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desertwanderer81
Jul 20, 2010, 10:00 PM
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CamCardon wrote: And what would that be? If there is no friction, then masses dont matter, they will both accelarate at 9.81m/s^2...
I'm sorry, but jrathfon is correct in this situation.
Just because the friction is not there, doesn't mean that Paul's fall is not being arrested by the force of gravity on Jay.
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desertwanderer81
Jul 20, 2010, 10:03 PM
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CamCardon wrote: adatesman wrote: Remember when you'd post sets of climbing related problems for people to calculate the answer to? Well, I was flipping through one of my old physics textbooks last night and happened to notice this: Obviously it needs more Red and Green Arrows, but not bad otherwise!  -a. Oooh, this looks like a fun one. a) The tension in the rope can be figured by adding the opposing forces on the rope, in this case, Paul and Jay. Pauls force on the rope is his full 52kg, but because Jay is on a slope, his will only be 47.565kg (74*sin40), Giving us 99.565kg of tension in the rope.
Ignoring the second part because it has already been addressed, you are making a very basic but common mistake. The tension on the rope is due to the force of gravity on Jay which resists his movement along the slope and the force which is required to accelerate Jay.
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CamCardon
Jul 20, 2010, 10:12 PM
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desertwanderer81 wrote: CamCardon wrote: desertwanderer81 wrote: CamCardon wrote: jrathfon wrote: CamCardon wrote: And what would that be? If there is no friction, then masses dont matter, they will both accelarate at 9.81m/s^2... um, jay has an opposing force.... so yes mass does matter. however as jay falls once the rope is cut, you are correct, his mass does not matter due to the frictionless surface. for bonus points we should consider that a person has a drag coefficient of Cd ~ 1.3 GO!! I realize he has an opposing force, but since Pauls is larger, and there is no friction, they will both be taken Pauls direction with the accelaration of gravity. And ive done enough math today to deal with adding in more crap into this problem... :p Except that you have to look at the vector components of Paul's weight. The force along the slope is less than Jay's weight. There are no components of Paul's weight, he is falling straight down. And the opposing component of Jay's weight is still less than Paul's total weight. ooops! I flipped their names around. Sorry. In reply to: Except that you have to look at the vector components of Jay's weight. The force along the slope is less than Paul's weight. I'm not sure I'm gathering at what you were getting at then..... You're stating "masses don't matter since there is no friction" The mass of each person does matter :p
Often in physics, the mass does matter. If we were factoring in friction, it would. But this is the same concept of which falls faster, a single brick or a pallet of bricks? Gravity has the same effect on all objects, ie making them fall at 9.81m/s^2, if there are no interfering forces.
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NoSoup4U
Jul 20, 2010, 10:15 PM
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Maybe the *engineers* here should stick to climbing...
Paul will deck at 3.72 m/s (that's 12.2 ft/s) assuming g=9.81m/s^2
As far as Jay is concerned, if you assume that the rope is cut instantaneously when Paul hits the ground (that is, when Jay also travels with an absolute velocity of 3.72 m/s on the icy incline), then Jay will deck at 21.0 m/s (that's 68.9 ft/s).
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desertwanderer81
Jul 20, 2010, 10:23 PM
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CamCardon wrote: desertwanderer81 wrote: CamCardon wrote: desertwanderer81 wrote: CamCardon wrote: jrathfon wrote: CamCardon wrote: And what would that be? If there is no friction, then masses dont matter, they will both accelarate at 9.81m/s^2... um, jay has an opposing force.... so yes mass does matter. however as jay falls once the rope is cut, you are correct, his mass does not matter due to the frictionless surface. for bonus points we should consider that a person has a drag coefficient of Cd ~ 1.3 GO!! I realize he has an opposing force, but since Pauls is larger, and there is no friction, they will both be taken Pauls direction with the accelaration of gravity. And ive done enough math today to deal with adding in more crap into this problem... :p Except that you have to look at the vector components of Paul's weight. The force along the slope is less than Jay's weight. There are no components of Paul's weight, he is falling straight down. And the opposing component of Jay's weight is still less than Paul's total weight. ooops! I flipped their names around. Sorry. In reply to: Except that you have to look at the vector components of Jay's weight. The force along the slope is less than Paul's weight. I'm not sure I'm gathering at what you were getting at then..... You're stating "masses don't matter since there is no friction" The mass of each person does matter :p Often in physics, the mass does matter. If we were factoring in friction, it would. But this is the same concept of which falls faster, a single brick or a pallet of bricks? Gravity has the same effect on all objects, ie making them fall at 9.81m/s^2, if there are no interfering forces.
YES! And there is an interfering force, namely a rope tied to the guy :p
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desertwanderer81
Jul 20, 2010, 10:31 PM
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NoSoup4U wrote: Maybe the *engineers* here should stick to climbing... Paul will deck at 3.72 m/s (that's 12.2 ft/s) assuming g=9.81m/s^2 As far as Jay is concerned, if you assume that the rope is cut instantaneously when Paul hits the ground (that is, when Jay also travels with an absolute velocity of 3.72 m/s on the icy incline), then Jay will deck at 21.0 m/s (that's 68.9 ft/s).
We have the same answer, except that I assumed that Ho for Paul was 25m as opposed to 20m since the problem states that he is starting from rest. It is impossible with our frictionless environment for him to start 5m down. Although the diagram does seem to present that as the situation. I'd go either way.
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desertwanderer81
Jul 20, 2010, 10:37 PM
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jrathfon wrote: 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.
You're an engineer? Did you ever even bother to take Dynamics, let alone Physics I? :p
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davidnn5
Jul 20, 2010, 11:04 PM
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Quick! Calculators at five paces!
In other words, I feel nerdy just READING this stuff (okay, skimming it while trying to keep my eyes open).
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majid_sabet
Jul 21, 2010, 12:08 AM
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chadnsc wrote: Notice how Midget didn't answer the question but instead is asking one of his own where he controls the parameters and wording of said question?
if you take a PR course (public relation) and IA (information officer) training, you will learn the skills necessary to take the question, rotate it, turn it around, move it for hours and then stick it back to where it started without people even noticing it and yet, you did not provide nothing.
that is an art by itself and you learn it when you learn politics.
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bill413
Jul 21, 2010, 12:28 AM
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kriso9tails wrote: donald949 wrote: 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.  If you look closely, the rope seems to be affixed to something inside of Jay's pants. The foot belay is sounding pretty reasonable right about now.
The foot belay sounds much more PG than "affixed to something inside of Jay's pants." The X belay?
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ddooddodo
Jul 21, 2010, 12:53 AM
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they are unsafe
first off-jay probably shouldn't have dropped his ice axe
no.2 you shouldn't climb frictionless surfaces incase you fall
no. 3 the belayer should have been anchored
no. 4 paul should be more careful
no. 5 the where climbing way to close to the end of the rope there should
have been extra rope trailing the belayer
lastly-the two people suck and shouldn't have been climbing together nor climbing at all
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chadnsc
Jul 21, 2010, 1:14 PM
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majid_sabet wrote: chadnsc wrote: Notice how Midget didn't answer the question but instead is asking one of his own where he controls the parameters and wording of said question? if you take a PR course (public relation) and IA (information officer) training, you will learn the skills necessary to take the question, rotate it, turn it around, move it for hours and then stick it back to where it started without people even noticing it and yet, you did not provide nothing. that is an art by itself and you learn it when you learn politics.
I assume you didn't take either classes.
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ajkclay
Jul 21, 2010, 2:01 PM
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Which "Jay" are we talking about?
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steple
Jul 21, 2010, 3:43 PM
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alpha = 40°
m_J = 74 kg
m_P = 52 kg
g = 9.806 m/s^2
While both are tied in, the force that pulls Paul downwards is
F_P = ( m_P - m_J * sin(alpha) ) * g
= 43.5 N .
a_P = F_P / m_P = ( 1 - m_J / m_P * sin(alpha) ) * g
= 0.836 m/s^2 .
v_P(t) = t * a_P
= t * 0.836 m/s^2 .
h_P(t) = h_P0 - 0.5 * t^2 * a_P
= 20 m - t^2 * 0.418 m/s^2 .
Paul grounds when h_P(t_G) = 0 m. The time of grounding t_G therefore is
t_G = sqrt( h_P0 * 2 / a_P )
= 6.92 s.
v_P(t_G) = t_G * a_P
= 5.78 m/s .
At the moment of Paul's grounding, Jay is sliding up the hill with v_J(t_G) = v_P(t_G) = 5.78 m/s.
Now I assume that the rope is being cut while there is no tension.
The force acting on Jay upwards in direction of the slope is
F_J = - m_J * g * sin(alpha)
= -466 N .
a_J = - g * sin(alpha)
= -6.30 m/s^2 .
v_J(t) = v_P(t_G) - g * sin(alpha) * ( t - t_G )
For simplicity, let us introduce a new time T = t - t_G. So Jay's speed will be
v_J(T) = v_P(t_G) - g * sin(alpha) * T.
s_J(T) = ( v_P(t_G) - 0.5 * g * sin(alpha) * T ) * T + s_J0
s_J0 = 1 / sin(alpha) * 25 m - 5 m .
T_G = ( v_P(t_G) + sqrt( 2 * g * sin(alpha) * s_J0 + v_P(t_G)^2 ) )/( g * sin(alpha) )
= 4.32 s .
v_J(T_G) = v_P(t_G) - g * sin(alpha) * T_G
= -21.5 m/s .
21.5 m/s * sin(alpha) = 13.8 m/s (edited)
21.5 m/s * cos(alpha) = 16.4 m/s . (edited)
Not only is Jay's impact velocity orthogonal to the ground higher, he also rolls horizontally at 16.4 m/s.
Bonus question: To minimize the impact velocity for both, at what time should they cut the rope?
(This post was edited by steple on Jul 21, 2010, 5:15 PM)
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bill413
Jul 21, 2010, 4:18 PM
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steple wrote: Bonus question: To minimize the impact velocity for both, at what time should they cut the rope?
When the moon is directly overhead, thus decreasing the acceleration due to earth's gravity.
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majid_sabet
Jul 21, 2010, 4:36 PM
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chadnsc wrote: majid_sabet wrote: chadnsc wrote: Notice how Midget didn't answer the question but instead is asking one of his own where he controls the parameters and wording of said question? if you take a PR course (public relation) and IA (information officer) training, you will learn the skills necessary to take the question, rotate it, turn it around, move it for hours and then stick it back to where it started without people even noticing it and yet, you did not provide nothing. that is an art by itself and you learn it when you learn politics. I assume you didn't take either classes.
if you read 8 years of MS postings in RC, you may find the answer
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jrathfon
Jul 21, 2010, 4:38 PM
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desertwanderer81 wrote: jrathfon wrote: 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. You're an engineer? Did you ever even bother to take Dynamics, let alone Physics I? :p
alright, i was wrong about the tension, but here you go:
http://www.sparknotes.com/...apter8section3.rhtml
my 20.7 and youse guy's 21.0 m/s for jay is just cause i used 9.8 instead of 9.81, sorry, i should have used my sig figs correct and it would have been 21.
yes, engineer. more of the chemical type (polymer science). i took phys 1 and 2 um.... 13 years ago, sorry i'm rusting on my free-body diagrams. did you take ochem 1 and 2, pchem 1 and 2, trans metal chem, inorganic chem? my penis is bigger than yours.
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jrathfon
Jul 21, 2010, 4:45 PM
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so would jay slide over the top of the ramp? that's be great, paul cuts the rope and jay flies over the hill and splats right on top of him.
actually no, we'd have to plot his trajectory, and take into account his body positioning so we can have an accurate take on his drag coefficient.
(This post was edited by jrathfon on Jul 21, 2010, 4:45 PM)
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LostinMaine
Jul 21, 2010, 4:47 PM
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jrathfon wrote: yes, engineer. more of the chemical type (polymer science). i took phys 1 and 2 um.... 13 years ago, sorry i'm rusting on my free-body diagrams. did you take ochem 1 and 2, pchem 1 and 2, trans metal chem, inorganic chem? my penis is bigger than yours.
must have been chemically enhanced.
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desertwanderer81
Jul 21, 2010, 4:54 PM
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jrathfon wrote: desertwanderer81 wrote: jrathfon wrote: 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. You're an engineer? Did you ever even bother to take Dynamics, let alone Physics I? :p alright, i was wrong about the tension, but here you go: http://www.sparknotes.com/...apter8section3.rhtml my 20.7 and youse guy's 21.0 m/s for jay is just cause i used 9.8 instead of 9.81, sorry, i should have used my sig figs correct and it would have been 21. yes, engineer. more of the chemical type (polymer science). i took phys 1 and 2 um.... 13 years ago, sorry i'm rusting on my free-body diagrams. did you take ochem 1 and 2, pchem 1 and 2, trans metal chem, inorganic chem? my penis is bigger than yours.
Ha, I didn't even solve for Jay's speed because it was stupidly easy to solve.
However, you attempted to solve Paul's speed fundamentally wrong. Actually LOOK at how I solved his velocity. Not only that, but there is additional tension from Jay's acceleration.
I have not taken those classes, however I am not pretending to understand how to solve those problems. Clearly you lack the fundamental tools to solve these problems. Oh, and they barely touch mechanics in Physics 2 so I don't really see the relevance of that :p
But hey, what do I know? I only tutor people regularly in mechanics type courses and have been an engineer who actually designs physical things for many years :p
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steple
Jul 21, 2010, 5:01 PM
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jrathfon wrote: so would jay slide over the top of the ramp? that's be great, paul cuts the rope and jay flies over the hill and splats right on top of him. actually no, we'd have to plot his trajectory, and take into account his body positioning so we can have an accurate take on his drag coefficient.
I was just thinking about that. Didn't consider that earlier. Let me check ...
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jrathfon
Jul 21, 2010, 5:01 PM
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desertwanderer81 wrote: jrathfon wrote: desertwanderer81 wrote: jrathfon wrote: 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. You're an engineer? Did you ever even bother to take Dynamics, let alone Physics I? :p alright, i was wrong about the tension, but here you go: http://www.sparknotes.com/...apter8section3.rhtml my 20.7 and youse guy's 21.0 m/s for jay is just cause i used 9.8 instead of 9.81, sorry, i should have used my sig figs correct and it would have been 21. yes, engineer. more of the chemical type (polymer science). i took phys 1 and 2 um.... 13 years ago, sorry i'm rusting on my free-body diagrams. did you take ochem 1 and 2, pchem 1 and 2, trans metal chem, inorganic chem? my penis is bigger than yours. Ha, I didn't even solve for Jay's speed because it was stupidly easy to solve. However, you attempted to solve Paul's speed fundamentally wrong. Actually LOOK at how I solved his velocity. Not only that, but there is additional tension from Jay's acceleration. I have not taken those classes, however I am not pretending to understand how to solve those problems. Clearly you lack the fundamental tools to solve these problems. Oh, and they barely touch mechanics in Physics 2 so I don't really see the relevance of that :p But hey, what do I know? I only tutor people regularly in mechanics type courses and have been an engineer who actually designs physical things for many years :p
salt in the wounds
and i admitted to being wrong. i'm sorry we can't all be mechanical engineers having daily practice with it. i have books and the interwebs to look up shit i understood 13 years ago.
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adatesman
Jul 21, 2010, 5:10 PM
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steple
Jul 21, 2010, 5:11 PM
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No, he doesn't make it to the top. He turns around after
T_turnaround = v_P(t_G) / g / sin(alpha)
s_J(T_turnaround) = ( v_P(t_G) - 0.5 * g * sin(alpha) * T_turnaround ) * T_turnaround + s_J0
= 36.5 m ,
l_slope = 25 / sin(alpha) = 38.9 m .
(This post was edited by steple on Jul 21, 2010, 5:17 PM)
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desertwanderer81
Jul 21, 2010, 5:19 PM
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ha, tit for tat ;)
It's more of the fact that you adding caveats and making excuses. Us engineers tend to believe we can solve any problem because we have such a broad background. The fact remains that while you are certainly very capable of understanding such problems, you lack the more in depth knowledge of the subject. Which is perfectly fine by the way! For instance I had to write a program in VB to compile and process a rather large amount of data. My program did what it needed to, however I am sure that if an actual programmer looked at it, he would have first smacked me upside the head for using VB then again when he actual saw my sloppy, terrible code!
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