Finding the magnitude of force

A 1889 kg car is pulling a(n) 483.584 kg trailer. Together the car and the trailer have an acceleration of 2.04 m/s^2 forward. The acceleration of gravity is 9.8 m/s^2 . Neglecting frictional forces acting on the trailer, find the magnitude of the force exerted on the car by the road. Answer in units of N.

I keep coming up with 4840 N but this is wrong. Any help would be great!

Actually the normal force is not the whole story. The car exerts 1889 kg * g vertically to support its weight. But it also exerts (1889 kg + 483.6 kg) * 2.04 m/sec^2 in the horizontal direction, in order to accelerate itself and the trailer.

The total force of the car on the road, or the road on the car, is the vector sum of those two perpendicular forces.

You've calculated the horizontal component. Now do F = sqrt(Fx^2 + Fy^2).

Using Newton Laws, you know that the force exerted by the road onto the car is equal and opposite by the force exerted by the car upon the road.

Using the Force = mass*acceleration relation, you know that in this case the acceleration will be gravity because the force is acting down. Therefore, the car is exerting a force on the road of 1889kg * 9.81 m/s^2, or approximately 18531.09 Newtons.

Since the force that the road exerts on the car is equal and opposite to this force and you are finding the *magnitude* of it, your answer will be 18531.09 Newtons.


Note: I'm assuming you mean the normal force and not the forces caused by friction between the wheels and the road. I don't see how you could solve for those.