Answer to Question #278460 in Physics for katie

Question

 Three horizontal ropes pull on a large stones stack in the ground, producing the vector forces A, B & C. Find the magnitude and direction of a fourth force on the stone that will make the vector sum of the four forces zero.

A = 150N 40o North of East

B = 90N 40o West of North

C = 50N 63o South of West 

Answer 1

Find the components of the forces in N and E directions:

$A_N=150\sin40°=96\text{ N},\\ A_E=150\cos40°=115\text{ N}.\\ B_N=90\cos40°=69\text{ N},\\ B_E=-90\sin40°=-58\text{ N}.\\ C_N=-50\sin63°=-45\text{ N},\\ C_E=-50\cos63°=-23\text{ N}.\\\space\\ R_N=A_N+B_N+C_N=120\text{ N},\\ R_E=A_E+B_E+C_E=34\text{ N}.$

Therefore, the balancing force must be 120 N directed South and 34 N directed W, or

$R=\sqrt{R_N^2+R_E^2}=125\text{ N}.\\\space\\ \theta=\arctan\frac{120}{34}=74°\text{ S of W}.$