Vector Application: Forces Acting on an Object in Equilibrium

by Dr. Carol JVF Burns (website creator)
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Forces are vectors.
Why? Forces have both size and direction.

For example, if you're pushing an object:

Depending on the definition being used, the weight of an object can be considered as either a vector or a scalar:

In the United States, a common unit of weight is the pound, abbreviated as ‘lb’.

A free-body diagram shows all the forces acting on an object.
A typical free-body 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.
a typical free-body diagram

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.

Example: Finding Tensions in Cables

Two cables are supporting a 200 pound object, as shown at right.
Find the tensions in the cables (as vectors).

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:

Slightly Different Approaches to Satisfy the Equilibrium Condition

Approach #1 is shortest, and is preferred by Dr. Burns.

Approach #1

  • The size of the components pointing left must equal the size of the components pointing right: $$ \cssId{s86}{T_1\cos 22.3^\circ = T_2\cos 41.5^\circ} $$
  • The sum of the sizes of the components pointing up must equal the size of the component pointing down: $$ \cssId{s88}{T_1\sin 22.3^\circ + T_2\sin 41.5^\circ = 200} $$
SYSTEM OF EQUATIONS TO BE SOLVED: $$\begin{gather} \cssId{s90}{T_1\cos 22.3^\circ = T_2\cos 41.5^\circ}\cr \cssId{s91}{T_1\sin 22.3^\circ + T_2\sin 41.5^\circ = 200} \end{gather} $$

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

  • Sum the ‘reference angle’ versions of the forces, and set equal to the zero vector: $$ \begin{gather} \cssId{s99}{\vec T_1 + \vec T_2 + \text{weight vector} = \vec 0}\cr\cr \cssId{s100}{\langle\ \color{red}{-T_1\cos 22.3^\circ}\,,\,\color{green}{T_1\sin 22.3^\circ} \ \rangle + \langle\ \color{red}{T_2\cos 41.5^\circ}\,,\,\color{green}{T_2\sin 41.5^\circ} \ \rangle + \langle \color{red}{0},\color{green}{-200}\rangle \ =\ \langle \color{red}{0},\color{green}{0}\rangle}\cr\cr \cssId{s101}{ \langle\ \color{red}{-T_1\cos 22.3^\circ + T_2\cos 41.5^\circ}\ ,\ \color{green}{T_1\sin 22.3^\circ + T_2\sin 41.5^\circ -200}\ \rangle \ =\ \langle \color{red}{0},\color{green}{0}\rangle} \end{gather} $$
  • For two vectors to be equal, their components must be equal.
SYSTEM OF EQUATIONS TO BE SOLVED: $$ \begin{gather} \cssId{s104}{\color{red}{-T_1\cos 22.3^\circ + T_2\cos 41.5^\circ} = \color{red}{0}}\cr \cssId{s105}{\color{green}{T_1\sin 22.3^\circ + T_2\sin 41.5^\circ -200} = \color{green}{0}} \end{gather} $$

Approach #3

  • Sum the ‘standard direction’ versions of the forces, and set equal to the zero vector: $$ \begin{gather} \cssId{s108}{\vec T_1 + \vec T_2 + \text{weight vector} = \vec 0}\cr\cr \cssId{s109}{\langle\ T_1\cos 157.7^\circ\,,\,T_1\sin 157.7^\circ \ \rangle + \langle\ T_2\cos 41.5^\circ\,,\,T_2\sin 41.5^\circ \ \rangle + \langle 0,-200\rangle \ =\ \langle 0,0\rangle}\cr\cr \cssId{s110}{\langle\ T_1\cos 157.7^\circ + T_2\cos 41.5^\circ\ ,\ T_1\sin 157.7^\circ + T_2\sin 41.5^\circ -200\ \rangle \ =\ \langle 0,0\rangle} \end{gather} $$
  • For two vectors to be equal, their components must be equal.
SYSTEM OF EQUATIONS TO BE SOLVED: $$ \begin{gather} \cssId{s113}{T_1\cos 157.7^\circ + T_2\cos 41.5^\circ = 0}\cr \cssId{s114}{T_1\sin 157.7^\circ + T_2\sin 41.5^\circ -200 = 0} \end{gather} $$

Solve the Resulting System of Equations in $\,T_1\,$ and $\,T_2\,$

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:

Find the Tensions in the Cables (as vectors)

These two observations should provide confidence in your answer:

Master the ideas from this section
by practicing the exercise at the bottom of this page.

When you're done practicing, move on to:
Introduction to Partial Fraction Expansion/Decomposition (PFE)

On this exercise, you will not key in your answer.
However, you can check to see if your answer is correct.
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(MAX is 23; there are 23 different problem types.)
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