Naked Science Forum
Non Life Sciences => Physics, Astronomy & Cosmology => Topic started by: bryanalter on 04/12/2007 22:44:51
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How do you quantify the additional force exerted by an object due to its motion? An object at rest on a surface exerts a downward force equal to it's weight: F=ma, where "a" is gravity. Now drop that same object from a height, and there would appear to a force greater than the object's weight exerted. In fact, the object may now cause damage (dent, crack, etc.) to the surface whereas when it was at rest no damage was caused. The object's momentum (mass x velocity) at impact can easily be calculated, but what relation does the momentum have to the additional force exerted by the object due to its motion. To illustrate this more graphically, envision holding a sheet of paper horizontally, with a 50 gm object resting on top of it. To start with, the object is supported by the sheet of paper. Gradually increase the weight of the object (while maintaining the same surface contact area with the paper), and you will eventually reach a weight at which the object breaks through, let's say that weight is 200 gm. Now, start again with the 50 gm object and a new (identical) piece of paper, and drop the object onto the paper at small increments of height, and eventually you will reach the minimum height at which the 50 gm object will break through the paper. Now, how do I relate the two scenarios??
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One way of calculating this force is via the impulse. The impulse in classical mechanics is a measurement of the force that acts (in period of time) on your mass to slow it down (or stop it) when it strikes something. If you know the force that's acting on your mass to slow it down, then the mass must exert an equal and opposite force back on the object that got in its way!
If the moving object has mass m, the amount it slows down by is Δv, and the time it takes to slow down is Δt, you can calculate the additional force it experiences.
F=m Δv/Δt.
In your paper experiment, you're increasing Δv slowly until eventually the forces involved are more than the bonds in the paper can take, and so it breaks.
(This is assuming a constant force applied over that time.)
There's more on the method here: http://en.wikipedia.org/wiki/Impulse , along with an explanation on how its related to momentum.
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How do you quantify the additional force exerted by an object due to its motion? An object at rest on a surface exerts a downward force equal to it's weight: F=ma, where "a" is gravity. Now drop that same object from a height, and there would appear to a force greater than the object's weight exerted. In fact, the object may now cause damage (dent, crack, etc.) to the surface whereas when it was at rest no damage was caused. The object's momentum (mass x velocity) at impact can easily be calculated, but what relation does the momentum have to the additional force exerted by the object due to its motion. To illustrate this more graphically, envision holding a sheet of paper horizontally, with a 50 gm object resting on top of it. To start with, the object is supported by the sheet of paper. Gradually increase the weight of the object (while maintaining the same surface contact area with the paper), and you will eventually reach a weight at which the object breaks through, let's say that weight is 200 gm. Now, start again with the 50 gm object and a new (identical) piece of paper, and drop the object onto the paper at small increments of height, and eventually you will reach the minimum height at which the 50 gm object will break through the paper. Now, how do I relate the two scenarios??
In addition to what jpetruccelli said: you can't quantify that force, since it depends on the time Δt during which the momentum is transferred, which, in turn, can vary with different materials and different experimental settings, for example how strong you take the paper and how is its surface: Δt decreases if you increase the paper's tension (in an unpredictable way) and increases if you increase the paper's surface (in quite unpredictable way). To make an exact computation you should exactly predict how the paper behaves microscopically instant by instant (quite impossible).
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One way of calculating this force is via the impulse.
Yes; A good way of approaching a lot of problems.
You can also consider the problem in terms of the energy needed to rip the paper. Whilst the breaking process is happening, there is a force acting in a direction and the work done is the force times the distance. (Again, this may not be uniform - but you can / could sum the effect over many small intervals). If there is enough KE at the start, the object can do the work needed to get through - else it stays there. The static weigh force would also be relevant, but would be a small proportion of the force, once the drop height was large. (The weight of the hammer, doesn't contribute much to driving a nail in! )
There have been many experiments, using high speed photography, to examine the speeds of bullets hitting armour (etc.) in which the progress of an object has been analysed as it goes through a substance.
I wouldn't be bothered enough, personally, to do the actual sums but you could make some reasonable estimates after some simple experiments in which you measured the depth of 'dimple' formed when the object doesn't quite get through, for a range of drop heights and weights.
A complex structure, like paper, would be very non-linear, though; metal is a bit better behaved.. You might be better to use thin sheets of foil and ball bearings. I could see an 'interesting' evening in the kitchen with some simple equipment, like a ruler and kitchen scales.
You could have a few cans of beer to take the pain away.
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One way of calculating this force is via the impulse.
Yes; A good way of approaching a lot of problems.
You can also consider the problem in terms of the energy needed to rip the paper. Whilst the breaking process is happening, there is a force acting in a direction and the work done is the force times the distance. (Again, this may not be uniform - but you can / could sum the effect over many small intervals). If there is enough KE at the start, the object can do the work needed to get through - else it stays there. The static weigh force would also be relevant, but would be a small proportion of the force, once the drop height was large. (The weight of the hammer, doesn't contribute much to driving a nail in! )
There have been many experiments, using high speed photography, to examine the speeds of bullets hitting armour (etc.) in which the progress of an object has been analysed as it goes through a substance.
I wouldn't be bothered enough, personally, to do the actual sums but you could make some reasonable estimates after some simple experiments in which you measured the depth of 'dimple' formed when the object doesn't quite get through, for a range of drop heights and weights.
A complex structure, like paper, would be very non-linear, though; metal is a bit better behaved.. You might be better to use thin sheets of foil and ball bearings. I could see an 'interesting' evening in the kitchen with some simple equipment, like a ruler and kitchen scales.
You could have a few cans of beer to take the pain away.
The problem is that he could never, in practice, be able to compute the force in that way, for the reasons I wrote.
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It depends on how dedicated he is and how much measuring he's prepared to do.
The minimum drop height for puncturing the paper and the amount of distortion would give you a very reasonable idea of the actual average force. You don't need to use the time of the event.
The principle is there but, personally, I'd rather not.
You don't have an objection to the basics? Possibly the accuracy would be poor but so is cosmology and people get very excited about those statistics.
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It depends on how dedicated he is and how much measuring he's prepared to do.
The minimum drop height for puncturing the paper and the amount of distortion would give you a very reasonable idea of the actual average force. You don't need to use the time of the event.
The principle is there but, personally, I'd rather not.
You don't have an objection to the basics? Possibly the accuracy would be poor but so is cosmology and people get very excited about those statistics.
Probably I didn't understand exactly what you intended. Ok, let's make an example: a spherical steel ball, mass = 10 grams, radius = 7 mm, breaks the paper when released from 0.5 m of height. Which is the force that the ball made on the paper?
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OK
(Edited to deal with silly late night arithmetical error.)
If the hole through the paper is, say, 5mm deep (0.005m) (the paper has been pushed that far, leaving burrs of 5mm) and this is the limiting case. By that I mean that any less height and the paper is not broken. So the ball will exit the hole at almost zero speed and have been exerting a force whilst it breaks through the paper.
The GPE of the ball, before release would be mgh.
that's 0.01*10*0.5 = 0.05J of energy
Work done in pushing the ball through the paper is
average force X distance = 0.005F
Equating these two energies.
0.005 F = 0.05
F = 10N, which is the weight of a 1000g mass. That seems a believable value for the force that would be needed to push a small ball through a strong piece of paper. Not having done the experiment, I can't be sure of the depth of hole, of course.
Failing that, you could use a bar with a hemispherical end and add weights to it until it broke through the paper. Of course, that would give the static force needed, rather than the dynamic force which, as in friction measurements, can be significantly different.
The argument is 'full of holes' as they say, because the force wouldn't be constant and would involve movement of the paper, whilst tearing, in all directions. It must be the sort of estimation that astronomers do when looking at crater sizes, only they are a bit better informed.
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If the hole through the paper is, say, 5mm deep (0.005m) (the paper has been pushed that far, leaving burrs of 5mm)
Are we talking about a sheet of paper or about, e.g., a block of lead? I don't understand what is that 5mm deep hole.
When the ball touch the paper, first it makes work against the elastic force due to the sheet of paper put in tension, that is, it makes work to stretch the paper linearly, then it makes anelastic work during the yielding phase, then maybe the strain hardening phase (usually for metals) then necking phase then fracture. How the heck can you evaluate the strain in each phase? Notice that the stress can vary a lot during these phases. And even if you want to consider just the average force, how can you evaluate the total strain, that is the total distance made by the ball during the interaction, without a very accurate analysis of many very high-speed films of the experiment? It doesn't seem as simple as you say, even to get an approximated value of the force.
(I would like to say however that I like your idea, even if I consider it quite impracticable; these kinds of experiments have always fascinated me.)
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Very naff diagram but do you see what I mean?
the distance over which the ball was exerting a force would be 5mm. It would be much easier if the ball fell through a soft, thick medium for which the distance would be better defined, of course, and the force would not be constant, at all.[diagram=296_0]
An incredibly impracticable thing to do but, if you were investigating a crime, for instance, it could yield some ball-park figures about the mass of a falling object or where it fell from, for instance. Elementary my dear Watson?
In any case - it's easier than bloody photons!
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Very naff diagram but do you see what I mean?
the distance over which the ball was exerting a force would be 5mm. It would be much easier if the ball fell through a soft, thick medium for which the distance would be better defined, of course, and the force would not be constant, at all.[diagram=296_0]
An incredibly impracticable thing to do but, if you were investigating a crime, for instance, it could yield some ball-park figures about the mass of a falling object or where it fell from, for instance.
"ball-park figures"? [???] Is it good to eat? [:)]
Elementary my dear Watson?
In any case - it's easier than bloody photons!
Ah, yes, agree completely!
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"ball park figures"
Getting a figure near enough "within the ball park".
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"ball park figures"
Getting a figure near enough "within the ball park".
Thank you.
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Pleasure.