Anyone able to help me out working some stuff out?

I'm trying to find the density of a metal cylinder which floats 50% in water. 1m dimeter 2m high made up of metal with a sheet thickness of 10mm.

Worked out the buoyant force of it fully submerged as rho g Volume to be 61606.8N..., so 50% of that 30803.4N

Now, I'm not sure how to work out the cylinders density from just the BF and its volume.

Any help? I assume waters density to be 1000 kg/m

PSN ID: ArranP
I believe this is a very simple problem, but proving it is trickier. If 50% of the cylinder is submerged, the density of the cylinder is 50% that of water, or 500 kg/m^3. I'm pretty sure that the volume of the cylinder doesn't matter in this case because the volume submerged = the volume of water displaced. The given units may be just to mislead. A proof wouldn't require calculations I THINK. I'll try to work something out.
Last edited by billytalent77 at Oct 14, 2013,
See I thought that too, but it seems too simple...?

I was thinking of it as a drum, although its not as described as a drum but hints to this as its made of 10mm thick metal...?

PSN ID: ArranP
1) D = m/V (where D is density, instead of rho) 2) Fb = Dw*g*Vw 3) Fg = Dc*g*Vc
Since 50% is submerged, Fb = ½ Fg
We can assume Vc = Vg due to the water displaced = volume submerged
Solve each equation for g
Insert Fb = ½ Fg into equation 2 and then solve both equations 2* and 3 for g
g = ½ Fg / (Dw * Vw) g = Fg / (Dc * Vc)
let the two equations be equal and cross out equal values (Vc = Vw), (Fg = Fg)
½ Fg / (Dw * Vw) = Fg / (Dc * Vc)
Rearrange into Dc = ½ Dw
Oh so it could be a hollow cylinder? Then you may be able to calculate the volume of the metal shell as a fraction of the entire cylinder, and use an air density of 1.225 kg/m^3. Then if the shell was say, 3% of the total volume, you could work backwards from a total density of 500kg/m^3 to find the density of just the metal. I'm not sure if that's thr best way to do it, but it's A way to do it?
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I was thinking calculating the volume hollow cylinder, with a thickness of 10mm, its height being 1.8m and two "lids" which would be 10mm thick, 0.5m in diameter at each end. This would the the total volume of the metal once added together.

Once acquiring this volume. Would I substitute this volume into Rho = Mass / volume... Rho being 500kg/m3 as the cylinder floats to calculate its mass.... Then once I have its mass its weight would be weight = mass x gravity?

Yea?

PSN ID: ArranP
Quote by apmaman
Anyone able to help me out working some stuff out?

I'm trying to find the density of a metal cylinder which floats 50% in water. 1m dimeter 2m high made up of metal with a sheet thickness of 10mm.

Worked out the buoyant force of it fully submerged as rho g Volume to be 61606.8N..., so 50% of that 30803.4N

Now, I'm not sure how to work out the cylinders density from just the BF and its volume.

Any help? I assume waters density to be 1000 kg/m

The force on the cylinder from the water is water density (p') times g times pi*R^2*h, where R is the radius of the entire cylinder. Divided by two, because only half of this volume is in the water. This is equal to the force from the cylinder, which is the density of the cylinder (p) times g times pi*(R^2-r^2)*h because you think of the volume as being a solid cylinder (radius 0.5m) minus another smaller solid cylinder (radius 0.49m), so you're left with just the shell which has thickness 0.5m-0.49m... 0.01m or 10mm which is what you want.

p'=1000kg/m^3, R=0.5m, r = 0.49m

F = p'g*pi*R^2*h/2 = p*g*pi*(R^2-r^2)*h)
p'*(R^2)/2 = p*(R^2-r^2)
p=p'*R^2/[2*(R^2-r^2)]
p=1000*0.5^2/[2*(0.5^2-0.49^2)]

12626 kg/m^3
Quote by apmaman

I'm trying to find the density of a metal cylinder which floats 50% in water. 1m diameter 2m high made up of metal with a sheet thickness of 10mm.

Worked out the buoyant force of it fully submerged as rho g Volume to be 61606.8N..., so 50% of that 30803.4N

Now, I'm not sure how to work out the cylinders density from just the BF and its volume.

Any help? I assume waters density to be 1000 kg/m

There is no way that number is right, 30803 N is a shit ton of force. And is the cylinder closed at the bottom or open?
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There is no way that number is right, 30803 N is a shit ton of force. And is the cylinder closed at the bottom or open?

closed.

looking at the working done i can see where iv went wrong (i think) in calculating its buoyant force.

i dispise fluid mechanics haha.

PSN ID: ArranP