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October 20, 2014

October 20, 2014

Posted by **Brian** on Sunday, November 3, 2013 at 5:56pm.

- Calculus II -
**Steve**, Sunday, November 3, 2013 at 8:22pmConsider the cable as a whole lot of little cable bits, each of which must be lifted a different distance. The bits at the top only have to be lifted a little way, the bits near the bottom have to be lifted a long way.

The 700 lb load has to go all the way up.

Each little piece of length dx weighs 1.5 dx and has to be lifted a distance of x feet.

The work done to lift each bit is its weight * the distance lifted. So, the whole job done is

∫[0,400] 1.5x + 700 dx

multiply the lot by a conversion factor if you want Newtons instead of ft-lbs.

- Calculus II -
**Brian**, Sunday, November 3, 2013 at 10:03pmWow this problem is easier than I thought it was. Thanks.

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