Volume of a Cavity in an Iron Casting

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Homework Help Overview

The problem involves determining the volume of cavities in an iron casting based on its weight in air and water. The subject area pertains to buoyancy and density calculations within the context of fluid mechanics and material properties.

Discussion Character

  • Exploratory, Assumption checking, Conceptual clarification

Approaches and Questions Raised

  • Participants discuss the calculation of mass from weight and the implications of buoyancy on effective weight. There are attempts to relate the mass of the casting to its volume and density, while questioning how buoyant force affects the weight in water.

Discussion Status

The discussion is ongoing, with participants providing guidance on the relationship between mass, volume, and buoyant force. There is exploration of different interpretations regarding the application of weight and buoyancy, but no consensus has been reached yet.

Contextual Notes

Participants note the need to clarify the relationship between the actual weight of the casting and the apparent weight when submerged in water, as well as the role of buoyant force in this context.

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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