We have already used vectors to solve problems.  This will continue, and we will need to extend our ability to use them in new situations.  Don't worry; we'll start out simple...
The drawings below are called "free body diagrams" and show all the force vectors acting on an object.
It's pretty easy after you've done it for a while, but I still leave things out by mistake sometimes...
 
 
This picture shows what we already know about the 3rd law: the forces are balanced.
mg is the wieght of the book
N is the table's upward force, it acts at 90° to the table surface, and is usually referred to as the "normal force".  I think it's a poor name, but oh well.
mg = N
Here a mass hangs from a string. 
mg is again the weight of the block.
T is short for tension.
If the system is not moving OR moving at a constant speed, then mg = T
Here, a hanging object was pulled to the side by q degrees.  It should be obvious which way the object will move if released.
Here, a hanging object was made to spin around an axis at q degrees from vertical.
It is similar to the above drawing, but when objects more in a circle, a force exists that acts toward the axis.  It's called centripetal force, Fc
This is a variation of the top drawing, but with a Tension in the string provided by a pulling force.
This one is almost identical to the one just above, except that the hand is pushing on the block itself, so we use "F" instead of "T"
Here is where people start to get confused, so I will try to go step by step...
1) mg acts straight down
2) N acts at 90° to surface. Since surface is tilted at q degrees, we need the X component (pretend the tilted surface is the X axis) of the usual "N", so it becomes mgcosq. N = mgcosq
3)mgsinq acts down the plane.  It uses the sin function because the vector comes from the side opposite the given angle q.
This one is a lot like the 2nd one from the top.
We can see from the drawing that there is no net force in the X plane (Sx = zero).

We can also see that mg must be opposed by T1y+T2y such that the sum = zero.

mg = T1sinq + T2sinq
in this case, T1 = T2 so a problem like this is pretty easy to solve.
This one incorporates parts of two different schemes: the inclined plane & the hanging mass.
The only thing to notice is that the two tension vectors are equal.  Why? Well, it's just one piece of string, so can only have one tension in it!
Here, we introduce the force of friction.
Fa is "force applied"
Fk is "force of kinetic friction" which means that the box is moving here.
If the motion is constant, Fa = Fk
Here we add friction to the inclined plane.
Fs is "force of static friction" and implies that the box is not moving.
The box will not move until
msN < mgsinq    is true.

This drawing also shows the vector "mgcosq" opposite of "N".   N = mgcosq