Buoyancy question about styrofoam

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To determine the volume of Styrofoam needed to keep a 0.137 kg solid platinum object floating just under the water surface, the buoyant force must balance the combined weight of the platinum and Styrofoam. The buoyant force is calculated using the equation for volume displaced, which incorporates the densities of both the fluid and Styrofoam. The final equation derived is 9800V + 0.0559421 = 1.3426 + 931V, leading to a required volume of 1.45E-4 m3 of Styrofoam. This approach effectively uses the previously calculated tension to simplify the problem. The discussion highlights the importance of equilibrium in buoyancy calculations.
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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.
 
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