### Re: is the density of a body equal its density in liquid+density of liquid?

Date: Fri Dec 26 11:25:17 2003
Posted By: Kenneth Beck, Staff, Chemistry and Physics of Complex Systems (C&PCS), Pacific Northwest National Laboratory
Area of science: Physics
ID: 1071754714.Ph
Message:
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Chris,

We need to understand a couple of concepts to answer your question.

First, let’s make sure we understand what is meant by  the “density of a
body“.  Density is the mass of a body divided by the volume occupied by
the body (d = m/V).  This is an accurate, physically rigorous
definition.  In a common approximation, sometimes people talk about
density as the “weight” of  an object divided by its volume.  At the
surface of the Earth when the force of gravity on an object (a body’s
weight) is more or less proportional to its mass, this common
approximation is OK too.

Second, there are two types of bodies we need to concern ourselves with
here.  Those are “rigid bodies” and “non-rigid bodies”.  Rigid bodies are
bodies whose shape stays the same and is unaltered by any forces acting
within our question (i.e., air pressure, liquid pressure, force of
gravity, etc.). Think of a steel chest or an oil tanker as a more or less
rigid body.  On the other hand, non-rigid bodies can have their shape
altered, compressed, expanded, or distorted by the forces acting within
our question.  Think of a balloon, or box made of paper or straw as a non-
rigid body.

If we measure the mass of a RIGID body and measure its volume in air, we
can calculate its density in air by dividing the mass by the volume.  If
we then put that rigid body in a liquid (like water or oil), we will find
its mass will not have changed.  Because it is a rigid body, its volume
will not have changed, either.  Therefore, its density will not have
changed.  The density of a rigid body is the same in air or in the liquid.

But, let's do this with a NON-RIGID body. If we measure the mass of a non-
rigid body and measure its volume in air, we can calculate its density in
air by dividing the mass by the volume. If we then put that non-rigid
body in a liquid, we will find its mass will not have changed.  However,
upon measuring its volume we may find that its volume has changed.  If it
is compressed (less volume) then its density will have INCREASED.  If the
body expanded (more volume), its density will have DECREASED.

As an example, we can fill a toy balloon (a more or less perfect non-
rigid body) with air. If we immerse it in oil by pushing down on it with
our hand, it will compress to the point where the pressure exerted by the
balloon plus the air inside it equals the pressure exerted by the oil in
which it is immersed.  The pressure exerted by a body (defined as Force
divided by Volume, p = F/V) is directly proportional to its density (
again, d = m/V).  Thus, the balloon plus the air will compress until its
density EQUALS the density of the oil in which it is immersed.  If we
then let it go, it will come to the top surface of the oil by expanding.
It will “float” on the oil surface because now - in air - its density is
less than that of the oil.

Dr. Ken Beck

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