\begin{align*}\mathrm{Common~Forces} \begin{cases} F_g = m g & \text{Gravity}\\ F_N & \text{Normal force: acts perpendicular to surfaces}\\ F_T & \text{Force of tension in strings and wires}\\ F_{sp}= -k \Delta x & \text{Force of spring stretched a distance } \Delta x \text{ from equilibrium}\end{cases}\end{align*}

### Normal Force

Often, objects experience a force that pushes them into another object, but once the objects are in contact they do not any move closer together. For instance, when you stand on the surface of the earth you are obviously not accelerating toward its center. According to Newton's Laws, there must be a force opposing the earth's gravity acting on you, so that the net force on you is zero. The same also applies for your gravity acting on the earth. We call such a force the **Normal Force**. The normal force acts between any two surfaces in contact, balancing what ever force is pushing the objects together. It is actually electromagnetic in nature (like other contact forces), and arises due to the repulsion of atoms in the two objects. Here is an illustration of the Normal force on a block sitting on earth:

### Tension

Another force that often opposes gravity is known as **tension**. This force is provided by wires and strings when they hold objects above the earth. Like the Normal Force, it is electromagnetic in nature and arises due to the intermolecular bonds in the wire or string:

If the object is in equilibrium, tension must be equal in magnitude and opposite in direction to gravity. This force transfers the gravity acting on the object to whatever the wire or string is attached to; in the end it is usually a Normal Force --- between the earth and whatever the wire is attached to --- that ends up balancing out the force of gravity on the object.

### Friction

Friction is a force that opposes motion. Any two objects in contact have what is called a mutual coefficient of friction. To find the force of friction between them, we multiply the normal force by this coefficient. Like the forces above, it arises due to electromagnetic interactions of atoms in two objects. There are actually two coefficients of friction: static and kinetic. Static friction will oppose *initial* motion of two objects relative to each other. Once the objects are moving, however, kinetic friction will oppose their continuing motion. Kinetic friction is lower than static friction, so it is easier to keep an object in motion than to set it in motion. \begin{align*}
f_s \le \mu_s | \vec{F_N}| && \mathrm{[5]~Static~friction~opposes~potential~motion~of~surfaces~in~contact}\\
f_k = \mu_k | \vec{F_N}| && \mathrm{[6]~Kinetic~frictions~opposes~motion~of~surfaces~in~contact}
\end{align*} There are some things about friction that are not very intuitive:

- The magnitude of the friction force does not depend on the surface areas in contact.
- The magnitude of kinetic friction does not depend on the relative velocity or acceleration of the two objects.
- Friction always points in the direction opposing motion. If the net force (not counting friction) on an object is lower than the maximum possible value of static friction, friction will be equal to the net force in magnitude and opposite in direction.

### Spring Force

Any spring has some equilibrium length, and if stretched in either direction it will push or pull with a force equal to: \begin{align*} \vec{F_{sp}} = -k \vec{\Delta x} && \mathrm{[7]~Force~of~spring~}\vec{\Delta x}\mathrm{~from~equilibrium} \end{align*}

### Examples

#### Example 1

A woman of mass 70.0 kg weighs herself in an elevator.

- a) If she wants to weigh less, should she weigh herself when accelerating upward or downward?
- b) When the elevator is not accelerating, what does the scale read (i.e., what is the normal force that the scale exerts on the woman)?
- c) When the elevator is accelerating upward at 2.00 m/s
^{2}, what does the scale read?

a) If she wants to weigh less, she has to decrease her force (her weight is the force) on the scale. We will use the equation \begin{align*} F=ma \end{align*} to determine in which situation she exerts less force on the scale.

If the elevator is accelerating upward then the acceleration would be greater. She would be pushed toward the floor of the elevator making her weight increase. Therefore, she should weigh herself when the elevator is going down.

b) When the elevator is not accelerating, the scale would read \begin{align*}70.0\mathrm{kg}\end{align*}.

c) If the elevator was accelerating upward at a speed of \begin{align*}2.00\mathrm{m/s^2}\end{align*}, then the scale would read \begin{align*} F=ma=70\mathrm{kg}\times (9.8\mathrm{m/s^2}+2\mathrm{m/s^2})=826\mathrm{N} \end{align*} which is \begin{align*}82.6\mathrm{kg}\end{align*}.

#### Example 2

A spring with a spring constant of \begin{align*}k=400\mathrm{N/m}\end{align*} has an uncompressed length of \begin{align*}.23\mathrm{m}\end{align*} and a fully compressed length of \begin{align*}.15\mathrm{m}\end{align*}. What is the force required to fully compress the spring?

We will use the equation\begin{align*}F=kx \end{align*} to solve this. We simply have to plug in the known value for the spring and the distance to solve for the force. \begin{align*} F=kx=(400\mathrm{N/m})(.23\mathrm{m}-.15\mathrm{m})=32\mathrm{N} \end{align*}