# Volume of a Cavity in an Iron Casting

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1. Apr 20, 2015

### Okazaki

1. The problem statement, all variables and given/known data
An iron casting weighs 300 N in air and 200 N in water. What is the volume of cavities in the casting, if the density of iron is 7800 kg/m3 ?

2. Relevant equations

d = m/V
Fg = mg

3. The attempt at a solution
I was not sure in the slightest of how to solve this problem. So I tried using an approach I tried in a similar problem.

Fg = mg
300 N = m * 9.8 m/s2
==> m = 30.6 kg

diron = 7800 kg/m3
dwater = 999.97 kg/m3

ddifference = -(6800.03) kg/m3

V = m/d
= 30.6 kg /6800.03 kg/m3
= 0.0045 m3

I know this is probably so far from the right answer, but I literally don't have any idea where to start.

2. Apr 20, 2015

### Staff: Mentor

Well, you did start.
The mass is 30.6 kg, good.
How much volume of iron does that correspond to? This is independent of the weight in water.

What is the effect that reduces the (effective) weight in water, and what does it tell you about the object? This is independent of its composition.

3. Apr 20, 2015

### Okazaki

Well, if you do it out:

d = m/v
v = m/d
= 3.92 x 10-3m3

And the buoyant force is the one that basically reduces the effective weight of an object in water. This force is equal to the displaced water, correct? So:

Fb = mfg
==> dwater*v*g

...We do use the density of water here, right?

=999.97 kg/m3 * 3.92 x 10-3m3 * 9.8
=38.21 N

And I just looked in the book, and the formula for weight vs. apparent weight is:

(apparent weight) = (actual weight) - (magnitude of buoyant force.)

...But, I'm confused. Would we put 300 N for actual weight? If we did, there's no way you could get 200 N by subtracting the buoyant force.

4. Apr 20, 2015

### Staff: Mentor

Yes, but you don't know the total volume yet.
You have to find the force by comparing the 300 N with the 100 N.