Previous 1
Topic: Here's a science/physics question
songbird274's photo
Thu 01/08/09 10:20 AM
Say you have a level parking lot that is fully covered with ice (you could pretty much ice skate on it). Parked in this parking lot are two cars: one the size of an Escort or Focus, and one an SUV or Minivan. Could a strong enough wind, such as a hurricane wind, make either or both of the cars move AT ALL on the ice? Another way to ask the question: Will either or both of the cars' tires slide on the ice, in any direction?

no photo
Thu 01/08/09 10:23 AM

Say you have a level parking lot that is fully covered with ice (you could pretty much ice skate on it). Parked in this parking lot are two cars: one the size of an Escort or Focus, and one an SUV or Minivan. Could a strong enough wind, such as a hurricane wind, make either or both of the cars move AT ALL on the ice? Another way to ask the question: Will either or both of the cars' tires slide on the ice, in any direction?


I would think the SUV/Minivan would be more apt to move, because of more surface area against the wind. JMO

Jill298's photo
Thu 01/08/09 10:24 AM
Edited by Jill298 on Thu 01/08/09 10:24 AM
I would think strong enough hurricane winds could move either of the cars regardless if there's ice or not.

no photo
Thu 01/08/09 10:27 AM

I would think strong enough hurricane winds could move either of the cars regardless if there's ice or not.
TRUE HURRICANE WINDS CAN DERAIL TRAINS.......

Shagwell's photo
Thu 01/08/09 10:28 AM
it is possible to predict that if couple of things are known to us, the weights of the 2 cars, the wind speed and the friction between the ice and the 2 cars noway

Jill298's photo
Thu 01/08/09 10:31 AM
Is this a theoritical question? Because I'm not seeing how this could actually happen in the first place?

no photo
Thu 01/08/09 10:35 AM
Edited by quiet_2008 on Thu 01/08/09 10:36 AM

Say you have a level parking lot that is fully covered with ice (you could pretty much ice skate on it). Parked in this parking lot are two cars: one the size of an Escort or Focus, and one an SUV or Minivan. Could a strong enough wind, such as a hurricane wind, make either or both of the cars move AT ALL on the ice? Another way to ask the question: Will either or both of the cars' tires slide on the ice, in any direction?


prolly not

the ice will have formed aroiund the tires not under them. The tires will still be resting on pavement

assuming the ice formed under the tires, the weight of the vehicles would still melt the ice under the tires and they still would be resting on pavement

EDIT: oh

a hurricane force wind? that would blow em away even if there was no ice

no photo
Thu 01/08/09 10:35 AM
Edited by Bushidobillyclub on Thu 01/08/09 10:57 AM
Everyone who has posted above has made valid points.

Depending on the situation and forces involved it is possible.

F=MA (Force = Mass X Acceleration),
or for this application A=F/M Acceleration = Force / Mass

But we must account for how friction effects the needed force.


Ff = μ FN

Ff = force due to friction (Newtons)
FN = normal force (Newtons)
μ = Greek letter “mu”, coefficient of friction between two surfaces (no units)
μs is static, μk is kinetic

When the vehicles are at rest the friction is static, when the force applied has overcome the friction and started the vehicle moving it is kinetic friction.


Each material has a different friction coefficient so we would need to measure the friction of the rubber on the ice.

s1owhand's photo
Thu 01/08/09 10:56 AM
i've parked on ice before after an ice storm and
pushed my car off the ice sideways from the curb
to get it back onto the street...

so...depending on the situation either car could
be moved.

drinker

no photo
Thu 01/08/09 10:59 AM
Edited by Bushidobillyclub on Thu 01/08/09 11:00 AM

i've parked on ice before after an ice storm and
pushed my car off the ice sideways from the curb
to get it back onto the street...

so...depending on the situation either car could
be moved.

drinker
What is interesting is we could put a force measurement device against the vehicle and have you push on the device, and stop once the car moves, this would be close enough to make our calculations and determine friction coefficient, but would defeat the purpose in that we would then not need to make the calculation because we could just find out how much wind speed would be needed to achieve the force based on surface area.

Not hip to that the wind speed calculation, id have to look it up lol.

drinker

songbird274's photo
Thu 01/08/09 11:13 AM

Is this a theoritical question? Because I'm not seeing how this could actually happen in the first place?


Yes, it is just a thought. I never tried this. It just came to me one day.

s1owhand's photo
Thu 01/08/09 11:27 AM
http://www.tuninglinx.com/html/aquaplaning.html


no photo
Thu 01/08/09 11:44 AM
Edited by Bushidobillyclub on Thu 01/08/09 12:16 PM
Excellent baseline resource!


Edit: Well questionable resource after looking at it Vs my text book but who knows not like I am any expert on this stuff I have extremely limited experience lol no really.

Ice to Ice friction coefficient is .1 so looks like either the tires are not better on ice, then ice on ice or this is a very generalized coefficient on this website.

Citizen_Joe's photo
Thu 01/08/09 10:01 PM

Say you have a level parking lot that is fully covered with ice (you could pretty much ice skate on it). Parked in this parking lot are two cars: one the size of an Escort or Focus, and one an SUV or Minivan. Could a strong enough wind, such as a hurricane wind, make either or both of the cars move AT ALL on the ice? Another way to ask the question: Will either or both of the cars' tires slide on the ice, in any direction?


How thick is the ice, and were the cars parked before or after it froze over? Without the ice, and without question, the wind could easily move both vehicles a noticable amount, and with the ice, both will move too, but the the likelyhood of being moved from their frozen position decreases.

s1owhand's photo
Sat 01/10/09 05:10 AM

Excellent baseline resource!


Edit: Well questionable resource after looking at it Vs my text book but who knows not like I am any expert on this stuff I have extremely limited experience lol no really.

Ice to Ice friction coefficient is .1 so looks like either the tires are not better on ice, then ice on ice or this is a very generalized coefficient on this website.


I found the reference citation...

Bauer, Horst, et al, Bosch Automotive Handbook 5th ed., Robert Bosch GmbH, 2000

laugh

maybe ice on ice is stickier than rubber on ice...
or vice versa, or maybe it depends on the compound
or flatness or temperature or pressure...etc..etc.

:wink:

no photo
Mon 01/12/09 02:58 PM
Edited by Bushidobillyclub on Mon 01/12/09 02:58 PM


Excellent baseline resource!


Edit: Well questionable resource after looking at it Vs my text book but who knows not like I am any expert on this stuff I have extremely limited experience lol no really.

Ice to Ice friction coefficient is .1 so looks like either the tires are not better on ice, then ice on ice or this is a very generalized coefficient on this website.


I found the reference citation...

Bauer, Horst, et al, Bosch Automotive Handbook 5th ed., Robert Bosch GmbH, 2000

laugh

maybe ice on ice is stickier than rubber on ice...
or vice versa, or maybe it depends on the compound
or flatness or temperature or pressure...etc..etc.

:wink:
I think your right, not only that, but really how good is rubber on ice? I live in Florida and have never driven on ice (no really) So really I guess it could be little better then ice on ice.

If you placed 2K lbs of a car shaped chunk of ice on top of a bigger chunk of flat ice I bet it would be pretty close to the same force required to move that chunk of ice as your car.

Ice kinda sucks for traction!

hansomguy's photo
Wed 01/14/09 03:54 AM
there really is only one way to know for sure, and that's to put the theory into practice and observe.

DeKLiNe0fMaN's photo
Sat 01/24/09 10:44 PM
Edited by DeKLiNe0fMaN on Sat 01/24/09 10:46 PM

DeKLiNe0fMaN's photo
Sat 01/24/09 10:45 PM

Everyone who has posted above has made valid points.

Depending on the situation and forces involved it is possible.

F=MA (Force = Mass X Acceleration),
or for this application A=F/M Acceleration = Force / Mass

But we must account for how friction effects the needed force.


Ff = μ FN

Ff = force due to friction (Newtons)
FN = normal force (Newtons)
μ = Greek letter “mu”, coefficient of friction between two surfaces (no units)
μs is static, μk is kinetic

When the vehicles are at rest the friction is static, when the force applied has overcome the friction and started the vehicle moving it is kinetic friction.


Each material has a different friction coefficient so we would need to measure the friction of the rubber on the ice.
well said my friend , ahhhh math i love it!

quickin0's photo
Sat 01/24/09 10:48 PM
true true

Previous 1