Pressure in water and different liquid

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Homework Help Overview

The discussion revolves around understanding pressure in fluids, specifically in relation to buoyant forces acting on submerged objects. Participants are examining the principles of pressure, buoyancy, and the relationship between weight and buoyant force in a fluid context.

Discussion Character

  • Exploratory, Conceptual clarification, Mathematical reasoning, Assumption checking

Approaches and Questions Raised

  • Participants explore the calculation of pressure and buoyant force, questioning how to apply the relevant equations. There is discussion about the relationship between the weight of the object and the buoyant force, as well as the volume of the object submerged in the liquid.

Discussion Status

Some participants have offered guidance on using buoyant force as a more sensible approach than pressure calculations. There is ongoing clarification about the definitions and roles of variables in the equations being discussed, with no explicit consensus reached on the final approach.

Contextual Notes

Participants are navigating assumptions about the submerged volume of the object and the densities involved, as well as the implications of the object being half submerged. There is a noted confusion regarding the application of pressure and buoyant force equations.

Scarlet_pat
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Homework Statement


consider the figures 1 and 2 which show two identical, rectangular wooden blocks

Weight of the block:



Homework Equations



Pressure = F/a
Pressure = Density x gravity x height



The Attempt at a Solution



The weight of the block,
i have tried to find the pressure of water of figure1

pressure = p g h = p = 1000 x 9.8 x 0.05 = 490 N/m^2

but i do not have any idea of what to do after.

thank you very much
 

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There is no need for finding pressure. Find the buoyant force
 


the buoyant force = V p g ...
the buoyant force = to the force exerted by the object.
therefore times the buoyant force by two, Because it is exact half submerged ?
 
Hi Scarlet_pat! :smile:

(have a rho: ρ :wink:)
Scarlet_pat said:
the buoyant force = V p g ...
the buoyant force = to the force exerted by the object.
therefore times the buoyant force by two, Because it is exact half submerged ?

Yes, but that's no way to prove it :redface:

you need to write an equation, weight + buoyant force = 0.

(btw, you could have used pressure … the upward force from the water is the pressure times the area of the base … but buoyant force is more sensible :wink:)
 


thank you very much for comfirming my answer :)
i was actually try to the method you have just suggested.
F= Pa = p g h * A

the the pressure there is always referring to Pressure of liquid right ?

however i think i have made a mistake ...
for F= V p g <-- the volume should be the volume of the object which submerged in the liquid right ?
 
Scarlet_pat said:
for F= V p g <-- the volume should be the volume of the object which submerged in the liquid right ?

(What happened to that ρ I gave you? :confused:)

I'm confused …

is ρ here the density of the block or of the water, and is this supposed to be the buoyant force or the weight?
 


lets just stick to the first equation ;)
F = V p g

and what I'm wondering is F = upthrust ,
V = volume of object submerged in liquid OR volume of the block?
p = density of liquid
g = acceleration due to gravity

am i right ?
 
The upthrust is the weight of the liquid displaced, which is the density of the liquid times the volume of the liquid displaced (ie the volume of the object submerged). :smile:
 


ok thanks ... thank you. ill have to time the upthrust by 2, because the object is only half submerged ... right ?
 
  • #10
The upthrust is the upthrust, you don't "times" it by anything, you add it to the weight to get zero.
 
  • #11


ok thanks. :)
 

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