Mass on a spring and Gravity..Pain in the neck


by saltrock
Tags: gravitypain, mass, neck, spring
saltrock
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#1
Dec11-04, 03:03 AM
P: 68
How can the gravity of earth be determined using mass ona spring?
Myworking:T=2pie squarerook of m/k
There no g in the question above.I have spend ages trying to figure out the link between the gravity and the mass on spring.
If you can give me some hint then i'l be able to work it out.
Any help would highly be appriciated
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cepheid
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#2
Dec11-04, 03:17 AM
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What pulls the mass downward? Gravity does. What restoring force acts on the mass, trying to pull it back up to its equilibrium position? The spring restoring force. If the spring is "Hookian", then this restoring force obey's Hooke's law: F = -ky. The displacement y from the equilibrium position is downward, and the farther down gravity pulls the mass, the stronger the restoring force (upward, because of the negative sign) becomes. The equilibrium position occurs when the two forces are equal (gravity has pulled the mass down far enough that the restoring force is now strong enough to exactly counteract gravity). The mass will hang suspended at this elongation y. That is all you need to answer this problem (equate the forces). If I put this into equation form for you, I'd be giving away the answer (in fact, I already have ).
saltrock
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#3
Dec11-04, 06:42 PM
P: 68
that means F=Kx=g.So,i have gotta find Kx and that will give me the gravity right??Extension at equilibrium position is zero so K times o gives me 0 gravity.....so i am not gettin any value fo gravity>>>..can you make me more clear plz..I know i am a bit thick..;)

kornwestheim
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#4
Dec11-04, 06:57 PM
P: 7

Mass on a spring and Gravity..Pain in the neck


Just remember to apply Newton second law. Which states that SUM F = ma.
Combining Hooke's law F = -kx and newton law you can get your acceleration or in this case the gravity.
cepheid
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#5
Dec12-04, 12:29 AM
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Quote Quote by saltrock
that means F=Kx=g.
You might want to rethink that. Is g a force?

Quote Quote by saltrock
So,i have gotta find Kx and that will give me the gravity right??Extension at equilibrium position is zero so K times o gives me 0 gravity.....so i am not gettin any value fo gravity>>>..can you make me more clear plz..I know i am a bit thick..;)
We are talking about a mass and spring that are suspended vertically. that's why I called the restoring force ky instead of kx. I guess I should have stated that clearly from the beginning my first post, although that would have been answering the question for you, of how to measure the value of g using this apparatus. (I also thought it was obvious, because if the mass/spring system were horizontal...why would gravity be affecting the length of the spring???) ...

I see the problem...I used the word "equilibrium" in my post to mean both the unstretched position, and later on to mean the position at which there was equilibrium between the two forces. Notice that this second, new, "equilibrium" extension of the spring is no longer "zero" ie. unstretched. On the contrary...it has some finite extension. Does that clear things up?
saltrock
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#6
Dec13-04, 07:12 AM
P: 68
I sort of like have understood it..can you plz give me a weblink that clearly explains this process please.Any help would highly be appriciated
cepheid
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#7
Dec13-04, 09:43 PM
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I really don't think you need a weblink for this man...it's really simple. I'll take one last stab at explaining...

1. Hang mass on a spring from the ceiling.

2. There are TWO forces acting on it:
i) gravity pulls it down -- it is falling
ii) the spring is trying to pull it back up.

3. When the two forces are equal, the mass will come to a rest.

4. How do we know when the two forces are equal?
i) the gravitational force is constant and equal to the mass' weight,which is mass, m, multiplied by acceleration due to gravity: g. -->mg

ii) the spring restoring force is NOT constant. it gets stronger the more the spring is stretched. In other words it is proportional to the distance the spring has been stretched from its unstretched length. Let's call this 'y' The force is --> -ky

5. So, based on #3, all you have to do is equate the forces and solve for g.
saltrock
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#8
Dec14-04, 10:26 AM
P: 68
so,g=-ky/m but i dont know the values for k and m and i am only allowed to use a stopwatch,millimetre rule and apparatus for supporting the spring.
cepheid
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#9
Dec14-04, 12:19 PM
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A problem I hadn't considered, but not to worry: a minor one.

Normally, the experiment I have just described is put forward as a means of measuring 'k' when g is assumed to be known beforehand.

You know neither g nor k. So what to do? First: measure k experimentally by some other method. Then use its value in the experiment I just described to get g.

What other method? Well, what else can we do with a mass-spring system other than hang it vertically? Well, we could set it oscillating horizontally in simple harmonic motion, just as you had intended orginially. If you do so, I think you will find that there is something you wrote in your very first post that you could measure experimentally. It wouldn't give separate values for m and k necessarily, but it would allow you to take care of both.

Sorry for not thinking about this before...
saltrock
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#10
Dec14-04, 05:46 PM
P: 68
T=2piesq.root of m over k
Regrranging this will give me k=2pieM/t^2

g=-kx/m
Putting the value of K in the above equation will give me

g=-4Pie^2x/T^2

I can measure T and i got rid of K and m(i didnt know what the mass was as i wasnt allowed to use any equipment to measure mass) :)

Thanks a lot cepheid for helping me out and responding really quickly.Cheers


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