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adatesman


Jul 19, 2010, 12:31 PM
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InDaDacks


Jul 19, 2010, 12:49 PM
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Re: [adatesman] Hey Majid!!! [In reply to]
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Ohh man, I just sprained a brain muscle


shimanilami


Jul 19, 2010, 1:05 PM
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Re: [adatesman] Hey Majid!!! [In reply to]
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ANSWER: Jay should cut the rope.


Carnage


Jul 19, 2010, 1:13 PM
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Re: [shimanilami] Hey Majid!!! [In reply to]
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shimanilami wrote:
ANSWER: Jay should cut the rope.

he should cut the rope, but in a way that he can then use the core strands to make a long enough rope to safely lower everyone to the ground.


clews


Jul 19, 2010, 1:41 PM
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Re: [adatesman] Hey Majid!!! [In reply to]
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who climbs with a 30m rope?


adatesman


Jul 19, 2010, 1:43 PM
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moose_droppings


Jul 19, 2010, 1:47 PM
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Re: [adatesman] Hey Majid!!! [In reply to]
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Jay weighs about 50lb's more than Paul, how the hell is Paul (even frictionless) going to pull Jay up that incline?


adatesman


Jul 19, 2010, 1:55 PM
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clews


Jul 19, 2010, 2:01 PM
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Re: [moose_droppings] Hey Majid!!! [In reply to]
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at a 40% incline much of jay's downward force is reduced

edited due to a brain fart


(This post was edited by clews on Jul 19, 2010, 2:24 PM)


moose_droppings


Jul 19, 2010, 2:24 PM
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Re: [clews] Hey Majid!!! [In reply to]
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clews wrote:
at a 40% incline, less than half of jay's mass is pulling downward.

Most of it will be sent horizontally

I kind of figured it had something to do with the frictionless incline, but they must also figure there is no friction in the bend of the rope going over the edge.

Way *over my head*, much like everything.


clews


Jul 19, 2010, 2:25 PM
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Re: [moose_droppings] Hey Majid!!! [In reply to]
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moose_droppings wrote:
clews wrote:
at a 40% incline, less than half of jay's mass is pulling downward.

Most of it will be sent horizontally

I kind of figured it had something to do with the frictionless incline, but they must also figure there is no friction in the bend of the rope going over the edge.

Way *over my head*, much like everything.


ahhh!! you quoted my stupidity before I could edit it!!


moose_droppings


Jul 19, 2010, 2:25 PM
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Re: [adatesman] Hey Majid!!! [In reply to]
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adatesman wrote:
I'll give you a hint... This was in the chapter dealing with vector components of forces.

There was a book???
Laugh

edit: Paul should lose his ice axe, crampons, all his gear and clothing to slow down his decent rate


(This post was edited by moose_droppings on Jul 19, 2010, 2:29 PM)


majid_sabet


Jul 19, 2010, 3:39 PM
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Re: [adatesman] Hey Majid!!! [In reply to]
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let me figure out the f2-f1 first.


(This post was edited by majid_sabet on Jul 19, 2010, 3:58 PM)


cruxstacean


Jul 19, 2010, 3:41 PM
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Re: [adatesman] Hey Majid!!! [In reply to]
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if the slope is so frictionless how did they get up their to begin with???


bill413


Jul 20, 2010, 5:43 AM
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Re: [cruxstacean] Hey Majid!!! [In reply to]
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cruxstacean wrote:
if the slope is so frictionless how did they get up their to begin with???

Using the ice axe which was then dropped.


bustloose


Jul 20, 2010, 7:25 AM
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they are both clearly noobs, with no business being out in the wild on their own, and deserve whatever fate befalls (heh) them. oh, and they should definitely have to pay for the rescue.


jrathfon


Jul 20, 2010, 8:11 AM
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Re: [adatesman] Hey Majid!!! [In reply to]
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aiight:

a) T = 976 N, and Paul decks at 6.47 m/s

b) assuming Paul chills before he cuts the rope and Jay is at rest, Jay craters at 20.7 m/s


jrathfon


Jul 20, 2010, 8:14 AM
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Re: [jrathfon] Hey Majid!!! [In reply to]
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jrathfon wrote:
aiight:

a) T = 976 N, and Paul decks at 6.47 m/s

b) assuming Paul chills before he cuts the rope and Jay is at rest, Jay craters at 20.7 m/s

so:

a) do i win a cookie

b) that's 14.5 and 46.3 mph respectively for all you 'mericans. aka Jay's getting screwed, why would Paul cut the rope? unless he's got some beef.

c) this all assumes they are in a vacuum.


donald949


Jul 20, 2010, 8:35 AM
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Re: [shimanilami] Hey Majid!!! [In reply to]
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shimanilami wrote:
ANSWER: Jay should cut the rope.
Exactly. Although he could save time by just letting go, as I noticed his belay technique is pretty bad.
And Paul has it coming, as clearly it is his plan to drop Jay later by untieing.
Paul you are a bastard, dropping Jay after he held you. I will never climb with you. Mad


cruxstacean


Jul 20, 2010, 8:50 AM
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Re: [bill413] Hey Majid!!! [In reply to]
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bill413 wrote:
cruxstacean wrote:
if the slope is so frictionless how did they get up their to begin with???

Using the ice axe which was then dropped.

So ice axes can hold in frictionless ice? this warrants further testing


majid_sabet


Jul 20, 2010, 9:09 AM
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jrathfon wrote:
jrathfon wrote:
aiight:

a) T = 976 N, and Paul decks at 6.47 m/s

b) assuming Paul chills before he cuts the rope and Jay is at rest, Jay craters at 20.7 m/s

so:

a) do i win a cookie

b) that's 14.5 and 46.3 mph respectively for all you 'mericans. aka Jay's getting screwed, why would Paul cut the rope? unless he's got some beef.

c) this all assumes they are in a vacuum.

not so fast

how did you get these #s?


adatesman


Jul 20, 2010, 9:16 AM
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bill413


Jul 20, 2010, 9:30 AM
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Re: [cruxstacean] Hey Majid!!! [In reply to]
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cruxstacean wrote:
bill413 wrote:
cruxstacean wrote:
if the slope is so frictionless how did they get up their to begin with???

Using the ice axe which was then dropped.

So ice axes can hold in frictionless ice? this warrants further testing

Yes. If they penetrate the surface of the ice and create a hole at an acute angle (or even perpendicular to) the surface, they'll hold by mechanical interlock, not by friction.


jrathfon


Jul 20, 2010, 11:02 AM
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Re: [adatesman] Hey Majid!!! [In reply to]
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adatesman wrote:
majid_sabet wrote:
jrathfon wrote:
jrathfon wrote:
aiight:

a) T = 976 N, and Paul decks at 6.47 m/s

b) assuming Paul chills before he cuts the rope and Jay is at rest, Jay craters at 20.7 m/s

so:

a) do i win a cookie

b) that's 14.5 and 46.3 mph respectively for all you 'mericans. aka Jay's getting screwed, why would Paul cut the rope? unless he's got some beef.

c) this all assumes they are in a vacuum.

not so fast

how did you get these #s?

Must have had a copy of the Teacher's edition. The Student edition only has answers to the odd-numbered questions in the back. Oh, the book is Physics for Scientists and Engineers, 3rd Edition by Tipler.

Um, no, I have no such thing. I just happen to be an engineer and know how to use sine... and the equations of motion.

p.s. Majid, I'm not going to let you cheat.

but, find the vector of force pulling on the rope from jay's weight, add that to paul's force (hint F=ma), there's your tension. subtract those two values to get the force paul is accelerating toward the ground with. calculate paul's acceleration and plug it into the EOM. voila time it takes paul to crate, thus you can back calculate his final velocity.

now, calculate the length of the slope, knowing it's 25 m tall at a 40deg incline (with sine). subtract 5 m from it, since there is still 5 m of rope left when paul crater's, now that's how far jay will slide. change that back into the vertical he will fall (again sine), since it's a frictionless slope. he will fall at 1G, again since it's a frictionless slope. now you know jay's cratering acceleration, EOM's, voila, dun.


jrathfon


Jul 20, 2010, 11:04 AM
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bill413 wrote:
cruxstacean wrote:
bill413 wrote:
cruxstacean wrote:
if the slope is so frictionless how did they get up their to begin with???

Using the ice axe which was then dropped.

So ice axes can hold in frictionless ice? this warrants further testing

Yes. If they penetrate the surface of the ice and create a hole at an acute angle (or even perpendicular to) the surface, they'll hold by mechanical interlock, not by friction.

mmmmmmm, you said interlock. is that the same as interlocken, that town in austria?

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