Archimedes principle and the density of a rock

[Megatronlol]

Hi, so I have a problem that I am working on and I think that I understand Archimedes principle but the density that I am calculating is just absurd. The problem is as follows:

"A geologist finds that a moon rock whose mass is 9.28 kg has an apparent mass of 6.18 kg when submerged in water. What is the density of the rock?"

I know that the apparent weight of an object when submerged in water is equal to the amount of displaced water. So the actual weight of 9.28 kg minus the apparent weight of 6.18 kg means that 9.28 kg - 9.18 kg = 3.1 kg of displaced water? Is this also true for volume then?

I know the density of water to be 1000kg/m3 and I have a mass of displaced water to be 3.1 kg, so rearranging p=m/v I get v=m/p to find the volume. Plugging in I get v = 3.1 kg/1000 kg/m3 which gives me a volume of displaced water of 3.1 x 10-3 m3 or 0.0031 m3 .

If a submerged object loses weight equal to the weight of the displaced water, does this mean that the volume of displaced water is equal to the volume of the rock? My current attempt assumes that this is true.

Assuming the volume of the displaced water = the volume of the rock then I have:

prock = 9.28 kg / 0.0031 m3 which equals 2993 kg/m3 which is ridiculous because 2993 kg is greater than the actual mass of 9.28 kg...

Any guidance is appreciated. I must be off by a few orders of magnitude for something? 2.9 kg/m3 seems reasonable.

 

[russ_watters]

Hi, so I have a problem that I am working on and I think that I understand Archimedes principle but the density that I am calculating is just absurd. The problem is as follows:
...
prock = 9.28 kg / 0.0031 m3 which equals 2993 kg/m3 which is ridiculous because 2993 kg is greater than the actual mass of 9.28 kg...

Any guidance is appreciated. I must be off by a few orders of magnitude for something? 2.9 kg/m3 seems reasonable.

Take another look at your units. Density is per CUBIC METER. Your rock is a lot smaller than a cubic meter so clearly the density is a much larger number than the mass (and the numbers aren't really comparable anyway, since they have different units). You're trying to find how much a cubic meter of it would weigh.

 

[Charles Link]

Water has a density of 1.00 gram/cm^3 which equates to 1000 kg/m^3 , so a density of 2993 kg/m^3 for a rock is quite reasonable. Your calculations to arrive at this number are in fact correct. ## \\ ## You computed the volume of the rock to be ## 3.1 \cdot 10^{-3} \, m^3 ##, which is 3.1 liters. And the 9.28 kg is about 20 pounds. Your numbers are good.

 

[Megatronlol]

Take another look at your units. Density is per CUBIC METER. Your rock is a lot smaller than a cubic meter so clearly the density is a much larger number than the mass (and the numbers aren't really comparable anyway, since they have different units). You're trying to find how much a cubic meter of it would weigh.

I think that I understand what you mean, but I am still struggling to grasp it a little.

Water has a density of 1.00 gram/cm^3 which equates to 1000 kg/m^3 , so a density of 2993 kg/m^3 for a rock is quite reasonable. Your calculations to arrive at this number are in fact correct. ## \\ ## You computed the volume of the rock to be ## 3.1 \cdot 10^{-3} \, m^3 ##, which is 3.1 liters. And the 9.28 kg is about 20 pounds. Your numbers are good.

Thank you for confirming.

 

[russ_watters]

I think that I understand what you mean, but I am still struggling to grasp it a little.

The use of density is to standardize a way to compare masses of different sized objects. It's like saying, "if I had a cubic meter of this, how much would its mass be?" Then you can compare it to the density of other objects.

 

[Charles Link]

I think that I understand what you mean, but I am still struggling to grasp it a little.
Thank you for confirming.

A cubic meter is 1 million cubic centimeters ## =1.0 \cdot 10^6 \, cm^3 ##, so that a cubic meter of water has a mass (weighs) one million grams which is 1000 kilograms. Thereby the density of water is ## \delta_{water}=1.00 \, g/cm^3=1000 \, kg/m^3 ##.

 

[sophiecentaur]

I know that the apparent weight of an object when submerged in water is equal to the amount of displaced water.

I don't think you can mean this. The apparent weight is the actual weight minus the weight of water displaced (buoyant force).
I think you got lost in your calculations by making them more complicated than needed.
If you use the formula
Specific Gravity (old term for ratio of density to that of water)
= Actual Weight of Rock/ Weight of water displaced
=W/(W -W_{under water} )
the whole thing is easy to see in one go.