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 Author Topic: Ski as example to illustrate the friction force (normal force...etc)  (Read 30447 times) 0 Members and 1 Guest are viewing this topic. Click to toggle author information(expand message area).
Fu-Kwun Hwang
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 « Embed this message on: April 16, 2009, 12:20:31 am »

This applet show normal force/friction force/net force for a ski case.
You can change angle of the slope,friction coeffieient mu, weight.
The Normal force, friction force and net force will be calculated and displayed.
If the Net force is not zero, you can click play (right triangle) to let go.
You can change the scale,too. But All the arrows and text messages will not be displayed.

For the following conditions: mass m (weight=m*g), angle θ. static friction coefficient μ
The normal force N= m*g*cosθ.
The component alone the slope =m*g*sinθ.
The maximum static friction force Frmax=μ*N=μ* m*g *cosθ.
If m*g*sinθ is smaller than μ* m*g *cosθ (μ>tanθ), then the static friction = -*m*g*sinθ
If m*g*sinθ is larger than  μ* m*g *cosθ (μk* m*g *cosθ
(uk is the coefficient of kinetic friction , which is a little smaller than coefficient of static friction μ)

The following are suggestions from received email message:
Quote
1.  The friction mu should have 0.1 as default (more realistic for snow) – it is approximately 0.05 at zero degrees Celsius, 0.2 at minus ten degrees
2.  Default slope inclination: 20 degrees
3.  Show two gravity components, one in the opposite direction of the normal force, one along the slope
4.  Show a speed reading (km/h) when the animation starts
5.  The “Drag me” text is not necessary (can be explained elsewhere)
6.  The ability to translate the apple by means of tags.
The applet has been modified according to the above suggestions.
Right click inside the simulation window and select GUI options ->translate , and select your language from the selection list. Then , translate all the strings, save the result.
Right  click -> GUI options -> language -> your language : to show it.
It should work later on when you run it at your computer.
You will find users/ntnu/fkh/ski_pkg/ski_xx.properties file, where xx the language code.
Please upload property file as attachment so that I can inclide it in the jar file.

Drag force due to air is included. F=(1/2) rho *Cd *Ap *V2, where rho=1.2 kg/m3,
You can change Cd*Ap with slider.
Cd*Ap=0.11 for an upright body, minimum frontal area
Cd*Ap=0.84 for a horizontal body,maximum frontal area
C*A=0.46 for a body in tuck position

Embed a running copy of this simulation

Embed a running copy link(show simulation in a popuped window)
Full screen applet or Problem viewing java?Add http://www.phy.ntnu.edu.tw/ to exception site list
• Please feel free to post your ideas about how to use the simulation for better teaching and learning.
• Post questions to be asked to help students to think, to explore.
• Upload worksheets as attached files to share with more users.
Let's work together. We can help more users understand physics conceptually and enjoy the fun of learning physics!
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Michael Seemann
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 « Embed this message Reply #1 on: April 17, 2009, 07:49:16 pm »

The simulation could be further improved by adding the air resistance, calculated as D=1/2 CpAv2. This means that the skier will only accelerate to a certain degree, when the net force equals D. On a 20 degree slope and an erect body position this would approximately occur at 80 km/h.
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Fu-Kwun Hwang
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 « Embed this message Reply #2 on: April 17, 2009, 09:07:17 pm »

If the air resistance is include, the net force will be F= mg sinθ - μ mg cosθ -(1/2) CpAv2
When the terminal velocity is reached, it requires F=0
So (1/2) CpAv2=mg sinθ - μ mg cosθ =mg (sinθ - μ cosθ )
v2=2 mg (sinθ - μ cosθ )/ (CpA)
For  μ=0.1 and 20 degree (θ=0.349),  (sinθ - μ cosθ )=0.248, air density p=1.2 kg/m3
which give us v2=4.05 * m/(C A)
The drag coefficient C for a skier is between 1.0-1.1 (http://en.wikipedia.org/wiki/Drag_coefficient)
The area is estimated to be A=0.5*1.7*cos(θ)=0.8 then we will have v2=5.06*m
For a skier with mass (80kg) it will give us v=20.1m/s=72.5 km/h.
It is very close to your value 80 km/h.
I will add this drag force to the simulation and update it soon!

C*A=0.11 for an upright body, minimum frontal area
C*A=0.84 for a horizontal body,maximum frontal area
C*A=0.46 for a body in tuck position
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Michael Seemann
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 « Embed this message Reply #3 on: April 18, 2009, 02:01:56 pm »

This looks great! Is it possible to update the net force bar during the animation, to show how this force eventually will vanish due the the air resistance? Perhaps even the air resistance could be illustrated with an arrow object?

vMax seem to be 8.0 km/h (rather than 72.5 km/h) for a 80 kg skier with Cd*Ap=0.84. Is there an error in the speed calculation?
 « Last Edit: April 18, 2009, 02:54:59 pm by Michael Seemann » Logged
Fu-Kwun Hwang
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 « Embed this message Reply #4 on: April 18, 2009, 06:53:38 pm »

I found the bug , change the GUI and updated the simulation!
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Michael Seemann
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 « Embed this message Reply #5 on: April 18, 2009, 07:32:44 pm »

The simulation is now close to perfection. However, the skier is displaced when the slope angle changes....
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Fu-Kwun Hwang
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 « Embed this message Reply #6 on: April 18, 2009, 09:29:38 pm »

Sorry! I did not notice that when I modified the view.
I think it is fixed now. Please check out updated applet.
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Michael Seemann
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 « Embed this message Reply #7 on: April 21, 2009, 01:22:18 pm »

This simulation illustrates the interesting fact that the acceleration down the slope and thereby the terminal velocity depends on the skier's mass. The reason for this is that the net force acting on the skier equals the sin(theta) gravity component minus friction minus air drag:

m*A = m*g*sin(theta) - mu*m*g*cos(theta) - (Cd*Ap*rho*V^2)/2

if we divide by mass, we get

A = g*sin(theta) - mu*g*cos(theta) - (Cd*Ap*rho*V^2)/(2*m)

The mass appears in the denominator of the last term, which explains the mass/speed relation (a heavier skier is faster).
 « Last Edit: April 21, 2009, 01:24:44 pm by Michael Seemann » Logged
Michael Seemann
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 « Embed this message Reply #8 on: May 04, 2009, 07:13:15 pm »

There might be an error in the terminal velocity generated by this simulation. Using the equation of motion we can describe the ulitmate speed of the skier, the speed at which the acceleration equals zero:

Fs - Ff = Fd

where

Fs = W*sin(theta), Ff=mu*Fn=mu*W*cos(theta) and Fd=(Cd*A*p*v^2)/2

hence

sin(theta) - mu*cos(theta)=(Cd*A*p*v^2)/2*W

By using the initial values of the simulation (theta=20, mu=0.1, W=60 and Cd*A*p=0.84), and solving the equation for v, we find that vmax = 19.4 m/s (53.9 km/h) where the simulation predicts 63.8 km/h.
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Fu-Kwun Hwang
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 « Embed this message Reply #9 on: May 04, 2009, 09:06:09 pm »

According to the formula from http://www.math.utah.edu/~eyre/rsbfaq/physics.html
, the air drag force is equal to Cd*Ap*rho*V2/2.
You forgot the air density term which is 1.2

I used g=10 m/s2 in the previous version. It will be changed to 9.8 m/s2

Another smaller difference was due to I added another cosθ in the air drag force:
dV/dt=(NetF=-fnet+airR*cs*V*V)/M;

I am going to remove the cosθ and change it to
dV/dt=(NetF=-fnet+airR*V*V)/M;

From the formula:
sqrt(60*9.8*(sin(c)-cos(c)*0.1)*2/0.84/1.2)*3.6=61.24 km/h

P.S. I am still working on another case, however, there are so many adjustable parameters that I have not found a good way to model the  new case.

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Michael Seemann
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 « Embed this message Reply #10 on: May 06, 2009, 12:37:57 am »

I have now tested the simulation with various initial conditions, and the terminal velocity seem to be in accordance with the equations

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teacher/web course designer
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 « Embed this message Reply #11 on: May 07, 2009, 03:47:21 pm »

It might be good to label the red belly vector "velocity" . I thought it was a force first and was comparing it to the "air drag".

I was also wondering it the air drag was static length, just showing direction. I could see by stopping (which also zooms in) that it wasn't. Since the beginning of the process is quite interesting and hard to see now, it would be interesting to have a zoomed in version that just showed the velocity with the red vector (maybe plus some

O
|
/\  )
markings on the slope
that would sweep by.)
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You cannot always have happiness but you can always give happiness. ..."Mother Teresa(1910-1997, Roman Catholic Missionary, 1979 Nobel Peace Prize)"

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