A 1 480-kg automobile has a wheel base (the distance between the axles) of 3.10 m. The automobile's center of mass is on the centerline at a point 1.25 m behind the front axle. Find the force exerted by the ground on each wheel.

Force exerted by ground on each back wheel = 2924.195 N

Force exerted by ground on each front wheel = 4327.81 N

Explanation:

We are given;

Mass of car; m = 1480 kg

Distance between front and back axle;d = 3.1 m

Now, let F_f and F_b be the force exerted by the level ground on the back and front wheels respectively.

So, At translational equilibrium, we have;

F_f + F_b = mg = 1480 × 9.8 = 14504 N

F_f + F_b = 14504 - - - - (1)

For rotational equilibrium, on taking the torque about the C.G., we have:

F_f(1.25) = F_b(3.1 - 1.25)

1.25F_f = 1.85F_b

F_f = (1.85/1.25)F_b

F_f = 1.48F_b - - - (2)

Put 1.48F_b for F_f in eq 1 to obtain;

1.48F_b + F_b = 14504

2.48F_b = 14504

F_b = 14504/2.48

F_b = 5848.39 N

So, F_f = 1.48 × 5848.39

F_f = 8655.62 N

Thus, force exerted by the ground on each back wheel = 5848.39/2 = 2924.195 N

And force exerted by the ground on each front wheel = 8655.62/2 = 4327.81 N

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