Rubber Band Forces I have a question regarding the force a band places on an object.  Say I have a rubber band wrapped around 2 pegs at a certain distance, and at that distance I know the pounds of force per inch it is pulling at.  For example, if the pegs are placed such that the band is stretched to 52 inches, the force the bands apply to the pegs are 57.6 lbs per inch.  
My question involves what happens with the addition of a third peg?  If I make a triangle with the pegs such that it makes a triangle with the same height as before, namely 52 inches, but the base is now 24 inches.  What would the force be pulling straight down (no longer along the band itself) from the high point of the triangle toward the 24 inch base? 
The free length of the band is 41 inches(peg to peg).  
 A: With two pegs, there are two strips of rubber working in parallel contributing to a total stiffness $K_{\rm total} = 57.6 \;\rm lbf/in$. So each strip is $K = 28.8\;\rm lbf/in$. 
With the three pegs you now have two strips at 26° apart, or 13° from vertical each. The effective spring constant in the vertical direction is thus $K_{eff} =2 K \cos^2(13^\circ) = 54.68\; \rm lbf/in$. One of the $\cos(13^\circ)$ comes from the force projection to vertical and the second from the displacement projection to vertical from along the side of the triangle.
Another way to get the same result if the base is $b$ and the height is $h$ is
$$ K_{eff} = \frac{8 h^2}{b^2+4 h^2} K $$
  $$  = \frac{8*52^2}{24^2+4*52^2} 28.8 = 54.68 $$
But also the stretch force value has changed. To get this you need the free length of the band which is not given and cannot be calculated from the given values. 
Edit 1
With the values given I came up with the following force on the top pin
$$ F = \frac{288 h \left( \sqrt{b^2+4 h^2}+b-82\right)}{5 \sqrt{b^2+4 h^2}} $$
So with two pins $b=0$, $h=52$ the force is $F=633.6\;\rm lbf$ with stiffness $K_{eff}=\frac{\partial F}{\partial h}=57.6$.
With three pins and $b=24$, $h=52$ the force is $F=1367.6\;\rm lbf$ and stiffness $K_{eff}=\frac{\partial F}{\partial h}=56.02$.
Why ?
The force on the pin is equal to two times the tension projected vertically (from a Free Body Diagram on the pin).
$$ F = 2 T \cos\left(\frac{\theta}{2}\right) $$ 
with $\cos\left(\frac{\theta}{2}\right) = \frac{h}{\sqrt{\left(\frac{b}{2}\right)^2+h^2}}$ and tension $T = K (L-L_0) $. The initial band length is $L_0 = 2\cdot 41 = 82\;\rm in$ and the stiffness of each strip is actually $K = 14.4\;\rm lbf/in$. This comes from $K_{eff}=4\,K$ when $b=0$ and with $K_{eff}=57.6\;\rm lbf/in$.
The length of the rubber band is the circumference of a triangle with base $b$ and height $h$
$$ L = b + 2 \sqrt{\left(\frac{b}{2}\right)^2 + h^2} $$
