Question

A 1550 kg automobile has a wheel base (distance between the axles) of 2.8 m. the automobiles center of mass is on the centerline at a point 1.00 m behind the front axle. Find the force exerted by the ground of each wheel.

Each front wheel

Each rear wheel

Answer 1

let
m = 1550 kg
let F_front is the force exerted by ground on each front wheel.

let F_rear is the force exerted by ground on each rear wheel.

As the automobile is in equilibrium net force and net torque acting on it must be zero.

Apply net torque about rear wheels = 0

2*F_front*2.8 - m*g*(2.8 - 1) = 0

2*F_front*2.8 = m*g*1.8

F_front = m*g*1.8/(2*2.8)

= 1550*9.8*1.8/(2*2.8)

= 4882 N <<<<<<<------------------------Answer

Apply net torque about front wheels = 0

2*F_rear*2.8 - m*g*1 = 0

2*F_rear*2.8 = m*g*1

F_rear = m*g*1/(2*2.8)

= 1550*9.8*1/(2*2.8)

= 2712 N <<<<<<<------------------------Answer