法向力

Let's suppose we're standing on the ground, and we're stationary.

We know that gravity is a force that acts on us.

Correct! Gravity cannot be the only force acting on us because otherwise our net force is towards the ground. We need another force acting on us to counteract the force of gravity.

Correct! By Newton's third law, since we are standing on the ground, the ground is pushing back on us.

In general, we call the force that a surface exerts on an object in contact with it the normal force.

The normal force is usually represented by the equation $F_{n}=mg$, as it must be equal to the force of gravity pulling down. The normal force is always perpendicular to the ground, even on an incline.

We know that gravity pulls us downward, toward the center of the Earth. However, most objects are clearly not accelerating at the rate of the gravitational acceleration. Why is this so?

The answer becomes simple when one uses Newton's Third Law of Motion. When an object lying on the ground experiences a downward force due to gravity, it pushes on the ground with some force. Then, by the Third Law, the ground exerts a force back upward on the object with the same magnitude. Thus, this force, called the normal force, balances with the gravitational force to stop objects from falling through the ground.

Let us say that the ground has an angle of inclination $\theta$, and suppose that the weight of the object is $W$.

We may decompose the weight into two parts: one in a direction parallel to the ground, and one in a direction perpendicular to the ground. Call these $W_{\parallel}$ and $W_{\perp}$, respectively.

Using simple trigonometry: $$W_{\parallel}=W\sin\theta$$ $$W_{\perp}=W\cos\theta$$ Only the component $W_{\perp}$ actually presses against the ground. Thus, the normal force $N$ equals $W\cos\theta$ for an angle of inclination $\theta$.

This also means that the gravitational force actually working to pull an object downward, when compared to free fall, is multiplied by the sine of the angle of inclination.