A-Consider four vectors F~ 1, F~ 2, F~ 3, and F~ 4 with magnitudes F1 = 42 N, F2
ID: 1576260 • Letter: A
Question
A-Consider four vectors F~ 1, F~ 2, F~ 3, and F~ 4 with magnitudes F1 = 42 N, F2 = 29 N, F3 = 25 N, and F4 = 52 N, 1 = 120 , 2 = 150 , 3 = 34 , and 4 = 59 , measured from the positive x axis with counterclockwise positive. What is the magnitude of the resultant vector F~ = F~ 1 + F~ 2 + F~ 3 + F~ 4? Answer in units of N. THE ANSWER IS (8.83) I am looking for answer B
B-What is the direction of this resultant vector F~ , within the limits of 180 and 180 as measured from the positive x axis with counterclockwise positive ? Answer in units of ***********************************************************************************************only answers (B)********************************************************************** .
Explanation / Answer
F1 = 42*cos120 i + 42*sin120 j
F2 = 29*cos(-150) i + 29*sin(-150) j
F3 = 25*cos34 i + 25*sin34 j
F4 = 52*cos(-59) i + 52*sin(-59) j
F = F1 + F2 + F3 + F4
F = (42cos(120))+(29cos(-150))+(25cos(34))+(52cos(-59)) i + (42sin(120))+(29sin(-150))+(25sin(34))+(52sin(-59)) j
F = 1.39 i - 8.72 j
magnitude = sqrt(1.39^2+8.72^2) = 8.83 N
part(B)
direction = tan^-1(-8.72/1.39) = -80.9o or -81.0 o