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{{context|date=August 2012}}
{{expert-subject|1=Mathematics|date=August 2012}}
[[File:Force Diagram.svg|thumb |alt=A diagram that shows the direction of several forces. |The load is pointing straight down, and is being held up by two forces pointing up at various angles <math>\alpha</math> and <math>\beta</math>.]]

A mathematical analysis of static load-sharing (also called load-distributing) 2-point anchor systems.

==Derivation==
Consider the node, where the two anchor legs join with the main line. At this node, the sum of all forces in the x-direction must equal zero since the system is in [[mechanical equilibrium]].
<br />
:<math>F_{x1}=F_{x2} \,</math>
<br />
:<math>F_{1}Sin(\alpha )=F_2Sin(\beta ) \,</math>
<br />
{{NumBlk|:|<math>F_{1}=F_2\frac{Sin(\beta )}{Sin(\alpha )} \,</math>|{{EquationRef|1}}}}
<br />
The net force in the y-direction must also sum to zero.
<br />
:<math>F_{y1}+F_{y2}=F_{load} \,</math>
<br />
{{NumBlk|:|<math>F_{1}Cos(\alpha )+F_2Cos(\beta )=F_{load} \,</math>|{{EquationRef|2}}}}
<br />
Substitute <math>F_{1}</math> from equation ({{EquationNote|1}}) into equation ({{EquationNote|2}}) and factor out <math>F_{2}</math>.
<br />
:<math>F_2\frac{Sin(\beta )}{Sin(\alpha )}Cos(\alpha )+F_2Cos(\beta )=F_{load} \,</math>
<br />
:<math>F_2\left [\frac{Sin(\beta )}{Sin(\alpha )}Cos(\alpha )+Cos(\beta )\right ]=F_{load} \,</math>
<br />
Solve for <math>F_{2}</math> and simplify.
:<math>F_2=\frac{F_{load}}{\left [\frac{Sin(\beta )}{Sin(\alpha )}Cos(\alpha )+Cos(\beta )\right ]} \,</math>
<br />
:<math>F_2=\frac{F_{load}}{\left [\frac{Sin(\beta )Cos(\alpha )}{Sin(\alpha )}+\frac{Cos(\beta )Sin(\alpha)}{Sin(\alpha)}\right ]} \,</math>
<br />
:<math>F_2=\frac{F_{load}}{\left [\frac{Cos(\alpha )Sin(\beta )+Sin(\alpha)Cos(\beta )}{Sin(\alpha )}\right ]} \,</math>
<br />
:<math>F_2=F_{load}\frac{Sin(\alpha )}{{Cos(\alpha )Sin(\beta )+Sin(\alpha)Cos(\beta )}} \,</math>
<br />
Use a [[Trig_Identity#Angle_sum_and_difference_identities|trigonometric identity]] to simplify more and arrive at our final solution for <math>F_{2}</math>.
<br />
{{NumBlk|:|<math>F_2=F_{load}\frac{Sin(\alpha )}{Sin(\alpha+\beta)} \,</math>|{{EquationRef|3}}}}
<br />
Then use <math>F_2</math> from equation ({{EquationNote|3}}) and substitute into ({{EquationNote|1}}) to solve for <math>F_1</math>
:<math>F_{1}=F_{load}\frac{Sin(\alpha )}{Sin(\alpha+\beta)}\frac{Sin(\beta )}{Sin(\alpha )} \,</math>
<br />
{{NumBlk|:|<math>F_{1}=F_{load}\frac{Sin(\beta )}{Sin(\alpha+\beta)} \,</math>|{{EquationRef|4}}}}

==Symmetrical Anchor - Special Case==
[[File:Symmetrical Anchor.svg|thumb|This diagram shows the specific case when the anchor is symmetrical. Notice <math>2\alpha = \theta</math>.]]

Let us now analyze a specific case in which the two anchors are "symmetrical" along the y-axis.

Start by noticing <math>\alpha</math> and <math>\beta</math> are the same. Let us start from equation ({{EquationNote|4}}) and substitute <math>\beta</math> for <math>\alpha</math> and simplify.

:<math>F_{eachAnchor}=F_{load}\frac{Sin(\alpha )}{Sin(\alpha+\alpha)} \,</math>
<br />
:<math>F_{eachAnchor}=F_{load}\frac{Sin(\alpha )}{Sin(2\alpha)} \,</math>
<br />
And using another [[Trig_Identity#Double-.2C_triple-.2C_and_half-angle_formulae|trigonometric identity]] we can simplify the denominator.

:<math>F_{eachAnchor}=F_{load}\frac{Sin(\alpha )}{2Sin(\alpha)Cos(\alpha)} \,</math>
<br />
:<math>F_{eachAnchor}=\frac{F_{load}}{2Cos(\alpha)} \,</math>
<br />
Remember that <math>\alpha</math> is HALF the angle between the two anchor points.  To use the entire angle <math>\theta</math>, you must substitute <math>\alpha=\theta/2</math> we can edit our equation thus.

:<math>F_{eachAnchor}=\frac{F_{load}}{2Cos(\frac{\theta}{2})} \,</math>

==References==
*"Non-Load Sharing Anchors." Ratref. Web. 11 February 2012. <http://ratref.com/content/non-load-sharing-anchors>.

[[Category:Mathematical proofs]]