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Let's say there is a bridge nearby the ocean (like Golden Gate bridge in San-Francisco). On the bridge there are two cars: one is moving along the bridge, the other one is at a standstill . A strong wind is blowing from the ocean (with an equal force on every part of the bridge). Which car will be blown away first? The one that is moving or the one that is at a standstill? Why?

2006-07-14 05:33:39 · 11 answers · asked by dolempap 2 in Science & Mathematics Physics

Important detail: the wind is blowing sideways, perpedicular to the side surface of both cars. Each car weighs, let's say, 700 kg

2006-07-14 05:44:01 · update #1

It seems I have to add more to get to the answer:
The two cars are totall the same - the same color, weight, trademark, the year, month and date of manufacture :). The windows are closed in both cars. The bridge is weird -- it has no lips on either side, so the wind is blowing freely on both cars. The speed of the wind is that one, which is enough to blow away a usual car. The air pressure is unknown, but let's say it's just a usual one.

2006-07-14 06:14:44 · update #2

11 answers

i would have to say the moving vehicle would blow away first.
why: well i would say that a moving vehicle becomes lighter because of aerodynamics, the force of wind under and around the vehicle, whereas the non moving vehicle is for the moment a more stable platform.

2006-07-14 05:41:41 · answer #1 · answered by Ranger 3 · 0 1

First off neither car has overcome static friction. Wheels that break static friction are sliding and not every effective. Secondly this would have to be one heck of a wind to blow a car away. The structure of the bridge also plays a role. For instance:

|| Car ||
-----------------
if the car is blocked by a lip on the side of the bridge or railings the wind may be diffracted. what's the surface area of the side of the cars. Are they the same? Are the windows open? What's coefficient of friction between the tires and the road? What's the speed of the wind? What's the air pressure? So many factors, so little information.

2006-07-14 05:55:40 · answer #2 · answered by Nick N 3 · 0 0

Not enough details to form an answer. Which direction is the wind blowing with respect to the direction in which the moving car is moving? What are the weights of each vehicle?

2006-07-14 05:39:03 · answer #3 · answered by Anonymous · 0 0

Ok a lot of physics talk and babble, but on the real world approach. Parked car wobbles and shakes in the wind but on the freeway at 65 the car jumps lanes and really unstable so I have to think the moving vehichle is in trouble......unless these are racecars and can generate significant down force with the forward movement.

2006-07-14 05:58:38 · answer #4 · answered by NVHSChemGuy 2 · 0 0

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2016-11-02 01:34:15 · answer #5 · answered by mcthay 4 · 0 0

the moving car has already overcome the coefficient of static friction, which is generally larger than the coefficient of kinetic friction.

there are two types of friction forces, static and kinetic. to change from static (or standing still) to kinetic (or moving) the force acting on the body must be greater than the force of static friction. To keep the body moving, the force must be greater than the force of kinetic friction.

in other words, the force due to friction on the stationary car is greater, and it takes more energy (force) to move it.

2006-07-14 05:38:19 · answer #6 · answered by jimvalentinojr 6 · 0 0

The moving car because less friction would be created by moving tires than stationary ones.

2006-07-14 05:37:10 · answer #7 · answered by Hunter S. Thompson 3 · 0 0

Somebody mention surface tension of a stopped vehical versus moving one? You need a couple "all else being euqal" statements & a few disregarding these factors statements.

2006-07-15 00:17:01 · answer #8 · answered by djack 5 · 0 0

the one standing still because it is getting hit by the wind the most

2006-07-14 05:37:22 · answer #9 · answered by Anonymous · 0 0

depends upon the direction of the wind.

2006-07-14 05:38:09 · answer #10 · answered by T_N_I_F 2 · 0 0

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