Pascal’s Principle (hydraulic
systems)
LARGE
force out
Small force
in
Pressure is the same at a
given depth, but force is
P*A
Archimedes’ Principle
As we saw in deriving the depth
dependence of pressure, the
forces acting on any subsection
of a fluid are equal to that
subsection’s weight, but those
forces from the rest of the fluid
also depend only on the shape of
that subsection. If we replace that
subsection of fluid with an object
of the same shape, the forces
from the fluid remain unchanged:
Fbuoy = weight of displace fluid
1.
Joey Soprano tasks two underlings to dispose of a large heavy
“package” in the bottom of a holding tank at a local water treatment
plant. The pair take the package out to the middle of the (large) tank
using a small boat, and drop it over the side (at which point it sinks to
the bottom of the tank). Does the water level (as measured at the
side of the tank) go up, go down, or stay the same after the package
reaches the bottom and while the two hoods are moving back to the
side of the tank (compared to its level when the package was still in
the boat)?
(DOWN: 3 UP: 11 SAME: 13 NO ANSWER: 27 Can’t tell: 1 )
•
The water level would go up because the package displaces the
water. [keep in mind, it must also displace water while in the boat!!]
•
i think the water level would increase because the boat is in the water
pushing the water down, and the boat is floating due to the force of
the water pushing it back, and to hold up a weight something must
give a little bit. in this case i think it is the water that is ˜bending˜ to
hold the boat up. When the package is dropped into the water it has a
density greater than the water because it sinks to the bottom. so in
this case the package is displacing the water pushing water up. [the
key is displacement of water, not direction of forces here]
•
The water level around the side of the tank should remain the same
after the ˜package˜ has sunk to the bottom of the tank. The ˜package˜
displaces the same amount of water in the boat as it does when it's
submerged. [In the boat it displaces its weight, in the water it just
displaces its volume!! Which is bigger?]
Chapter 14
problems
Equation of Continuity
(mass in must = mass out)
A1v1=A2v2
Assuming that r is constant
(i.e. an incompressible fluid)
E.G. with the Equation of Continuity
(mass in must = mass out)
What is the flow through the unmarked pipe?
E.G. with the Equation of Continuity
(mass in must = mass out)
13
What is the flow through the unmarked pipe?
(4+8+5+4 = 21 in; 13+2+6 = 21 out)
Bernoulli’s Principle
P + ½rv2 + rgy = const
Rank the pressures at positions 1, 2, 3, 4 ?
• Why does the curtain crowd you in the shower?
• (Bernoulli: 10; Pressure: 7; Temperature: 4; other 5;
no answer: 26)
• The answer, of course, has to do with a pressure
difference (lower pressure inside the shower
curtain), but the question is where does that
difference come from? The water creates a flow of
air on the inside=> lower pressure (Bernoulli effect).
[research indicates that there is an additional effect
from vorteces created in the air by the slowing down
of the air, but Bernoulli has a lot to do with it]
When:
Who:
April 19, 2009
Teams of 3*
*Must be all male/all female
Where:
DeVault Alumni
Center
jills-house.org/events.php
for a registration form
Find us on Facebook...Hoosiers for
$10/person, includes t-shirt
Questions Contact
[email protected]
Chapter 14
problems
•
A basketball is dropped onto a good hardwood floor and
is observed to bounce several times. In your opinion, is
such motion oscillatory? Is this motion simple harmonic
motion? Please explain (briefly) why or why not in each
case.
Most (20) said yes to osc. But NO to SHM (16) (changing
amplitude [9] or freq. [7]
•
The motion is oscillatory because it continues to move
about the ground in an up and down pattern; however,
the motion is not simple harmonic motion because it
does not bounce to a constant height ad infinitum.
[Close, but the argument against SHO is more restrictive]
•
Yes the motion is oscillatory because it repeats itself as
the ball bounces up and down. This is simple harmonic
motion because it is periodic and the ball's displacement
is a sinusoidal function of time. [Right emphasis, but
wrong conclusion; it is not sinusoidal!]
The restoring force is not linearly proportional to
displacement from equilibrium [This is the key; see
the next question]
•
Explain as compactly as you can manage the
essential requirement on a force if it is to be able
to produce simple harmonic motion. [14 answers
were confused or confusing]
•
The forces must resemble a spring in that a displacement
from that point causes a force to be produced in a
direction back toward the origin. [It has to be more than
just that]
Must be a force that is proportional to the displacement
but opposite in sign. ( 13 Answered like this; but few
applied this to the previous Q!!).
The force must be restorative (i.e., the force must be
proportional to the displacement at a given point x, but
opposite in sign of the displacement). The magnitude of
the force is determined by Hooke's law. F=-kx [This is the
key, but keep in mind that a conventional “spring” need
not necessarily be involved]
•
•
Simple Harmonic Motion
Consider any situation where a system is near a stable equilibrium point, and the
restoring force is directly proportional to the system’s displacement from equilibrium.
A mass on a spring satisfies these conditions, but there are many others that do as
well (e.g. atoms in molecules, torsional oscillators, pendula, some electrical circuits,
…
F = - k Dx = ma = m d2x/dt2
Chapter 15 problems
(The vertical scale mark is at
t=0.)
What is the maximum speed of the blocks at this amplitude?
Connection to Circular
motion
Angular separation between
Jupiter and Callisto as viewed
from the Earth by Galileo in 1610!
Here there is a real angle associated with the
phase*; in most SHM case the phase angle
is just a way of keeping track of the motion.
* What is that angle in the case of the Callisto observations?
Simple Harmonic Motion
T = 2p/w
Chapter 15
problems
DONE in
DISCUSSION in 2009
1.
•
•
•
A 30kg child is swinging on a swing whose seat is a
distance of 5.0m from the pivot point. Estimate the
optimal time between “pumps” that the child should
execute to increase her swinging amplitude. How does
this time change for a 15.0 kg child?
(9 got this right, 3 made a slight error, 32 didn’t
answer and 10 didn’t know what to do).
I know that the problem probably involves T = 2 (pi)
(L/g)^(1/2), but I don't know how to proceed (for
someone who did not know what to, this is VERY
GOOD!)
Since this is just an estimation, I used the equation T =
2 x pi (L/g)^1/2 where the mass is all concentrated at
the end of the pendulum. So with L=5.0m, the optimal
time is 4.5 seconds. Since the mass is not in this
equation, the weight of the child doesn't matter. (This
is the period; depending on how you pump this is the
answer, or half this time, or some multiple of this time).
Physical Pendulum
I a = - mghsin(q) but if q<<1 then:
a= - (mgh/I) q = d2q/dt2
Be careful to distinguish the physical
angle from the phase angle here!