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# A Geometry question.

Posted on 2011-09-26
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Visualize a 50 foot pole that is perpendicular to the ground.
The pole breaks, 17 feet from the ground. The 33 foot top section, begins it's fall in the due west direction. There is no wind or other external force, other than gravity.
The top 33 feet begins it's arc downward, and it's arcing motion creates part of a semi-circle, with a radius of 33 feet (the length of the section of pole).

An obstruction is resting on the ground, and it is 36 feet from the base of the pole, and sitting outside the radius of the 33 foot arc.

If the obstruction was to be raised off the ground, completely perpendicular to the ground level, how many feet would it have to be raised for the obstruction to touch the outside edge of the 33 foot arc?

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Question by:nickg5
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Accepted Solution

Rich Weissler earned 90 total points
ID: 36601919
Hmm... drawing out my interpretation of what you are asking...

Either there is no answer, or I've misinterpreted the problem.
Drawing1.jpg
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Expert Comment

ID: 36601938
I get the same answer. It's not going to hit the object no matter how tall the object is. If the object was less that 33 feet from the pole, then it might matter how tall it was. Razmus' excellent figure shows that quite nicely.
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Expert Comment

ID: 36601989
Are you sure the break acts like a hinge?
How do you insure that the top of the pole falls in a particular direction?
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Author Comment

ID: 36602016
I've added red numbers to Razmus diagram to show the 3 measurements.

Drawing1.jpg
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ID: 36602052
d-glitch is right to point out that it depends how the break operates.
For example, if it were to act initially as a hinge and then shear, you might find that the top of the pole lands more than 36' from the base of the pole, as it would have some momentum leftwards (in the figure above).

Based on a quick experient with a straw, it seems to me unlikely it would get as far as 36'.  And besides, even if it did, making the obstruction taller would not increase the chances of it hitting the obstruction.
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Expert Comment

ID: 36602053
From Razmus' excellent figure:  If the break acts like hinge, there is no way the top of the pole can reach out 36 feet.
If the break does not maintain it integrity, the bottom of the falling section will kick to the right.  Even less of a problem.
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ID: 36602093
Try it by holding a straw or a chopstick vertically on the edge of your table and letting go.  See where it lands on the floor.  For me it kicks to the left!
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Expert Comment

ID: 36602103
From the question "The top 33 feet begins it's arc downward, and it's arcing motion creates part of a semi-circle, with a radius of 33 feet (the length of the section of pole)."

So it's not a real life problem and we can assume perfect circular motion.
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Assisted Solution

d-glitch earned 75 total points
ID: 36602114
Maybe this is more complicated than we thought.
The stump provide the torque to rotate the pole, and it does cause the center of mass to move horizontally as well.

If the hinge shears, the pole goes into free fall for a while:  the horizontal and rotational velocities will remain constant
until the next collision.
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ID: 36602121
I think most of the problems that nickg5 brings here are real, like cutting down trees.
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ID: 36602139
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Author Comment

ID: 36602177
If the 33 foot section of pole falls due west, the direction of Razmus diagram, if the "thing" that is 36 feet away from the base of the pole, when the "thing" is on the ground, then:

Could the "thing" be a 2 story building, instead, and the arcing, falling section of pole, would never hit the building?
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Expert Comment

ID: 36602189
As long as the pole falls in a semicircular arc and the hinge remains, the pole will never hit the obstruction, no matter what it is or how tall it is.
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Author Comment

ID: 36602395
jan24 is bringing up a good point:

"d-glitch is right to point out that it depends how the break operates.
For example, if it were to act initially as a hinge and then shear, you might find that the top of the pole lands more than 36' from the base of the pole, as it would have some momentum leftwards (in the figure above)."

That is the unknown, the hinge may very well shear.

And d-glitch has located the variables.
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Assisted Solution

deighton earned 75 total points
ID: 36707907
it never hits it.

Arguments about the broken pole bouncing off the ground and the whole thing eventually reaching the obstruction are not unreasonable in the real world, but any calculation would depend on numerous factors, such as the flexibility of the pole and ground, friction, and so-on - all of which are totally unknown - also any such calculation would be very difficult
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Assisted Solution

jan24 earned 75 total points
ID: 36708203
Agreed that a full solution in the real world is very difficult.  But could we not simplify it by taking a worst case:
- ignore atmospheric friction
- assume the break acts as a perfect hinge until the moment the top section is horizontal, and then snaps cleanly
- the lower section is entirely rigid
I would suggest that to the extent that any of the above do not hold, it would reduce the likelihood of hitting the obstruction, not increase it.  And therefore to solve the simplified problem is enough to be sure that the obstruction is safe from being hit.

After that, it becomes a Newtonian dynamics problem which can definitely be solved numerically, if not algebraically.  What I think we'll find is that whether it hits depends not on the height of the obstruction but rather on the force of gravity.  It might hit on the moon, but not on Earth.

The one exception to my argument is the possibility of wind blowing in the wrong direction of course.  A sufficiently strong wind will always lead the pole to hit the obstruction irrespective of its height.  And I guess there is a correlation between the chances of the pole breaking and wind occurring...
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Expert Comment

ID: 36708370
The question as I understood it isn't the pole hitting the obstruction... it's the obstruction hitting the arc traced out by the pole.  Unless that part of the question is changed, the two can not intersect.  Since the only object that changes is the obstruction, and it isn't specified that the obstruction is perpendicular to the ground, I suppose it is possible for the obstruction to intersect the arc if it does LEAN into it.
At that point, for the minimum angle of the lean... the obstruction would have to seventeen and a quarter foot tall to just touch the arc where the falling rod were parallel to the ground.  *shrug*
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Expert Comment

ID: 36708493
so maybe the question is, on hitting the ground and the hinge breaking, is there enough momentum for the wood to upright itself, then topple into the building?

I guess the common sense answer is that it looks like there is, based on angular momentum that would be gained in the fall.

The height of the building still seems irrelevant though, since the closest point is at height zero from the ground
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Expert Comment

ID: 36708881
> At that point, for the minimum angle of the lean (17.26' - in green on the diagram)

And the minimum possible height for the obstruction to be able to touch the arc SHOULD be the point where the intersection is perpendicular to a tangent, and would be 6.8'... but the obstruction is leaning WAY over at that point.  (red in the diagram.)
Drawing1a.JPG
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Author Comment

ID: 36709222
Razmus:
And the minimum possible height for the obstruction to be able to touch the arc SHOULD be the point where the intersection is perpendicular to a tangent, and would be 6.8'.
------------------
The obstruction is actually a wire, horizontal to the ground, and 12 feet high.
I knew the distance from the base of the pole, to a point directly under the wire, was 36 feet. I was not sure about the arc of the falling section, as it starts to arc, and it's top falls toward, then near, and then past the wire, as it heads to the ground.

Once the top hits the ground there is no factor that could cause it to bounce as high as 12 feet. It would be the risk of hitting the wire, as the arc passes near the wire.
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ID: 36709225
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ID: 36709274
I see.
If (the broken part of the pole remains at least slightly attached to the bottom part of the pole) AND (the bottom part of the pole does not deflect in the wind) the 33' section of the pole can not bridge the 36' distance and touch the wire.
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Expert Comment

ID: 36710264
so maybe the question is, on hitting the ground and the hinge breaking, is there enough momentum for the wood to upright itself, then topple into the building?

Certainly not. The angular momentum will not be anywhere near that high. The pole is hitting the ground at roughly a 30 degree angle to the ground. You would have a perpetual motion machine if it could stand itself back up from there.

The only way it could possibly hit the object is if the hinge breaks when the horizontal velocity of the center of mass is high enough to carry it the remaining 3 feet (which intuitively seems possible).

Someone get the physics book out and see what the maximum horizontal velocity of the center of mass would be while the pole is rotating. I'm at work and don't have time to do it.
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Author Comment

ID: 36710985
What is the maximum horizontal velocity of the center of mass would be while the pole is rotating?

That is an interesting question.

It seems like the chance of movement is away from the wire, as the top falls within it's arc. So, the person on the cutting end, needs to beware of a backward thrust.
------------------------------------------------------------

Previous discussions on another thread are non conclusive on whether there should be a hinge or not. Two articles on "how," state it differently.
There could be:
1. A hinge that does not break at all
2. One that breaks when the top hits the ground.
3. One that break before the top hits the ground.
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Expert Comment

ID: 36711095
Take a pencil (or a 36 inch dowel).
Hold it vertically on a mark on the table, and then let it fall over.

Sometimes the bottom kicks back (but only a little).
Some times it skid forward (toward the phone line) after it lands.
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Assisted Solution

TommySzalapski earned 120 total points
ID: 36711194
1. A hinge that does not break at all
Excellent. Misses the object easily.

2. One that breaks when the top hits the ground.
Very good. The part that was connected at the hinge will now fall on the other side and land about 5 feet on the right side of the pole.

3. One that break before the top hits the ground.
This could give backward thrust or carry the piece farther toward the pole.

Sometimes the bottom kicks back (but only a little).
Some times it skid forward (toward the phone line) after it lands.

This is because you are dropping it on a flat surface similar to if you cut the pole at the very base.
Set a short box on the table and drop the pencil off the edge of the box or just balance the pencil on the edge of the table and let it fall. You will see a bit of backward thrust just before the pencil drops below the table and you will also see some horizontal velocity.
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ID: 36711198
"toward the pole" should read "toward the object"
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ID: 36712267
On this one:
One (hinge) that breaks when the top hits the ground.

It was concluded that the top hitting the ground, at such an angle, would cause a spring like action and force the butt, backwards, an unknown distance, toward the guy with the saw, the ladder, the house, the windows, the fence, the gas meter.

And is was concluded that if the hinge broke before the top hit the ground, the weight of the butt end would cause it to catch up to the falling top, and hopefully the whole thing falls parallel to the ground, which is the safest result.

But, the above two conclusions could be wrong.
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ID: 36712491
30 degrees is a shallow enough angle that I would fully expect the top of the pole to stay where it was when it landed. Then the butt would fall in a similarly shaped arc as before but its 'hinge' would be where the top has landed.

the weight of the butt end would cause it to catch up to the falling top
Things fall at the same rate regardless of weight. Since the piece would be rotating before the hinge broke, it would keep rotating after it broke, but the whole thing would be falling. So it would appear that the top fell faster than the butt (although in reality, the system is falling as a whole and rotating within itself).

Anyway, the angle that it lands at would depend on when the hinge broke. What you can guarantee is that the butt will not angle up more than 30 degrees above the top so it will land more horizontal than if the hinge didn't break.

Or if the hinge breaks very early in the process, the top will still be above the butt. The butt will land and the top will finish landing soon after.
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Assisted Solution

paultomasi earned 65 total points
ID: 36713239
To ensure the tree falls away from the house, cut away as much as you can reach off the side of the tree nearest the house. This will 'lighten' one side of the tree (the tree will fall towards the heavier side).

Have 2 helpers using guide ropes to gently tug at the tree spaced apart forming a vee such that the tree will fall between them. The lover part of the tree is heavier than the upper part so the rope need only be fixed 1/3rd up from the cut (28 feet).

Cut a 60 degree vee 2/3rds in 17' up on the heavy side. Cut downward 30 degrees on the light side so that the bottom of the 'line-of-cut' is in line with a point 3 inches higher than the vee crotch of the previous vee cut.

There is no need to cut all the way to the vee. At ths point, both helpers should gently tug the ropes. When you notice the tree toppling, let gravity do the rest.

tree.JPG
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TommySzalapski earned 120 total points
ID: 36717909
Okay, I have some time. The only concern is if the hinge breaks mid fall.

Using 17 feet up as our 0 of evergy (yes, this allows negative but if the hinge breaks after the 33 foot piece is below the horizonal it will move to the right, farther from the object).

So the initial kinetic energy of the pole is 0 and the initial potential energy of the pole (assuming mass is evenly distributed which may not be the case) is

maxE = mgL/2
(m is mass, g is acceleration due to gravity and L is 33 ft = 10.06 meters)

The formula for potential and kinetic energy at any point during the fall is:
pE = mgLcos(a)/2
kE = (Iw^2)/2

Also pE + kE = maxE at all times (conservation of energy)

a is the angle between the falling piece and the vertical, w is angular velocity, and I is moment of inertia (mL^2)/2

So kE = (mL^2*w^2)/2

So our whole equation is
kE + pE = maxE or

Someone check my math and then after my next class I'll see if I can maximize the horizontal component of the velocity of the center of mass unless someone beats me to it.

By the way, the 'back thrust' is just the fact that it's still rotating after the hinge breaks.
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Expert Comment

ID: 36717920
The one thing I'm not sure on yet is how the angular velocity will change when it breaks. Some of the motion will become vertical and horizonal and some will still be rotational. (and of course all the acceleration will be vertical)
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Author Comment

ID: 36717985
paultomasi:
To ensure the tree falls away from the house, cut away as much as you can reach off the side of the tree nearest the house. This will 'lighten' one side of the tree (the tree will fall towards the heavier side).
--------------------------------------------------------------------------
The issue is the tree is leaning slightly to the "northwest" direction, when the 90 degree landing area is "south to southwest to west."
Northwest would cause it to fall over the corner of the house.
The two remaining limbs, one directly over the house, and one on the fall side, are both high up. The one over the house is the heaviest. I'm trying to figure out how to get it cut. Not only is the tree leaning toward the house, that is also the heaviest side.

I began cutting off all limbs from the ground up. I was not thinking which side would be heaviest or not. I've about reached my limit. The limb over the house is about 12 feet above my head when I am standing on the roof and I can not use a ladder. I'm planning to try to get a rope over the limb, make a slip knot, throw the other end over the limb that is on the fall side, and tie it off. Then I can use my rope chain saw to cut the limb so it won't fall so fast and so heavy.

TommySzalap:
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Expert Comment

ID: 36718755
The limb is not starting out vertical then, so my formula would need some adjusting. Also, the center of mass is probably closer to the base of the limb than the tip. So the real life max horizontal velocity would come out lower than my formula would which makes it even less likely to hit the house.

If you anchor the base of the limb with a rope or something, then we are back into the cases where the hinge doesn't break (the rope being the new hinge) so again, we won't hit the house.
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Author Comment

ID: 36719023
TommySzalap:

The tree makes a fork about 26 feet up.
In my question about geometry, I wanted the solution to assume that the tree is 100% vertical, and the two limbs on each fork are the same weight.

There will not be a rope to anchor anything.

A rope may be used to cut the existing limb on the north side of the tree, so that the center of gravity is closer to the center of the tree.

The word rope should not affect your formulas above as the rope is not being used to anchor the 33 foot piece to be cut.
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ID: 36719761
I'm still stuck on that one point. The rest of the equation is relatively easy. I just can't figure out how much of the kinetic energy goes into rotation and how much into displacement.
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Author Comment

ID: 36719989
Hold the phone. The guy came over and cut the small tree down. It was 24 feet tall. he made his first cut, parallel to the ground. Then he cut down, at some angle, maybe 45 degrees, to make the wedge. Then he went to the back and began a cut downward. The hinge did not break until the tree "rotated" and a pallet helped buffer the house, as the tree missed the edge of the roof by maybe 6 inches. It broke the pallet in half but the pallet was just resting there and was not put in place for any reason.

With the topping of the large tree being from the roof top, and the unexpected rotation on the smaller one, the plan is the cable company will come tomorrow and lower the cable line, so the guy can cut the tree by standing on the ground. A rope will be put over the fork in the tree, and two of us will pull in the direction of the fall.

Here are photos of the tree and the safety measures taken for the two windows and the gas line.

treeeeee-001.JPG
treeeeee-002.JPG
treeeeee-003.JPG
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ID: 36720049
The smaller tree only had maybe 10 feet exposed above the rooftop.
So, the fact that the 24 foot (already topped) tree, did rotate on the hinge and came many, many degrees closer to the house, than expected, the large one will be cut from the ground.

The small tree was clearly leaning "southwest" and it fell closer to "west" than "southwest" and just did miss the roof.

He just came over and said let's do it.
His chain saw is only a 14 inch, it seems. He says he has cut dozens of trees, but if those were on his vast amount of vacant land, well.......

Why did it rotate? I was not close enough to see how level his back side cut was as compared the the wedge. If he was cutting at a different level, when it started to fall, and since his backside cut was not exactly as the wedge, then it rotated due to the weight. Then he realized the larger one, needs to be done the same way, but with two guys on a rope, etc.
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Expert Comment

ID: 36815532
err, we cannot be responsible for loss due to our answers, they are provided for academic learning purposes only, not real world tree surgery :)

good luck!
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Author Comment

ID: 36816220

Any neighbors that thought the large one, was going to provide any "excitement" (roof, ambulance, police, etc.) were disappointed.
It came down fine.
stumps-001.JPG
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ID: 36816292
Blimey! I didn't realise it was THAT close!! No wonder you were apprehensive!
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Author Closing Comment

ID: 36816570
The answer was no, it would not hit the wire-pole.
Razmus got that with his photos, etc.

Thanks to TommySzalapski for formulas that reminded me why I did not major in math.
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Author Comment

ID: 36816592
Location near house was the reason the guy wanted to cut the large tree from the roof (top it).
But, how to cut the wedge on the fall side?
All that added up to a ladder, man standing at edge of roof, butt of the top could rotate, etc.
So, it was decided to let the cable co. lower the wire, no charge, and cut it at ground level. I counted at least 100 green black walnuts after the cut.
I've picked up over 250, the last couple months after wind, rain, etc. They hit my roof and got in the neighbor's grass = undesired trees.
I gave the wood to someone who wants to make furniture with it.
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