3.4.2 Tilted Axes and Arbitrary Directions

As we’ve noted, the coordinate system is entirely your choice. It is a grid that you impose on the problem in a manner that will make the problem easiest to solve. As you saw in Chapter 2, it is often convenient to tilt the axes of the coordinate system, such as those shown in FIGURE 3.20. The axes are perpendicular, and the y-axis is oriented correctly with respect to the x-axis, so this is a legitimate coordinate system. There is no requirement that the x-axis has to be horizontal.

The figure illustrates the components of a displacement vector labeled vector C on the x y coordinate system with tilted axes.

FIGURE 3.20 Full Alternative Text

Finding components with tilted axes is no harder than what we have done so far. Vector $\vec{C}$ in FIGURE 3.20 can be decomposed into $\vec{C} = C_{x} î + C_{y} ĵ$ , where $C_{x} =$ $C cos θ$ and $C_{y} = C sin θ$ . Note that the unit vectors $î$ and $ĵ$ correspond to the axes, not to “horizontal” and “vertical,” so they are also tilted.

Tilted axes are useful if you need to determine component vectors “parallel to” and “perpendicular to” an arbitrary line or surface. This is illustrated in the following example.

EXAMPLE 3.7 Muscle and bone

The deltoid—the rounded muscle across the top of your upper arm—allows you to lift your arm away from your side. It does so by pulling on an attachment point on the humerus, the upper arm bone, at an angle of $15 °$ with respect to the humerus. If you hold your arm at an angle $30 °$ below horizontal, the deltoid must pull with a force of 720 N to support the weight of your arm, as shown in FIGURE 3.21a. (You’ll learn in Chapter 5 that force is a vector quantity measured in units of newtons, abbreviated N.) What are the components of the muscle force parallel to and perpendicular to the bone?

The figure consists of two parts labeled ("a") and (b). The first part illustrates the force vector acting on a deltoid muscle of a human arm. The second part illustrates the components of that force on the x y coordinate plane.

FIGURE 3.21 Full Alternative Text

VISUALIZE FIGURE 3.21b shows a tilted coordinate system with the x-axis parallel to the humerus. The force $\vec{F}$ is shown $15 °$ from the x-axis. The component of force parallel to the bone, which we can denote $F_{‖}$ , is equivalent to the x-component: $F_{‖} = F_{x}$ . Similarly, the component of force perpendicular to the bone is $F_{⊥} = F_{y}$ .

SOLVE From the geometry of Figure 3.21b, we see that

\begin{array}{l} F_{‖} = F cos 15 ° = (720 N) cos 15 ° = 695 N \\ F_{⊥} = F sin 15 ° = (720 N) sin 15 ° = 186 N \end{array}

REVIEW The muscle pulls nearly parallel to the bone, so we expected $F_{‖} \approx 720 N$ and $F_{⊥} ≪ F_{‖}$ . Thus our results seem reasonable.

Chapter 3 CHALLENGE EXAMPLE Finding the net force

FIGURE 3.22 shows three forces acting at one point. What is the net force $\vec{F}_{net} = \vec{F}_{1} + \vec{F}_{2} + \vec{F}_{3}$ ?

VISUALIZE FIGURE 3.22 shows the forces and a tilted coordinate system.

The figure illustrates three force vector arrows labeled vector F subscript 1, vector F subscript 2, and vector F subscript 3 on the x y coordinate system with tilted axes.

FIGURE 3.22 Full Alternative Text

SOLVE The vector equation $\vec{F}_{net} = \vec{F}_{1} + \vec{F}_{2} + \vec{F}_{3}$ is really two simultaneous equations:

\begin{array}{l} (F_{net})_{x} = F_{1 x} + F_{2 x} + F_{3 x} \\ (F_{net})_{y} = F_{1 y} + F_{2 y} + F_{3 y} \end{array}

The components of the forces are determined with respect to the axes. Thus

\begin{array}{l} F_{1 x} = F_{1} cos 45 ° = (50 N) cos 45 ° = 35 N \\ F_{1 y} = F_{1} sin 45 ° = (50 N) sin 45 ° = 35 N \end{array}

$\vec{F}_{2}$ is easier. It is pointing along the y-axis, so $F_{2 x} = 0 N$ and $F_{2 y} = 20 N$ . To find the components of $\vec{F}_{3}$ , we need to recognize—because $\vec{F}_{3}$ points straight down—that the angle between $\vec{F}_{3}$ and the x-axis is $75 °$ . Thus

\begin{array}{l} F_{3 x} = F_{3} cos 75 ° = (57 N) cos 75 ° = 15 N \\ F_{3 y} = - F_{3} sin 75 ° = - (57 N) sin 75 ° = - 55 N \end{array}

The minus sign in $F_{3 y}$ is critical, and it appears not from some formula but because we recognized—from the figure—that the y-component of $\vec{F}_{3}$ points in the $- y$ -direction. Combining the pieces, we have

\begin{array}{l} (F_{net})_{x} = 35 N + 0 N + 15 N = 50 N \\ (F_{net})_{y} = 35 N + 20 N + (- 55 N) = 0 N \end{array}

Thus the net force is $\vec{F}_{net} = 50 î N$ . It points along the x-axis of the tilted coordinate system.

REVIEW Notice that all work was done with reference to the axes of the coordinate system, not with respect to vertical or horizontal.

3.4.2 Tilted Axes and Arbitrary Directions

FIGURE 3.20 A coordinate system with tilted axes.

EXAMPLE 3.7 Muscle and bone

FIGURE 3.21 Finding the components of force parallel and perpendicular to the humerus.

Chapter 3 CHALLENGE EXAMPLE Finding the net force

FIGURE 3.22 Three forces.