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# Answer to Question #119193 in Mechanics | Relativity for glyy

Question #119193
three forces pull on object: f₁ = 50-N towards the +x; F₂ = 50-N toward the +y and F₃ = 70-N, 30° below -x. what force will balance the three? [Analythical method]
1
2020-06-01T14:26:36-0400

Solution.

"F_1=50N;"

"F_2=50N;"

"F_3=70N;"

"\\phi=30^o;"

"F-?;"

"\\overrightarrow{F}=\\overrightarrow{F_1}+\\overrightarrow{F_2}+\\overrightarrow{F_3};"

"\\overrightarrow{F_{12}}=\\overrightarrow{F_1}+\\overrightarrow{F_2};"

Since the forces F1 and F2 are perpendicular, the equivalent of these forces is found by Pythagoras' theorem.

"F_{12}=\\sqrt{F_1^2+F_2^2};"

"F_{12}=\\sqrt{(50N)^2+(50N)^2}=50\\sqrt{2}N;"

By the cosine theorem, we find the equivalent module:

"F^2=F_{12}^2+F_3^2-2F_{12}F_3cos\\theta" ;

"F^2=(50\\sqrt{2}N)^2+(70N)^2-2\\sdot50\\sqrt{2}N\\sdot70N\\sdot cos15^o="

"=339N^2;"

"F=18.4N;"

"\\dfrac{F_3}{sin\\alpha}=\\dfrac{F}{sin\\theta};\\implies sin\\alpha=\\dfrac{F_3sin\\theta}{F};"

"sin\\alpha=\\dfrac{70N\\sdot 0.2588}{18.4N}=0.9845;"

"\\alpha=80^o"

Equivalent force is directed at an angle of 55o above -x.

Answer: "F=18.4N" at an angle of 55o above -x.

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