# GCSE physics problem

Dick
Homework Helper
i need to get rid of the x ,x is the length of the chain hanging over the edge before the chain starts to move, so when t=O x=b

this gives arccosh(1/1) = 0*sqrt[g/a]
which gives 0 = 0

which doesnt help, then i thought when i intergrated the RHS i should had a constant.

t*sqrt[g/a] + C

so
arccosh(x/b) = t*sqrt[g/a] + C

i dont see how either of these option help me to progress to get
$$\\T= \sqrt{\frac{a}{g}} \\ln \frac{a+{\sqrt{ a^2-b^2 }}}{b}$$
Nah. It just says that when t=0 there is b length of chain hanging over the edge. That is the correct initial condition. So putting C=0 for your constant is GOOD. You now want to know when ALL of the chain is hanging over then edge. Then x=?. (Next you'll need a formula for arccosh - then the formula will start looking like you want. You might want to Google 'arccosh' and try to find it!).

hmmm ive been doing sum research and found that the intergral of the LHS is

log[x+[sqrt[x^2-b^2]
is this not correct??

if it is we can substitue x to be a, when all of x is hanging over the edge;

log[a+[sqrt[a^2-b^2]

or using the arccosh approach the formula for arccosh is

arccosh = In [(x/b) + sqrt[(x/b)+1] * sqrt[(x/b)-1]

im a bit unsure as to which path to go down?
thanks dick and cristo.
peace

Dick
Homework Helper
I would just put x=a in the arccosh formula. Then combine and simplify the square root product and you are done.

YES!!!!! got it thanks a lot Dick. Im very happy to have finally understood that :D cheers

peace

Dick