A floor pushes up on whatever rests on it. That push, perpendicular to the surface, is the normal force $F_N$. With the second law along an axis you can find it in any situation, and it is often not $mg$.
Key idea
The normal force is whatever size the surface must push to keep the object from moving into it. Find it from $\Sigma F_y = ma_y$. A bathroom scale reads the normal force, your apparent weight.
Why it works
A surface is like a very stiff spring: pressed harder, it squeezes slightly more and pushes back harder. It has no fixed value. Stand still and it balances gravity. Ride an elevator that accelerates upward, and the floor must push harder than $mg$ to speed you up; accelerate downward, and it pushes less.
Worked examples
Worked example: An accelerating elevator
A $60\ \text{kg}$ person rides an elevator that is moving up and speeding up at $2\ \text{m/s}^2$. What does the scale read?
- Speeding up while moving up means $a$ points up: $a = +2\ \text{m/s}^2$.
- $F_N - mg = ma$, so $$F_N = m(g + a) = 60(9.8 + 2) = 708\ \text{N}.$$
Answer: $708\ \text{N}$, more than $mg = 588\ \text{N}$
Worked example: Slowing on the way down
The same elevator moves down and slows down at $2\ \text{m/s}^2$. What does the scale read?
- Moving down and slowing down: the acceleration points opposite to the velocity, so up: $a = +2\ \text{m/s}^2$.
- $F_N = 60(9.8 + 2) = 708\ \text{N}$ again.
Answer: $708\ \text{N}$
Worked example: Pressing down
You press down on a $10\ \text{kg}$ box with $30\ \text{N}$. What is the normal force from the floor?
- At rest: up $=$ down, so $F_N = mg + 30 = 98 + 30 = 128\ \text{N}$.
Answer: $128\ \text{N}$
Common mistake
Assuming the normal force always equals $mg$. That is true only on a level surface with no other vertical forces and no vertical acceleration. Check: write $\Sigma F_y = ma_y$ every time and solve for $F_N$; extra pushes, rope pulls, ramps and accelerations all change it.
Try it in the Science Lab: Forces at Work includes the elevator scale.