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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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missnola2a
#1
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.
N
(b) Find the the volume of the chuck of steel.
m3
(c) What is the mass of the chuck of steel?

2. Relevant equations


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

d=m/v

Fb=gp(f)V

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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Sakha
#2
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
rl.bhat
#3
Jul27-09, 12:46 AM
HW Helper
P: 4,439
Weight of the displaced water = 39 N = m*g = density of water*volume*g.

missnola2a
#4
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??
missnola2a
#5
Jul27-09, 01:05 AM
P: 13
-38.96 (which cant be right) *rather than /38.96
rl.bhat
#6
Jul27-09, 01:11 AM
HW Helper
P: 4,439
Quote Quote by missnola2a View Post
-38.96 (which cant be right) *rather than /38.96
Fw(in air ) = density of steel*volume*g.
missnola2a
#7
Jul27-09, 01:20 AM
P: 13
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) ?
rl.bhat
#8
Jul27-09, 01:33 AM
HW Helper
P: 4,439
then to get weight in water I subgract (39N/9.81) ?
No.
The weight in water = F(air) - 39 N.


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