Suppose a pilot flies 40.0 km in a direction 60º north of east and then flies 30.0 km in a direction 15º north of east as shown in Figure 3.61. Find her total distance R from the starting point and the direction θ of the straight-line path to the final position. Discuss qualitatively how this flight would be altered by a wind from the north and how the effect of the wind would depend on both wind speed and the speed of the plane relative to the air mass.
Solution:
The pilot’s displacement is characterized by 2 vectors, A and B, as depicted in Figure 3.61. To determine her total displacement R from the starting point, we need to add the two given vectors. To do this, we individually get the x and y components of each vector. This is presented in the table that follows:
Vector
x-component
y-component
A
40\:\cos 60^{\circ} =20\:\text{km}
40\:\sin 60^{\circ} =34.6410\:\text{km}
B
30\:\cos 15^{\circ} =28.9778\:\text{km}
30\:\sin 15^{\circ} =7.7646\:\text{km}
Sum
48.9778\: \text{km}
42.4056 \:\text{km}
The table above indicates east and north as positive components, while west and south indicate negative components. The last row is the sum of the components. These are also the x and y components of the resultant vector.
To calculate the magnitude of the resultant, we simply use the Pythagorean Theorem as follows:
Therefore, the pilot’s resultant displacement is about 64.8 km directed 40.9° North of East from the starting island.
Discussion:
If the wind speed is less than the speed of the plane, it is possible to travel to the northeast, but she will travel more to the east than without the wind. If the wind speed is greater than the speed of the plane, then it is no longer possible for the plane to travel to the northeast, it will end up traveling southeast.
In an attempt to escape his island, Gilligan builds a raft and sets to sea. The wind shifts a great deal during the day, and he is blown along the following straight lines: 2.50 km 45.0º north of west; then 4.70 km 60.0º south of east; then 1.30 km 25.0º south of west; then 5.10 km straight east; then 1.70 km 5.00º east of north; then 7.20 km 55.0º south of west; and finally 2.80 km 10.0º north of east. What is his final position relative to the island?
Solution:
Gilligan’s displacement is characterized by 7 vectors. To determine his final position relative to the starting point, we simply need to add the vectors. To do this, we individually get the x and y components of each vector. This is presented in the table that follows:
Vector
X-Component
Y-Component
(1)
-2.5\:\cos 45^{\circ} =-1.7678\:\text{km}
+2.5\:\sin 45^{\circ} =+1.7678\:\text{km}
(2)
+4.70\:\cos 60^{\circ} =+2.3500\:\text{km}
-4.70\:\sin 60^{\circ} =-4.0703\:\text{km}
(3)
-1.30\:\cos 25^{\circ} =-1.1782\:\text{km}
-1.30\:\sin 25^{\circ} =-0.5494\:\text{km}
(4)
+5.1000\:\text{km}
0
(5)
+1.70\:\sin 5^{\circ} =+0.1482\:\text{km}
+1.70\:\cos 5^{\circ} =+1.6935\:\text{km}
(6)
-7.20\:\cos 55^{\circ} =-4.1298\:\text{km}
-7.20\:\sin 55^{\circ} =-5.8979\:\text{km}
(7)
+2.80\:\cos 10^{\circ} =+2.7575\:\text{km}
+2.80\:\sin 10^{\circ} =+0.4862\:\text{km}
Sum
3.2799\:\text{km}
-6.5701\:\text{km}
The table above indicates east and north as positive components, while west and south indicate negative components. The last row is the sum of the components. This is also the x and y components of the resultant vector.
To calculate the magnitude of the resultant, we simply use the Pythagorean Theorem as follows:
Find the components of vtot along a set of perpendicular axes rotated 30º counterclockwise relative to those in Figure 3.55.
Solution:
By isolating the vtot from the rest of the other vectors, we come up with the following figure.
The resultant velocity has a magnitude of 6.72 m/s and is directed 49° from the positive x-axis. Now, we shall create another set of axes rotated at 30° counterclockwise. We call the axes x’ and y’ axes. The figure is shown below.
From the figure, we can see that the resultant velocity is 19° from the x’ axis. Therefore, the x’ and y’ components are:
Find the components of vtot along the x- and y-axes in Figure 3.55.
Solution:
By isolating the vtot from the rest of the other vectors, we come up with the following figure. Also, the x and y-components are shown.
The resultant velocity has a magnitude of 6.72 m/s and is directed 49° from the positive x-axis. To solve for the x and y components, we just need to solve the legs of the right triangle formed by the three vectors. That is,
Find the magnitudes of velocities vA and vB in Figure 3.55
Solution:
Basically, we are given an oblique triangle. First, we shall determine the value of the interior angle at the intersection of vA and vB. We can solve this knowing that the sum of the interior angles of a triangle is 180°.
Show that the order of addition of three vectors does not affect their sum. Show this property by choosing any three vectors A, B, and C, all having different lengths and directions. Find the sum A + B + C then find their sum when added in a different order and show the result is the same. (There are five other orders in which A, B, and C can be added; choose only one.)
Solution:
Consider the three vectors shown in the figures below:
Vector A
Vector B
Vector C
First, we shall add them A+B+C. Using the head-tail or graphical method of vector addition, we have the figure shown below.
Now, let us try to find the sum of the three vectors by reordering vectors A, B, and C. Let us try to find the sum of C+B+A in that order. The result is shown below.
We can see that the resultant is the same directed from the origin upward. This proves that the resultant must be the same even if the vectors are added in different order.
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