Mr. Jones Posted November 8, 2007 Report Share Posted November 8, 2007 Riddle me this riddler, How long does it take a Mod to lock a fucking stupid thread? There have been far worse things posted. Be paitent newb. Quote Link to comment Share on other sites More sharing options...
Cougar1647545494 Posted November 8, 2007 Author Report Share Posted November 8, 2007 You said force on the brakes, not the energy converted. the force of friction on the brakes slowing the car down cannot exceed the force of static friction between the tires and the road. once the force of friction on the brakes reaches the maximum static friction of the tires on the road, the tires will begin to slide. since the force of static friction is defined as F=KN where K is the coefficient of static friction between the tires and the road and N= the normal force of the road on the car. since the car is not flying into the air the normal force is equal to the weight of the car. if the two cars have the same tires, then the coefficient of friction is the same. therefore if you increase the weight of the car, the maximum force of static friction will increase. since the force of friction on the brakes has to be equal and opposite to the force of friction, the force will be greater on the car with a greater mass. this is assuming that both cars braking systems are strong enough to lock up the wheels and overcome the static friction of the tires on the road. ughhhh kinetics..... Quote Link to comment Share on other sites More sharing options...
dan93z28 Posted November 8, 2007 Report Share Posted November 8, 2007 and if you want to think about it in terms of energy, then consider that kinetic energy represented by: K=(1/2)mv^2 where m= mass and V= velocity. therefore if the velocity is the same between the two cars, then the heavier car will have a greater initial kinetic energy than the lighter car, and thus will have more energy in the form of heat after it has stopped. Quote Link to comment Share on other sites More sharing options...
Draco-REX Posted November 8, 2007 Report Share Posted November 8, 2007 I think I know what he's trying to say.. ASSUMING that both cars are identical except for weight.. ASSUMING they both start from 0mph. ASSUMING they both accellerate at WOT for the exact same amount of time, then the heavier car will be traveling at a much slower rate. Then ASSUMING both cars have unlimited grip. THEN it is concieveable that the brakes will be see the same amount of force needed to stop the cars. In a perfectly closed system, energy in must equal energy out. However, there's FAR too many things that must be assumed in order to arrive at that answer. So: http://i165.photobucket.com/albums/u59/wolfclown1/fail.jpg Quote Link to comment Share on other sites More sharing options...
dan93z28 Posted November 8, 2007 Report Share Posted November 8, 2007 question is WAAAAAAAY too vague Quote Link to comment Share on other sites More sharing options...
Cougar1647545494 Posted November 9, 2007 Author Report Share Posted November 9, 2007 Who comes to a dead stop in club racing? Question is a BIRDS EYE concept. Quote Link to comment Share on other sites More sharing options...
Cougar1647545494 Posted November 9, 2007 Author Report Share Posted November 9, 2007 HP only matters if your foot is on the brake and the gas at the same time. Again, closed system..Ins and outs...racing... HP dictates braking on a road course (closed system) Quote Link to comment Share on other sites More sharing options...
TheHaze Posted November 9, 2007 Report Share Posted November 9, 2007 Riddle me this, riddle me that, who's afraid of the big black bat? http://www.geocities.com/torontodreaming/film/images/riddler_sm.jpg Quote Link to comment Share on other sites More sharing options...
dan93z28 Posted November 9, 2007 Report Share Posted November 9, 2007 ohhhh so your saying like throughout a road race? i thought you meant a single brake test scenario like i described. Quote Link to comment Share on other sites More sharing options...
El Karacho1647545492 Posted November 9, 2007 Report Share Posted November 9, 2007 F=ma, since a isn't stated and I'm assuming it's 0, F is equal. fucking physics Quote Link to comment Share on other sites More sharing options...
fattyona12 Posted November 9, 2007 Report Share Posted November 9, 2007 D, cause they aren't moving? Quote Link to comment Share on other sites More sharing options...
SPLN SUX Posted November 9, 2007 Report Share Posted November 9, 2007 Why does one have to be smart to answer this? The bigger question... how could you answer this wrong? You should stop going to what ever college it is youre attending, and immediatly go to Milford HighSchool in Cincinnati, because i learned that physics as a freshman, almost a decade ago. You Fail. Quote Link to comment Share on other sites More sharing options...
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