Question 1.8: Vector addition helps win a Porsche As in Example 1.7, the v...
Vector addition helps win a Porsche
As in Example 1.7, the vectors in this example are specified in terms of magnitude and direction. The first thing we need to do is to find their components. Consider four finalists in a treasure hunt who are brought to the center of a large, flat field. Each is given a meterstick, a compass, a calculator, a shovel, and (in a different order for each contestant) these three displacements:
72.4 m, 32.0° east of north;
57.3 m, 36.0° south of west;
17.8 m straight south.
The three displacements lead to the point where the keys to a new Porsche are buried. The finder gets the Porsche. Three of the four contestants start measuring immediately, but the winner first calculates where to go. What does he calculate?
SET UP The situation is shown in Figure 1.21. We have chosen the x axis as east and the y axis as north. This is the usual choice for maps. Let \overset{\rightarrow }{\pmb{A}} , be the first displacement, \overset{\rightarrow }{\pmb{B}} , the second, and \overset{\rightarrow }{\pmb{C}} , the third. We can estimate from the diagram that the vector sum \overset{\rightarrow }{\pmb{R}} , is about 10 m, 40° west of north.
The angles of the vectors, measured counterclockwise from the +x axis, are 58°, 216°, and 270°. We have to find the components of each.
SOLVE The components of \overset{\rightarrow }{\pmb{A}} , are
A_x=A\cos\theta =(72.4 m)(\cos 58°)=38.37 m,
A_y=A\sin \theta =(72.4 m)(\sin 58°)=61.40 m.
The table shows the components of all the displacements, the addition of components, and the other calculations. To find the magnitude R and direction θ, use the Pythagorean theorem R=\sqrt{R^2_x+R_y^2} and the definition of the inverse tangent, \theta =\tan^{-1}\frac{R_y}{R_x}. Always arrange your component calculations systematically as in the table. Note that we have kept one too many significant figures in the components; we’ll wait until the end to round to the correct number of significant figures.
| Distance | Angle | x component | y component |
| A = 72.4 m | 58° | 38.37 m | 61.40 m |
| B = 57.3 m | 216° | -46.36 m | -33.68 m |
| C = 17.8 m | 270° | 0.00 m
|
-17.80 m |
| R_x = -7.99 m | R_y = 9.92 m |
R=\sqrt{(-7.99 m)^2+(9.92 m)^2}=12.7 m,
\theta =\tan^{-1}\frac{9.92 m}{-7.99 m}=129° west of north
REFLECT The losers try to measure three angles and three distances totaling 147.5 m, 1 meter at a time. The winner measures only one angle and one much shorter distance.
Practice Problem: In a second contest, the three displacements given are 63.5 m, 52° west of north; 12.6 m straight north; and 81.9 m, 41° south of east. What magnitude and direction does the winner calculate for the displacement to find the keys to the Porsche? Answers: R = 11.9 m and θ = 9.8° south of east.
