An object in equilibrium has three forces exerted on it. A 36 N force acts at 90° from the x axis and a 42 N force acts at 60°. What are the magnitude and direction of the third force?

Answer 1

ΣF(x) =F1(x) +F2(x) +F3(x) =0
0+42•cos60°+F3(x) =0
F3(x) = -42•cos60°= -21 N
ΣF(y) =F1(y) +F2(y) +F3(y) =0
36+42•sin60°+F3(y) =0
F3(y) =-36 -42•sin60°= -72.4 N
F=sqrt{F(x)²+F(y)²}
tan θ=F(y)/F(x)