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Engineering Statics: Open and Interactive

Section 3.1 Equilibrium

Engineering statics is the study of rigid bodies in equilibrium so it’s appropriate to begin by defining what we mean by rigid bodies and what we mean by equilibrium.
A body is an object, possibly made up of many parts, which may be examined as a unit. In statics, we consider the forces acting on the object as a whole and also examine it in greater detail by studying each of its parts, which are bodies in their own right. The choice of the body is an engineering decision based on what we are interested in finding out. We might, for example, consider an entire high-rise building as a body for the purpose of designing the building’s foundation, and later consider each column and beam of the structure to ensure that they are strong enough to perform their individual roles.
A rigid body is a body that doesn’t deform under load, that is to say, an object which doesn’t bend, stretch, or twist when forces are applied to it. It is an idealization or approximation because no objects in the real world behave this way; however, this simplification still produces valuable information. You will drop the rigid body assumption and study deformation, stress, and strain in a later course called Strength of Materials or Mechanics of Materials. In that course, you will perform analysis of non-rigid bodies, but each problem you do there will begin with the rigid body analysis you will learn to do here.
A body in equilibrium is not accelerating. As you learned in physics, acceleration is velocity’s time rate of change and is a vector quantity. For linear motion,
\begin{equation*} \vec{a} = \frac{d\vec{v}}{dt}\text{.} \end{equation*}
For an object in equilibrium \(\vec{a} = 0\) which implies that the body is either stationary or moving with a constant velocity
\begin{equation*} \vec{a} = 0 \implies \begin{cases}\vec{v}=0\\\vec{v}=C\end{cases}\text{.} \end{equation*}
The acceleration of an object is related to the net force acting on it by Newton’s Second Law
\begin{equation*} \Sigma\vec{ F} = m \vec{a}\text{.} \end{equation*}
So for the special case of static equilibrium Newton’s Law becomes
\begin{equation} \Sigma\vec{F} = 0\text{.}\tag{3.1.1} \end{equation}
This simple equation is one of the two foundations of engineering statics.
There are several ways to think about this equation. Reading it from left to right it says that if all the forces acting on a body sum to zero, then the body will be in equilibrium. If you read it from right to left it says that if a body is in equilibrium, then all the forces acting on the body must sum to zero. Both interpretations are equally valid but we will be using the second one more often. In a typical problem equilibrium of a body implies that the forces sum to zero, and we use that fact to find the unknown forces which make it so. Remember that we are talking about vector addition here, so the sums of the forces must be calculated using the rules of vector addition; you won’t get correct answers if you can’t add vectors!
We’ll be using all of the different vector addition techniques introduced in Section 2.6, which may lead to some confusion. It doesn’t matter, mathematically, which technique you use but part of the challenge and reward of statics is learning to select the best tool for the job at hand; to select the simplest, easiest, fastest, or clearest way to get to the solution. You’ll do best in this course if can use multiple approaches to solve the same problem.
In Chapter 5 we will add another requirement for equilibrium, namely equilibrium equation (5.3.2) which says the forces which cause rotational motion and angular acceleration \(\boldsymbol{\alpha}\) also must sum to zero, but for the problems of this chapter the only condition we’ll need for equilibrium is \(\Sigma\vec{F} = 0\text{.}\)