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Author Topic: Friction force vs external force  (Read 8919 times)
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Fu-Kwun Hwang
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on: June 25, 2013, 02:54:21 pm » posted from:Taipei,T'ai-pei,Taiwan

A block with mass m is put on top of a surface. The friction coefficient between block and the surface is \mu.
The normal force is m g, so the maximum static friction F_\mu\le m g \mu.
It means when the external force is less than m g, the friction force is - m g.
However, if the external is larger than the maximum static friction force then the friction force stay constant - mg \mu.
And the block starts to move.
The following simulation let you play with the above case.
You can drag the spring horizontally to change the external force.

CLick the following image to display the simulation.

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Press the Alt key and the left mouse button to drag the applet off the browser and onto the desktop. This work is licensed under a Creative Commons Attribution 2.5 Taiwan License
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Let's work together. We can help more users understand physics conceptually and enjoy the fun of learning physics!


* springFriction20130.jpg (25.48 KB, 762x290 - viewed 333 times.)
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Fu-Kwun Hwang
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Reply #1 on: June 26, 2013, 08:04:59 am » posted from:,,Satellite Provider

Here is another more complicated version.

Bock A with mass m is put on top of a surface. There is another block (B) with mass m2 on top of block A.
Friction coefficient between block A and surface is \mu, between block A and block B is \mu_2.

The normal force between block A and surface is (m+m_2) g, between block A and block B is m_2 g
The external force is F:

1. F\le (m+m_2)g \mu: both block would not move
  Friction force between block A and surface is -F, between block A and block B is 0.
2. (m+m_2)g\mu<F\le (m+m_2) g (\mu+ \mu_2) : block A and block B move together
   (\frac{F-(m+m_2)g\mu}{m+m2}m_2< m_2 g\mu_2 Force on block B less than maximum static friction force)
 Friction force between block A and surface is (m+m_2)g \mu , acceleration of both blocks is \frac{F-(m+m_2) g\mu}{m+m2}=\frac{F}{m+m_2}-g\mu
 friction force between block A and block B is \frac{m_2}{m+m_2} (F-(m+m_2) g\mu)=\frac{m_2}{m+m_2}F-m_2 g \mu
3. F>(m+m_2) g (\mu+\mu_2): block A and block B move separately.
 Friction force between block A and surface is (m+m_2)g \mu , acceleration of blocks A is \frac{F-(m+m_2) g\mu-m_2 g\mu_2}{m}
 friction force between block A and block B is m_2 g \mu_2, acceleration of block B is g \mu_2

Enjoy the simulation!
CLick the following image to display the simulation.

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
Press the Alt key and the left mouse button to drag the applet off the browser and onto the desktop. This work is licensed under a Creative Commons Attribution 2.5 Taiwan License
  • 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!


* springFriction2013.jpg (7.12 KB, 798x202 - viewed 312 times.)
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RUPESH BABU
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Reply #2 on: July 24, 2013, 08:39:23 am » posted from:Pondicherry,Puducherry,India

please, help me..!
differentiate between
1. Body at rest which has no net force
2. Body moving at constant velocity which has no net force Sad Sad
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Reply #3 on: July 24, 2013, 08:51:52 am » posted from:Taipei,T'ai-pei,Taiwan

A person sit on a car moving with constant velocity.

For observer in the same car: that person is at rest which has no net force.
For observer at rest relative to the ground: that person is moving at constant velocity(the same as the car) which has no force.
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Reply #4 on: July 25, 2013, 09:22:40 am » posted from:Pondicherry,Puducherry,India

 parameters :
velocity = 0 , final velocity = v ;
acceleration = 0 , acc_mid = a [from 0 to v] , final acceleration = 0 ;

concept:
1. Initially ,velocity is zero.
2. Force is applied to reach constant velocity (i.e) to overcome friction and to accelerate , producing net force.
3. After reaching velocity 'v' , net force is zero. (i.e) force is applied only to overcome friction.

My understandings says that, at CONSTANT VELOCITY,
FORCES are there , only NET FORCE is zero.
 If my concept is right then acknowledge me else help me with CORRECT concept...

SIR, many questions are with me. I want to get clarified one by one. Here after ,you can see my posts frequently. Huh -->  Smiley
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Fu-Kwun Hwang
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Reply #5 on: July 25, 2013, 06:24:21 pm » posted from:,,Satellite Provider

Yes. Net force is zero when an object move with constant velocity.
The external is required if there are friction force or other kind of force resist the motion of the object.
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Reply #6 on: July 26, 2013, 09:44:55 pm » posted from:Pondicherry,Puducherry,India

What is moment of inertia ?
- polar moment of inertia ?
Please explain the term briefly with its importance in dynamics with real time examples.
please brief brief..... Smiley
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Fu-Kwun Hwang
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Reply #7 on: July 27, 2013, 05:48:46 pm » posted from:,,Satellite Provider

Please check out wikipedia first.

You are welcomed to post in more detail  what you do not understand and we will try to help.
But you should learn to check out information by yourself.
 
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RUPESH BABU
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Reply #8 on: September 12, 2013, 09:08:36 am » posted from:Pondicherry,Puducherry,India

"Moment of  Inertia or (Polar I) is the tendency to resist the rotational motion".
this is the fact I grabbed...
But I need some  insight meaning related statics as well as dynamics...(ie. about its role in these areas).
PLease be brief.! that will help me a lot Smiley
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Youe can not help men permanently by doing for them what they could and should do for themselves. ..."Abraham Lincoln(1809-1865, US President 1861-1865"
 
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