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

by missnola2a
Tags: difference, object, water, weight, weights
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Jul27-09, 12:39 AM
P: 13
1. The problem statement, all variables and given/known data

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.
(b) Find the the volume of the chuck of steel.
(c) What is the mass of the chuck of steel?

2. Relevant equations

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



BUT I dont know how to find the volume.

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

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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Jul27-09, 12:45 AM
P: 295
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
Jul27-09, 12:46 AM
HW Helper
P: 4,433
Weight of the displaced water = 39 N = m*g = density of water*volume*g.

Jul27-09, 01:04 AM
P: 13
Object Weight given only difference in weights in and out of water

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??
Jul27-09, 01:05 AM
P: 13
-38.96 (which cant be right) *rather than /38.96
Jul27-09, 01:11 AM
HW Helper
P: 4,433
Quote Quote by missnola2a View Post
-38.96 (which cant be right) *rather than /38.96
Fw(in air ) = density of steel*volume*g.
Jul27-09, 01:20 AM
P: 13

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


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) ?
Jul27-09, 01:33 AM
HW Helper
P: 4,433
then to get weight in water I subgract (39N/9.81) ?
The weight in water = F(air) - 39 N.

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