Jump to content

Smart? answer this...


Cougar1647545494

Recommended Posts

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)

Link to comment
Share on other sites

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.

Link to comment
Share on other sites

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...

Link to comment
Share on other sites

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.
Link to comment
Share on other sites

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

Link to comment
Share on other sites

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?

Link to comment
Share on other sites

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.

Link to comment
Share on other sites

Join the conversation

You can post now and register later. If you have an account, sign in now to post with your account.

Guest
Reply to this topic...

×   Pasted as rich text.   Paste as plain text instead

  Only 75 emoji are allowed.

×   Your link has been automatically embedded.   Display as a link instead

×   Your previous content has been restored.   Clear editor

×   You cannot paste images directly. Upload or insert images from URL.

×
×
  • Create New...