Answer to Question #123754 in Analytic Geometry for Emmanuel Kojo Mensa Kent

Question

Two equal concurrent forces, each of F N have a resultant of 6 N. When the magnitude of one of the forces is doubled, the resultant becomes 11 N. Calculate the value of F and the angle between the forces.

Answer 1

Let $F_1=F_2=x,$ $\ \ \alpha$ - angle between these forces.

$\vec{F_3}=\vec{F_1}+\vec{F_2},\ \ F_3=6;$

$\vec{F_4}=2\vec{F_1}+\vec{F_2}=\vec{F_3}+\vec{F_1},\ \ F_4=11;$

We can find $F_3$ using formula: $F_3=2F_2\cos(\alpha /2), \ \ 3=x\cos(\alpha /2)$ .

($AB=BC, \ BD\perp AC\ \Rightarrow AC=AD= AB\cos(\alpha/2))$

And we can find $F_4$ using law of cosines for $\Delta ECA:$ $11^2=x^2+6^2-2\times x\times 6\times \cos (\pi-\alpha/2)=x^2+36+12x\cos(\alpha/2)=x^2+36+12\times 3,$

$121=72+x^2, \ \ x^2=49, \ \ x=7.$

$3=x\cos(\alpha/2)=7\cos (\alpha/2), \ \ \cos(\alpha/2)=3/7, \ \ \alpha/2=\arccos 3/7, \ \ \alpha=2\arccos 3/7.$

Answer: $F=7, \ \ \alpha =2\arccos 3/7.$