calc + applications in physics


by ecoli
Tags: applications, calc, physics
ecoli
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#1
Oct6-05, 10:08 PM
P: 9
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?
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Tom Mattson
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#2
Oct6-05, 10:11 PM
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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
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#3
Oct6-05, 10:12 PM
P: 9
you're right, I forgot to copy the first part of the message... namely the problem. I'm an idiot... sorry guys.

amcavoy
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#4
Oct6-05, 10:22 PM
P: 669

calc + applications in physics


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
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#5
Oct6-05, 10:27 PM
P: 9
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
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#6
Oct6-05, 10:29 PM
P: 669
Quote Quote by ecoli
Where did you get the 500 from... because won't that change when you pull the cable upwards?
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
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#7
Oct6-05, 10:31 PM
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Quote Quote by ecoli
you're right, I forgot to copy the first part of the message... namely the problem. I'm an idiot... sorry guys.
No problem.

Quote Quote by ecoli
[tex] \int_{0 ft}^{500 ft} (800lbs + 2x) x\ dx [/tex]
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
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#8
Oct6-05, 10:31 PM
P: 9
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
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#9
Oct6-05, 10:32 PM
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If you follow the advice of either apmcavoy or myself, you will get the correct answer of 650,000 ft-lbs.
ecoli
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#10
Oct6-05, 10:34 PM
P: 9
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
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#11
Oct6-05, 10:35 PM
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Want to post your steps and have us check them?
ecoli
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#12
Oct6-05, 10:47 PM
P: 9
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
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#13
Oct6-05, 10:51 PM
P: 669
Quote Quote by 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
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
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#14
Oct6-05, 10:52 PM
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Quote Quote by ecoli
Fd = (800 + 2x)
[tex] \int_{0}^{500} 800 + 2x dx [/tex]
Right.

then I pull out the 800 and get [tex] 800 \int_{0}^{500} 2x\ dx [/tex]
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
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#15
Oct6-05, 11:00 PM
P: 9
ahhh... thanks guys, all clear now.


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