How Is the Density of the Rock Determined in a Buoyancy Problem?

[xgoodtimesx]

One end of a light thin string is attached to a cork with density ρC = 0.24 g/cm3. The other end is attached to a rock. When the arrangement is placed in liquid 1 with density ρ1 = 0.80 g/cm3, a fraction f1 = 0.75 of the cork becomes submerged. When the arrangement is placed in liquid 2 with density ρ2 = 0.89 g/cm3, a fraction f2 = 0.67 of the cork becomes submerged. What is the density of the rock?


Please derive a equation to solve this problem

 

[tiny-tim]

Welcome to PF!

Hi xgoodtimesx! Welcome to PF!

Show us what you've tried, and where you're stuck, and then we'll know how to help!

 

[nlightner]

and then use the equation to find the density of the rock.

To solve this problem, we can use the principle of buoyancy, which states that the buoyant force acting on an object is equal to the weight of the fluid it displaces.

First, we can set up an equation for the buoyant force acting on the cork in both liquids 1 and 2:

Fb1 = ρ1Vg
Fb2 = ρ2Vg

Where ρ1 and ρ2 are the densities of the liquids, V is the volume of the cork, and g is the acceleration due to gravity.

Next, we can set up an equation for the weight of the cork:

W = ρCVg

Where ρC is the density of the cork.

Since the buoyant force is equal to the weight of the cork, we can set Fb1 = W and Fb2 = W and solve for V:

ρ1Vg = ρCVg
ρ2Vg = ρCVg

Solving for V and setting the two equations equal to each other, we get:

ρ1V = ρC
ρ2V = ρC

Dividing the two equations, we get:

ρ1/ρ2 = V/V

Substituting in the given values for f1 and f2, we get:

0.75/0.67 = V/V

Solving for V, we get:

V = 0.75/0.67 = 1.12 cm3

Now, we can solve for the density of the rock by using the equation for weight:

W = ρRVg

Where ρR is the density of the rock.

Substituting in the given values for f1 and f2, and the calculated value for V, we get:

0.75ρ1Vg = 0.24ρRVg
0.67ρ2Vg = 0.24ρRVg

Dividing the two equations and solving for ρR, we get:

ρR = 0.24(0.67/0.75)ρ1 + 0.24(0.67/0.67)ρ2 = 0.215 g/cm3

Therefore, the density of the rock is 0.215 g/cm3.