MATH 117
Adding Forces
Solution
1. Let F1 be a force of 50 pounds in the direction 30º West of South.
Let F2 be a force of 60 pounds in the direction 40º East of South.
Compute the magnitude and heading of the resultant F1 + F2 .
Solution.
F1 = ( 50 cos240º , 50sin 240º )
F2 = ( 60 cos310º , 60sin 310º )
So the resultant force is
F1 + F2 = ( 50 cos240º + 60 cos310º , 50sin 240º + 60sin 310º )
≈ (13.56725658, –89.26393678)
Then F1 + F2 has magnitude
F1 + F2 = 13.567256582 + 89.263936782 ≈ 90.289 pounds .
Its angle in the 4th Quadrant is tan−1(−89.26393678 / 13.56725658) ≈ −81.3577381 , which
(by adding 90º) gives a heading of 8.64226º East of South.
2. (a) A small plane flies at 240 mph in a direction 20º North of West. It encounters a
35 mph wind heading 30º East of North. Find the resulting speed and heading of the
plane.
Plane: P = ( 240 cos160º , 240sin160º )
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Wind: W = ( 35cos60º , 35sin 60º )
Result: R = P + W = ( 240 cos160º + 35cos60º , 240sin160º + 35sin 60º )
≈ (–208.026229, 112.3957235) = (A, B)
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Resulting Speed:
P + W ≈ A 2 + B 2 ≈ 236.448 mph
Angle: tan−1(B / A) ≈ −28.382º ; thus, the resulting heading is 28.382º North of West.
(b) If the plane encounters the wind described in Part (a), what speed and heading
should it take in order to stay on course its desired course of 20º North of West at 240
mph?
Desired result is R = ( 240 cos160º , 240sin160º )
Wind: W = ( 35cos60º , 35sin 60º )
R = P + W , so
€ P = R − W = ( 240 cos160º – 35cos60º ,€240sin160º – 35sin 60º )
≈ (–243.026229, 51.77394527) = (A, B)
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The plane’s speed should be
A 2 + B 2 ≈ 248.48 mph
And tan−1(B / A) ≈ –12.0264 , so the plane should fly 12.0264º North of West.