
I am kind of worried why FF is such an expert on things falling from the sky and deforming?
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I can't remember what FF's degree was - but maybe it was physics or maths? And then he got interested/involved in flying at Uni? Pretty sure his degree was technical. Or maybe he just did physics or applied maths at A level?
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>> I guess we can assume that it has negligible deformation.
But it is the deformation that determines the g-force.
If the object does not deform and the floor does not deform then there is a force of infinite g.
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perhaps if dave were say what he is dropping, and why he wishes to know?
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Should we be advising how high a balcony an eight and a half stone (ish) person should be dropped from?
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I was wondering something similar. Bridge in my case...
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Ha, I thought it may be more complicated than it seems!
I am considering crash testing my dog boxes. There is a swedish test place (like the old BSI) that has developed a series of tests designed to help protect the dog and the back seat passengers in the event of a crash. Only one manufacturer has so far carried out the tests on their boxes, and because of this, and the swedish peoples' desire for all things safety, they are able to charge a hefty premium on their boxes, and are selling lots of them as a result. So I want to get in on the action and increase my profit substantially.
One of the tests is to run a sledge carrying a Volvo V70 'tub' down the track into a deforming metal bar, such that the car decelerates at something like 28g to simulate a 50kph impact. In the back of the car (against the rear seat in the boot) is the test box containing a 'dog' that weighs 35kg. The dog is a canvas bag containing sand and marbles. The aim of this test (there are 2 others) is to see if the box falls apart/splits open, the doors open, or deforms with sharp edges. After the test the door must be able to be opened without tools or excessive force.
So before I pay the money (ca. £1k), I would like to get an idea of how my boxes will hold up - the pop rivets, welds, door locks, hinges etc. So I thought it may be an idea to try and replicate it by dropping a box (with the back of it hitting first) onto the floor with a 'dog in it.
Here's the test:- www.youtube.com/watch?v=CrTQ0qsJGRw
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35kg? you got fat dogs out there!
dont forget to put a sticker on the box
"No dogs were harmed in testing this product"
Do you give the boxes funny name, like Ikea?
"Dented" the dog box?
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>> Do you give the boxes funny name, like Ikea?
>>
Well the funny thing is, most of those names do actually mean something in Swedish.
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"that the car decelerates at something like 28g"
Goes in a Great Dane, comes out a bulldog.
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>>Goes in a Great Dane, comes out a bulldog.
Sounds like a typical Friday night.
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I wonder if the OP could build me a box for the mother-in-lard.
200kg.
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Ooh dear, can you see your future Dave....
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Yeah, I need a box for the wife to keep her away from the fridge.
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To get a drop test to simulate a particular acceleration level is quite tricky. The test needs to be carried out a large number of times, while varying the stiffness of the local contact to tune the peak deceleration. As an example, in one shock test I did, where we had to obtain a specified shock pulse to obtain qualification, we obtained the final tuning to get the acceleration level just right by changing the number of paper towels in a stack between the underside of the shock table and the landing "stop". I wouldn't recommend this route unless you absolutely have to.
The valid comparison with the drop test on the video is simply to replicate the height and the orientation. The deceleration measured during such a test is simply a characteristic of the test object rather than the most meaningful way to specify the test.
However, in this case, I wouldn't be spending any effort on tests which won't lead directly to any official certification. As you know the acceleration levels which have been obtained by other people's tests, you can reduce your risk by carrying out a static calculation - as an example, what force is there across the hinge or across the latch when the dog and door is subject to 28g? what stress results from this force? How much force does the heaviest dog inside your cage impart when decelerating at 28g? what's the worst location and direction to apply this load to your structure? If you can demonstrate that all parts remain below the yield stress of the material under that loading, then, you are very unlikely to be caught out when you do the full scale test - as the test allows some limited plastic deformation.
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>> Of course, it may be that at terminal velocity (10m/s), 28g can't be achieved.
To add to all the "not enough information" comments, terminal velocity is a product of downward force (gravity and mass) vs. air resistance (drag). When the two cancel out, you have terminal velocity. Thus without knowing the drag factor of the 55kg object it is not possible to determine its terminal velocity.
Think of a skydiver. The mass of the bloke and his 'chute does not change when the 'chute is opened, yet his terminal velocity decreases dramatically. The only change is to the drag factor of the assembly.
Incidently, before he opens the parachute his terminal velocity is about 56m/s in the lower atmosphere. Going higher gives you thinner air and a higher terminal velocity up to the point where, if you're Austrian and barking, you can go supersonic.
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"The only change is to the drag factor of the assembly"
Wearing a dress?
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For the maths and physics experts out there. - Fursty Ferret
