Answer to Question #161284 in Molecular Physics | Thermodynamics for Kyle

Question

a set of forces are acting on an object. The first force, with a magnitude of 24.70N, is directed at -258 degrees and the second force, with a magnitude of 53.20B, is directed at 37 degrees north of east. The third force with a magnitude of 166.50N is directed at 270, and the fourth force of 49.92N is directed at 50 degrees east of south. Solve the resultant force and exact direction analytically.

Answer 1

$F_x=-F_1cos78°+F_2cos37°-F_3+F_4sin50°=-80.64~N.$

$F_y=F_1sin78°+F_2sin37°-F_4cos50°=24.09~N.$

$F=\sqrt{F_x^2+F_y^2}=84.16~N$ directed at $arctan\frac{F_x}{F_y}=73.37°$ west of north.