# Homework Help: Object Weight given only difference in weights in and out of water

1. Jul 27, 2009

### missnola2a

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)

2. Jul 27, 2009

### Sakha

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

3. Jul 27, 2009

### rl.bhat

Weight of the displaced water = 39 N = m*g = density of water*volume*g.

4. Jul 27, 2009

### missnola2a

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??

5. Jul 27, 2009

### missnola2a

-38.96 (which cant be right) *rather than /38.96

6. Jul 27, 2009

### rl.bhat

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

7. Jul 27, 2009

### missnola2a

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) ?

8. Jul 27, 2009

### rl.bhat

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

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