Cougar1647545494 Posted November 8, 2007 Report Share Posted November 8, 2007 LOL got a few fellow M.E.s` with this one: 2 cars, one weighs roughly twice the other, same average bhp. Force on the brakes is: A)More on the heavier car B)Less on the lighter car C)More on the lighter car D)Equal Quote Link to comment Share on other sites More sharing options...
thorne Posted November 8, 2007 Report Share Posted November 8, 2007 I pose question. Brake Pedal? Brake Pads ? Brake System ? Quote Link to comment Share on other sites More sharing options...
smokin5s Posted November 8, 2007 Report Share Posted November 8, 2007 what kind of cars are they? and what type of brake system are they running... my guess would be equal.... but I'd really need to get more info to know for sure. Quote Link to comment Share on other sites More sharing options...
Cougar1647545494 Posted November 8, 2007 Author Report Share Posted November 8, 2007 I pose question. Brake Pedal? Brake Pads ? Brake System ? Generalized system on a track setting Quote Link to comment Share on other sites More sharing options...
Cougar1647545494 Posted November 8, 2007 Author Report Share Posted November 8, 2007 what kind of cars are they? and what type of brake system are they running... my guess would be equal.... but I'd really need to get more info to know for sure. same make/model/etc its a big picture question, both cars run the same system (same pad,rotor,etc) Quote Link to comment Share on other sites More sharing options...
Art Posted November 8, 2007 Report Share Posted November 8, 2007 D. Quote Link to comment Share on other sites More sharing options...
Mallard Posted November 8, 2007 Report Share Posted November 8, 2007 Uhh, A and B are the same answer. I'm not sure what you mean by "the force on the brakes," but if you're talking about the amount of energy the brakes will have to dissipate, then the heavier car will be harder on the brakes. You're trying to slow down a larger inertia, so you will have to dissipate more energy, the brakes are going to get hotter, fade quicker, the car won't stop as fast, etc. Anyone that's towed anything will know what I'm talking about. But if you're talking about the opposing forces the caliper sees when it clamps the rotor, then I would answer D. Quote Link to comment Share on other sites More sharing options...
Cougar1647545494 Posted November 8, 2007 Author Report Share Posted November 8, 2007 Uhh, A and B are the same answer. I'm not sure what you mean by "the force on the brakes," but if you're talking about the amount of energy the brakes will have to dissipate, then the heavier car will be harder on the brakes. You're trying to slow down a larger inertia, so you will have to dissipate more energy, the brakes are going to get hotter, fade quicker, the car won't stop as fast, etc. Anyone that's towed anything will know what I'm talking about. But if you're talking about the opposing forces the caliper sees when it clamps the rotor, then I would answer D. speaking in Kinetics... Quote Link to comment Share on other sites More sharing options...
hpfiend Posted November 8, 2007 Report Share Posted November 8, 2007 Force is equal assuming same mastercylinder/caliper diameters and same applied pedal pressure. Stopping distance is a different thing. Quote Link to comment Share on other sites More sharing options...
dan93z28 Posted November 8, 2007 Report Share Posted November 8, 2007 do they have the same type and size of tire? if so then the answer would be A. this is because the maximum force of braking has to be equal to the maximum static friction of the tire on the road. since static friction is equal to the coefficient of friction times the normal force of the system(in this case normal force is equal to weight) the heavier car would have more friction, and thus would have more breaking force. however braking force does not equate to stopping distance or rate of deceleration. Quote Link to comment Share on other sites More sharing options...
ansonivan Posted November 8, 2007 Report Share Posted November 8, 2007 The force is equal, you never stated that the brakes were being used. Quote Link to comment Share on other sites More sharing options...
copperhead Posted November 8, 2007 Report Share Posted November 8, 2007 I'd like to point out that horsepower has nothing to do with the question, I don't know why it's there. Beyond that, I would agree with Mallard. Quote Link to comment Share on other sites More sharing options...
SpaceGhost Posted November 8, 2007 Report Share Posted November 8, 2007 My answer is "I'm stupid" so this question is irrelevant... Quote Link to comment Share on other sites More sharing options...
Draco-REX Posted November 8, 2007 Report Share Posted November 8, 2007 Are we trying to stop in the same distance, or just come to a stop regardless of distance? Quote Link to comment Share on other sites More sharing options...
excell Posted November 8, 2007 Report Share Posted November 8, 2007 You fucking failed with this thread. Quote Link to comment Share on other sites More sharing options...
Cougar1647545494 Posted November 8, 2007 Author Report Share Posted November 8, 2007 I'd like to point out that horsepower has nothing to do with the question, I don't know why it's there. Beyond that, I would agree with Mallard. Horespower has everything to do with it... the car is a closed system on the track, and the energy going in is provided by the engine, and the energy going out is friction(brakes, tires, aero drag etc.). lighter car may have more aero drag due to higher speeds, and a bit more cooling since they have more total pressure to work with, but its basically braking requirements being dictated by engine HP Quote Link to comment Share on other sites More sharing options...
Cougar1647545494 Posted November 8, 2007 Author Report Share Posted November 8, 2007 The force is equal, you never stated that the brakes were being used. They are! Quote Link to comment Share on other sites More sharing options...
Cougar1647545494 Posted November 8, 2007 Author Report Share Posted November 8, 2007 do they have the same type and size of tire? if so then the answer would be A. this is because the maximum force of braking has to be equal to the maximum static friction of the tire on the road. since static friction is equal to the coefficient of friction times the normal force of the system(in this case normal force is equal to weight) the heavier car would have more friction, and thus would have more breaking force. however braking force does not equate to stopping distance or rate of deceleration. brakes are just converting the cars kinetic energy into heat energy. where is the heavy car getting extra energy to be harder on brakes? Quote Link to comment Share on other sites More sharing options...
Mr. Jones Posted November 8, 2007 Report Share Posted November 8, 2007 You have left out an infinite amount of details, and have failed miserably in this thread. Quote Link to comment Share on other sites More sharing options...
Kevin R. Posted November 8, 2007 Report Share Posted November 8, 2007 Yeah. This thread is retarded. Nice effort at sounding smart though. Quote Link to comment Share on other sites More sharing options...
turbomark Posted November 8, 2007 Report Share Posted November 8, 2007 depends on how hard you push the pedal Quote Link to comment Share on other sites More sharing options...
black00ws6 Posted November 8, 2007 Report Share Posted November 8, 2007 The heavier one has more force... IF We have the speed of each car weight the distance they have to stop So if car A weighs 4,000 lbs, going 60mph and has 200 feet to stop in car B weighs 2,000, the only difference. Quote Link to comment Share on other sites More sharing options...
Mensan Posted November 8, 2007 Report Share Posted November 8, 2007 HP only matters if your foot is on the brake and the gas at the same time. Quote Link to comment Share on other sites More sharing options...
dan93z28 Posted November 8, 2007 Report Share Posted November 8, 2007 brakes are just converting the cars kinetic energy into heat energy. where is the heavy car getting extra energy to be harder on brakes? 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. Quote Link to comment Share on other sites More sharing options...
SpaceGhost 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? Quote Link to comment Share on other sites More sharing options...
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