User:Eml4500.f08.lulz.abcd/HW3

Solve for AC and AB
As given in class, AC and AB are dependent on their elemental P's and K's. In HW 2 we solved for both elemental K values and one of the P values: that work is reproduced as appropriate in the proof below

$$AC = \frac{|P^{(1)}_2|}{K^{(1)}} = \frac{\sqrt{\left(f_3^{(1)}\right)^2 + \left(f_4^{(1)}\right)^2}}{\left(\frac{EA}L\right)} = \frac{\sqrt{(4.4378)^2 + (2.5622)^2}} = \frac{5.1243}{0.75} = 6.8324$$

$$AB = \frac{|P^{(2)}_1|}{K^{(2)}} = \frac{\sqrt{\left(f_1^{(2)}\right)^2 + \left(f_2^{(2)}\right)^2}}{\left(\frac{EA}L\right)} = \frac{\sqrt{(-4.438)^2 + (4.438)^2}} = \frac{6.2763}5 = 1.2553$$