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Old Jan 8th 2014, 04:07 PM   #1
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spring weight noob problem

i have two plates and between it their is leaf spring inside, and in top plate i have a 100lb weight if i put a scale that measures the weight in between the actual weight and the first plate would it be less than 100lb? would it be absorb by the leaf spring? this is stationary
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Old Jan 8th 2014, 07:39 PM   #2
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Originally Posted by angelulloa
i have two plates and between it their is leaf spring inside, ...

inside what? Do you mean that there is a leaf spring in-between the two plates?

Originally Posted by angelulloa
… and in top plate i have a 100lb weight…

Here again you use the term inside. Inside what? I can’t even see what it’d be in-between this time.

Originally Posted by angelulloa
… if i put a scale that measures the weight..

The weight of what?

Originally Posted by angelulloa
.. in between the actual weight and the first plate would it be less than 100lb? would it be absorb by the leaf spring? this is stationary


This problem is too poorly phrased to be able to be clearly understood. Please think it over and try describing this again please.
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Old Jan 9th 2014, 04:32 AM   #3
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is it possible to have a weight reduction on the first plate because of the spring?

http://imgur.com/J9niE1f
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Old Jan 9th 2014, 04:32 AM   #4
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Static Spring.

The spring will initially compress to a certain extent (determined by the weight).
The more a spring is squashed the more it resists being squashed any further.
Once it has been compressed just enough, the push back from the spring will exactly match the push down from the weight.

Once this equilibrum point has been reached, the spring can be viewed as a solid structure.
Only while the system is changing will the springyness of the spring make any difference.
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Old Jan 9th 2014, 04:42 AM   #5
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Originally Posted by MBW View Post
The spring will initially compress to a certain extent (determined by the weight).
The more a spring is squashed the more it resists being squashed any further.
Once it has been compressed just enough, the push back from the spring will exactly match the push down from the weight.

Once this equilibrum point has been reached, the spring can be viewed as a solid structure.
Only while the system is changing will the springyness of the spring make any difference.
(most beautiful description)

is their any machine that can reduce the weight of something, dont know if its the right question?
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Old Jan 9th 2014, 06:45 AM   #6
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Originally Posted by angelulloa87 View Post
is their any machine that can reduce the weight of something, dont know if its the right question?
Yes, there is such a machine - it's called a rocket ship. Let me explain:

An object's weight is equal to the force of gravity acting upon it, and the general equation for this force is:

F = GMm/R^2

Here G is the gravitational constant, 'M' is the mass of the earth, 'm' is the mass of the object under consideration and R is the distance from the center of the Earth to the object. For object's at or near the surface of the earth the value of GM/r^2 is typically shortened to 'g,' which yields:

F = mg

The value of g is about 9.8 m/s^2.

So, to reduce an object's weight you can:

1. Reduce the object's mass. Not practical, other than through chemical or nuclear reactions, but I doubt that's something you want to consider.

2. Increase the value of 'R,' which means you move away from the Earth. If you place the object in a rocket and launch it into space its weight (i.e. gravitational attraction to Earth) is reduced.

3. Decrease the value of M/R^2. If you place the object on a different planet that has smaller value of M/R^2 the gravitational force is reduced. For example objects placed on the surface of the moon are said to weigh about 1/6 what they do on earth.

Hence the machine you use to reduce weight is a rocket ship, so you can put the object into space or on another planet.

Another thing to consider is that if you remain near the Earth's surface an object's weight is pretty much constant, but the net force required to suspend or support it (as measured by the scale in your diagram) can be reduced using dynamic means. For example - if you place the object in an elevator and accelerate the elevator downward at rate 'a' the force of the object on the scale is reduced by the amount of acceleration:

F= m(g-a)

If you make a=g, then F becomes zero. This is precisely what happend for an object in earth orbit - its rate of downward acceleration as it orbits is exactly equal to the acceleration due to gravity, and we refer to the object as being "weightless." Strictly speaking the object's weight hasn't changed at all, but the downward force as measured by the scale has - hence the scale is no longer an accurate measure of the weght.

MBW's earlier reply about the dynamic motion of the spring is along these same lines - as the spring/mass system oscillates the net force on the object changes. But once the spring comes to rest (no more dynamic movement) the weight of the object as measured by the scale settles to F=mg.

Last edited by ChipB; Jan 9th 2014 at 09:49 AM.
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Old Jan 9th 2014, 02:15 PM   #7
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About Weightlessness

I've looked at "weightless" as a condition which might happen to a mass marked by a "vanishment of weight." A critical condition I ask is does "weightlessness" mean zero force-of-Earth on the mass?

If I understand mass (of a system and Earth) and gravity via Newton's force description (force toward Earth, mostly), I see no circumstance... I don't understand, I say, the idea that a mass might ever be "weightless," might ever have zero magnitude force of gravity (from Earth and all else).

I wrote an "armchair" explanation. I'm usually wrong... What is the correct way?

http://www.thermospokenhere.com/wp/0...tationary.html
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Old Jan 9th 2014, 03:33 PM   #8
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Thermo: as usual your explanation is very thorough, but I feel you tend to use very complicated equations where a simpler explanation would suffice. To wit:

The force due to gravity is F=GMm/r^2

For an object of mass m, from F=ma you get that a = GM/r^2.

For an object in circular orbit the acceleration due to gravity must equal centripetal acceleration:

GM/r^2 = w^2r, where w = rotational velocity. So:

r^3 = GM/w^2.

Set w to equal one revolution/day, and you get the altitude of the geostationary satellite, r.

As for whether anything can be truly weightless, as you say the answer is no. However, given large values of r the force becomes vanishingly small. For example, the contribution to your weight on Earth due to the gravity of the star Alpha Centauri is too small to measure.
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