When forces acting on an object which is at rest are balanced, then we say that the object is in a state of static equilibrium.
The resultant of these forces equals zero. That is, the vector sum of the forces adds to zero.
Example #1
Suppose two dogs are struggling for the same shoe as shown in the diagram below. The left dog's force is shown by the yellow/black arrow while the right dog's force is shown by the teal arrow. Notice below that when the forces are added headtotail, the resultant force, shown in grey, acts straight up the yaxis.
Force_{right dog} + Force_{left dog} = Resultant
F_{r} +
F_{l} =
R
To place the shoe into a state of static equilibrium, a third force (shown in red) would have to be added to the diagram. Then the forces exerted on the shoe would equal zero.
Force_{right dog} + Force_{left dog} + "Equilibrium" vector = 0
F_{r} +
F_{l} + E = 0
F_{r} +
F_{l} =
E
By comparing the vector equations presented above
F_{r} +
F_{l} = R
F_{r} +
F_{l} = E we notice that the third force required for equilibrium to be established is a vector that has the same magnitude as the resultant, but points in 180º the opposite direction.
R = E



Each dog's force has been resolved into its x and ycomponents.

A vector diagram showing only the components of each dog's force.

The addition of a third force would place the shoe into a state of static equilibrium.

Example #2
Suppose you are asked to calculate the tensions in the three ropes (A, B, C) that are supporting the 5kg mass shown below. Since the system is at rest, we will work the problem using the properties of static equilibrium.
Let's begin with a freebody diagram showing the forces acting on the knot. Since these forces belong to three separate ropes, all three tensions can be different.

The next step will be to build an xy chart showing the components of each force.
If θ were to equal 37º then 
Since we do not have any knowledge of any of the three tensions, we must now do a similar analysis using a freebody diagram of the 5kg mass.

The xy chart for the 5kg mass would look as follows:
Since we know that the 5kg mass is in static equilibrium, we know that the sum of the forces in each column equals zero. We only need to write the equation for the ycolumn since there are no nonzero forces in the xcolumn.
y: C + (mg) = 0
C  5(9.8) = 0 C = 49 N

Now, knowing the value for C, we can solve for the tensions in ropes A and B by setting up the following equations from each column in our xy chart for the knot:
x:


A cos(37º) + B = 0

y:


A sin(37º)  C = 0




x:


0.8A + B = 0

y:


0.6A  C = 0




Since we know the value for C we will next solve for A and then for B.

y:


0.6A  49 = 0 A = 49/0.6 A = 81.7 N




x:


0.8(81.7) + B = 0 B = 0.8(81.7) B = 65.3 N

Example #3
This procedure can also be applied to the following situation in which a mother is just at the instant of releasing her child in a swing.
