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Section 5.3 Equations of Equilibrium

Now that you have thoroughly learned how to draw accurate free-body diagrams, it is time to bring in some equations so that we can solve problems. Recall that Newton’s 2nd law tells us that Newton’s second law states that a particle accelerates in the direction of and at a rate proportional to the net applied force. For translational motion, Newton’s 2nd law can be represented by the equation:

\begin{equation} \Sigma\vec{F}=m\vec{a}\text{, where: }\tag{5.3.1} \end{equation}
\begin{align*} \Sigma \vec{F} \amp\equiv\text{sum of forces on the body}\\ m\amp\equiv \text{mass of the body}\\ \vec{a} \amp\equiv \text{linear acceleration of the body} \end{align*}

Rotational motion has a parallel equation which is

\begin{equation} \Sigma\vec{M}_o=I_o \boldsymbol{\alpha}\text{, where: }\tag{5.3.2} \end{equation}
\begin{align*} \Sigma \vec{M}_o \amp\equiv \text{sum of moments around point }o\\ I_o \amp\equiv \text{mass moment of inertia about point }o\\ \boldsymbol{\alpha} \amp\equiv \text{angular acceleration of the body} \end{align*}

In statics, we focus on systems which are not accelerating, thus both the linear acceleration \(\vec{a}\) and angular acceleration \(\vec{\alpha}\) must equal zero. As described in Chapter 3, this means that the linear velocity must either be constant or zero. The same argument holds for the angular velocity. Thus we are left with the simplified equations:

\begin{equation} \Sigma\vec{F}=0 \text{ in }\textbf{all}\text{ directions}\tag{5.3.3} \end{equation}
\begin{equation} \Sigma\vec{M}=0 \text{ about }\textbf{all} \text{ axes of rotation}\label{sigmaM}\tag{5.3.4} \end{equation}

Let us examine how we can put this knowledge to use in a two-dimensional problem. For the \(\Sigma\vec{F}=0\) equation, there are infinite directions in the \(xy\) plane of a two-dimensional problem, so it is best to be strategic on which directions you select. You should recall from Chapter 2 the reason we use the \(xy\) orthogonal coordinate system is that the \(x\) and \(y\) axes are independent of each other because they are perpendicular. Thus, picking any two orthogonal directions to solve the equation \(\Sigma\vec{F}=0\) gives us independent equations to solve for two of the unknowns, which also takes care of two of the degrees of freedom. We are then left with the final rotation degree of freedom. Setting the sum of moments about any of the infinite points in the \(xy\) plane equal to zero ensures no rotational motion is allowed. Often we sum moments around a point on the body, but the equation \(\Sigma\vec{M}_o=0\) works equally as well for points off the body.

Recognize that equations of equilibrium are the mathematical representation of a free-body diagram. Hence, drawing an incorrect free-body diagram will cause you to write the wrong equations of equilibrium and not be able to solve the problem correctly!