Volume of a Cavity in an Iron Casting

AI Thread Summary
The problem involves calculating the volume of cavities in an iron casting weighing 300 N in air and 200 N in water, with the density of iron given as 7800 kg/m3. The effective weight reduction in water is attributed to the buoyant force, which equals the weight of the displaced water. The mass of the casting is calculated to be 30.6 kg, leading to a volume of iron of approximately 3.92 x 10^-3 m3. The discussion emphasizes the need to understand the relationship between actual weight, apparent weight, and buoyant force to solve for the volume of cavities accurately. The confusion arises around using the correct values for weight and buoyant force in the calculations.
Okazaki
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Homework Statement


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 ?

Homework Equations



d = m/V
Fg = mg

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.
 
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Okazaki said:
but I literally don't have any idea where to start.
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.
 
mfb said:
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.

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.
 
Okazaki said:
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?
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.
 
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