Weight Problem
A chain of length A cm and mass B kg is suspended vertically by one end with the other end immediately above a scale. The chain is released and falls. At the instant the entire chain has fallen onto the scale what does the scale read?
Labels: funphysics
posted by Rajesh Lal at
8:37 AM
58 Comments:
I'm pretty sure it's 3B kg.
The weight!
Actually I think it would be more than the weight of the chain, because of the force required to stop the chain's fall.
The weight!
Actually I think it would be more than the weight of the chain, because of the force required to stop the chain's fall.
The weight!
Actually I think it would be more than the weight of the chain, because of the force required to stop the chain's fall.
That's right. 3B is 3 times the mass of the chain. I actually calculated the force (apparent weight) and divided g to convert to kg as that is what I assumed the scales would be calibrated as.
How I did it, will wait until I think the time is right to explain the perhaps surprising answer.
i believe the result is 2mg because F=mg+F1(F1 being the force required to stop the chain)F1/t=p.
t=sqrt(2*h/g) and p=m*V where V=sqrt(2hg) and then F1=mg and F=2mg.
Hi milos. You made a typo - should have read F1=p/t, but you did it right when you substituted.
Unfortunately, F1=p/t is inapplicable as p is not a constant. Neither can you use p=mV as the whole of mass m doesn't have speed V.
You are definitely in the right direction though.
PS sorry if I seem to be hogging the place, but I'm stuck at home with a broken leg and am desparate for entertainment.
i have another explanation... i believe the last one wasnt quite good.lets name the mass of the last chain m1 and its length h1 then the accelaration required to stop the chain is a.it is the acceleration after this part of the chain has fallen.the force required is Fthe force required to stop it.V=sqrt(2*g*(h-h1)) and V^2=0^2 + 2a1h1 and a=a1+g(because this part is also affected by gravity wich gives acceleration)using these equations we get the a1=g*(h-h1)/h1 and a=g*h/h1.therefore F=m1*a.now because of m1/m=h1/h we get F=m*g and the scale shows F+mg=2mg.if u have another way of solving i would be quite glad to see it
The starting pont is Newton's Second law: F = dp/dt where p = mv.
So F = m dv/dt + v dm/dt
Now apply this to the chain at the very last instant of its fall. It is
essential to interpret his meaning of the equation clearly. m dv/dt = 0
because m is not accelerating (it is resting on the scales). For the very
last bit of the chain, v = sqrt(2gh) where h is the length of the chain in
metres. (I'm not going to use A as it's in cm).
Now, dm/dt = (B/h)v. So v dm/dt = (B/h)v^2 = (B/h)2gh = 2Bg.
NB B/h is the mass per unit length of the chain as is the same everywhere.
Putting this altogether =>
F = Bg + 2Bg = 2Bg. (The Bg term is the weight of the resting chain).
I assumed the scales on calibrated in kg, so the reading is 3B kg.
Unfortunately your second attempt isn't easily rescued as it refers to some bizarre
acceleration "a" which can be chosen to be arbitrarily large by choosing "h1"
arbitrarily small.
Although, for simplicity, I have focused on the last moment, it is not
hard to generate an equation for the pan reading as a function of time
for the entire fall of the chain.
Just before posting this reply, I did some Googling and found that my
answer is the "accepted" one, but some of the proofs are awfully messsy.
Anyone, How do you remove a posting? I want to delete my last one as the formatting is dreadful.
The starting pont is Newton's Second law: F = dp/dt where p = mv.
So F = m dv/dt + v dm/dt
i believe there is something wrong with this or u just didnt type it quite good:S...also on your previous post u said u cannot use F=p/t witch u used here:S...i will google it myself to find some other answers.and u can delete just by signin in and then going for post a new comment and then u go to see previous given the possibility to delete yours.
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ok i used now the principer of conservation of energy.and u get
m1gh1+m1V^2/2=A=F'*h1.the thing is that after calculating u get F'=mg but that force isnt constant.it starts from zero and when the whole chain has fallen it becomes highest so (F+0)/2=F' and u get F=2mg.i couldnt find anything googling it but i believe this is a good explanation...u can use this principe on the whole chain because the whole chain will have the speed of v=sqrt(2*g*h/2) and then mgh/2+mV^2/2=A and u again get the F'=mg but again the force isnt constant so F=2*F'=2mg.and dont forge the mass of chain so all in all it gives the 3m on scale
also my way with the acceleration a can be used because it isnt constant also:d but i still dont quite understand your solution and i believe that the conservation of energy is the best way to solve this
I should have said that dp/dt is the rate of change of momentum, dv/dt is the rate of change of speed and dm/dt is the rate of change of mass. But its such a standard differential equation that I didn't think it necessary.
I accept that the v dm/dt part is not usually included, but it is the only part of interest to e.g. the rocket equation.
Aren't you familiar with calculus? It would explain a lot about your methods. That's a compliment in case that sounds ambiguous.
Looking at your energy method, you have F h1 = m1 g h1 + m1 v^2/2. You then correctly say that this yields F = mg (m=B here) [that is a problem]. So is this: you now seem to forget that you defined h1 to be the last bit of the chain (if you've changed your mind about the definition of h1, then the original energy equation is no longer true and then the whole of your argument must be reworked) and take some kind of average value of F and several other averages - none of which have physical meaning as far as I can tell, and use them to fiddle in factors of 2.
I am surprised that your technique didn't pay off though. I thought you had a truly original solution for a moment. I'm going to see if what went wrong there.
I think I've seen what went wrong. The force F' (sorry I hadn't noticed the ' before) not only has to stop the chain, it also has to account for the energy in the generated heat and sound. Perhaps that explains the needed factor of 2. However, my guts are saying that your F' h1 term isn't a proper interpretation of work = force times distance.
You have nowhere, clearly, used the law of conservation of energy.
The "a" acceleration is a nonsense. It can be made arbitrarily large by letting h1 become arbitrarily small.
You assert that some things = A, what is A (it's not the length of the chain as it has the dimensions of energy.
I'm sorry, but, there are so many mistakes in your current and previous solutions, that it would take too much time for me to cover them all. This web page is already very long, and I don't have any desire to start a war. This stuff is meant to be fun.
Here's a pretty good version: http://answers.yahoo.com/question/index?qid=20061108225109AADuW5Y
The equation F=P/t used there is not the same as your one as t is defined differently. I have only used such a relation for infinitesimal quantities, in a correct manner.
I Googled with "chain falling force". I'm startled at how bad some of the answers are.
PS I notice you have pulled some posts - please tell me how to do that.
Oops, I'm getting tired. I just noticed you had explained how to delete a posting. Thanks
i believe u dont really understand the meaning of F=2*F'.we dont have to use the h1 and m1 just ordinary m and h.u will still get the force of mg.and it is clear that the force isnt constant.it increases with time obviously starting from zero to some max value.that is why we have the F'=(0+F)/2.maybe this is not best explained.i have the proper way but it would be too long with a lot of complex math...
Let the chain have dropped through a distance y. Then the speed of
the moving part of the chain is then V = sqrt(2gy). I now focus on
the physics at that instant.
Let m1 and h1 correspond to a very small element of the chain that
is just about to come to rest. Also m1/h1 = B/h => m1 = (B/h) h1
Substituting into F' h1 = m1 g h1 + m1 V^2/2 =>
F' h1 = (B/h) h1 (g h1 + 2gy/2) = (B/h) h1 g (h1 + y)
Divide throughout by h1 =>
F' = (B/h) g (h1 + y). Now take the limit h1 -> 0 =>
F' = (B/h) g y and this is the actual instantaneous "force".
For the total scale reading we must of course add the force due to
the mass at rest, which also = (B/h) g y. On this we are agreed.
So as y varies from 0 to h, F' varies from 0 to Bg uniformly (with
respect to distance fallen by the chain). So at the last moment,
F' = Bg (and cannot be fiddled into 2Bg through some weird
averaging process).
The fact that m1 and h1 are infinitesimals is crucial. If you were
familiar with the methods of calculus you would know that my
reasoning is unshakeable.
I nevertheless, remain slightly stunned that the method didn't work.
And I can understand why you want to rescue it. I am convinced any
rescue will have to be contrived to fit the facts. The defining
expression for F' h1 is not physics (too my reducing surprise).
I believe I've implicitly covered why you can't take m (=B) and h
as a whole (you must work with infinitesimals). Also be aware that
the force goes up with the square of the time that the chain has
been falling.
I do understand what F = 2 F' means, that's why I say it's nonsense (sorry).
milos, I believe that if I can't explain something, it's almost certainly because I don't really understand it. You would be wiser if you adpopted that attitude.
It is important (for you) to thoroughly understand my last post. If you think I have made a mistake, I'd be happy to hear about it, but recommend that you assume that I haven't. Take particular care to note that my variables and descriptions are very precise. I have left no room for sophistry.
In your last post, I could only agree with "it is clear that ... some max value". i.e. most of that post is wrong. Even that troubled me as you are not clear about which force you are referring to.
I'm reposting the proper solution, but now making it more useful.
I've also fixed a couple of typos.
Newton's Second law is: F = dp/dt where p = mv.
So F = m dv/dt + v dm/dt. Assume the chain is made from infinitely
many, infinitesimally small links and that we apply the above
equation to one link at a time. i.e. m = 0. So we have F = v dm/dt.
Now consider the chain after it has fallen a distance y. Length y
will be on the scales and length (h-y) will be falling at speed
v = sqrt(2gy)
Now, dm/dt = (B/h)v. So v dm/dt = (B/h)v^2 = (B/h)2gy
NB B/h is the mass per unit length of the chain and is a constant.
So the reading on the scales is (B/h)gy + 2(B/h)gy = 3(B/h)gy.
The first term is the weight of the resting chain.
So the reading on the scale starts at 0 and rises uniformly (with y)
to 3(B/h)gh = 3Bg. It then drops back to Bg.
No more from me unless you indicate you want more.
although i believe that my way of solving this is good i would be quite glad if u could try and solve it using the method of energy conservatio.i agree that your solution is accurate but i want to search for some other solutions
Hi milos. The CoE approach is what I tried in my Aug 14, 9:31 am post.
The only way I can think of to get the factor of 2 is to propose that F' must also be responsible for providing the energy lost as sound and heat. Bear in mind we are dealing with an (totally) inelastic collision process, CoE usually isn't helpful then.
Unless you find some physics to say that "of course F' is twice as big as you calculated because...", then you are doomed to failure. I'm prety sure that if there was a neat explanation I'd have known about it for many years.
Are you completely satisfied that the F=2F' is wrong and that my other criticisms are well observed?
i still believe that my reason for F=2*F' is correct and totally disagree with yout proposal that it must also be responsible for providing the energy lost as sound and heat.but nevertheless i agree with your solution and will try to solve this in other ways...
i still believe that my reason for F=2*F' is correct and totally disagree with yout proposal that it must also be responsible for providing the energy lost as sound and heat.but nevertheless i agree with your solution and will try to solve this in other ways...
milos, I guarantee the your belief in the F = 2*F' is because you are not careful in your definition of the problem and because it, unfortunately, gives the right result. I can't understand why you can't see that F = 2F' is a misunderstanding, and hence proceed to find out what really has to be done to get a CoE based solution.
You introduce variables, like A and a1 without defining what they are or remembering what you meant them to be. You then change their meanings and conditions of use (e.g. h1 was associated was the last bit of the chain at the final speed, and suddenly it is any small length at an intermediate speed) and don't even know that you are doing so, even though I have warned you of it several times. You are simply playing about with letters and that doesn't make for good physics.
That's why I wrote the 14 Aug, 9:31 post - it is a model of clarity. You cannot have understood it. If you had, you would have been able to that there is no way to fiddle in the F = 2F'. I recommend that you re-read it.
The reason that it fails is because F' h1 = m1 g h1 + m1 V^2/2 is incorrect. It is dimensionally OK and very appealing, but that's all.
I'm not happy with F' being responsible for producing the heat and sound either, the suggestion was the prodct of desperation.
I have not given up though. But I have moved onto other problems and put this one is background mode.
milos, I want to assure you that I am totally intrigued by the failure of the CoE method. If I can't understand what's wrong, it also makes me concerned that I necessarily don't know why the momentum method does work. That disturbs me profoundly.
That is very much why I'm happy to labour away at it. I've GOT to know what's going on.
Our discussion is marred by the F = 2F' mistake. I hope this puts
examination puts it to rest.
m1gh1+m1V^2/2 = A = F'*h1 => F'= Bg (I'm not using m as it hasn't
been defined). [What is A? - you never say.]
But you only got F'=Bg because you did the calculation at the very
last instant of the chain falling => V = sqrt(2gh).
If you had tweaked the calculation to examine it when the chain
was half way (exactly) then you would have got F' = Bg/2 (as V then
would have been sqrt(2gh/2) = sqrt(gh). As F' is varying uniformly,
with respect to distance fallen, I would then have accepted that
the maximum force = F = 2F' => F = 2 (Bg/2) = Bg.
I hope we can now lay the F = 2F' mistake to rest.
milos, yeehah. I have a sort of solution for you. I don't like it much as it uses too microscopic a view of a force law which in turn is simply the simplest law I can come up with. It also makes gross simplications.
The pan can only exert a force over a very short range up the chain (which i prefer to think of as a perfectly smooth rope). I assume the following force versus height profile. The force will be a maximum (Fm) at the surface of the pan and fall uniform with height to 0 at X (a very small distance above the pan). So the force, Fm = Fm - x Fm/X where x is the height above the pan. Integrating this force over height => Work = Integral[Fm - x Fm/X, x from 0 to X]
So Work = [Fm x - x^2 Fm/(2X)] {x=0 to X}
=> Work = Fm X - X^2 Fm/(2X)
=> Work = Fm X/2
i.e. The effective (energy reducing) force is half the pan force. Bingo, we've now got the desired pan force vs stopping force relationship.
I know it looks like your F = 2 F', but it is a completely different interpretation (but the same maths).
Despite this result, I suspect that if I had chosen a different force/height profile, I wouldn't have got the need factor of 2.
I'm dropping this now, unless you give me reason to see a better answer forthcoming.
Cheers
I lied, I'm back. I suspect that whatever the force law is, it must be such that we can extract the factor 2. It can't be a function in the normal sense, it's a distribution.
As you see, this is a tortuous explanation. Whereas the momentum method was extremely straightforward.
Rats, I've just spotted a typo, I wrote Fm = Fm -x Fm/X. The LHS Fm should have been F.
It doesn't affect anything else.
I should have made it clearer that Fm is the force at the pan (i.e. At 0 height above the pan).
I feel the solution is a last sad act of deperation.
i dont have a solution at the moment...i will try to finish my new idea asap
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Is that all you've got to say? Did you even bother to read it? I put a fair amount of effort into that "solution" (dated Aug 17, 5:43 pm).
Surely it (along with the few small posts after it) is exactly what you're looking for.
It is now very clear to me that the energy approach requires a (contrived) quasi-knowledge of the force distribution applied to the chain by the pan. The force distribution is short range - it cannot extend up the chain.
The result that the force law is a distribution that gives the factor of 2 is perversely interesting. But it is a consequence of the law of conservation of momentum, and in turn CoM is the most straightforward approach.
I strongly recommend that you learn calculus, it's not very difficult. It's power is extraordinary. It is an essential pre-requisite if you want to make real progress with maths or physics.
ok i have a solution wich u are probably not going to like but here it goes:if we take the very last part of the chain with length h1 and mass m1 then it has a speed of v=sqrt(2gh) and its Ek=mV^2/2=mgh therefore A=Ek-0=Ek and F*h=mgh.F=mg.but as the force of interaction between the scale and the chain.there is also the force of gravity.the resulting force must be then 2mg.i find your solution of CoM good as i stated but i dont really know what is calculus
hi milos, I've read your last post. It is clearer than some of the others. I still don't know what "A" is. Although you say m1 is the mass of the last bit of the chain, you only use "m" in your calculations. Initially you use it for the last bit of the chain, then later you use it to mean the whole of the chain. You also then just add mg "for luck" as far as I can see.
Back to the problem. What I have learnt is that the CoE approach isn't a sensible one to use, for THIS problem, CoM is. But for other problems it may be the other way round. In general you need to pick and choose each as required.
I can understand why you don't fully understand what I've been writing. It uses calculus freely.
It is too big a topic to discuss here. Use Google to find out more about it. A few hours ago, I gave you a hint about it for the "bug on a band" problem.
I cannot over-emphasize how amazingly powerful calculus is. Mathematics and physics, would be nowhere nearly as advanced as they are without it. Most of today's technology would not exist either.
Even back in the days of steam power, calculus was very important.
its my mistake its m1 and not m:P...also A is the work we need to do to reduce its energy...its how we explain it...as i stated english is not my mother language so i dont know how u do it...i will definitelly try to search for calculus explanation.
have searched and the fact is i am 15 and we havent started with any differential or integral calculations so its no surprise for me to have no idea what calculus is
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milos, for a 15 year old, without knowledge of calculus, I think you're doing very well.
Unlike you, I can only do it in english, more credit to you.
I admire your determination.
Calculus is considered to be advanced mathematics. You probably won't be taught it until you're a few years older. However, the basic ideas are very simple and very easy to learn.
Hi milos. OK I can fully understand why you aren't able to follow what I'm doing. I have used calculus freely in my answers.
Without calculus and at your age, and with English not being your first language, I've got to say that I'm very impressed by your attempts at this problem and by the determination that you've shown. You have a lot of potential.
milos, I'm not going mad. The first comment didn't appear. So I wre-wrote it. Then they both appeared together. I don't want to delete either of them.
i will definitely try searching for some calculus explanations:D.in my country it is translated something like mathematical analysis
Calculus free method. But uses infinitesimal elements in an ordinary way.
Let m1 be a small element of the chain that is just about to hit
the pan. It's length is h1. Let h = total length of the chain. If
the chain has dropped through distance y it will have a speed
V=sqrt(2gy). The element is carrying momentum m1*V. The rate at
which momentum is being delivered to the pan is therefore
m1*V*(V/h1) = (m1/h1)*V^2 = (B/h)V^2 = (B/h)2gy = force applied.
Add the usual (B/h)gy for the resting mass. When y=h, F=3Bg.
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CoE Just for completeness - trying to avoid calculus
Let m1 be a small element of the chain that is just about to hit
the pan. It's length is h1. Let h = total length of the chain. If
the chain has fallen distance y it will have a speed V=sqrt(2gy).
The element is carrying KE, m1*V^2/2 = m1*gy = (B/h)gy*h1
We now have to assume a force law acting on the chain, which is a
max at the pan and quickly reduces with with height above the pan.
Now imagine that F versus height is a triangle shape, the height
of the triangle being Fpan, the width being h1, then the area of
the triangle is (1/2)Fpan*h1 = Faverage*h1
Faverage*h1 = (B/h)gy*h1 => Fpan = 2(B/h)gy, then add the (B/h)gy
for the resting mass. DO NOT DO IT WITH CoE in an exam.
These may be of help:
http://en.wikipedia.org/wiki/Calculus
http://en.wikipedia.org/wiki/Derivative
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Quickie introduction to calculus.
Let y = x^3, let Dx = change in x (Dx is not D*x, it is a
single variable). There will be a corresponding change Dy in y.
So y+Dy = (x + Dx)^3 = x^3 + 3x^2 Dx + 3x(Dx)^2 + (Dx)^3
Now subtract y = x^3 => Dy = 3x^2 Dx + 3x(Dx)^2 + (Dx)^3
Divide bthroughout by Dx=> Dy/Dx = 3x^2 + 3x(Dx) + (Dx)^2
Define the derivative of y with respect to x as:
dy/dx = limit Dx->0 Dy/Dx = 3 x^2
When you first learn calculus you will be taught that dy/dx is an
indivisible symbol. That's no longer true in advanced maths, where
you will learn that dy and dx can be separated.
On a graph, dy/dx is the gradient or slope at point (x,y).
More examples:
y = a*x^n => dy/dx = a*n*x^(n-1) [a, n constants]
y = cos(x) => dy/dx = -sin(x) [x in radians]
y = sin(x) => dy/dx = cos(x) [x in radians]
y = e^x => dy/dx = e^x [and that's what makes e very special]
y = ln(x) => dy/dx = 1/x [ln is logarithm base e]
It is very common to abbreviate this, e.g. d(cos(x))/dx = -sin(x)
Advanced maths: you can say e.g. for y = cos(x), dy = -sin(x) dx
if have two functions f(x) and g(x), then
d(fg)/dx = f dg/dx + g df/dx [chain rule]
or d(f(g(x))/dx = df/dg * dg/dx [function of a function]
Lagrange used another notation.
If y = g(x), then dy/dx = g'(x). [ there is a ' after the g]
This can be really convenient sometimes. It helps to emphasise
that derivatives are themselves functions.
Integrals, sometimes called antiderivatives.
e.g. integral of ln(x) is 1/x.
Integration corresponds to finding the area under a graph.
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