| Topic: |
Science > Physics |
| User: |
"Dylan Buffum" |
| Date: |
03 Apr 2006 05:56:26 PM |
| Object: |
Basic arithmetic |
I'm a rock climber just visiting over from <i>frictionlist</i>.
Perhaps you can answer a simple, high-school physics question for me:
How can a 60kg mass falling at 9.8m/ss generate in excess of 12-24 kn?
Here's the context:
Rock climbing ropes are rated to 12 kilonewtons. A great deal of the
other gear is rated upwards of 20 kn. Sometimes this gear does fail.
But to my mind, ten years removed from any education in physics,
someone of my body size only generates just over a 1/2 kn. F=ma F=60
x 9.8 = 588n. The distance that I fall shouldn't make a difference,
right, since accelleration has already accounted for that.
So where am I wrong?
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| User: "Llanzlan Klazmon" |
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| Title: Re: Basic arithmetic |
03 Apr 2006 07:23:55 PM |
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"Dylan Buffum" <DJCBUFFUM@HOTMAIL.COM> wrote in
news:1144104986.230873.241140@i39g2000cwa.googlegroups.com:
I'm a rock climber just visiting over from <i>frictionlist</i>.
Perhaps you can answer a simple, high-school physics question for me:
How can a 60kg mass falling at 9.8m/ss generate in excess of 12-24 kn?
How quickly is the 60kg mass brought to rest by the rope? That is the
accelleration you have to worry about, not the accelleration due to
gravity. That of course depends on the type of rope. A bungy stops you with
much less accelleration than say a nylon rope.
Klazmon
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| User: "Hexenmeister" |
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| Title: Re: Basic arithmetic |
04 Apr 2006 01:55:46 PM |
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"Llanzlan Klazmon" <Klazmon@llurdiaxorb.govt> wrote in message
news:Xns979B7E2023F44Klazmonllurdiaxorbgo@203.97.37.6...
| "Dylan Buffum" <DJCBUFFUM@HOTMAIL.COM> wrote in
| news:1144104986.230873.241140@i39g2000cwa.googlegroups.com:
|
| > I'm a rock climber just visiting over from <i>frictionlist</i>.
| > Perhaps you can answer a simple, high-school physics question for me:
| >
| > How can a 60kg mass falling at 9.8m/ss generate in excess of 12-24 kn?
|
| How quickly is the 60kg mass brought to rest by the rope? That is the
| accelleration you have to worry about, not the accelleration due to
| gravity. That of course depends on the type of rope. A bungy stops you
with
| much less accelleration than say a nylon rope.
|
| Klazmon
|
Perhaps he's a "jerk". :-)
Androcles.
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| User: "Dylan Buffum" |
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| Title: Re: Basic arithmetic |
04 Apr 2006 05:06:58 PM |
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That helps. Here's some more.
Ropes are 60, meters, but you never fall that distance. Say the
climber has ascended 30 meters, and is five meters above the last
anchor point. When he falls, he descends 10 meters before the rope
begins to catch. A particular 9.8mm nylon climbing rope, under a force
of 8.8 kn, has a dynamic elongation of 26.4%.
What kind of force is a 60kg mass generating at the moment the rope
begins to slow the fall, and at the moment the descent is complete? I
guess the result will be a curve, steepening as the rope gets towards
it's maximum elongation, yeah?
Thanks.
PS. frictionlist is a googlegroup of climbers in Wisconsin.
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| User: "Llanzlan Klazmon" |
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| Title: Re: Basic arithmetic |
04 Apr 2006 06:15:49 PM |
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"Dylan Buffum" <DJCBUFFUM@HOTMAIL.COM> wrote in
news:1144188418.157377.170380@u72g2000cwu.googlegroups.com:
That helps. Here's some more.
Ropes are 60, meters, but you never fall that distance. Say the
climber has ascended 30 meters, and is five meters above the last
anchor point. When he falls, he descends 10 meters before the rope
begins to catch. A particular 9.8mm nylon climbing rope, under a force
of 8.8 kn, has a dynamic elongation of 26.4%.
What kind of force is a 60kg mass generating at the moment the rope
begins to slow the fall, and at the moment the descent is complete? I
guess the result will be a curve, steepening as the rope gets towards
it's maximum elongation, yeah?
Have a look at Gregory Hansen's reply.
Klazmon.
Thanks.
PS. frictionlist is a googlegroup of climbers in Wisconsin.
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| User: "Gregory L. Hansen" |
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| Title: Re: Basic arithmetic |
03 Apr 2006 07:38:20 PM |
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In article <1144104986.230873.241140@i39g2000cwa.googlegroups.com>,
Dylan Buffum <DJCBUFFUM@HOTMAIL.COM> wrote:
I'm a rock climber just visiting over from <i>frictionlist</i>.
Perhaps you can answer a simple, high-school physics question for me:
How can a 60kg mass falling at 9.8m/ss generate in excess of 12-24 kn?
Here's the context:
Rock climbing ropes are rated to 12 kilonewtons. A great deal of the
other gear is rated upwards of 20 kn. Sometimes this gear does fail.
But to my mind, ten years removed from any education in physics,
someone of my body size only generates just over a 1/2 kn. F=ma F=60
x 9.8 = 588n. The distance that I fall shouldn't make a difference,
right, since accelleration has already accounted for that.
So where am I wrong?
The force of a falling object at the end of a rope depends on the
elasticity of the rope-- the more it stretches, the gentler the stop will
be and the smaller the maximum force will be. Think of falling onto a
trampoline versus falling onto concrete.
The easiest way to analyze it is to recall the equations of motion for a
simple harmonic oscillator,
x(t) = A sin(wt)
v(t) = -A w cos(wt)
a(t) = -A w^2 sin(wt)
where the angular frequency w=sqrt(k/m) and k is the spring constant, m
the mass of the object. Give the boundary conditions x(0)=0, v(0)=v_0.
Plugging v(0) into the second equation,
v_max = -A w cos(w*0) = -A w
That gives us our coefficient, A=-v_0/w. Plug that into the third
equation,
a(t) = v_0 w sin(wt)
Sine is a maximum at sin(wt)=sin(pi/2)=1, so
a_max = v_0 w
a_max = v_0 sqrt(k/m)
F=ma, so F_max=m*a_max, or
F_max = v_0 sqrt(k*m)
Neglecting air resistance (and you'd hope you don't fall so far that you
need to correct for it!), velocity after falling some height h can be
found by conservation of energy-- kinetic energy plus gravitational
potential energy.
1/2 m v^2 = mgh
v = sqrt(2gh)
Or...
F_max = sqrt(2ghkm)
--
"Out of the way, you swine, a physicist is coming!"
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| User: "" |
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| Title: Re: Basic arithmetic |
04 Apr 2006 12:45:16 AM |
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I'm a rock climber just visiting over from <i>frictionlist</i>.
What part of the wolrd is <i>frictionlist</i> in??????
************************
"Out of the way, you swine, a physicist is coming!"
"Yeah, wooden want you pigs to get a physicist's ***** on you!!!!"
- Don
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| User: "John C. Polasek" |
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| Title: Re: Basic arithmetic |
03 Apr 2006 07:32:37 PM |
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On 3 Apr 2006 15:56:26 -0700, "Dylan Buffum" <DJCBUFFUM@HOTMAIL.COM>
wrote:
I'm a rock climber just visiting over from <i>frictionlist</i>.
Perhaps you can answer a simple, high-school physics question for me:
How can a 60kg mass falling at 9.8m/ss generate in excess of 12-24 kn?
Here's the context:
Rock climbing ropes are rated to 12 kilonewtons. A great deal of the
other gear is rated upwards of 20 kn. Sometimes this gear does fail.
But to my mind, ten years removed from any education in physics,
someone of my body size only generates just over a 1/2 kn. F=ma F=60
x 9.8 = 588n. The distance that I fall shouldn't make a difference,
right, since accelleration has already accounted for that.
So where am I wrong?
It's how fast you stop with the simple impulse formula
F*dT = M*dV
But you have to know the distance to get dV
dV = sqrt(g*H)
Or put it this way; fall 20 units, stop in 2 units and the force is 10
G's. The force is 10 times your weight.
John Polasek
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