Tire Pressure

Imagine you've lost a bet. Your “friend” is now going to run over your thumb with his car*, a 1996 Honda Accord. (This, by the way, was the most stolen car in 2012 (8637 cars according to the NICB Hot Wheels report)). He (I assume it’s a he) goes inside to get his video camera (of course). What can you do to the car in the seconds he’s gone to help this go better for you? 

One idea would be to let as much air out of the tire as possible. If the rubber itself is very flexible**, the force your thumb will see is limited by the pressure the tire can apply to the area of your thumb. The less tire pressure, the less force. The force the tire supports overall is the same (about ¼ of the car), but it’s supported over more area as the tire flattens and is in contact with more road. So if it’s flat enough hopefully more of the tire is born by the road and less by your digit.

That car weighs about 3000 lbs. The tire width is 7.28”. Normal tire pressure is about 35 psi. So normally (3000 lbs) / (4 tires) / (35 psi) = 21 square inches of tire is in contact with the road (tire width is about 7 in, so that’s about 3 in of contact length). If your thumb is 0.625 in wide by 2.5 in long its area is 1.6 square inches. That’s 8% of the contact area, so you’ll get something like 8% of the wheel load, or 57 lbs. If you can reduce the tire pressure to 10 psi, more like 75 square inches will be on the road, but your thumb is the same 1.6 sqin, so you only take 2% of the load, or 16 lbs: much better.

*   Do not do this.
** Actually, the tire rubber is not perfectly flexible, so those forces will be higher (so do not try this), but the idea of less pressure meaning less force under a given area of the tire is still valid. Also, if too much air is let out the steel rim could come into play and the result would be very bad (so do not try this). Also, we’re not even talking about the pinching and abriasian that could occur as the tire goes up one side and down the other (so dont’ try this)

Second Law

You have a spool of rope laying on it's side. The rope is passing under the central axle and to the right toward you. You pull slowly on the rope directly to the right. What will happen? Will the spool roll away from you to the left, or toward you to the right?

Despite the common intuition that such pulling will cause the spool to spin counterclockwise and roll to the left, in fact it will roll to the right (and wind up the rope you are pulling). Try it.

One way to think of why this must be true is through Newton's Second Law of Motion. Newton's second law is one of the most simply stated yet powerfully predictive ideas in physics. It is:

F = M * A

Where F is the force exerted on an object. This is a vector, so it is a direction and a magnitude.
M is the mass of the object.
And A is how the object accelerates, also a vector.

This means that if you pull on the spool to the right (F), and nothing else is pushing or pulling on it*, the acceleration vector (A) must be in the same direction and just scaled by the mass, M.

*Gravity is pulling it down, but the ground is pushing it up just the same, and our pulling is not at all in the same direction, so it shouldn't affect anything. Also friction is pushing left, but by its nature friction can't be greater than our force, F. Since we're just talking about the direction and not magnitude of the motion, it's fine to ignore it.