- #1

Ocata

- 198

- 5

## Homework Statement

Hello, I know how to solve this problem, I just don't understand an aspect of the conclusion.

A car is traveling at some constant velocity forward in a horizontal direction. The ground has a coefficient of static friction of .6. What acceleration in the forward direction would cause the vehicle to skid? Approximate gravity to 10m/s^2.

## Homework Equations

[itex][F_{net} = ma] = [F_{friction} = ma] = [μ(F_{N}) = ma] = [μ(F_{G}) = ma] = [μ(mg) = ma]

[μ(g) = a][/itex]

## The Attempt at a Solution

.6(10) = 6m/s^2

This is fine, but I am not really understanding why m would cancel as though it does not matter.

To me, the heavier the object, the less likely it is to skid. That is, the greater the mass of an object, the greater its weight, and there for the greater the normal force.

So on a particular surface of μ = .6, then:

Force of Friction = [itex]μ(F_{N}) = μmg = .6(10kg)(10m/s^{2}) = 60N[/itex].

However, if I increase the mass from 10kg to 100kg, then:

Force of Friction = [itex]μ(F_{N}) = μmg = .6(100kg)(10m/s^{2}) = 600N[/itex].

So, clearly, mass has an impact on force of friction. If this is in fact true, why does it cancel from the formula above in the part 2 (relevant equations) section?