Density of a Bar of Soap/ buoyancy

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SUMMARY

The density of a rectangular bar of soap floating in water can be calculated using Archimedes' Principle. Given that 3.5 cm of the soap is submerged and 1.5 cm is above water, the total thickness of the soap is 5 cm (0.05 m). The volume of the soap is expressed as 0.05A m³, where A is the cross-sectional area. By equating the buoyancy force, which is the weight of the displaced water (0.035A m³), to the weight of the soap, the density of the soap can be determined using the formula: density = (density of water * 0.035A) / (0.05A).

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Homework Statement



A rectangular bar of soap floats with 3.5 cm extending below the water surface and 1.5 cm above. what is its density?

Homework Equations


p(rho)=m/v



The Attempt at a Solution


I don't know how to compute the volume since we are given only two lengths.
Athough, i think a better way to approach this problem will be with buoyancy.

To start,can we consider this block to be in equilibrium since it is floating? I am confused because it is also partially submerged.
 
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Yes. This question is solved using the idea of buoyancy and Archimedes Principle.
Here's a way of starting...
Let the area of the top and bottom surface be A m²
The volume of the soap is therefore 0.05A m³ [0.05 is the height/thickness of the soap]

Now because the soap is floating, its weight is exactly balanced by the upthrust due to the weight of the volume of water it is displacing.
Weight = mass x g
Mass = volume x density

Does this help you to get started?
 
how did you get .05A?

So if it is floating, mg=Buoyancy.
mg=density x volume x g

how do i get the mass if i don't know the density?
Sorry, there are two unknowns still.. density and mass. i don't know how to get either with just the volume given
 
If there is 3.5cm of soap below water and 1.5 above, then the thickness of the soap is 3.5 + 1.5 = 5cm
5cm = 0.05m

Volume is area of cross section times length/thickness
If the area of cross section (of the top and bottom of the soap) is A then the volume of the soap is 0.05A

When the soap is floating, only 3.5cm (0.035m) is below the water.
This means that the volume under water is 0.035A
The volume of water displaced is V=0.035A
The mass of water is volume V times density of water ρw
The weight of water is mass times g
w x 0.035A x g
This equals the buoyancy force.

This force must be equal to the weight of the soap.
The weight of the soap is its volume (0.05A) times its density.
If you equate the buoyancy force to the weight of the soap you will get a value for the density of the soap.
A and g cancel out. You need a value for the density of water.
 

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