User:Trevorolson/sandbox



The forces in the x-direction must equal zero since the system is in equilibrium.


 * $$F_{x1}=F_{x2} \,$$


 * $$F_{1}Sin(\alpha )=F_2Sin(\beta ) \,$$

The forces in the y-direction must equal zero since the system is in equilibrium.


 * $$F_{y1}+F_{y2}=F_{load} \,$$

Substitue $$F_{1}$$ from equation ($$) into equation ($$) and factor out $$F_{2}$$.


 * $$F_2\frac{Sin(\beta )}{Sin(\alpha )}Cos(\alpha )+F_2Cos(\beta )=F_{load} \,$$


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

Solve for $$F_{2}$$ and simplify.
 * $$F_2=\frac{F_{load}}{\left [\frac{Sin(\beta )}{Sin(\alpha )}Cos(\alpha )+Cos(\beta )\right ]} \,$$


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


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


 * $$F_2=F_{load}\frac{Sin(\alpha )} \,$$

Use a trigonometric identity to simplify more and arrive at our final solution for $$F_{2}$$.

Then use $$F_2$$ from equation ($$) and substitute into ($$) to solve for $$F_1$$
 * $$F_{1}=F_{load}\frac{Sin(\alpha )}{Sin(\alpha+\beta)}\frac{Sin(\beta )}{Sin(\alpha )} \,$$

Symmetrical Anchor - Special Case


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 ($$) and substitute $$\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 in between your two anchors. Simply put, if you want to measure your entire angle $$\theta$$, you must remember that $$\alpha=\theta/2$$ so we can edit our equation thus.


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