Buoyancy question about styrofoam

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HadManySons
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Homework Statement


The first part of the question states this: If we want to buoy a 0.137 kg solid platinum object in water with an attached string from above, what is the tension required to support the piece of platinum?

The correct answer is: 1.28 Newtons. I got this by subtracting the buoyant force, which was

(.137kg/(21450kg/m^3)) * (9.8m/s^2) * (1000 kg/m^3) = .062593N
the volume of platinum g density of water

from the weight of the platinum: 1.3426 Newtons


What I'm trying to find now is this, "If the object is now attached to a piece of Styrofoam, what volume of Styrofoam (density rho_s = 95 kg/m3) should be used to have the Styrofoam and the metal floating just under the water surface?". But I've hit a stopping point and can't seem to get the right answer

Homework Equations



density = mass/volume

buoyant force = volume displaced * gravity * density of fluid

The Attempt at a Solution



I had set everything in equilibrium, meaning:
Buoyant force + Weight of Styrofoam + weight of the platinum = 0

But how do I know the volume of the water displaced if I don't already know the volume of Styrofoam that I need?
 
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HadManySons said:
I had set everything in equilibrium, meaning:
Buoyant force + Weight of Styrofoam + weight of the platinum = 0
That's fine, but it might be easier to just look at forces on the Styrofoam piece. (Since you calculated the tension in the string, why not use it?)

But how do I know the volume of the water displaced if I don't already know the volume of Styrofoam that I need?
Express the forces in terms of the volume. Then you can solve for it.
 
Well I finally figured it out. The equation ended up being:

9.8 m/s2 * 1000 kg/m3 * (VolumeStyrofoam + 5.67E-6The volume of platinum) = 1.3426 Nweight of platinum + (9.8m/ss * 95kg/m3density of Styrofoam * VolumeStyrofoam)

Which is then:
9800V + .0559421= 1.3426 + 931V
8869V = 1.2870058
V = 1.45E-4 m3 of Styrofoam.
 
Good. Making use of the results from the first part (the tension), you could use:
ρwVg = ρsVg + T

Solving for V gives you the same result.
 
Doc Al said:
Good. Making use of the results from the first part (the tension), you could use:
ρwVg = ρsVg + T

Solving for V gives you the same result.

I had wondered how to go about that, since you had mentioned it earlier. Thanks for pointing that out.