Car Center of Mass Problem

[Jacob87411]

A 1470 kg automobile has a wheel base (the distance between the axles) of 2.70 m. The center of mass of the automobile is on the center line at a point 1.20 m behind the front axle. Find the force exerted by the ground on each wheel.

A bit confused here. Were going to use the sum of the torque = 0, using the center of mass (1.2m behind the front axle and 1.5m infront of the back) as our starting point. We need to find the force in kN on each wheel. So between the 4 wheels they must support 14,406 N (1470*9.8), i know this much but am confused where to go from here

[Doc Al]

This is a basic equilibrium problem, so you'll have to use an equilibrium condition such as: The torques about any point must add to zero. (Choose points where the tires touch the ground as your pivot point.)

Write this mathematically and see what it tells you. (Realize that the same force is exerted on each of the two front tires since the center of mass is on the center line; the same logic applies to the back tires)

[Jacob87411]

Yes but what is the force..for example if you use the back wheels as the point of rotation the torque is zero as you said. Isn't the torque equation then:

0=(2.7)(Ff)
The 2.7 is the distance to the front wheels, the Ff is the force on the front wheels. The torque from the back wheels is canceled out because the distance is 0?

[Gladi8or2]

You can choose the center of torque as you wish.. the result will be the same. It's best to choose the center of torque in a point where a force acts, so this force will cancel out. This simplifies your equations.

And don't forget... there's also a force caused by gravity. You forgot it in your previous post.

[Doc Al]

Don't forget the torque due to the car's weight.

[Jacob87411]

Ahhhh

t=(1.5)(14,406)-(2.7)(Ff)

Got it now, thanks

[Doc Al]

Right. And realize that the force you call Ff is distributed over two tires.