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GCSE physics problem

  • Thread starter reidy
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    gcse
  • #26
Dick
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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
[tex]\\T= \sqrt{\frac{a}{g}} \\ln \frac{a+{\sqrt{ a^2-b^2 }}}{b}[/tex]
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!).
 
  • #27
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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
 
  • #28
Dick
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I would just put x=a in the arccosh formula. Then combine and simplify the square root product and you are done.
 
  • #29
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YES!!!!! got it thanks a lot Dick. Im very happy to have finally understood that :D cheers

peace
 
  • #30
Dick
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Congratulations!!!
 

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