This isn’t exactly a high school problem, but I wasn’t sure where else to post it.

The problem is calculating how much more lift force exerted by the ground on the front tires of a car ( ) than the life force on the rear tires () when braking.


Here’s the remaining problem example:
is the force of gravity that pulls the car toward the center of the Earth. This is the weight of the car; weight is just another word for the force of gravity.

If the car were standing still or coasting, and its weight distribution were 50-50, then would be the same as . It is always the case that plus equals , the weight of the car. Why? Because of Newton's first law. The car is not changing its motion in the vertical direction, at least as long as it doesn't get airborne, so the total sum of all forces in the vertical direction must be zero. points down and counteracts the sum of and , which point up.

By how much does exceed ? The braking torque is proportional to the sum of the braking forces and to the height of the CG. Let's say that height is 20 inches. The counterbalancing torque resisting the braking torque is proportional to and half the wheelbase (in a car with 50-50 weight distribution), minus times half the wheelbase since is helping the braking forces upend the car. has a lot of work to do: it must resist the torques of both the braking forces and the lift on the rear tires. Let's say the wheelbase is 100 inches. Since we are braking at one g, the braking forces equal , say, 3200 pounds. All this is summarized in the following equations:


With the help of a little algebra, we can find out that:



Thus, by braking at one g in our example car, we add 640 pounds of load to the front tires and take 640 pounds off the rears! This is very pronounced weight transfer.

This all makes sense. The problem I have is when I attempt the same calculations using different variables. The new variables are as follows:
G (weight of car) – 2886
Wheelbase – 104.3 inches
Weight distribution – 61 (front) / 39 (rear)
Height – 20 inches

Following the example's formula:
2886 times 20 inches = times 63.623 inches - times 40.677 inches.
+ = 2886 lbs

The problem is the next part, I could not figure out how the following section was calculated. From my guess, it may be 1760.46 (.61 x 2886) + 2886 /x = ? lbs and vice-versa for the rear tires. The only possibility I came up with how the ‘5’ was conceived was by dividing the total wheelbase of 100 by the height of 20 (100/20=5). Please let me know, sorry for the long explanation. Thanks for reading.