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.

 

 

 

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 head-to-tail, the resultant force, shown in grey, acts straight up the y-axis.

 

 

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.

 

 

Each dog's force has been resolved into its x- and y-components.	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.

 

 

 

Suppose you are asked to calculate the tensions in the three ropes (A, B, C) that are supporting the 5-kg 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 x|y 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 5-kg mass.

 

The x|y chart for the 5-kg mass would look as follows:     Since we know that the 5-kg 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 y-column since there are no non-zero forces in the x-column. 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 x|y 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

 

 

 

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.

 

 

 

In cases where the rope passes over a massless, frictionless pulley the tension on either side of the pulley is the same. This term really means that we are not considering the pulley to have any rotational inertia; that is, the rope slides over the pulley without causing it to rotate. When pulleys with rotational inertia are used, we have to assign different tensions to the ropes on each side and calculate the torque produced on the pulley.

 

 

 

Consider this next system in which both suspended blocks have a mass of 1.5 kg and the pulley has a mass of 1.0 kg.

 

 

Since all of the forces are vertical, we can easily identify the tensions in each of the three ropes by setting the "sum of the magnitudes of the downward forces" equal to the "sum of the magnitudes of the upward forces."

 

Isolating each block and then the pulley we can generate the following freebody diagrams and their equations.

 

T2 = mg	T3 = mg	T1 = T2 + mg + T3
T2 = (1.5)(9.8) T2 = 14.7 N	T3 = (1.5)(9.8) T3 = 14.7 N	T1 = 14.7 + 1(9.8) + 14.7 T1 = 39.2 N

 

If the masses had been unequal, then the system would have accelerated in the direction of the greater mass. Yet, even during that period of acceleration, the tensions T₂ and T₃ would have remained equal.

   

 

 

Let's use all of the properties learned on this page to now determine the mass of the monkey. In the process, also calculate T₁, T₂, T₃, T₄, T₅, T₆, and W₂ if W₁ equals 350 N. If you need assistance getting started, reference the information in the hints located to the left of each question.

 

 

left knot

T1   T3   T4

right knot

T5   T2   W2

pulley

T6   massmonkey