Key Questions
- What is the definition of static equilibrium?
- How do I choose which are the most efficient equations to solve two-dimensional equilibrium problems?
Section 5.3 Equations of Equilibrium
In statics, our focus is on systems where both linear acceleration \(\vec{a}\) and angular acceleration \(\mathbf{\alpha}\) are zero. These systems are frequently stationary, but could be moving with constant velocity.
Under these conditions Newton’s Second Law for translation reduces to
\begin{equation}
\sum \vec{F} = 0\text{,}\tag{5.3.1}
\end{equation}
and, Newton’s second law for rotation gives the similar equation
\begin{equation}
\sum \vec{M} = 0\text{.}\tag{5.3.2}
\end{equation}
The first of these equations requires that all forces acting on an object balance and cancel each other out, and the second requires that all moments balance as well. Together, these two equations are the mathematical basis of this course and are sufficient to evaluate equilibrium for systems with up to six degrees of freedom.
These are vector equations; hidden within each are three independent scalar equations, one for each coordinate direction.
\begin{align}
\sum\vec{F} \amp= 0 \implies \begin{cases}\Sigma F_x \amp= 0\\ \Sigma F_y \amp= 0 \\ \Sigma F_z \amp= 0\end{cases} \amp \sum\vec{M} \amp= 0 \implies \begin{cases}\Sigma M_x \amp= 0\\ \Sigma M_y \amp= 0 \\ \Sigma M_z \amp= 0\end{cases}\tag{5.3.3}
\end{align}
Working with these scalar equations is often easier than using their vector equivalents, particularly in two-dimensional problems.
In many cases we do not need all six equations. We saw in Chapter 3 that particle equilibrium problems can be solved using the force equilibrium equation alone, because particles have, at most, three degrees of freedom and are not subject to any rotation.
To analyze rigid bodies, which can rotate as well as translate, the moment equations are needed to address the additional degrees of freedom. Two-dimensional rigid bodies have only one degree of rotational freedom, so they can be solved using just one moment equilibrium equation, but to solve three-dimensional rigid bodies, which have six degrees of freedom, all three moment equations and all three force equations are required.
