04_39 Buoyant forces and densities

[AznBoi]

A spring scale calibrated in kilograms is used to determine the density of a rock specimen. The reading on the spring scale is 0.45 kg when the specimen is suspended in air and 0.36 kg when the specimen is fully submerged in water. If the density of water is 1000 kg/m^3, the density of the rock specimen is

A) 2.0 x 10^2
B) 8.0 x 10^2
C) 1.25 x 10 ^3
D) 4.0 x 10^3
E) 5.0 x 10^3

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Solution of book:
The buoyant force is equal to the weight of the water displaced and is the difference between the apparent weight and actual weight. So the mass of the water displaced is the diff between the actual and apparent masses... or 0.45 - 0.36 = 0.09 kg is mass of water displaced.

the water displaced and the rock have the same volume, so the ratio of their densities is the same as the ratio of their masses. Ratio of masses is

0.45 / 0.09 = 5.0 the rock is five times the mass, so must be five times denser than water.

Density of water is 1000 kg/m3 so density of rock is 5000 kg/m3


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I've used a method to get the answer, and since the specimen is complete submerged in the water the volumes are the same. (am I right so far?)

So then you can combine the volumes and solve for the actual density of the specimen: = 5000kg/m^3
Is this a logical/correct way of solving this problem?

I know that using ratios and such is much more efficient in a lot of situations where the problem gives no values for unknowns except for ratios. How do you know what ratio to use and how do you get better at using them? I'm confused about the solution's method and would like to learn it, thanks!

 

[Mindscrape]

I would use Newton's second law. You know that the buoyancy force is $$F_b = \rho V g$$. You also know, by Newton's second law and conservation laws, that the force of gravity minus the buoyant force will be the displaced force $$F_g - F_b = F_a$$. At which point the density of the object will be the ratio of force of gravity to the buoyancy force, which will be the difference of the force of gravity and the apparent force of gravity (apparent weight). So $$\rho_{rock} = \frac{F_g}{F_b} = \frac{F_g}{F_g-F_a}$$. Your textbook just factored out the gravity term since it appears in every term.

 

[m2003]

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Yes, your method is also correct. By combining the volumes and using the formula for density (density= mass/volume), we can solve for the density of the rock specimen. In this case, the volume of the water displaced is equal to the volume of the rock specimen, so we can simply divide the mass of the rock (0.45 kg) by the volume of the rock (which is equal to the volume of the water displaced). This gives us a density of 5000 kg/m^3, which is the same as the solution provided in the book.

As for using ratios, it is important to understand the relationship between different quantities in a problem. In this case, we know that the ratio of the mass of the rock to the mass of the water displaced is 5:1, because the rock is five times denser than the water. This means that for every 1 kg of water displaced, there is 5 kg of rock. By setting up a ratio and solving for the unknown quantity (in this case, the density of the rock), we can easily find the answer. Practice and understanding the concepts behind ratios will help you become better at using them in problem solving.