Question #228416

The resultant of two forces in a plane is 400 N at 120 degrees. If one of the forces is 200 lbs at 20 degrees what is the other force?


Expert's answer

Given:

R=400 NR=400\:\rm N

θ=120∘\theta=120^{\circ}

F1=200 lbs=890 NF_1=200\:\rm lbs=890\: N

α=20∘\alpha=20^{\circ}




The resultant force

R=F1+F2\bf R=F_1+F_2

Hence,

F2=R−F1\bf F_2=R-F_1

F2=R2+F12−2RF1cos⁡(120∘−20∘)F_2=\sqrt{R^2+F_1^2-2RF_1\cos(120^{\circ}-20^{\circ})}

F2=4002+8902−2∗400∗890cos⁡(100∘)F2=1037 NF_2=\sqrt{400^2+890^2-2*400*890\cos(100^{\circ})}\\ F_2=1037\:\rm N


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