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Is friction a constant or a fraction of travelling speed? (1 Viewer)

zenger69

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I have an arguement over a foolish maths matter. Over whether friction will actually stop something. Let teacher be Mr P

I said to him that friction is constant and
ACTUAL SPEED = APPLIED SPEED - FRICTION so when a person takes their foot off the accelerator, the car wil lose speed and eventually reach 0. This occurs both mathematically and in real life.

But Mr P reckons it's a fraction. He's example is that "A car is travelling along a road and travels at x kilometres/second. Friction causes it to lose 1/4 of it's speed each second. Mr P says it will approach zero and never get to zero (the frog leaping thing or the limiting sum). Therefore mathematically the car would be moving forever. " but he can't explain why this doesn't happen in real life.

Therefore I'm arguing that the vehicle loses speed due to friction is constant if the cars are on the same surface, whereas Mr P is arguing the speed lost to friction is a fraction of it's speed and therefore friction depends on the speed the vehicle is travelling at.

To me it sounded unlogically and a bit foolish, because no one inlcuding Mr P can defy the laws of physics. But then my knowledge of physics is very basic.

So could someone judge who is right with evidence?
 
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physician

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nick1048 said:
rofl the good old frog in a well theory... I think your explanation is more legit, however I'm no physician.
I think you mean 'physicist'


zenger69 said:
friction depends on the speed the vehicle is travelling at

well if ur looking at it in terms of a vehicle... the friction on a fast car is diffrent to that on a slow car... u need to keep in mind the friction between the rubber tyres and the ground depends on the speed of the vehicle... speeding up results in redced friction... but if us low down friction will eventually lead u to a halt... therefore u need to accelerate or remian at a constant speed so that the force (acceleraton) offers an opposing force to the friction between the road and the tyre...

look i'm proably not making any sense here... but couldn't u just get ur physics teacher to solve the dispute?
 

rama_v

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physician said:
I think you mean 'physicist'

speeding up results in redced friction...
This is true only for the tyres. But there are more air particles smashing into te front of teh car creating more friction (i.e. wind resistance) at higher speeds, I think
 
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Jumbo Cactuar

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zenger69 said:
I have an arguement over a foolish (or i think is foolish) maths teacher. Over whether friction will actually stop something.

I said to him (the fool) that friction is constant and
ACTUAL SPEED = APPLIED SPEED - FRICTION so when a person takes their foot off the accelerator, the car wil lose speed and reach 0. This occurs both mathematically and in real life.

But the fool reckons it's a fraction. He's example is that "A car is travelling along a road and travels at x kilometres/second. Friction causes it to lose 1/4 of it's speed each second. The fool says it will approach zero and never get to zero. Therefore mathematically the car would be moving forever. " but he can't explain why this doesn't happen in real life.

Therefore I'm arguing that the vehicle loses speed due to friction is constant, whereas the fool is arguing the speed lost to friction is a fraction of it's speed and therefore friction depends on the speed the vehicle is travelling at.

Can someone judge who is right with evidence?
So here's how friction works;

Friction is proportional to weight (in our example) such that;

Initial condition of movement;
F = F(thrust) - F(fric) = F(thrust) - mu(s) * W > 0

Friction during movement;
F(fric) = mu(d) * W

where mu(s) >= mu(d).
and 0 <= mu(s or d) <=1.

mu is a function of the two surfaces involved. (hence constant for two surfaces)

Frictionless surface interactions: mu = 0

There was a question in 2003 4u Maths HSC about friction proportional to speed, which I scoffed at when I read.

In the above example, say a person took their foot of the accelerator:
F = F(thrust) - F(fric) = 0 - mu(d) * M * g
a = -mu(d) * g
v = - 9.8 * mu(d) *t + v(0)
when v=0, t = v(0)/(9.8 * mu(d))
which has a solution for mu(d) >< 0 and all v(0).

QED your teacher is fool. :cool:

excuse any mistakes, I am a bit rusty
 

zenger69

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alrite thanks, maybe i should print out these links and see.

plus he taught the four unit course, so it's a bit scary.

Otherwise he is actually a really good teacher. Thanks guys.

Also I gotta edit fool outta each post not from me.
 

zenger69

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Also anyone else able to explain it in plain english is more than welcome.

Maybe we need to step away from maths.
 

Jumbo Cactuar

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Friction is a deceleration which is constant, with the exception, i think, of aerodynamic friction.

However, friction force can't cause an object to then accelerate the other way.


If you don't understand say specifically what has you stumped.
 

Budz

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Friction is constant proportional to mass.
Speed does not affect the amount of friction, only increases the forward momentum.
 

Budz

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If u apply or take ur foot of the accelarator u
accelerate constantly
not exponetially
 

Xayma

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Jumbo Cactuar said:
QED your teacher is fool. :cool:

excuse any mistakes, I am a bit rusty
Yes Kinetic energy may be a constant (its value given by μsN (Car tyres experience static friction and rotate) however you must also consider (if you want to get really into it) drag which is given by 1/2*CDSρv2 where:

CD is the coefficent of Drag, for a Monaro this is a value of about 0.31

S is the refrence area (in this case I think the top area of the car but not 100% sure (CD will change depending on the refrence area used))

ρ is the density of the medium (Air has a density of 1.225kg*m-3 in standard conditions).

and

v is the velocity.

So because it is determined by velocity, we can conclude that due to the frictional force due to the ground while moving being a constant that there is a quadratic relationship to the total frictional forces. Where there will always be a frictional force (which would stop the car moving). However zenger if it was losing 1/4 of its speed it wouldnt matter it would come to a point where the loss is negligble and unnoticeable.
 

M-turkey

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What did your teacher mean when they said that an object would never stop?

You can observe things coming to rest, and it is because of friction.


There is a change of friction between an object and something it is in contact with (i.i the ground) however, between an object at rest and when it is moving. Friction is higher when an object is at rest and more force is required to create initial movement. Once an obeject is moving, it's friction with the ground is reduced. Less force is needed to maintain an objects velocity.
 

DAAVE

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Everything in classical mechanics comes back to the good old 2nd law of newton:

F = m*a

More generally, it can be said that the sum of the forces acting on a body will result in an accelleration.

Now on a car moving along the ground you have several forces acting - weight, friction, drag. Now the car obviously isn't accellerating downwards so we can conclude that the forces in the vertical direction must equal zero, so we must have an upwards force that exactly equals the weight, which we call the normal force (ie, W + n = 0, or W = -n)

Now lets neglect air drag and say the car is in a vacuum. Experiment shows that the cars decelleration will be proportional to the normal force, and more generally that (little f for friction here):

f = constant * n

this constant we denote mu subscript s (s for static friction), so we have:

f = mu_s*n

Now adding all the forces on the car along the horizontal direction we have (since we neglected drag, and the car isn't using its engine):

F = mu_s*n = m*a

since n = (-mg):

mu_s*(-mg) = m*a

a = -g*mu_s

so the accelleration is not related to the velocity of the vehicle and will always be constant.

You could always just prove him wrong, take any object, give it a push along the floor and watch friction do its thing (actually its a little more complicated than that for a rolling object, the above is basically correct for a body without wheels moving along the ground, the conclusion is correct though.)
 
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gordo

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friction is a second order system and alot more difficult to understand than you think

go do some googling :)
 

Jumbo Cactuar

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DAAVE said:
Now the car obviously isn't accellerating downwards so we can conclude that the forces in the vertical direction must equal zero.
Actually, that is an assumption of your model.
 

DAAVE

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Jumbo Cactuar said:
Actually, that is an assumption of your model.
OK, mr pedantic:

Assume the car is on a flat, level surface.

Happy?
 

Jumbo Cactuar

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DAAVE said:
OK, mr pedantic:

Assume the car is on a flat, level surface.

Happy?
Inhomogeneities in the road material means subtle variations elastic modulus of the material. Different elastic modulus means more of less give for your load.

As it gives more or less the car is accelerated up or down very quickly; just not for very far .. in the realm of micrometers :rolleyes:

How that sir is pedantic!! :)
 

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