Two vector, 6 and 9 units long, from an angle of (a) 0°, (b) 60°, (c) 90°, (d) 150°, and (e) 180°. Find the magnitude and direction of their resultant with respect to the shorter vector?
1 Answer
For part b, the length is 13 units and the angle is
Explanation:
Start with part b.
I will recommend that you use the "tip to tail" graphical method (also called the "head to tail" method) to sketch the resultant. Unfortunately, this does cause one complication. I will try to help you deal with that complication when we get there. Open this link if you need a refresher on the "tip to tail" method.
Start with the short side using the "tip to tail" method.. (Drawing it carefully to scale is not necessary since we will switch to the Law of cosines once the triangle is sketched.) Using the "tip to tail" method will give you a triangle.
Now we will use the Law of Cosines. You know the values of 2 sides and 1 angle. Label the 6 unit long side, side a; and the 9 unit long side, side b. Label the angle between those 2 sides angle
Before we plug in data, there is one complication. The angle between the directions of the 2 vectors of part b is given as 60 degrees. That would be the angle you see if you drew them both originating from the same spot.
In the "tip to tail" method, you slide the 2nd vector out to the end of the 1st vector, so the angle between them,
Next is to determine the angle between side a and the resultant (side c). We label that angle
Plugging in the data,
9^2 = 6^2 + 13^2 - 2613*cosbeta#
Solving that for
For parts a, c, d, and e, subtract the given angle from 180 and use the result along with the 6 and 9 units for the lengths as above. And then execute the Laws of Cosines, as above.
I hope this helps,
Steve