Quick Question

[Haftred]

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

[da_willem]

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!

[Haftred]

I still can't figure it out :mad:

[dextercioby]

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 wanna write:
\rho_{body}-\rho_{fluid}=\frac{F_g-F_{buoyance}}{g V_{displaced}}
From the previous equation,the density of the body is immediate.
:surprised

[Haftred]

ok, so my original two formulas are correct then, right?

[dextercioby]

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.