calc + applications in physics

ecoli

I think I found a way to do this, but it's not working.

The problem gives a cable weighing 2lb/fot, that is used to lift 800 lb of coal up a mineshaft 500ft deep. find the work

I looked up the answer, which 650,000 ft-lbs. The weight is the force, so there is no need for the acceleration due to gravity. I tried something that I thought was right, but it didn't yield the correct answer. Am I not integrating correctly or is my set-up wrong?

[tex] \int_{0 ft}^{500 ft} (800lbs + 2x) x\ dx [/tex]

I distributed the x and integrated, ending up with [tex] ({800/2})x^2 + ({2/3})x^3 [/tex] from 0 to 500.

That doesn't come up with the rgith answer, though. Any ideas?

Tom Mattson

ecoli, are you continuing a thread here that you started at SFN? Because if you are, I don't think that anyone except me is going to be able to follow you!

ecoli

you're right, I forgot to copy the first part of the message... namely the problem. I'm an idiot... sorry guys.

amcavoy

I couldn't tell from your previous post if you already have an answer, but here's how I would set it up:

The 800 is constant from the bottom to the top, only the 2x will change. You can write it out as:

[tex]W=800*500+\int_{0}^{500}2x\,dx[/tex]

Do you see why?

Alex

ecoli

Where did you get the 500 from... because won't that change when you pull the cable upwards?

The book gives the answer as 650,000 ft-lbs, so I'm not sure your answer is right...

amcavoy

Remember for a constant force, the work is W=Fd. You can split the work up into the work for the coal (constant weight=constant force) + work for the cable. The force on the cable is the only thing that changes.

Alex

Tom Mattson

No problem.

You've got 1 major problem that is screwing the entire thing up, and that is that you don't seem to really get the definition of work.

It is (for 1D forces and displacements):

[tex]W=\int_{x_1}^{x_2}Fdx[/tex]

Now, your integral seems to be taking the definition as:

[tex]W=\int_{x_1}^{x_2}Fxdx[/tex]

which is not the same thing. So, drop that extra "x" in the integrand, and you'll be fine.

ecoli

the weight of the cable changes. I understand that is equal to the force.

I see why your answer is correct, but it's not the same answer the book gives... unless I'm integrating incorrectly.

Tom Mattson

If you follow the advice of either apmcavoy or myself, you will get the correct answer of 650,000 ft-lbs.

ecoli

I am following your advice, I must be integrating incorrectly. That's making the answer come out wrong, so I thought it was the setup that was wrong. Of course that's not the case.

Tom Mattson

Want to post your steps and have us check them?

ecoli

ok, thanks.

first off, I have
Fd = (800 + 2x)
[tex] \int_{0}^{500} 800 + 2x dx [/tex]

then I pull out the 800 and get [tex] 800 \int_{0}^{500} 2x\ dx [/tex]

I integrate and get [tex] 800 * x^2 [/tex] from 0 to 500

which is 200 million, and not the correct answer

amcavoy

There's your problem, you pulled out the 800! Remember this?

[tex]\int\left(f(x)+g(x)\right)\,dx=\int f(x)\,dx+\int g(x)\,dx[/tex]

Alex

Tom Mattson

Right.

Wrong.

You're pulling out the 800 as if it were a factor of the integrand. It's not, it's a term of the integrand. So the correct way to split it up would be:

[tex] \int_0^{500} (800 + 2x) dx = \int_0^{500} 800 dx + \int_0^{500} 2x dx[/tex]

Now, you can pull 800 out of the first integral and 2 out of the second.

ecoli

ahhh... thanks guys, all clear now.