# calc + applications in physics

by ecoli
Tags: applications, calc, physics
 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? $$\int_{0 ft}^{500 ft} (800lbs + 2x) x\ dx$$ I distributed the x and integrated, ending up with $$({800/2})x^2 + ({2/3})x^3$$ from 0 to 500. That doesn't come up with the rgith answer, though. Any ideas?
 PF Patron Sci Advisor Emeritus P: 5,539 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!
 P: 9 you're right, I forgot to copy the first part of the message... namely the problem. I'm an idiot... sorry guys.
P: 668

## 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:

$$W=800*500+\int_{0}^{500}2x\,dx$$

Do you see why?

Alex
 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...
P: 668
 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
PF Patron
Emeritus
P: 5,539
 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 by ecoli $$\int_{0 ft}^{500 ft} (800lbs + 2x) x\ dx$$
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):

$$W=\int_{x_1}^{x_2}Fdx$$

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

$$W=\int_{x_1}^{x_2}Fxdx$$

which is not the same thing. So, drop that extra "x" in the integrand, and you'll be fine.
 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.
 PF Patron Sci Advisor Emeritus P: 5,539 If you follow the advice of either apmcavoy or myself, you will get the correct answer of 650,000 ft-lbs.
 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.
 PF Patron Sci Advisor Emeritus P: 5,539 Want to post your steps and have us check them?
 P: 9 ok, thanks. first off, I have Fd = (800 + 2x) $$\int_{0}^{500} 800 + 2x dx$$ then I pull out the 800 and get $$800 \int_{0}^{500} 2x\ dx$$ I integrate and get $$800 * x^2$$ from 0 to 500 which is 200 million, and not the correct answer
P: 668
 Quote by ecoli ok, thanks. first off, I have Fd = (800 + 2x) $$\int_{0}^{500} 800 + 2x dx$$ then I pull out the 800 and get $$800 \int_{0}^{500} 2x\ dx$$ I integrate and get $$800 * x^2$$ from 0 to 500 which is 200 million, and not the correct answer
There's your problem, you pulled out the 800! Remember this?

$$\int\left(f(x)+g(x)\right)\,dx=\int f(x)\,dx+\int g(x)\,dx$$

Alex
PF Patron
Emeritus
P: 5,539
 Quote by ecoli Fd = (800 + 2x) $$\int_{0}^{500} 800 + 2x dx$$
Right.

 then I pull out the 800 and get $$800 \int_{0}^{500} 2x\ dx$$
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:

$$\int_0^{500} (800 + 2x) dx = \int_0^{500} 800 dx + \int_0^{500} 2x dx$$

Now, you can pull 800 out of the first integral and 2 out of the second.
 P: 9 ahhh... thanks guys, all clear now.

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