Re: [Phys-L] Physics, Errors and Differenet Teaching Styles

[Jeff Bigler <jcb@mit.edu>]

On 6/26/2012 6:15 PM, Jeffrey Schnick wrote:
> > #3. "A boat carrying a large rock Is floating on a lake. The 
> > boulder is thrown overboard and sinks. The water level in the 
> > lake (with respect to the
> > shore) 
> >
> > 1) rises 
> >
> >  2) drops 
> >
> >  3) remains same"
> 2) The water level in the lake, relative to its banks, drops.  (I am
> assuming that the density of the rock is greater than that of the water
> so the rock sinks to and comes to rest on the bottom where, in
> equilibrium the bottom is exerting a non-zero normal force upward on the
> rock--e.g. we are not talking about a pummice rock that might sink down
> below the surface just after being dropped but would then bob back up to
> the surface.)  The gravitational force that would act on that amount of
> water that would fit in the space occupied by the boat+rock below the
> surface of the water (let's call this the displaced water) is now less
> than the gravitational force on the boat+rock because the buoyant force
> is no longer equal in magnitude to the gravitational force on the
> boat+rock.  Rather, the magnitude of the net upward force which is equal
> to the magnitude of the total buoyant force on the boat+rock plus the
> magnitude of the normal force exerted on the rock by the bottom of the
> lake minus the magnitude of the gravitational force on the boat+rock is
> zero, so the magnitude of the buoyant force on the boat+rock is equal to
> the magnitude of the gravitational force on the boat+rock minus the
> magnitude of the normal force exerted on the rock by the bottom of the
> lake. Now if the gravitational force that would act on the displaced
> water is less, that means the volume of the displaced water is less.
> The net change is equivalent to taking a thin layer of water from the
> top of the water in the lake and using it to fill in some of the
> original displacement volume, just enough to reduce the displacement
> volume to the new value.  The new water level in the lake is reduced by
> the thickness of that layer.

The buoyant force on the rock does not change when it is resting on the
bottom.  The normal force from the bottom of the lake plus the buoyant
force must equal the weight of the rock, which means the normal force
from the bottom must be less than the weight of the rock.

The boat's mass would be reduced by the mass of the rock, so the change
in the amount of water displaced by the rock must be the negative of the
change in the water displaced by the boat.  If we have an ideal lake and
an ideal rock (all of the water has the same temperature, and the water
and rock are completely incompressible), the level of the lake water
should remain unchanged.

If we want to look at very small effects, the density of water at the
bottom of the lake is slightly greater than at the surface, partly due
to compression from the water above and mostly due to the fact that the
water in the lake probably is not uniform in temperature, and the denser
water sinks to the bottom.  (As an example, the water in Seneca Lake,
the largest of the Finger Lakes in New York State, has a bottom
temperature of approximately 4°C year round, which would have a density
of 999.9 kg/m^3. In the summer, the water at the surface has a
temperature of about 20°C, which means a density of 998.2 kg/m^3.)  The
density difference means the buoyant force at the bottom of the lake
will be slightly greater than the buoyant force at the top.  (Of course,
the rock is also slightly smaller due to thermal contraction and
compression from the additional pressure, so we would need to figure
that in too, but I believe these effects are insignificant compared with
the thermal effects.)  However, the simplicity of the above problem
suggests that these effects are intended to be neglected.

-- 
Jeff Bigler
Lynn English HS; Lynn, MA, USA
"Magic" is what we call Science before we understand it.