9.3.2 Work as a Dot Product of Two Vectors

There’s something different about the quantity $F (Δ r) cos θ$ in Equation 9.17. We’ve spent many chapters adding vectors, but this is the first time we’ve multiplied two vectors. Multiplying vectors is not like multiplying scalars. In fact, there is more than one way to multiply vectors. We will introduce one way now, the dot product.

FIGURE 9.8 shows two vectors, $\vec{A}$ and $\vec{B}$ , with angle $α$ between them. We define the dot product of $\vec{A}$ and $\vec{B}$ as

\vec{A} \cdot \vec{B} = A B cos α

(9.18)

A dot product must have the dot symbol $\cdot$ between the vectors. The notation $\vec{A} \vec{B}$ , without the dot, is not the same thing as $\vec{A} \cdot \vec{B}$ . The dot product is also called the scalar product because the value is a scalar. Later, when we need it, we’ll introduce a different way to multiply vectors called the cross product.

The figure illustrates the method to calculate the dot product using two vectors and the angle between them.

FIGURE 9.8 Full Alternative Text

The dot product depends on the orientation of the vectors. FIGURE 9.9 shows five different situations, including the three “special cases” where $α = 0 °, 90 °$ , and $180 °$ .

The figure consists of five parts, and illustrate the different orientations of a pair of vectors to calculate their dot product.

FIGURE 9.9 Full Alternative Text

NOTE The dot product of a vector with itself is well defined. If $\vec{B} = \vec{A}$ (i.e., $\vec{B}$ is a copy of $\vec{A}$ ), then $α = 0 °$ . Thus $\vec{A} \cdot \vec{A} = A^{2}$ .

EXAMPLE 9.4 Calculating a dot product

Compute the dot product of the two vectors in FIGURE 9.10.

The figure illustrates two vectors, and the angle between them, to calculate the dot product.

FIGURE 9.10 Full Alternative Text

SOLVE The angle between the vectors is $α = 30 °$ , so

\vec{A} \cdot \vec{B} = A B cos α = (3) (4) cos 30 ° = 10.4

Like vector addition and subtraction, calculating the dot product of two vectors is often performed most easily using vector components. FIGURE 9.11 reminds you of the unit vectors $î$ and $ĵ$ that point in the positive x-direction and positive y-direction. The two unit vectors are perpendicular to each other, so their dot product is $î \cdot ĵ = 0$ . Furthermore, because the magnitudes of $î$ and $ĵ$ are 1, $î \cdot î = 1$ and $ĵ \cdot ĵ = 1$ .

The figure illustrates two unit vectors labeled i hat and j hat on an x y coordinate plane.

FIGURE 9.11 Full Alternative Text

In terms of components, we can write the dot product of vectors $\vec{A}$ and $\vec{B}$ as

\vec{A} \cdot \vec{B} = (A_{x} î + A_{y} ĵ) \cdot (B_{x} î + B_{y} ĵ)

Multiplying this out, and using the results for the dot products of the unit vectors:

\begin{array}{l} \vec{A} \cdot \vec{B} & = A_{x} B_{x} î \cdot î + (A_{x} B_{y} + A_{y} B_{x}) î \cdot \hat{j} + A_{y} B_{y} \hat{j} \cdot \hat{j} \\ = A_{x} B_{x} + A_{y} B_{y} \end{array}

(9.19)

That is, the dot product is the sum of the products of the components.

EXAMPLE 9.5 Calculating a dot product using components

Compute the dot product of $\vec{A} = 3 î + 3 ĵ$ and $\vec{B} = 4 î - ĵ$ .

SOLVE FIGURE 9.12 shows vectors $\vec{A}$ and $\vec{B}$ . We could calculate the dot product by first doing the geometry needed to find the angle between the vectors and then using Equation 9.18. But calculating the dot product from the vector components is much easier. It is

\vec{A} \cdot \vec{B} = A_{x} B_{x} + A_{y} B_{y} = (3) (4) + (3) (- 1) = 9

The figure illustrates two vectors drawn in the x y coordinate plane, to calculate the dot product using components of the vectors.

FIGURE 9.12 Full Alternative Text

Looking at Equation 9.17, the work done by a constant force, you should recognize that it is the dot product of the force vector and the displacement vector:

W = \vec{F} \cdot Δ \vec{r} (work done by a constant force)

(9.20)

This definition of work is valid for a constant force.

EXAMPLE 9.6 Calculating work using the dot product

A 70 kg skier is gliding at $2.0 m / s$ when he starts down a very slippery 50-m-long, $10 °$ slope. What is his speed at the bottom?

MODEL Model the skier as a particle and interpret “very slippery” to mean frictionless. Use the energy principle to find his final speed.

VISUALIZE FIGURE 9.13 shows a pictorial representation.

The figure illustrates the initial and final position of a skier of mass 70 kilograms, skiing on a slope of 10 degrees.

FIGURE 9.13 Full Alternative Text

SOLVE The only forces on the skier are $\vec{F}_{G}$ and $\vec{n}$ . The normal force is perpendicular to the motion and thus does no work. The work done by gravity is easily calculated as a dot product:

\begin{array}{l} W & = {\vec{F}}_{G} \cdot Δ \vec{r} = m g (Δ r) cos α \\ = (70 kg)(9 .8 {m/s}^{2})(50 m)cos 80 ° = 5960 J \end{array}

Notice that the angle between the vectors is $80 °$ , not $10 °$ . Then, from the energy principle, we find

\begin{array}{l} Δ K = \frac{1}{2} m v_{1}^{2} - \frac{1}{2} m v_{0}^{2} = W \\ v_{1} = \sqrt{v_{0}^{2} + \frac{2 W}{m}} = \sqrt{{(2.0 m/s)}^{2} + \frac{2 (5960 J)}{70 kg}} = 13 m/s \end{array}

NOTE While in the midst of the mathematics of calculating work, do not lose sight of what the energy principle is all about. It is a statement about energy transfer: Work causes a particle’s kinetic energy to either increase or decrease.

9.3.2 Work as a Dot Product of Two Vectors

FIGURE 9.8 Vectors $\vec{A}$ and $\vec{B}$ , with angle $α$ between them. Their dot product is $\vec{A} \cdot \vec{B} = A B cos α$ .

FIGURE 9.9 The dot product $\vec{A} \cdot \vec{B}$ as $α$ ranges from 0 $°$ to 180 $°$ .

EXAMPLE 9.4 Calculating a dot product

FIGURE 9.10 Vectors $\vec{A}$ and $\vec{B}$ of Example 9.4.

FIGURE 9.11 The unit vectors $î$ and $ĵ$ .

EXAMPLE 9.5 Calculating a dot product using components

FIGURE 9.12 Vectors $\vec{A}$ and $\vec{B}$ .

EXAMPLE 9.6 Calculating work using the dot product

FIGURE 9.13 Pictorial representation of the skier.