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The load is pointing straight down, and is being held up by two forces pointing up at various angles $\alpha$ and $\beta$.

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.

$F_{x1}=F_{x2}$

$F_{1}Sin(\alpha )=F_{2}Sin(\beta )$

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The net force in the y-direction must also sum to zero.

$F_{y1}+F_{y2}=F_{load}$

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Substitute $F_1$ from equation (Template:EquationNote) into equation (Template:EquationNote) and factor out $F_2$.

$F_2\frac{Sin(\beta )}{Sin(\alpha )}Cos(\alpha )+F_{2}Cos(\beta )=F_{load}$

$F_2[\frac{Sin(\beta )}{Sin(\alpha )}Cos(\alpha )+Cos(\beta )]=F_{load}$

Solve for $F_2$ and simplify.

$F_2=\frac{F_{load}}{[\frac{Sin(\beta )}{Sin(\alpha )}Cos(\alpha )+Cos(\beta )]}$

$F_2=\frac{F_{load}}{[\frac{Sin(\beta )Cos(\alpha )}{Sin(\alpha )}+\frac{Cos(\beta )Sin(\alpha )}{Sin(\alpha )}]}$

$F_2=\frac{F_{load}}{\left[\frac{Cos(\alpha )Sin(\beta )+Sin(\alpha )Cos(\beta )}{Sin(\alpha )}\right]}$

$F_2=F_{load}\frac{Sin(\alpha )}{Cos(\alpha )Sin(\beta )+Sin(\alpha )Cos(\beta )}$

Use a trigonometric identity to simplify more and arrive at our final solution for $F_2$.
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Then use $F_2$ from equation (Template:EquationNote) and substitute into (Template:EquationNote) to solve for $F_1$

$F_1=F_{load}\frac{Sin(\alpha )}{Sin(\alpha +\beta )}\frac{Sin(\beta )}{Sin(\alpha )}$

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Symmetrical Anchor - Special Case

This diagram shows the specific case when the anchor is symmetrical. Notice $2\alpha =\theta$.

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

Start by noticing $\alpha$ and $\beta$ are the same. Let us start from equation (Template:EquationNote) and substitute $\beta$ for $\alpha$ and simplify.

$F_{eachAnchor}=F_{load}\frac{Sin(\alpha )}{Sin(\alpha +\alpha )}$

$F_{eachAnchor}=F_{load}\frac{Sin(\alpha )}{Sin(2\alpha )}$

And using another trigonometric identity we can simplify the denominator.

$F_{eachAnchor}=F_{load}\frac{Sin(\alpha )}{2Sin(\alpha )Cos(\alpha )}$

$F_{eachAnchor}=\frac{F_{load}}{2Cos(\alpha )}$

Remember that $\alpha$ is HALF the angle between the two anchor points. To use the entire angle $\theta$, you must substitute $\alpha =\theta /2$ we can edit our equation thus.

$F_{eachAnchor}=\frac{F_{load}}{2Cos(\frac{\theta }{2})}$

References

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