What is the density of the body in Archimedes' Principle problem?

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The discussion focuses on calculating the density of a body using Archimedes' Principle, given that the body weighs half as much in water as it does in air. The density of water is stated as 1000 kg/m^3. The user derives the equation by equating the buoyant force in water to half the weight of the body, leading to the conclusion that the density of the body is twice that of water. The final calculation confirms that the density of the body is 2000 kg/m^3, which aligns with the principles discussed. The solution is validated by other participants in the thread.
chmate
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Hello all!

Inside the water, body weights two times less than in air. Density of water is 1000kg/m^3. Find the density of body.


1. Well I tried this solution and I'm not sure if it's correct. I started from this principle Fa=W/2 (inside water force equals weight of the body divided by two). Then:


\rho 1 -> density of water
\rho 2 -> density of body
V -> volume

\rho 1 gV=\dfrac{\rho 2gV}{2}

\rho 1=\dfrac{\rho 2}{2}
\rho 2=2\rho 1

Anything wrong?

Thank you.
 
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Looks good to me.
 

Thread 'Magnitude of buoyant force in fluids of different densities'

The book claims the answer is that all the magnitudes are the same because "the gravitational force on the penguin is the same". I'm having trouble understanding this. I thought the buoyant force was equal to the weight of the fluid displaced. Weight depends on mass which depends on density. Therefore, due to the differing densities the buoyant force will be different in each case? Is this incorrect?
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