F = 365N[228o]CCW+224N[334o]CCW+488N [102o]CCW.
Fx=365*Cos228 + 224*Cos334 + 488*cos102
= -144.4 N.
Fy=365*sin228 + 224*sin334 + 488*sin102
= 107.9 N.
Tan Ar = Fy/Fx = 107.9/-144.4 = -0.74718
Ar = -36.8o = Reference angle.
A = -36.8 + 180 = 143.2o CCW = 36.8o N.
of W.
F = Fx/Cos A = -144.4/Cos143.2 = 180.3 N
[143.2]CCW = 180.3N[36.8o] North of West.
⦁ Find the resultant vector (magnitude and direction) for the sum of:
vector A which has magnitude 365 N at 48° south of west,
vector B which has magnitude 224 N at 26° south of east, and
vector C which has magnitude 488 N at 12° west of north.
Express the direction of the resultant relative to due west.
2 answers
fire