Question

It is proposed to dispose of nuclear wastes in drums with weight W=640 lbs and volume 8ft^3 by droping them into the ocean v0=0. the force equation of a drum falling in the water is m(dv/dt)=-W+B+Fr where the buoyant foce B is equal to the weight (at 6.25 lb/ft^3) of the volume of water displaced by the drum and Fr is the force of water resistance found empiraclly to be 1 lb for each foot per second of the velocity of a drum. If the drums are likely to burst upon impact of more than 75 ft/s what is the maximum depth to which they can be dropped in the ocean without likelihood of bursting?

Answer 1

What I have started off with is:
\[\huge mx''-F_rx'=-W+B\]
We have to divide by m, and then choose our integrating factor to be -F/m and wind up with (Please check this work)
\[\huge x''+{-F_r \over m}x'=-W+B\] \[\huge e^{(-F_rt/m)}x'={-W+B \over m} \int\limits e^{(-F_rt/m)} dt\]
\[\huge x' = {W-B \over F} + Ce^{F_rt/m}\]

This is were we're not sure if we are correct. Continuing further with another derivative, yields an answer that doesn't make sense when taking the limit of t->infinity (maximizing to find the depth).

Anyone have some pointers or can help us figure out where to go with the problem?

Answer 2

solve for t with x' = 75 ft/s
integrate once more to find an expression for x
plug in the time you found to get the max depth

Answer 3

if i solve for x'(t)=75, what is t? is t=0? I ask because i'm not sure how to get that constant figured out since t and c are both unknown.

Answer 4

"by droping (sic) them into the ocean v0=0."
when t=0 x' =0
find C
solve for t when x'=75

Answer 5

Unless i did my calculations incorrect and/or fubared the derivatives up, i couldn't yield the right answer doing the above. Oh well, thanks for the help though.