Recent content by reidy

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

hmmm I've 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...

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 doesn't help, then i thought when i intergrated the RHS i should had a constant. t*sqrt[g/a] +...

sooo does that leave me with; arccosh(x/b) = t*sqrt[g/a] which gives; t= arccosh(x/b) / sqrt[g/a]

thanks a lot cristo, i did go a bit crazy with the \\

haha sorry that doesn't help. i don't have a clue about trig functions either. I am only 16 and my knowledge of maths is no where near that need to complete this question. although i have read up outside the syllabus and would consider myself to be very good at math, it would be good to have a...

noo, I am trying to research them but any hints would be appreciated. thanks dick

how do i intergrate dx/sqrt((x^2)-(b^2))

ARGHHHH I GIVE UP WITH LATEX! sorry to spam this topic peace

Having worked out acceleration and intergrating to find velocity then rearranging the formula, i am left with; \\int \\frac{dx}{\\sqrt{x^2 - b^2}} = \\int \\sqrt{\\frac {g}{a}} dt how do i intergrate the left hand side? im soo frustrated this question has been annoying me for ages, i...

Im sorry the formula should be \\T= \sqrt{\frac{a}{g}} \\ln \frac{\a + sqrt{ a^2-b^2 }}{b} Having worked out acceleration and intergrating to find velocity then rearranging the formula, i am left with; \\int frac{dx}{\\sqrt{x^2 - b^2}} = \\int \\sqrt{frac {g}{a}} dt how do i...

The force exerted by the dangling chain is enough to create a resultant force allowing the rest of the chain to be accelerated, and hence be pulled off the table. The mass of the dangling chain increases(as more of the chain is pulled of the table), and the force exerted by the dangling chain...

I am currently studying for my GCSE's but have a keen interest in physics and maths. Today at school i noticed a question; A chain of length a is placed on a frictional-less table with length b of the chain dangling off the edge of the table. Using \\F = ma derive the formula: \\T=...

T is the time it takes for the entire chain to fall off the table im sorry i shoudl have made myself clear, but using LaTex for the first time is canny confusing

sorry that should be \\T= \sqrt{\frac{a}{g}} \\In \frac{\sqrt{ a^2-b^2 }}{b}