# Why does the contact force of an object on a surface equal its weight (even though it's not because of Newton's third law)?

E.g. take a scenario like this: Where the block $$A$$ is lying on a surface $$S$$ at rest. A common mistake to make when first studying Newton's laws of motion is to think that, because $$R_{S \: on \: A} = - W_{Earth \: on \: A}$$, they are a pair of forces that are equal and opposite due to Newton's third law. However, I know that this is not the case because they are exerted on the same object and are different types of force (contact vs gravitational).

This still leaves the question of why $$R_{S \: on \: A} = - W_{Earth \: on \: A}$$. I know why $$R_{S \: on \: A}$$ has to equal $$-W_{Earth \: on \: A}$$ for the object to be at rest and not accelerating up or down. Yet, I don't know why it does, even if I know it's not because they are Newton's third law pairs. What relation, and there surely must be some, does the weight of an object have on the contact force it exerts on a surface, and thus on the reaction force exerted on it? Is it always equal and, if so, why?

## 2 Answers

What relation, and there surely must be some, does the weight of an object have on the contact force it exerts on a surface, and thus on the reaction force exerted on it? Is it always equal and, if so, why?

As you have correctly guessed, they are not always equal. For example, when you are jumping up the surface force is substantially larger than your weight. And when you are falling the surface force is zero though your weight is unchanged.

So, Newton’s gravitational law determines the weight, but what determines the surface force?

To a good first approximation you can use Hooke’s law. As you stand on the floor you cause a small deflection or compression in the material. Per Hooke’s law, this deflection is proportional to the force: the more you deflect the surface the greater the force.

Now, of course, Hooke’s law is primarily intended to describe springs, and many floor surfaces don’t feel very “springy”. That is because Hooke’s law is a good first approximation, but many materials deviate substantially. Importantly, many materials have a part of the force that is proportional to the velocity of the deflection which brings in damping.

So in the case of an object resting on the ground the Hooke’s law term determines the amount of deflection so that the surface force is equal to the weight and the damping term dissipates any initial kinetic energy so that the object doesn’t bounce.

It is not in general. The normal force is independent from weight unless the rest state of the body needs that.

For example, if you take a rock and press it against a wall, you have that the normal force is orthogonal to the direction of the weight.

• So, you're saying that, whenever an object is resting on surface, the contact forces just happen to be equal to the right? I guess it's possible, but it seems fairly unlikely to me. Is it not like friction, where, up until a certain point, it's equal to the weight up until a certain maximum value determined by factors like the sturdiness of the surface etc.? – 0.5772156649 Mar 10 '20 at 16:58
• The contact force fairly complicated force, since is given by very complex interaction between the elements, but the zero sum forces condition on a resting element is what allow you to balance the forces (like gravity with surface force or the force that apply your hand to a chalk when you write being equal to the surface force). – Grego_gc Mar 11 '20 at 12:36