Forces are vectors.
Why? Forces have both size and direction.
For example, if you're pushing an object:
A freebody diagram shows all the forces acting on an object. A typical freebody diagram is shown at right. By definition, an object is in equilibrium if the sum of the forces acting on the object is the zero vector. The zero vector is denoted by ‘$\,\vec 0\,$’. If an object is at rest (not moving), then it is in equilibrium.
Note about the other direction:
If an object is in equilibrium, then its velocity is not changing—it is either at rest, or moving at a constant velocity. 

The notation $\,\\vec F\\,$ for the length of a vector $\,\vec F\,$ can get tedious.
Sometimes, a simpler notation is used: let $\,F\,$ (without an arrow) denote the length of $\,\vec F\,$.
This simpler notation will be used in this section.
Problem: Two cables are supporting a 200 pound object, as shown at right. Find the tensions in the cables (as vectors). Solution: Let $\,\vec T_1\,$ and $\,\vec T_2\,$ denote the cable tensions, as shown. Let $\,T_1\,$ and $\,T_2\,$ denote the lengths of $\,\vec T_1\,$ and $\,\vec T_2\,$, respectively. The suspended object causes a 200 lb force pointing straight down. 
Since the object is at rest (assume the cables don't break), it is in equilibrium.
There are different
ways to write down the equilibrium condition and then solve for the desired tensions.
Take a look at the variations shown below, and decide which you like best!
The author (Dr. Burns) prefers
approach #1.
The methods shown in the table below will use some/all of the following information:
vector  size of horizontal component  size of vertical component 
$\,\vec T_1\,$  $T_1\cos 22.3^\circ$  $T_1\sin 22.3^\circ$ 
$\,\vec T_2\,$  $T_2\cos 41.5^\circ$  $T_2\sin 41.5^\circ$ 
weight vector  0  200 
vector  sign of horizontal component  sign of vertical component 
$\,\vec T_1\,$  negative (points to the left)  positive (points up) 
$\,\vec T_2\,$  positive (points to the right)  positive (points up) 
weight vector  (0 has no sign)  negative (points down) 
Approach #1 is shortest, and is preferred by Dr. Burns.
Approach #1
SYSTEM OF EQUATIONS TO BE SOLVED:
$$\begin{gather}
T_1\cos 22.3^\circ = T_2\cos 41.5^\circ\cr
T_1\sin 22.3^\circ + T_2\sin 41.5^\circ = 200
\end{gather}
$$

Approaches #2 and #3 both use the same idea: for an object in equilibrium,
the sum of the forces is the zero vector.
The only difference is the naming of the forces:
Approach #2
SYSTEM OF EQUATIONS TO BE SOLVED:
$$
\begin{gather}
\color{red}{T_1\cos 22.3^\circ + T_2\cos 41.5^\circ} = \color{red}{0}\cr
\color{green}{T_1\sin 22.3^\circ + T_2\sin 41.5^\circ 200} = \color{green}{0}
\end{gather}
$$

Approach #3
SYSTEM OF EQUATIONS TO BE SOLVED:
$$
\begin{gather}
T_1\cos 157.7^\circ + T_2\cos 41.5^\circ = 0\cr
T_1\sin 157.7^\circ + T_2\sin 41.5^\circ 200 = 0
\end{gather}
$$

Since $\,\cos 157.7^\circ = \cos 22.3^\circ\,$, a moment's inspection shows that the three systems of equations are equivalent.
Use your favorite technique (perhaps substitution or elimination) to solve the system of your choice.
For your convenience, the system in Approach #1 is solved here:
x*cos(22.3 deg) = y*cos(41.5 deg), x*sin(22.3 deg) + y*sin(41.5 deg) = 200
These two observations should provide confidence in your answer:
On this exercise, you will not key in your answer. However, you can check to see if your answer is correct. 
PROBLEM TYPES:
IN PROGRESS 