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• 0 ## Question OK, this isn't so much a puzzle, just a calculus quiz.

A tree trimmer has shimmed up a tree, removing all the limbs on the way up.

The tree is 2 foot diameter at the base, and after topping the tree 100 feet up, the top is 3 inches in diameter. (he is a really brave tree trimmer.)

On his way down, he cuts the trunk into 2 foot long pieces (50 pieces in all.)

Each piece lands on a squarely on a bug.

Assuming equal density throughout the tree, which piece will squish the bug the flattest?

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• 0 Anything to do with Gravity has to be guided by Newton's Laws

Maximum effect achieved for the log with the biggest momentum. i.e. mass x velocity.

The mass depends on the volume which is a function decreasing with height from ground. (inversely proportional to square of height).Biggest mass at the bottom

Velocity at point of hitting ground depends on the height of the log piece from which it falls. (directly proportional to the Square root of height)

Reckon that 39th piece from top (12th from the bottom) will have the max effect on the corresponding bug

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• 0 The 50th piece.

Force exerted by each piece is proportional to the mass of the piece since acceleration due to gravity being constant.

So the 50th piece will have the maximum mass.(largest base)

So it will flatten the bug most.(spring action)

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• 0

If log 1 is the base of the tree, then log 13 transfers the greatest impulse.

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• 0

The 50th piece.

Force exerted by each piece is proportional to the mass of the piece since acceleration due to gravity being constant.

So the 50th piece will have the maximum mass.(largest base)

So it will flatten the bug most.(spring action)

no.. i suppose u are takin 50 as the base piece... then even though it has a large mass, there is no momentum bcoz of zero velocity.... it is stated that each log is landing on the bug

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• 0 The 12th piece (counting from the base as 1)

V = pi/12 * h * (D^2 + D*d + d^2)

h = the height of the frustrum

D = the Diameter of the large end

d = the diameter of the small end

Piece number Top Diameter Bottom Diameter Height (in) volume Height from ground(ft) velocity (ft/s) impulse

13 18.54 18.96 24 6627.074 24 39.19184 259727.2

12 18.96 19.38 24 6927.28 22 37.52333 259934.6

11 19.38 19.8 24 7234.135 20 35.77709 258816.3

Edited by tonyd
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• 0 The first piece will reduce the bug's original height by the largest percentage, from 100% height to flat. After that, each piece will only reduce the height of the bug by a very small percentage, resulting in a lower "flatness factor". Therefore, the first piece will have the greatest effect on the flatness of the bug, so the first piece will smash the bug the flattest.

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• 0 Any of the segments will flatten a bug. It is not necessary to calculate the amount of force exerted. The last and largest piece will do the trick though the why depends on how you read the puzzle.

If all logs fall on a separate bug then the flattest will be the one squished by the log with the largest diameter and therefore flattest surface. The last segment wins.

If all logs fall on the same bug then each segment will flatten the bug just a bit more than the one before. The first one will have the largest effect, but the question asks which one will "squish the bug the flattest." Each time the bug gets a little more flat so the last segment wins.

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• 0 My 1st post was incorrect! I think that the piece with maximum potential energy will flatten the bug most.

PE=mgh

so we need to find the maximum value for mxh.

density being constant>> m is proportional to volume.

so calc. for Vxh.

Solving for maximum value, the answer i got was the 20th piece from bottom.

So i think that the 31st piece to fall will flatten the bug most.

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• 0 I got the same answer as swapnil.

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• 0 My 1st post was incorrect! I think that the piece with maximum potential energy will flatten the bug most.

PE=mgh

so we need to find the maximum value for mxh.

density being constant>> m is proportional to volume.

so calc. for Vxh.

Solving for maximum value, the answer i got was the 20th piece from bottom.

So i think that the 31st piece to fall will flatten the bug most.

Would tend to agree. Squashing the bug involves Work. And the log with the maximum energy can do the most work. The kinetic energy of the log at impact will get transmitted to the bugs innards and make them fly in all directions. So the log with the most energy (in this case the 31st from top and 20th from bottom) will make the bug particles fly the maximum distance and cause the flattest bug

My earlier hypothesis that the log with maximum momentum (impulse), which incidentally is the 12th from bottom and 39th from the top will cause the maximum force to be applied on the bug stands corrected. The log with the max energy will transfer the max force on the bug to squash it.

Instead of the squashing bugs if the logs were to fall on nails or spikes stuck into the ground, then a similar poser would have been which log drives the nail the deepest. This would be the principle on which industrial pile drivers are designed. And these take into account the Kinetic energy transferred by the Hammer to the pile head.&nbsp;</p>

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• 0

Whichever model is a better independent measure, impulse (force is roughly constant and effective throughout crunch) or energy (all energy is transferred into squashing distance and no energy lost to earth), I think all of you are calculating volume wrong.

If log 1 is the base of the tree, I got 13th log and 23rd log respectively.

Personally I find the latter assumption too much.

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• 0 Everyone seems to be making several assumptions which may or may not have been intended by Cool Javelin but are unrealistic in the real world.

First, the assumption that the piece will start to fall unhindered immediately after being cut. In the real world an unrealistic assumption. Lumberjacks will push the piece tor use wedges so it fulcrums on the last section of the cut preventing the section from the full effects of gravity in a downward fashion until that last splinter breaks free. Next the assumption that the segment would land in an exact vertical position. Then the question of the surface of the ground. If the ground is soft, then the bug is pressed into the ground by the impact, reducing squashing effect more on larger diameter pieces than small diameter pieces. If the log lands on the side, then larger diameter would be a better squishing surface. Also going with these same assumptions, my answer would be the segment with the greatest kinetic energy at instance of impact. This is exactly equal to the potential energy before the segment begins to fall. Assuming it falls in an exact vertical position, ignoring the effect of air resistance, the answer would be 20th section by my calculations.

Whichever model is a better independent measure, impulse (force is roughly constant and effective throughout crunch) or energy (all energy is transferred into squashing distance and no energy lost to earth), I think all of you are calculating volume wrong. If log 1 is the base of the tree, I got 13th log and 23rd log respectively. Personally I find the latter assumption too much.
I did a complete calculation of the volume (V=((pi) r square)H where (pi)=3.14159..., r = average radius of log segment and H= height = 2ft. However, since H and (pi) are constant and we are only comparing relative amounts, it is only necessary to use r square , just as we can ignore exact density, another factor in the mass.
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• 0

Everyone seems to be making several assumptions which may or may not have been intended by Cool Javelin but are unrealistic in the real world.

Assumptions are intended; he said it's a calculus problem.

I did a complete calculation of the volume (V=((pi) r square)H where (pi)=3.14159..., r = average radius of log segment and H= height = 2ft. However, since H and (pi) are constant and we are only comparing relative amounts, it is only necessary to use r square , just as we can ignore exact density, another factor in the mass.

An example of why I'm accusing everyone of calculating volume incorrectly. It's not a standard cylinder.

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