# Density/Buoyancy question

Discussion in 'MCAT Study Question Q&A' started by aint settling, Aug 11, 2011.

1. ### aint settling

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The true mass of a rock is 10g. When submerged in water (D=1000kg/m3) the apparent mass of the rock is 5g.

What is the density of the rock?

2. ### geeyouknit 2+ Year Member

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Let's say force = F; b= buoyancy, w=weight

On the object, the overall force is Fnet;
Fnet = Fb - Fw;
Fw= mass * g = 0.01 kg * g (pointing down)
Fnet is related to the apparent mass = 0.005 kg* gravity (pointing down)
Fb = the force of buoyancy is the weight of the water displaced =
= mass of water * g (pointing up)
Cancel out gravity in all terms ==> -0.005 kg = Masswater - 0.01 kg
Mass of water is 0.005 kg; Use density to get vol of water, 5e-6
which gives volume of rock; 0.01 kg / vol of rock = density

Nevermind: I did most of it right, I just forgot that the net force is pointing down, so Fnet should be negative.

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3. ### inaccensa 7+ Year Member

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is the answer 5*10^-6 m3? oh yeah, sorry I wrote the volume...oopss hope not to do that on the test, density =10^10-3/5*10^-6 = 2000kg/m3

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4. ### costales 2+ Year Member

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This object-in-water problem is common. You can just take a ratio and multiply by the water density:
specific gravity
= true mass/(true mass-apparent mass)
So it's 10/(10-5) = 2, and the answer is 2000 kg/m3.

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5. ### docntrainin

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I'm confused as well, I thought the problem would be solved like this:

Fb = pfluid x g x Vsubmerged
(10 g - 5 g/ 1000) (10) = 1000 x 10 x Vsubmerged (aka volume of rock)
5/1000 = 1000 x V submerged
V submerged (Volume of Rock) = 5e-6

Density = 2000 kg/m3 (AHHHH STUPID MISTAKES LIKE THIS!!!!)

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6. ### inaccensa 7+ Year Member

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Well, I did
Weight air - Apparent weight = weight of the displaced fluid, although costales's formula is the easiest way, which i was not aware of.

There is something similar density object/ density fluid = fraction submerged ---> floating liquid

7. ### Majik

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Fnet = Normal Force + Bouyant Force - Weight

Note: Normal Force & Bouyant Force both act upwards; Weight acts down.
Provided the object is touching the sea floor, the net force is zero.
The apparent weight is equal to the normal weight.

Therefore, rearranging we get:

Weight - Bouyant Force = Normal Force

Weight = 0.1N
Normal Force (Apparent Weight) = 0.05N

We can solve for Bouyant Force now:
Weight - Normal Force = Bouyant Force
0.05N = Bouyant Force = pVg (p=density of water;V=volume of object)
0.05N / (1000kg/m^3)(10m/s^2) = Volume
5x10^-6 m^3 = Volume

Density of the object equals: mass / volume
Density: 10x10^-3 kg / 5x10^-6 m^3
Density: 2000 kg/m^3

8. ### muhali3 7+ Year Member

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If an object is floating, the relative density is the % submerged.

If sunken/submerged, the relative density is the ratio of the weight to the buoyant force (W/B).

9. ### Swagster 7+ Year Member

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Forgive me BerkTeach for stealing your method, but I have to say it's magic.

First, the object sinks, so we know that the weight/buoyant force ratio is equal to the relative densities (rock over water).

Second, the weight is 10 and the buoyant force (apparent weight loss) is 5, so the W/B ratio is 2. This means that the rock is twice as dense as the water.

For the exact breakdown of why W/B = (object density)/(water density) you need to read the page on buoyancy from the TBR book. Their approach is pure genius.

10. ### Swagster 7+ Year Member

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Did you take the TBR class or just use the books?

11. ### BerkReviewTeach Company Rep & Bad Singer Vendor 10+ Year Member

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It's not my method. That's all BR on that one. Gotta agree though, it's pretty genius the way they do certain math tricks.

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12. ### muhali3 7+ Year Member

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relative density questions also seem to come up pretty often, so those tricks really do save a lot of time.

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