GCSE Physics Problem: Deriving T Using F = ma

[reidy]

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:

$$\\T= \sqrt{\frac{a}{g}} \\In \frac{\sqrt{ a^2- b^2 }}{b}$$

 

[denverdoc]

What exactly is T? I have some guesses, but better to state it explicitly.

 

[reidy]

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

 

[reidy]

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

 

[reidy]

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:

$$\\T= \sqrt{\frac{a}{g}} \\In \frac{\sqrt{ a^2-b^2 }}{b}$$

where T= time taken for the chain to slip off the table
and the acceleration due to gravity is normal

any solutions would be greatly appreciated as i have gave the question a lot of thought and have some idea of how it should be solved. however i doubt my knowledge of physics/maths will allow me to complete this question.

cheers in advance

 

[denverdoc]

The forum rules ask what efforts you have made, so maybe here at least a description of your thoughts regarding the acceleration of the chain--is it constant, variable, and why?

 

[reidy]

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 increases, which means that a bigger force is pulling the chain on the table towards the ground. As F increase and mass stays constant doesn't this mean that the rate of aceleration must increase as F increases.

Im really quite unsure as to how to reach the final equation from F=ma.

Thanks in advance devendoc.

 

[denverdoc]

Thats how I see it. As the dangling chain gets longer, the purely inertial component on the table decreases. So far starters I would try to set up both as a function of time, maybe let x(t) be the change in length of the supported section, so that A(t) length falling vertical = A+x(t) while the other segment B-x(t). Assume that mass is proportional to length by inserting some constant, then you should have masses of both as a function of time, and from that acceleration as a function of time. Then if my crystal ball is working the math will get a little messy. See what you can come up with.

 

[Dick]

$$\\T= \sqrt{\frac{a}{g}} \\ln \frac{\sqrt{ a^2-b^2 }}{b}$$

Yes, it is kind of messy - involves setting up and solving an ODE with boundary conditions. The answer as a function of time comes out like b(t)=b*cosh(k*t) where k=sqrt(a/g). Inverting and solving for t involves solving a quadratic. And then finally, you notice that the above answer is wrong. If you take b very close to a, this formula would starting giving you negative values of T. Add a/b to the argument of the ln to correct it. Is this really all worth it?

 

[denverdoc]

Not to me it wasn't. Just understanding the concepts of the problem I thought was, which is not so hard to get to the ODE part, assuming the OP knows a(t)=d'' x(t)/dt^2. From there I planned to go running to my copy of Taylor. ;-D

 

[reidy]

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 intergrate the left hand side?
im soo frustrated this question has been annoying me for ages, i would love to complete it! thank you for any help in advance

peace

 

[reidy]

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 would love to complete it! thank you for any help in advance

peace

 

[reidy]

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

peace

 

[reidy]

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

 

[Dick]

Try the substitution x=b*cosh(u). Are you ok with hyperbolic functions?

 

[reidy]

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

 

[Dick]

They are a lot like trig functions, but even easier. Here's what you need to know.

cosh(x)=(e^x+e^(-x))/2
sinh(x)=(e^x-e^(-x))/2

Now as an exercise, you prove (which you'll need to do the problem):

d(cosh(x))=sinh(x)*dx
cosh(x)^2-1=sinh(x)^2

That makes the LHS integral easy.

 

[cristo]

Having worked out acceleration and intergrating to find velocity then rearranging the formula, i am left with;

$$\int \frac{dx}{\sqrt{x^2 - b^2}} = \int \sqrt{\frac{g}{a}} dt$$

As an aside; here's your LaTex corrected. You just put too many \'s in it!

 

[reidy]

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 bit more focused step by step advice. I've got this far through the question and it would be a shame to have to wait a few years before i had the knowledge to finish it.
thanks a lot Dick for the help and any future help.
peace

 

[reidy]

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

 

[cristo]

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.

Well, as Dick says, the trigonometric functions are defined as $$\cosh x=\frac{e^x+e^{-x}}{2} \hspace{1cm} \sinh x=\frac{e^x-e^{-x}}{2}$$ where e^x is the exponential function; are you familiar with this? Do you know what the derivative of the exponential function is? If so, you should be able to work out the derivative of the hyperbolic functions.

Also for this question, you will need to know $$\cosh^2x-\sinh^2x=1$$; this is a standard identity.

So now you can integrate the LHS using substitution. Do you know how to do this?

 

[Dick]

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 bit more focused step by step advice. I've got this far through the question and it would be a shame to have to wait a few years before i had the knowledge to finish it.
thanks a lot Dick for the help and any future help.
peace

Well, lack of experience with elementary functions and calculus doesn't necessarily have to stop you. It doesn't stop engineers. They use integral tables. Besides, you did such a good job of setting up the integral. The integral of the RHS is arccosh(x/b) (where arccosh is the inverse of the cosh function). You can do the integral of the LHS, right?

 

[reidy]

sooo does that leave me with;

arccosh(x/b) = t*sqrt[g/a]
which gives;

t= arccosh(x/b) / sqrt[g/a]

 

[Dick]

sooo does that leave me with;

arccosh(x/b) = t*sqrt[g/a]
which gives;

t= arccosh(x/b) / sqrt[g/a]

Seems ok so far. So what value of x do you want to put in?

 

[reidy]

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] + C

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

i don't see how either of these option help me to progress to get
$$\\T= \sqrt{\frac{g}{a}} \\ln \frac{a+{\sqrt{ a^2-b^2 }}}{b}$$

 

[Dick]

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] + C

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

i don't 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!).

 

[reidy]

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

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]

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

 

[reidy]

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

peace

 

[Dick]

Congratulations!