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





-----Original Message-----
From: phys-l-bounces@mail.phys-l.org
[mailto:phys-l-bounces@mail.phys-l.org] On Behalf Of Jeff Bigler
Sent: Tuesday, June 26, 2012 11:50 PM
To: Phys-L@Phys-L.org
Subject: Re: [Phys-L] Physics, Errors and Differenet Teaching Styles

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
boat+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.

In more colloquial terms than I used before: With the rock in the boat, the buoyant force is great enough to hold them both up, but with the rock on the bottom the buoyant force holds the boat up by itself but gets help from the normal force in holding the rock up so the total buoyant force is smaller meaning less water is displaced meaning the lake level is lower.
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.

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