# Normal Forces

"Normal forces" are best understood as the weight carried by each wheel. Unless the car is perfectly symmetrical from front to back, the weight carried by the rear wheels is not the same as the weight carried by the front wheels. That is, the weight distribution must be accounted for.

It is important to know the normal force on the drive wheels, because they determine the amount of traction that the wheel can provide. If the propulsive force exerted by the wheel is greater than the wheel's traction, the wheel will skid, and the car may "spin out". In order to compute the normal forces, it is necessary to know the weight of the vehicle (w), the x-location of the center of mass (Cx) relative to one of the wheel contact points, and the distance between the wheels (Lwb).

The calculation is done by first assuming that the car is resting on the ground in equilibrium -- that is, it is not rotating. It is clear that this is true, because the only forces acting on it are the force of gravity and the force of the racetrack surface pushing back.

If the car is not rotating, then the sum of all moments acting about any point must be zero. A moment is simply a force acting on an object, in either the x- or y- direction, multipled by its distance from some point of reference. In the drawing above, suppose that the point of reference is at point "b". The two moments acting on the car are:

(W * Cx), clockwise

(Ra * Lwb), counterclockwise

Because we know that the sum of moments is zero, we can write the following equation (counterclockwise moments are given a negative sign):

(Clockwise moments) - (counterclockwise moments) = 0

(W * Cx) - (Ra * Lwb) = 0

Ra = (W * Cx) / Lwb

Suppose that W = 1.0 lb, Cx = 1 inch, and Lwb = 4 inches. Then,

Ra = (1.0 lb) * (1 inch) / (4 inches) = 0.25 lb

What, then, is the normal force on the drive wheel, known as Rb? The sum of Rb and Ra (Ra + Rb) must equal the total weight (W) of the car. This means:

Rb = W - Ra

Rb = 1.0 lb - 0.25 lb

Rb = 0.75 lb

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