Archimedes Principle

  • Thread starter funcosed
  • Start date
  • #1
funcosed
37
0

Homework Statement


trying to make sense of this derivation...


Homework Equations


Buoyant force is not an independent force but is derived from difference in pressure from the bottom of an object to its top.

Lets say an object of density "ρ'" is located "d" meters down the surface of a liquid of density "ρ" relative to its upper surface so the upper surface experience a force of water pressure equal to

P1= ρ*g*d

Also assuming that the object itself is "h" meters in heights then the lower part is h +d meters down the surface and experiences a Pressure of

P2 = ρ*g* (d+h)

obviously P2 > P1

so ΔP = P2 - P1 = ρ*g* (d + h - d) = ρ*g*h

and then,

ΔP = ΔF / A = ρ*g*h

so we have

ΔF = ρ*g*h*A = ρ*g*V ; here V is the volume of the object that is submerged in the liquid, g is the gravity constant, and ρ is the density of the liquid.

Also since we have:

ΔF = ρ*g*V and ρ' = m / V so V = m / ρ' we have,

ΔF = ΔF = ρ*g*Vρ*g* (m / ρ') = (ρ * ρ')(mg) = W(object)* (ρ * ρ')

The Attempt at a Solution


I can follow most of it but not the last line. It looks like an equals sign has been left out
ΔF = ΔF = ρ*g*V = ρ*g* (m / ρ') ??
but then where does the (ρ * ρ') term come from??
 

Answers and Replies

  • #2
tiny-tim
Science Advisor
Homework Helper
25,838
256
hi funcosed! :wink:
I can follow most of it but not the last line. It looks like an equals sign has been left out
ΔF = ΔF = ρ*g*V = ρ*g* (m / ρ') ??
but then where does the (ρ * ρ') term come from??

yes, it should be ΔF = ρ*g*V = ρ*g* (m / ρ') = mg(ρ/ρ') …

that last * is a misprint :smile:
 
  • #3
funcosed
37
0
thanks
 

Suggested for: Archimedes Principle

Replies
22
Views
482
Replies
14
Views
2K
Replies
13
Views
870
  • Last Post
Replies
6
Views
660
Replies
13
Views
359
  • Last Post
Replies
9
Views
357
Replies
5
Views
739
Replies
1
Views
2K
  • Last Post
Replies
1
Views
678
Top