Reply #67: The force of friction is dependent on the mass of the truck [View All]

and the coefficient of friction between the tires and the road. F=Nµ_k, which in the case of a flat road reduces to F=mgµ_k. m is the mass of the truck, g is earth's gravity, m*g is the normal force (assuming a flat road) and µ_k is the coefficient of kinetic friction.

In theory, if Truck A weights twice that of Truck B, it will have twice as mush kineticn energy and momentum. It will also have twice as much braking force to act on it's twice as much mass, so the decceleration should be the same.

However, as the tires heat up they melt and lose traction, not to mention that it will take a little longer to bring the tires up to full braking force, time that transforms into distance. Panic-stopping twice as much mass will heat the tires up faster, causing less stopping friction and thus less decceleration.

Of course, I'm assuming wheel lockup by using the µ_k. If the big rig has anti-lock brakes, the coefficient of static friction, µ_s is used instead, which is significantly higher and thus imports more braking force is the driver or the ABS is able to hold the braking force right at impending lockup. This is why anti-lock brakes are such a critical safety features.

Likewise, if the following car has ABS, it will be able to stop much faster than, say, my old beater does.