Question

Need help setting up the Work/Force Integral:

Crane lifts a bucket of sand originally weighing 125 lb at a constant rate. Sand is lost from the bucket at a constant rate of 0.5lb/ft. How much work is done in lifting the sand 90 ft?

Answer 1

. . . . .

Answer 2

work is being done against gravity only.
therefore -ve of work done by gravity = work done by crane.
hence,
work done by gravity = \[\int\limits_{0}^{h} mgdx\]
At any instant,
At a height x above the ground,
mass = 125 - 0.5x
therefore work done = \[\int\limits_{0}^{h} [125 - 0.5x]gdx\]

taking g as constant try integrating it now.
remember the units.
i hope this is correct.

Answer 3

hermeezy, that is not the expression for force. i think.

Answer 4

Here's what i did without setting up the integral:

losing 0.5ft/lb in 90 ft is total of 45 lb of sand lost.
125-(45/2) = 102.5 <----average amount of sand in bucket
W = fd = 102.5(90) = 9225 lb/ft

Since i don't know how to set up a proper integral, this is the best i can come up with.

Answer 5

why is average amount of sand in the bucket equal to your force? force is mass times acceleration.

Answer 6

haha. ;)

Answer 7

@siddhantsharan what is g in your equation? gravity ? 9.8?

Answer 8

yep. gravitational acceleration it is. and when you find the answer, remember to convert it into ft/s^2. 9.8 is in si units.

Answer 9

Is it any difference based on your equation if the question says "neglect the weight of the bucket" ? sorry should have include that in there.

Answer 10

i neglected it in my equation. thought that if it was there you wouldve given it. :)

Answer 11

you understood the solution?

Answer 12

Still working it out, did you end up with an answer to your equation by any chance?

Answer 13

umm
\[-296805.15\]\[ft ^{2} lb s ^{-2}\]

Answer 14

that ll be a positive, sorry.

Answer 15

whoa...okay , i'll work on it. thanks!

Answer 16

cheers. :)