A car has 2 axles.
Weight on front axle = 6kn.
weight on rear axle = 4kn.
distance between axles =250 meters.

Calculate centre of gravity using the information provided.

part 2 -
A weight of 200kg is placed 0.5 meters in front of rear axle.
taking the value of gravity as 9.8, calculate the weight front and rear axles now support.
simple formula for a dummie would be great.thanks for looking.

Sorry... distance between axles is 2.5 meters

a) Ok, use the principle of moments to find this.

This is similar to:

You have a plank of length 2.5m with negligible mass. Two weights, one 4kN (placed at the left of the plank) and the other 6kN (placed at the right of the plank) are placed at opposite ends of the plank. Where do you think the pivot will be, if the system is in equilibrium

With a sketch, the anticlockwise moment is given by (4kN*x) where x is the distance from the rear axle to the pivot, or the centre of gravity. The clockwise moment is (6kN*[2.5-x]).

Since the system is in equilibrium, both moments are equal, so:

4kN*x = 6kN*[2.5-x]

Solve for x, the distance from the rear to the centre of gravity :)

b) Make use of your sketch again.

Use the principle of moments again:

Take the rear axle as the pivot. The moment due to the weight is 200*0.5 clockwise. The axle has to provide the same turning moment so that the system is in equilibrium. The moment about the front axle is given by F*2.5, where F is the force acting on it.

So,

200*0.5 = F*2.5

Solve for F to have the force on the front axle.

You can do the opposite for the rear axle. However, there is a shortcut. Since you have 200 kg, and you just found the weight that is on the front axle, the rest of that weight rests on the rear axle. So, subtract your answer to the force F from 200.

I hope it helped! :)