Hydrostatics/Equilibrium: An object hanging by a light cord in a bucket of water

[LunarJK]

Homework Statement

A rock with a mass of 2.8 kg is suspended from the roof of an elevator by a light cord. The rock is totally immersed in a bucket of water that sits on the floor of the elevator, but the rock doesn't touch the bottom or sides of the bucket.
When the elevator is at rest the tension of the cord is 19.9 N. Calculate the volume of the rock in cubic centimeters (cm^3).

Homework Equations

Pascal's Law: P = Po + gρh
V = Ah or V = m/ρ
P = F/A
ΣF = 0

The Attempt at a Solution

I am a little confused about how to go about this problem as I'm just starting into fluid dynamics... but my initial step into this was to assume that if the rock is not touching the bottom of the bucket, and is also not floating up, that it is in a state of equilibrium.

So I started by trying to recognize all forces acting upon the rock: T-tension on the cord upwards, mg-weight of the rock due to gravity, PoA- force of the Atmospheric pressure acting down on the rock, and PA-pressure of the water acting up on it. From this i stated that:
T + PA - mg - PoA = 0 thus,
T + PA = mg - PoA

However since i know the mass of the rock, tension of the cord, and I'm assuming gravity to be 9.81 m/s2, this equation can be simplified as far as:
A(P-Po) = 7.56N

Now the Po will equal just the standard atmospheric pressure of 101.345 kPa, but that still leaves 2 unknowns that I'm not too sure how to deal with.

Since i am looking for Volume of the rock, i can relate volume in only two equations i can think of, which are V = Ah, and V = m/density. Since I am already working with a formula that involves the area, I am thinking my next step would be to somehow use pascal's law to find the depth the rock is submerged into find the volume at the end?

Anyways, any tips would be nice. Let me know if I'm on the right track, or if I'm way overlooking a simpler solution.

 

[rl.bhat]

You have to use Archimedes' Principle, not Pascacl's law.

 

[thomate1]

You are on the right track in recognizing that the rock is in a state of equilibrium. From this, we can set up an equation of forces as you have done, with the tension of the cord, weight of the rock, atmospheric pressure, and water pressure all acting on the rock. However, we can simplify this equation by noting that the water pressure is equal to the weight of the water displaced by the rock, which in turn is equal to the weight of the rock. This can be represented as P = ρgh, where ρ is the density of water, g is the acceleration due to gravity, and h is the depth of the rock in the water.

Using this equation, we can rewrite our original equation as:

T + ρgh - mg - PoA = 0

Now, we can solve for the depth of the rock in terms of the known quantities:

h = (T + PoA - mg)/ρg

From here, we can use the equation V = Ah to find the volume of the rock. We know the area of the rock (since it is a solid object) and we have just solved for the depth, so we can plug these values into the equation and solve for V.

Finally, since the density of the rock is given as 2.8 kg, we can use the equation V = m/ρ to find the volume in cubic centimeters. This will give us the final answer for the volume of the rock in the bucket of water.

Overall, you were on the right track in recognizing the forces at play and using Pascal's law to find the depth of the rock. Keep in mind that in fluid dynamics, we often use the concept of pressure to relate different quantities, so it is important to always consider the pressure of the fluid when solving these types of problems.