Object Weight given only difference in weights in and out of water

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Homework Help Overview

The problem involves a chunk of carbon steel submerged in fresh water, with a focus on understanding the relationship between its weight in air and water, given a specific difference in weights. The subject area includes concepts of buoyancy, density, and weight calculations.

Discussion Character

  • Exploratory, Conceptual clarification, Mathematical reasoning, Assumption checking

Approaches and Questions Raised

  • Participants discuss the buoyant force and its relationship to the weights in air and water, questioning how to derive volume from the given weight difference. There are attempts to clarify the equations related to buoyancy and weight, with some participants exploring the conversion of density units.

Discussion Status

The discussion is active, with participants sharing their reasoning and calculations. Some guidance has been offered regarding the relationships between the weights and the buoyant force, but there is no explicit consensus on the correct approach or calculations yet.

Contextual Notes

Participants are working under the constraints of a homework assignment, which may limit the information they can use or the methods they can apply. There are also questions about unit conversions and the implications of the density values provided.

missnola2a
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Homework Statement




A chunk of carbon steel with density, ρ = 7.84 g/cm3 , is completely submerged in fresh water. The chuck of steel weighs 39 N more in air than in water. Please answer the following:
(a) Find the buoyant force acting on the chuck of steel.
N
(b) Find the the volume of the chuck of steel.
m3
(c) What is the mass of the chuck of steel?

Homework Equations




Fw(air)=Fw(water) + 39 N

d=m/v

Fb=gp(f)V

BUT I don't know how to find the volume.

density of air is 1.16 kg/m3
density of water is 1.0x10^3


The Attempt at a Solution



once I get the VOLUME I can find the relative weights by using rho*g*v
and Fw(water)+ 39N = Fw(air)
 
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The buoyant force (upward direction) in the air is less than in the water, therefore
Fw(air)=Fw(water) -39 N
And you know that (mg-Fw(air))/g = m
 
Weight of the displaced water = 39 N = m*g = density of water*volume*g.
 
ok, a little hazy. Fw(water) is 39N=dens water*v*g? than V = .00398

if so, Fw(water)-39N=Fw (air) than Fw(air) is /38.96??
 
-38.96 (which can't be right) *rather than /38.96
 
missnola2a said:
-38.96 (which can't be right) *rather than /38.96

Fw(in air ) = density of steel*volume*g.
 
OHH,...

so what you are saying is that I use previously acquired volume of .00398 in this eq.

I am sorry if I am frustrating. I mean well!

OH no... do I convert 7.84 g/cm^3 to kg/m^3 ??

if so 7.84 g/cm^3 (100cm*100cm*100cm) * 1kg = 7840 kg/m^3
(1m^3) 1000 G


Fw(air) = 7.84 g/cm3

OR

Fw (air) = 7840 kg/m^3 * .00398 m^3 * 9.81 m/s

is that right?

then to get weight in water I subgract (39N/9.81) ?
 
then to get weight in water I subgract (39N/9.81) ?
No.
The weight in water = F(air) - 39 N.
 

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