Work and integration

[Allanon]

Hi I can't seem to get the right answer for this problem, and I'm wondering how you're suppose to do it.

A storage tank in the shape of a hemisphere of radius 10 ft is full of oil weighing 50 lb/ft. How much work is done in pumping the oil out through a hole in the top of the tank?

Thanks

[DavidHooper]

Who reckons a volume integral over z dxdydz for a hemisphere would work?

That would (I think) give the total distance that each oil element must move. Then multiply by 50g.

I must say, lb and ft are confusing. What is g in ft?!

Oh, and BTW, the units on the weight of the oil are wrong.

[twanvl]

Theoraticly:
W=F*s
F=m*g
s (distance) is diferent for each 'layer' of oil in the tank, range 0-10ft
m(s) = pi*r^2*50
r(s) = sqrt((5ft)^2 - (s-5)^2) = sqrt(10s - s^2)
(sqrt is square root, x^0.5)

W=integral(0...10, 50*g*s*sqrt(10s-s^2) ds)
=50g*integral(0...10, s*sqrt(10s-s^2))

Unfortunatly this is a function you can't integrate (as far as I know), So you have to do with a non exact anser. The answer is 50*g*196.35. I don't know the value of g for imperial units, for SI it is 9.81.

In reality you also have to consider all kinds of friction forces, which make the equation more dificult

[teriaki]

Its been a while but as i recall, you use a volume integral and the height of any crossectional area of oil. To find this you must follow this procedure:

project a radius from the center of the base of the hemisphere to a given point on the surface (but not on the base).
project a vertical line from from that point on the surface to a point on the base. this is h.
connect the point on the base to the center of the base. this is x. there is now a triangle.

notice that each crossection can be defined as a circle of radius r*cos(theta) and height of r*sin(theta). you can integrate x and h over theta to get the force required to lift each crossection through h.

also, it is important to note that the orrientation of the hemisphere is critical to solving the problem.

[kedaman]

Originally posted by twanvl
Theoraticly:
W=F*s
F=m*g
s (distance) is diferent for each 'layer' of oil in the tank, range 0-10ft
m(s) = pi*r^2*50
r(s) = sqrt((5ft)^2 - (s-5)^2) = sqrt(10s - s^2)
(sqrt is square root, x^0.5)

W=integral(0...10, 50*g*s*sqrt(10s-s^2) ds)
=50g*integral(0...10, s*sqrt(10s-s^2))

Unfortunatly this is a function you can't integrate (as far as I know), So you have to do with a non exact anser. The answer is 50*g*196.35. I don't know the value of g for imperial units, for SI it is 9.81.

In reality you also have to consider all kinds of friction forces, which make the equation more dificult

substituting r in m(s) with sqrt(10s - s^2) should give you
pi*50*(10s - s^2), should be easy from there..