Calculating Object Density with Formulae

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The discussion focuses on the correct formulas for calculating the density of an object using Archimedes' principle. Two proposed formulas are evaluated, both of which are confirmed to be valid for determining density. The conversation emphasizes that buoyant forces can complicate calculations, suggesting a simpler approach of using the mass of the object and the volume of displaced fluid. It is noted that Archimedes' law applies primarily to incompressible fluids, allowing for straightforward density calculations. Ultimately, the consensus is that the density can be easily derived from the object's mass and the volume of fluid it displaces.
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Can I use either of these formulae to find the desity of an object?

a) ((Weightapparent/Forcebuoyant) + 1)(Densityfluid) = Densityobj

b) (Weightobject/Forcebuoyant *Densityfluid) = Densityobj

Thanks
 
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Archimedes law tells you the buoyance force is equal to the gravitational force on the displaced mass:

F_{buoyance}=g \rho_{fluid} V_{displaced}

The apparent weight of the object can be found by evaluating the net downward force F_g-F_{buoyance}=g[m-\rho_{fluid} V_{displaced}] Where the term between brackets is the apparent weight.

With \rho density, m mass V volume and g the gravitational acceleration. With this knowledge you can probably work out the answer yourself!
 
I still can't figure it out :mad:
 
da_willem said:
Archimedes law tells you the buoyance force is equal to the gravitational force on the displaced mass:

F_{buoyance}=g \rho_{fluid} V_{displaced}

The apparent weight of the object can be found by evaluating the net downward force F_g-F_{buoyance}=g[m-\rho_{fluid} V_{displaced}] Where the term between brackets is the apparent weight.

With \rho density, m mass V volume and g the gravitational acceleration. With this knowledge you can probably work out the answer yourself!

What's to figure out than applying what Da-willem said??

You may want to write:
\rho_{body}-\rho_{fluid}=\frac{F_g-F_{buoyance}}{g V_{displaced}}
From the previous equation,the density of the body is immediate.
 
ok, so my original two formulas are correct then, right?
 
Haftred said:
ok, so my original two formulas are correct then, right?

Yes,they are true.But why complicate with calculating buoyant forces,when u can simply say that:
\rho_{object}=\frac{m_{object}}{V_{displaced}}
It's Archimede's law that states that,if the fluid is incompressible (Archimedes didn't think of compressible fluids),then the volume of the body is equal to the volume of the displaced fluid.Since u can easily determine one's body mass,the density is immediate.

Daniel.
 
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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