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.


1
Expert's answer
2021-02-04T17:13:26-0500

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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