Consider three force vectors F~1 with magnitude 69 N and direction 278?, F~2 wit
ID: 1953637 • Letter: C
Question
Consider three force vectors F~1 with magnitude 69 N and direction 278?, F~2 with magnitude 64 N and direction 320?, and F~3 with magnitude 94 N and direction 155?. All direction angles ? are measured from the positive x axis: counter-clockwise for ? > 0 and clockwise for ? < 0. What is the magnitude of the resultant vector F~, where F~ = F~1 + F~2 + F~3 ?What is the direction of F~as an angle between the limits of -180? and +180? from
the positive x axis with counterclockwise as the positive angular direction?
Answer in units of ?.
Explanation / Answer
F1 = 69(cos(278)i + sin(278)j)
F1 = 9.60294397i -68.3284967j
F2 = 64(cos(320)i + sin(320)j)
F2 = 49.0268444i - 41.138407j
F3 = 94(cos(155)i + sin(155)j)
F3 = -85.192932i + 39.7261166j
F1 + F2 + F3 = -26.5631436i -69.7407871j = Ai + Bj
The magnitude is:
sqrt(A^2 + B^2) = 74.6282653 = 74.63 N
The angle is given by:
arctan(B/A) = 69.148968 degrees
But since both components were negative then the angle needs to be in the third quadrant and since you want a value between -180 and 180, just take away 180 degrees to get:
-110.851032 = -110.85 degrees