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

AI Thread Summary
The discussion revolves around calculating the buoyant force, volume, and mass of a submerged chunk of carbon steel given its weight difference in air and water. The buoyant force is derived from the weight difference, with the equation Fw(air) = Fw(water) + 39 N indicating that the weight in air is greater due to buoyancy. Participants clarify the need to convert the density of steel from g/cm³ to kg/m³ for accurate calculations. The volume of the steel is calculated using the buoyant force and the density of water. The conversation emphasizes the importance of correctly applying the principles of buoyancy and density to solve the problem effectively.
missnola2a
Messages
13
Reaction score
0

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)
 
Physics news on Phys.org
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.
 
Back
Top