Riddles with a Prize, by Mantus Anchors - Page 8 - SailNet Community
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post #71 of 94 Old 11-20-2012
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Re: Riddles with a Prize, by Mantus Anchors

m-M=0
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Re: Riddles with a Prize, by Mantus Anchors

Guys you are so close.....

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Re: Riddles with a Prize, by Mantus Anchors

(m+M)squared/32
let me try this again
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Re: Riddles with a Prize, by Mantus Anchors

If the weight of the spring is negligible, yet by pushing the head down the spring causes the head to pop up, the weight of the head has to be negligible. Both now have to be nearly the same weight if the head was pushed down to completely compress the spring. And there's the big unknown. How far does the spring have to be compressed to cause the head to pop up? Of course you have to define the word "pop" when using it here. Does that mean pop up like a jack in the box, or does it merely mean making the head rise back up, regardless how quickly?

For the box to be lifted, enough force has to be created through the compression of the spring to cause both the head and the box to be lifted off the surface the box is sitting on. The force of the spring pushes the head upward at a velocity sufficient enough to stretch the spring beyond it's natural state and in doing so create sufficient stretch in the spring that the box will be lifted off the surface before the velocity of the head stops and begins to succumb to gravity and the forces of the spring returning to its natural state.

Since the spring's weight in negligible, the weight of the box and the head would have to be negligible, barring any super-spring, space-age steel that the spring might be composed of.

The weight of all three would have to be negligible relative to the weight of any two, in other words, all are almost zero. Unless you state the head compressed the spring half way to cause it to pop up like a jack in the box, which would allow you to surmise that fully compressing the spring could cause the spring to decompress at such a rate as to lift the box off the surface as described above.

This is an answer that could be theorized in a calculus equation but I forgot most of the calculus symbols. It's been a while.
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post #75 of 94 Old 11-21-2012
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Re: Riddles with a Prize, by Mantus Anchors

2m+M

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Re: Riddles with a Prize, by Mantus Anchors

The formula should be expressed in terms of fig newtons per glass of milk. Unfortunately the variables get out of hand; temp of milk, % butterfat and whether it's really fig or those ersatz raspberry filled concoctions
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Re: Riddles with a Prize, by Mantus Anchors

F = mass * acceleration (F= m*a), and weight = mass * gravity

In order for the box to "jump" off the surface it's sitting on (assuming earth with g=9.8m/s^2), the upward force, F(u), would need to be greater than the weight of the box which is (M+m)*9.8 m/s^2.

Whatever downward force, F(d), is applied to the Jack will be returned with the Jack is released. Therefore, the force applied to just overcome the weight of the Box+Jack would be F(d) > (M+m)*9.8 m/s^2

If this isn't the answer, then all I can say is F(u). ;-)

Last edited by Capt-T; 11-22-2012 at 12:30 AM. Reason: change = to >
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Re: Riddles with a Prize, by Mantus Anchors

The force must be greater than Mg+mg, g being = to 9.80665 meters per second squared.One must be careful to run out of cookies and milk simultaneously ; as in life or maths.
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Re: Riddles with a Prize, by Mantus Anchors

If Jack is in the box, the spring is already compressed with enough potential energy to jump both. The only force necessary is to release the catch on the spring.


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Re: Riddles with a Prize, by Mantus Anchors

M-m
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