Inertia, Weight, and Friction


by 6/4 Blues
Tags: friction, inertia, weight
6/4 Blues
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#1
Oct26-07, 10:35 PM
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I want to become a better driver in the ice and snow. Understanding that a vehicle with more mass will resist change in motion more than one with less mass, what would be the relationship between weight and friction? Will a heavier truck have more friction?(assuming these two vehicles had the same tires) Ultimately, would a smaller vehicle with less inertia be easier to stop (in ice and snow) than a much heavier vehicle with more inertia but greater fiction?
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stewartcs
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Oct27-07, 12:34 AM
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The frictional force between two objects is equal to the coefficient of friction times the normal force. So by increasing the normal force (weight of the vehicle) you will increase the frictional force between the tires and the road.
stewartcs
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#3
Oct27-07, 12:45 AM
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Quote Quote by 6/4 Blues View Post
Ultimately, would a smaller vehicle with less inertia be easier to stop (in ice and snow) than a much heavier vehicle with more inertia but greater fiction?
Seems like it would be a trade off. Not sure of the exact numbers though. I would guess that it would be harder to stop the momentum of the heavier vehicle (i.e. take longer to stop it without sliding) even with greater frictional force, due to the fact that you are on ice and the coefficient of friction, and thus frictional force would be significantly smaller in relation to the time required to stop the smaller vehicle's momentum.

Here's some more information on the subject...

http://www.gi.alaska.edu/ScienceForum/ASF8/843.html

Gelsamel Epsilon
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#4
Oct28-07, 02:09 AM
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Inertia, Weight, and Friction


You get more friction, and thus more resistive force, but since f = ma

and f = xmg (where x is the coefficient and g is the acceleration due to gravity, hence mg is weight force which is also = to normal force)

thus ma = xmg

and a = xmg/m
a = xg
deacceleration (choosing the deacceleration direction as positive) = x (coefficient of friction) * g (acceleration due to gravity)

So in either case the friction provided gives an equal deacceleration in each case assuming the same coefficient (the same surface).

Tell me if I did something wrong.

Edit: If I wasn't clear this shows deceleration due to friction (on the ground) is not dependant on mass, thus if both vehicles have the same of amount of breaking force the one with the higher inertia will stop last.
pixel01
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#5
Oct28-07, 09:37 AM
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Epsilon did it right. The deceleration does't depend on the mass of the vehicle if the friction coeficient is equal.
But I think friction coeficient in this case have some other aspect. Normally when we identify friction coeficient, the surfaces must be smooth and dry and are not deformed. In the case of driving on ice or snow,, the surfaces are different and I may think heavier vehicles decelerate better for they can deform the ice/snow more than smaller ones do.
pixel01
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Oct28-07, 09:38 AM
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Gelsamel Epsilon
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#7
Oct28-07, 10:16 PM
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Quote Quote by pixel01 View Post
Epsilon did it right. The deceleration does't depend on the mass of the vehicle if the friction coeficient is equal.
But I think friction coeficient in this case have some other aspect. Normally when we identify friction coeficient, the surfaces must be smooth and dry and are not deformed. In the case of driving on ice or snow,, the surfaces are different and I may think heavier vehicles decelerate better for they can deform the ice/snow more than smaller ones do.
Actually you're right, ice would probably be close enough to be equal but snow...

Imagine a small toy car on the snow, it's coefficient would be much smaller then a truck trying to plow through the snow. I'm not sure which would stop first. I think the larger one would still stop later...


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