Given the forces Fa= 2.3 N [S 35° W], Fb= 3.6N [N 14° W] and Fc= 4.2N [S 24°E] calculate the following:

Question

Fa + Fb + Fc

Answer 1

Fa = 2.3N[235o],CCW.
Fb = 3.6N[104o],CCW.
Fc = 4.2N[294o],CCW.

X=2.3*cos235+3.6*cos104+4.2*cos294 =
-0.482 N.
Y=2.3*sin235+3.6*sin104+4.2*sin294=
-2.28 N.

tan Ar = Y/X = -2.28/-0.482 = 4.62215
Ar = 77.8o = Reference angle.
A = 77.8 + 180 = 257.8o CCW = 12.2o W of
S.

Fr = Y/sinA = -2.28/sin257.8 = 2.33 N.=
Resultant force.

Answer 2

Fa + Fb + Fc = 2.33N[S12.2W]