Calculating Object Density with Formulae

  • Thread starter Thread starter Haftred
  • Start date Start date
  • Tags Tags
    Density Formulae
Click For Summary

Homework Help Overview

The discussion revolves around calculating the density of an object using various formulae related to buoyancy and Archimedes' principle. Participants are exploring the relationship between apparent weight, buoyant force, and fluid density in the context of density calculations.

Discussion Character

  • Conceptual clarification, Mathematical reasoning, Assumption checking

Approaches and Questions Raised

  • The original poster questions the validity of two different formulae for calculating density. Some participants elaborate on Archimedes' law and the concept of buoyant force, while others express confusion about the application of these principles.

Discussion Status

There is ongoing exploration of the formulae and their implications. Some participants affirm the correctness of the original formulas, while others suggest simpler approaches to calculating density without involving buoyant forces. The discussion reflects a mix of interpretations and attempts to clarify the concepts involved.

Contextual Notes

Participants are navigating the complexities of buoyancy and density calculations, with references to Archimedes' law and the assumptions regarding fluid incompressibility. There is a noted frustration from the original poster regarding the clarity of the problem.

Haftred
Messages
53
Reaction score
0
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
 
Physics news on Phys.org
Archimedes law tells you the buoyance force is equal to the gravitational force on the displaced mass:

[tex]F_{buoyance}=g \rho_{fluid} V_{displaced}[/tex]

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

With [itex]\rho[/itex] density, [itex]m[/itex] mass [itex]V[/itex] 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:

[tex]F_{buoyance}=g \rho_{fluid} V_{displaced}[/tex]

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

With [itex]\rho[/itex] density, [itex]m[/itex] mass [itex]V[/itex] 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:
[tex]\rho_{body}-\rho_{fluid}=\frac{F_g-F_{buoyance}}{g V_{displaced}}[/tex]
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:
[tex]\rho_{object}=\frac{m_{object}}{V_{displaced}}[/tex]
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.
 

Similar threads

  • · Replies 24 ·
Replies
24
Views
2K
Replies
3
Views
2K
Replies
1
Views
1K
  • · Replies 4 ·
Replies
4
Views
2K
  • · Replies 9 ·
Replies
9
Views
1K
  • · Replies 2 ·
Replies
2
Views
2K
Replies
2
Views
2K
  • · Replies 23 ·
Replies
23
Views
3K
  • · Replies 10 ·
Replies
10
Views
3K
  • · Replies 2 ·
Replies
2
Views
1K