Question

There are four forces acting on the point O as represented below. F1=20 N, F2= 30 N, F3= 50N, F4= 25 N, and the forces F1, and F2 are acting along the x- and y-axises, respectively. If Q1=30, and Q2= 40, what is the net force at the point O?

F3 3 Q2 F FA

Answer 1

sum of X- component of all forces Fx =F1cos0 + F2cos90 -F3cos(90-Q1)-F4cos(Q2)

Fx = 20+0-50cos(90-30)-25cos(40) = -24.15

sum of Y- component of all forces Fy =F1sin90+F2sin90+F3sin(90-Q1)-F4sin(Q2)

Fy = 0+30+50sin60 - 25sin40 =46.07

magnitude of resultant F = $\sqrt{Fx^{2}+Fy^{2}}$ =$\sqrt{24.15^{2}+46.07^{2}}$

magnitude of net resultant F =52.016

angle dirrection= Q = tan^-1 (Fy/Fx )=-62.34⁰ (62.34⁰ degree with -X axis)