Question #161284

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.


Expert's answer

Fx=F1cos78°+F2cos37°F3+F4sin50°=80.64 N.F_x=-F_1cos78°+F_2cos37°-F_3+F_4sin50°=-80.64~N.

Fy=F1sin78°+F2sin37°F4cos50°=24.09 N.F_y=F_1sin78°+F_2sin37°-F_4cos50°=24.09~N.

F=Fx2+Fy2=84.16 NF=\sqrt{F_x^2+F_y^2}=84.16~N directed at arctanFxFy=73.37°arctan\frac{F_x}{F_y}=73.37° west of north.


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