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Occulis
RealPoor Jedi
Joined: 11 Oct 2002
Posts: 13293
Location: Moral Relativity Central
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 Posted: 10/30/03 - 13:24 Post subject: Okay you fat geeks, here's a problem!
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how much of a single-pulse force will i need to lift the wood to an upright position?
Edit!
Alternatively, answer this problem:
How much force would you have to strike a 200lb brick with to shoot it 6 feet in the air?
Last edited by Occulis on 10/30/03 - 13:50; edited 2 times in total
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Tura
RealPoor Guru
Joined: 29 Oct 2003
Posts: 4866
Location: Raleigh, NC
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Posted: 10/30/03 - 13:25 Post subject:
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I only made it to algebra2 <---spell?
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Occulis
RealPoor Jedi
Joined: 11 Oct 2002
Posts: 13293
Location: Moral Relativity Central
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Posted: 10/30/03 - 13:25 Post subject:
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Cmon 1 of you star-trek convention-going shitballz has to know the answer to this.
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Occulis
RealPoor Jedi
Joined: 11 Oct 2002
Posts: 13293
Location: Moral Relativity Central
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Posted: 10/30/03 - 13:26 Post subject:
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| Tura wrote: | | I only made it to algebra2 <---spell? |
Priceless
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WheresNWS
RealPoor Master of Posts
Joined: 19 Nov 2002
Posts: 6448
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Posted: 10/30/03 - 13:26 Post subject:
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Do we neglect friction?
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Tura
RealPoor Guru
Joined: 29 Oct 2003
Posts: 4866
Location: Raleigh, NC
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Posted: 10/30/03 - 13:27 Post subject:
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Hey, it got me through high school ;D
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Renaij
Luke Warm
Joined: 11 Oct 2002
Posts: 251
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Posted: 10/30/03 - 13:27 Post subject:
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put yourself under it so that your wee wee is under it, and then think of hot men, i mean women.
it shall work! a mans penis rising in an erection generates 12.39 kilajules or energy!
the problem is teh solved
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Fabulez
Luke Warm
Joined: 11 Nov 2002
Posts: 437
Location: up in here
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Posted: 10/30/03 - 13:29 Post subject: Re: Okay you fat geeks, here's a problem!
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| Dunn wrote: | how much of a single-pulse force will i need to lift the wood to an upright position?
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A lot.
But not too much! Just enough.
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Occulis
RealPoor Jedi
Joined: 11 Oct 2002
Posts: 13293
Location: Moral Relativity Central
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Posted: 10/30/03 - 13:33 Post subject: Re: Okay you fat geeks, here's a problem!
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| Fabulez wrote: | A lot.
But not too much! Just enough. |
finally, someone who thinks exactly like me!
so uh... you forgot all your physics education too, huh!
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atarom
Dalai Lama of RealPoor
Joined: 11 Oct 2002
Posts: 16398
Location: 375th st. Y
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Posted: 10/30/03 - 13:34 Post subject:
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we need to know the distance between the end and the pivot point, dont we?
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scoresia
RealPoor Sensei
Joined: 10 Mar 2003
Posts: 1500
Location: Florida
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Posted: 10/30/03 - 13:35 Post subject:
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Depends on what the darn thins weighs, what the dimensions are, Also, are you allowed to use anything else for counter balance
too many unknowns
I suppose
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Zapper
RealPoor Sensei
Joined: 11 Oct 2002
Posts: 1512
Location: Connecticut
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Posted: 10/30/03 - 13:36 Post subject:
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1 megawatt per hour per square foot ...times ummmm..... 1 joule per linear...... f**k it........
must goatse..............
nah
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Occulis
RealPoor Jedi
Joined: 11 Oct 2002
Posts: 13293
Location: Moral Relativity Central
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Posted: 10/30/03 - 13:39 Post subject:
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| scoresia wrote: | Depends on what the darn thins weighs, what the dimensions are, Also, are you allowed to use anything else for counter balance
too many unknowns
I suppose |
In the initial post the whole problem is given:
200lbs
6 feet long
Height doesn't really matter, but let's say 1 foot thick
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Occulis
RealPoor Jedi
Joined: 11 Oct 2002
Posts: 13293
Location: Moral Relativity Central
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Posted: 10/30/03 - 13:40 Post subject:
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| atarom wrote: | | we need to know the distance between the end and the pivot point, dont we? |
Let's say between 0 and 1 inches!
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scoresia
RealPoor Sensei
Joined: 10 Mar 2003
Posts: 1500
Location: Florida
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Posted: 10/30/03 - 13:41 Post subject:
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Hmm...quick calculation ... 100.6lbs of pressure(force) (lb ft per second) WHATEVER LEAST I TRIED!
Edited (thx Fabulez)
Last edited by scoresia on 10/30/03 - 13:44; edited 2 times in total
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atarom
Dalai Lama of RealPoor
Joined: 11 Oct 2002
Posts: 16398
Location: 375th st. Y
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Posted: 10/30/03 - 13:41 Post subject:
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er i was looking at it wrong, we need to know at what point the impact is hitting the board, wouldnt we?
if you were closer to the pivot point it would take more force than if you lifted from the very end.... am i wrong?
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Zapper
RealPoor Sensei
Joined: 11 Oct 2002
Posts: 1512
Location: Connecticut
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Posted: 10/30/03 - 13:43 Post subject:
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Hmmm I found Dunns tech support number... Can I call and ask for hepl?
My mouose dosen't work. I think it suffocated in the plastic bag it came in and it's dead. They should have put holes in the bag!!!!!
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Krumble
Toomuchtimeonhands
Joined: 11 Oct 2002
Posts: 771
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Posted: 10/30/03 - 13:44 Post subject:
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287,000 Newtons
No, not really.
All I know is, if you consider the thickness of the board in all places it's a really, very, extremely, incredibly hard problem.
Atarom, I would guess it's 3 inches from the bottom of the board. Since that's all that would be needed to touch the ground when standing upright.
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Occulis
RealPoor Jedi
Joined: 11 Oct 2002
Posts: 13293
Location: Moral Relativity Central
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Posted: 10/30/03 - 13:44 Post subject:
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| atarom wrote: | er i was looking at it wrong, we need to know at what point the impact is hitting the board, wouldnt we?
if you were closer to the pivot point it would take more force than if you lifted from the very end.... am i wrong? |
d00d i drew the arrow to it!
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atarom
Dalai Lama of RealPoor
Joined: 11 Oct 2002
Posts: 16398
Location: 375th st. Y
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Posted: 10/30/03 - 13:47 Post subject:
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yeah but dont we need something a ltitle more precise than a s****y photoshop drawing? like a number?
yeah krumble i figured that out, when i first saw the problem i thought we were pushing down on the short side, like a lever. then i saw the big IMPACT POINT sign =/
i just f****n woke up.
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Occulis
RealPoor Jedi
Joined: 11 Oct 2002
Posts: 13293
Location: Moral Relativity Central
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Posted: 10/30/03 - 13:49 Post subject:
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| atarom wrote: | yeah but dont we need something a ltitle more precise than a s****y photoshop drawing? like a number?
yeah krumble i figured that out, when i first saw the problem i thought we were pushing down on the short side, like a lever. then i saw the big IMPACT POINT sign =/
i just f****n woke up. |
lozlze
It's for an approximation, not a space shuttle launch!
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FealornTM
Luke Warm
Joined: 15 Nov 2002
Posts: 375
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Posted: 10/30/03 - 13:55 Post subject:
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Is this on earth?
Looks like a statics/dynamics 101 problem really
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Guest
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Posted: 10/30/03 - 13:58 Post subject: Re: Okay you fat geeks, here's a problem!
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| Dunn wrote: | how much of a single-pulse force will i need to lift the wood to an upright position?
Edit!
Alternatively, answer this problem:
How much force would you have to strike a 200lb brick with to shoot it 6 feet in the air? |
The answer is.
One 85 dollar a day single-pulse force mexican.
He could stand that slab up and lay it down, all day long.
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Tura
RealPoor Guru
Joined: 29 Oct 2003
Posts: 4866
Location: Raleigh, NC
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Posted: 10/30/03 - 14:06 Post subject:
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| Quote: | The answer is.
One 85 dollar a day single-pulse force mexican.
He could stand that slab up and lay it down, all day long. |
(teh win)[/quote]
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Occulis
RealPoor Jedi
Joined: 11 Oct 2002
Posts: 13293
Location: Moral Relativity Central
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Posted: 10/30/03 - 14:07 Post subject:
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| FealornTM wrote: | Is this on earth?
Looks like a statics/dynamics 101 problem really |
well man i can call it a "remedial physics" problem or even a "consumer ethics" problem but that doesn't put a number and a unit on the screen, either!
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FealornTM
Luke Warm
Joined: 15 Nov 2002
Posts: 375
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Posted: 10/30/03 - 14:10 Post subject:
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Well, I could probably answer this, however:
You need to define your distance from point of rotation to the point at which you're applying force. Unless you want the answer in the form of
Force times X with X being the distance.
I'm not fat.
I'm not a geek, just paid like one.
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atarom
Dalai Lama of RealPoor
Joined: 11 Oct 2002
Posts: 16398
Location: 375th st. Y
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Posted: 10/30/03 - 14:14 Post subject:
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| FealornTM wrote: | Well, I could probably answer this, however:
You need to define your distance from point of rotation to the point at which you're applying force. Unless you want the answer in the form of
Force times X with X being the distance.
I'm not fat.
I'm not a geek, just paid like one. |
this is what i was bringing up.
not that i ever took physics, or even remember how to solve this problem...
but logic insisted that i point it out.
nice avatar feal
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Krumble
Toomuchtimeonhands
Joined: 11 Oct 2002
Posts: 771
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Posted: 10/30/03 - 14:14 Post subject: Re: Okay you fat geeks, here's a problem!
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| Dunn wrote: | Alternatively, answer this problem:
How much force would you have to strike a 200lb brick with to shoot it 6 feet in the air? |
Working on it. I can tell you that it would take 1.224 seconds to get there if you hit it with the absolute minimum force. Haven't taken Engr. Physics yet so this next part should be fun.
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Guest
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Posted: 10/30/03 - 14:15 Post subject:
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| FealornTM wrote: | Well, I could probably answer this, however:
You need to define your distance from point of rotation to the point at which you're applying force. Unless you want the answer in the form of
Force times X with X being the distance.
I'm not fat.
I'm not a geek, just paid like one. |
The Reagan thingie is wrong.
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WheresNWS
RealPoor Master of Posts
Joined: 19 Nov 2002
Posts: 6448
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Posted: 10/30/03 - 14:18 Post subject:
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| FealornTM wrote: | Well, I could probably answer this, however:
You need to define your distance from point of rotation to the point at which you're applying force. Unless you want the answer in the form of
Force times X with X being the distance.
I'm not fat.
I'm not a geek, just paid like one. |
It's not that simple. I could have answered this a couple years ago, but I've forgotten a bunch of math. I think you have to calculate the entire amount of energy needed by integrating the necessary vertical force over the rotation of the block. The component force that contributes to its rotation is the vector perpendicular to the block, which you get with ol' Pythagoras. M / X (I think; M=mass, X=distance from pivot) will give you the total moment force which, as I said, needs to be integrated across the total distance rotation while using the perpendicular vector only. Like I said, I forgot all this stuff and don't have a physics book, anymore.
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