A 1 510-kg automobile has a wheel base (the distance between the axles) of 3.30 m. The automobile's center of mass is on the centerline at a point 1.05 m behind the front axle. Find the force exerted by the ground on each wheel. each front wheel kN each rear wheel kN

Force exerted on each back wheel = 2,356.63 N

Force exerted on each front wheel = 5,004.4N

Explanation:

We are given that;

Mass of automobile =1510 kg

Distance between axles (d) = 3.3m

Centre of mass of automobile is 1.05m behind the front axle.

Now,

If we imagine this system,

It is clear that when sum of forces about a system is considered , the sum of the upward forces would be equal to the sum of the downward forces. In this case, the the weight of the automobile is acting downwards and thus would be equal to the sum of the forces on the front and back wheel acting upwards.

Since the mass of the automobile is 1510kg,it's weight will be W = mg = 1510 x 9.81 = 14813.1 N

Thus;

F₁ + F₂ = 14,813.1 N

where;

F₁ = The force exerted by the ground on front two wheels

F₂ = Force exerted by the ground on back two wheels

Now, to continue this, let's derive the formula for torque ;

The formula to calculate torque is given as;

τ = F x d

Where;

F = the force action on the point

d = the distance of the point

So, taking equilibrium torque about the front axle, we have;

1.05 (14,813.1) - (F_2•3.3) = 0

15553.755 = 3.3F2

F2 = 15553.755/3.3 = 4713.26

This is the total force exerted on the 2 back wheels. So the force exerted on one back wheel = (1/2)F2 = 4713.26/2 = 2356.63 N

Now, for the front wheel;

Let's take equilibrium torque about the rear wheel;

(3.3 - 1.05)(14,813.1) - (F_1•3.3) = 0

(2.25 x 14,813.1) = 3.33F1

F1 = 33329.475/3.33 = 10,008.85 N

So force exerted on one front wheel = 10008.85/2 = 5004.4N

neither accurate nor precise.