Friction, physics, and racing

jimberg · 05-19-2001, 01:58 AM

Okay, here's what I read and how I understand it. First of all, there's more forces involved in a car than just basic friction, but I can start with the basic friction.

Say you have a block of rubber 10" square. That's 100 square inches. You have a force of 100 pounds pushing it down against the pavement. This force is distributed at 1 pound per square inch. Now, make the block of rubber 20" square. That's 400 square inches and only 1/4 pound of force per square inch. No matter how much you change the contact area, the amount of friction will remain the same as long as the same force is being applied. If, using this example, you would apply 400 pounds of force with the 400 square inch piece of rubber, you would quadruple the friction.

That's all they mean as far as friction not depending on surface area.

How does this work with dragsters? Let's say that a dragster has a 50/50 weight distribution, which wouldn't be good, but let's just say it does. The car weights 3000 pounds. This means that each tire is responsible for holding up 750 lbs. You've got skinnies on the front and big old slicks on the back. Let's assume that tire compounds have the same coefficient of friction. If you locked up all the wheels and tried to drag the car across the pavement, all the tires would be producing the same level of friction. That's what I explained in the previous paragraph.

Now, let's think of the car launching. Force is only being applied to the slicks and not the skinnies. This force is rotational, also. As the rear tires plant, the front end will go up and decrease the amount of force being applied between the front tires and the pavement. Because of this, the force that is being taken from the front is now being distributed to the rear. The wheels and the tires in the front are small to save weight. To save on friction, they can also use a harder compound.

Now, think of a real dragster with true weight distribution and how it applies.

I'm no physics major or anything, but I know the second paragraph is true and the next one is my extrapolation of that fact to the problem on hand.

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351W 89 Mustang GT Convertible

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05-19-2001, 01:58 AM	#4
jimberg Registered Member Join Date: Oct 1998 Location: Rogers, MN Posts: 2,089	Okay, here's what I read and how I understand it. First of all, there's more forces involved in a car than just basic friction, but I can start with the basic friction. Say you have a block of rubber 10" square. That's 100 square inches. You have a force of 100 pounds pushing it down against the pavement. This force is distributed at 1 pound per square inch. Now, make the block of rubber 20" square. That's 400 square inches and only 1/4 pound of force per square inch. No matter how much you change the contact area, the amount of friction will remain the same as long as the same force is being applied. If, using this example, you would apply 400 pounds of force with the 400 square inch piece of rubber, you would quadruple the friction. That's all they mean as far as friction not depending on surface area. How does this work with dragsters? Let's say that a dragster has a 50/50 weight distribution, which wouldn't be good, but let's just say it does. The car weights 3000 pounds. This means that each tire is responsible for holding up 750 lbs. You've got skinnies on the front and big old slicks on the back. Let's assume that tire compounds have the same coefficient of friction. If you locked up all the wheels and tried to drag the car across the pavement, all the tires would be producing the same level of friction. That's what I explained in the previous paragraph. Now, let's think of the car launching. Force is only being applied to the slicks and not the skinnies. This force is rotational, also. As the rear tires plant, the front end will go up and decrease the amount of force being applied between the front tires and the pavement. Because of this, the force that is being taken from the front is now being distributed to the rear. The wheels and the tires in the front are small to save weight. To save on friction, they can also use a harder compound. Now, think of a real dragster with true weight distribution and how it applies. I'm no physics major or anything, but I know the second paragraph is true and the next one is my extrapolation of that fact to the problem on hand. ------------------ 351W 89 Mustang GT Convertible