# Calc + applications in physics

1. Oct 6, 2005

### 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?

$$\int_{0 ft}^{500 ft} (800lbs + 2x) x\ dx$$

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?

Last edited: Oct 6, 2005
2. Oct 6, 2005

### Tom Mattson

Staff Emeritus
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!

3. Oct 6, 2005

### ecoli

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

4. Oct 6, 2005

### 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:

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

Do you see why?

Alex

5. Oct 6, 2005

### 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...

6. Oct 6, 2005

### 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

7. Oct 6, 2005

### Tom Mattson

Staff Emeritus
:rofl: 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):

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

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.

8. Oct 6, 2005

### 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.

9. Oct 6, 2005

### Tom Mattson

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

10. Oct 6, 2005

### 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.

11. Oct 6, 2005

### Tom Mattson

Staff Emeritus
Want to post your steps and have us check them?

12. Oct 6, 2005

### 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

13. Oct 6, 2005

### amcavoy

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

14. Oct 6, 2005

### Tom Mattson

Staff Emeritus
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:

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

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

15. Oct 6, 2005

### ecoli

ahhh... thanks guys, all clear now.