A tough physics/fluids question?

27m?

It’s been long enough that I don’t remember the formulas. If you can post all the fluid dynamics/mechanics formula here, I could give it a shot.

p = m/v
p – density
m – mass
v – volume

And this wiki page seems to have everything you would probably need to know.
http://en.wikipedia.org/wiki/Buoyancy

It’s been a while but…

by comparing the area of a dome using 4 pi r2 and the constant that it has a mass of 9/10 mass of water, and the whole mass (above and below) is
V=(4/3)X pi X R3.

Then recalculating (by trial & error) until the mass of the dome is 1/10 the mass of the sphere and the hight of the dome =3m…

I came up with a total radius of the ice at approx 26.7125m less the exposed 3m leaves 23.7125m below the surface.

It has been a long time, so don’t use this for your homework answer.

@drbill the probelms with your work is that the distance above the water is not the radius of the dome above the water

we worked it out last night and we got a depth of 15.3 m

it was a ridiculous amount of work

I had played around with the number using various numbers for r of the whole sphere, and used the same radius for the dome, and continued till I got close to the dome having 10% the mass of the whole.

I have been out of school for many years, that’s why I was not sure if I did it right.

I’m really not sure what to do – it seems like a trick question with the “trick” answer being 27m, but I can’t imagine the volume of a parabolic shape changing at a rate that isn’t in an exponential relationship with the height (I’m trying to put the diagram in my head into words – sorry if it’s confusing). I couldn’t get anywhere new with the numbers, sorry mate.

volume of sphere = (pi*d^3)/6
volume of exposed cap of sphere = (pi*h^2*r) – (pi*h^3)/3

we already know the h (height) of the cap is 3, so making one equation for the volume of the sphere under water we get(and substituting d/2 for r):

Volume of water displaced = (volume of sphere) – (volume of cap of sphere)
Volume of water displaced = (pi*d^3)/6 – (pi*d*9)/2 + pi*9

we know the density ratio is this:

density of the ice = (9/10)density of water ===> substituting density equations

(mass of ice)/(volume of ice) = (9/10)(mass of water)/(volume of water)

since the glacier is at rest and not bobbing, we can assume that the mass of the water displaced is equal to the total mass of the spherical glacier. if you divide both sides of the equation by this equal mass, they go away and your new equation is this:

Volume of water displaced = (9/10)*(Volume of ice sphere)

using the equations from above:

(9/10)(pi*d^3)/6 = (pi*d^3)/6 – (pi*d*9)/2 + pi*9

pi can be divided out and your only unknown is the diameter of the sphere. My solution is a formula that I’m not sure how to get the solution to, and I’m not sure my math was good. This is what I got:

d^3 – 180*d + 540 = 0

anyone else know the right answer for sure?

it was 12.3 m