Components of Vectors — Question 9

Question 9

Horizontal span and vertical rise

A straight cable rises from a lower anchor to an upper anchor 8 m higher. Its horizontal projection is 15 m long and points east. Ignore sag. Let positive $x$ be east and positive $y$ upward.

1. Find the cable length and its inclination above horizontal.

2. A technician writes $L\sin\theta=15$ m and $L\cos\theta=8$ m. What angle would make those equations correct?

3. If the same 17 m cable instead has a 15 m vertical rise, find its horizontal span and inclination. Is this the original geometry?

Question 9 - Solution

Horizontal span and vertical rise

(a) The cable is the hypotenuse.

$$\boxed{L=\sqrt{15^2+8^2}=17\,\mathrm m},\quad\theta=\tan^{-1}(8/15)=28.072^\circ$$ $L=\sqrt{15^2+8^2}=17$ m. The inclination above horizontal is $\theta=\tan^{-1}(8/15)\approx28.07^\circ$. The 15 m horizontal span is a projection, not the cable length.

(b) Identify the reference axis.

$$\phi=90^\circ-\theta=61.928^\circ,\quad L\sin\phi=15\,\mathrm m,\quad L\cos\phi=8\,\mathrm m$$ The equations are correct for an angle measured from vertical, $\phi=\tan^{-1}(15/8)\approx61.93^\circ$. Then $L\sin\phi=15$ m and $L\cos\phi=8$ m. For the horizontal angle, they must instead read $L\cos\theta=15$ m and $L\sin\theta=8$ m.

(c) Exchange the legs deliberately.

$$x'=\sqrt{17^2-15^2}=8\,\mathrm m,\quad\theta'=\tan^{-1}(15/8)=61.928^\circ$$ The new horizontal span is $\sqrt{17^2-15^2}=8$ m and the inclination is $\tan^{-1}(15/8)\approx61.93^\circ$. It uses the same length but a different direction and different components. Complementary angles describe the same arrow only when the reference axis also changes.