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Section 3.4 An Application of the Dot Product: Work

Consider the scenario in which a force \(\mathbf F\) causes an object to move from point \(P\) to point \(Q\text{,}\) as shown in Figure 3.18. The vector \(\vec{PQ}\) is often denoted \(\mathbf D\) and is referred to as the displacement vector. The work done by the force is then

Definition 3.16. Work.

\begin{equation} \text{Work} = \mathbf F \cdot \mathbf D = |\mathbf D||\mathbf F|\cos\theta,\tag{3.11} \end{equation}
where \(\left|\text{proj}_{\mathbf D}\mathbf F\right| = |\mathbf F|\cos\theta\) is the magnitude of the force in the direction of motion.
Figure 3.17. A constant force \(\mathbf F\) acting on an object causing a displacement from \(A\) to \(B\text{.}\)
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Figure 3.18. A force \(\mathbf F\) displaces an object from \(P\) to \(Q\) along the displacement vector \(\mathbf D\text{.}\) The reference segment shows the magnitude of the force in the direction of motion, \(|\mathbf F|\cos\theta\text{.}\)

Example 3.19. Computing work done by a force.

A force is given by the vector \(\mathbf F = 8\mathbf i - 6\mathbf j + 9\mathbf k\) and moves a particle from the point \(A(0,10,8)\) to the point \(B(6,12,20)\text{.}\) Find the work done.
Hint.
Consider using the formula for the dot product: \(\mathbf F \cdot \mathbf D \text{.}\) For more help, watch the video below.
Figure 3.20. Set up the work computation: build the displacement vector first, then take the dot product. What do you get?
Solution.
The displacement vector is \(\mathbf D = \vec{AB} = \langle 6,2,12\rangle\text{,}\) and the work done is
\begin{equation*} \text{Work} = \mathbf F \cdot \mathbf D = \langle 8,-6,9\rangle \cdot \langle 6,2,12\rangle = 48 - 12 + 108 = 144. \end{equation*}

Example 3.21.

Suppose a mass of \(10\text{ kg}\) is resting on an inclined plane. Gravity exerts a force equivalent to \(mg\) on the object, where \(g = 9.8\text{ m/s}^2\) is the gravitational acceleration. Suppose the incline is tilted at a \(30^{\circ}\) angle. Compute the components of the force that are parallel and perpendicular to the inclined plane.
Solution.
Figure 3.22. Decomposing gravity on a \(30^{\circ}\) incline: the parallel component \(\vec{F}_{\parallel} = \operatorname{proj}_{\vec{v}}\vec{F}\) pulls the mass down the slope, and the perpendicular component \(\vec{F}_{\perp} = \vec{F} - \vec{F}_{\parallel}\) presses it into the surface.
First, we determine the total gravitational force vector \(\vec{F}\) acting on the mass. Using \(g = 9.8\text{ m/s}^2\text{:}\)
\begin{equation*} |\vec{F}| = (10)(9.8) = 98\text{ N} \end{equation*}
Acting entirely in the negative vertical direction, this gives:
\begin{equation*} \vec{F} = \langle 0, -98 \rangle = -98\hat{j} \end{equation*}
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Figure 3.23. Forces acting on a mass on a \(30^{\circ}\) incline: the weight \(\vec{F}\) resolves into a component \(\vec{F}_{\parallel}\) down the slope and a component \(\vec{F}_{\perp}\) into the surface, with \(\vec{F} = \vec{F}_{\parallel} + \vec{F}_{\perp}\text{.}\)
Next, we define a unit vector \(\vec{v}\) pointing down the surface of the inclined plane[cite: 1]. Since the plane is tilted at \(30^{\circ}\) (or \(\frac{\pi}{6}\) radians), a unit vector pointing down the slope is directed at an angle of \(-\frac{\pi}{6}\) relative to the horizontal[cite: 1]:
\begin{equation*} \vec{v} = \cos\left(\frac{\pi}{6}\right)\hat{i} - \sin\left(\frac{\pi}{6}\right)\hat{j} = \left\langle \frac{\sqrt{3}}{2}, -\frac{1}{2} \right\rangle \end{equation*}
where \(|\vec{v}| = 1\)[cite: 1].
The component of the force parallel to the incline, \(\vec{F}_{\parallel}\text{,}\) is the projection of \(\vec{F}\) onto \(\vec{v}\)[cite: 1]:
\begin{equation*} \vec{F}_{\parallel} = \operatorname{proj}_{\vec{v}}\vec{F} = \left(\frac{\vec{F} \cdot \vec{v}}{|\vec{v}|^2}\right)\vec{v} = (\vec{F} \cdot \vec{v})\vec{v} \end{equation*}
Computing the dot product[cite: 1]:
\begin{equation*} \vec{F} \cdot \vec{v} = \langle 0, -98 \rangle \cdot \left\langle \frac{\sqrt{3}}{2}, -\frac{1}{2} \right\rangle = 0 + 49 = 49 \end{equation*}
Thus, the parallel force vector is[cite: 1]:
\begin{equation*} \vec{F}_{\parallel} = 49 \left\langle \frac{\sqrt{3}}{2}, -\frac{1}{2} \right\rangle \end{equation*}
Since \(\vec{F} = \vec{F}_{\parallel} + \vec{F}_{\perp}\text{,}\) the perpendicular force component is found by subtraction[cite: 1]:
\begin{align*} \vec{F}_{\perp} \amp = \vec{F} - \vec{F}_{\parallel}\\ \amp = \langle 0, -98 \rangle - 49 \left\langle \frac{\sqrt{3}}{2}, -\frac{1}{2} \right\rangle\\ \amp = 49 \left( \langle 0, -2 \rangle - \left\langle \frac{\sqrt{3}}{2}, -\frac{1}{2} \right\rangle \right)\\ \amp = 49 \left\langle -\frac{\sqrt{3}}{2}, -\frac{3}{2} \right\rangle \end{align*}
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