I am having trouble with the following question:
A more general equation for a catenary is $y = a \cosh(x/b)$. Find $a$ and $b$ to match the following characteristics of a hanging cable. The ends are $40 m$ apart and have a height of $y = 20 m$. The height in the middle is $y = 10 m$.
I can't seem to figure it out due to the two unknowns. Could someone help me? Thanks in advance!
Solve one unknown after the other. The requirement that "the height in the middle is $y=10$" gives us, since the middle is $x=0$,
$$a\cosh\left(\frac 0b\right)=10$$
You can easily find $a=10$ from that. Then use the other fact to get
$$10\cosh\left(\frac{20}b\right)=20$$
You can easily find $b$ from that: I'll leave that to you.
Assuming the cable is symmetrical, from the data you have $a=10$ and $20=10\cosh(\frac{20}{b})\Rightarrow b=\frac{20}{\operatorname{arcosh}2}$
