Drag racing acceleration and friction

[andro]

In a drag race starting from rest, assume we have constant acceleration and the acceleration is limited by tire friction. I understand we have $v(t) = at$ and I have read elsewhere on this forum that $a = \mu_sg$ where $\mu_s$ is the static friction involved, and $g$ is the acceleration due to gravity.

But how is this expression for $a$ derived? How is it independent of the mass of the car?

 

[Quantum Defect]

In a drag race starting from rest, assume we have constant acceleration and the acceleration is limited by tire friction. I understand we have $v(t) = at$ and I have read elsewhere on this forum that $a = \mu_sg$ where $\mu_s$ is the static friction involved, and $g$ is the acceleration due to gravity.

But how is this expression for $a$ derived? How is it independent of the mass of the car?

The frictional force is given by

F_friction = mu * mass * g

If the acceleration is limited by the Friction, as you have stated, then

mass * a <= mu * mass * g

Mass cancels.

As the mass of the car increases, the Normal force increases, as does the Frictional force. But as the mass increases, the inertia increases, as well -- the car is harder to accelerate.

You see the same thing (somewhat counterintuitive) with the speed of falling objects. As the mass of an object increases, the weight (force pulling down) increases. But as the mass increases, so does the inertia (resistance to acceleration). Since both weight and inertia are proportional to mass, the acceleration of falling bodies is independent of their mass.

 

[GiuseppeR7]

where did you find that problem?. I think it is difficult to solve. The friction created by the tire is a not so simple function of the normal force acting on the tire itself. Even the acceleration is a strong function of the normal force since it rule the grip of the tire.