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PHYSICS 231
Lecture 24: Walking on water & other
‘magic’
Remco Zegers
Walk-in hour: Thursday 11:30-13:30 am
Helproom
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P0
Pressure at depth h
P
=
P0+ fluidgh
h: distance between liquid surface
and the point where you measure P
h
P
Buoyant force for submerged object
B = fluidVobjectg = Mfluidg = wfluid
The buoyant force equals the weight of the
amount of water that can be put in the
volume taken by the object.
If object is not moving: B=wobject object= fluid
Buoyant force for floating object
The buoyant force equals the weight of the amount
of water that can be put in the part of the volume
of the object that is under water.
objectVobject= waterVdisplaced h= objectVobject/(waterA)
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B h
w
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Bernoulli’s equation
P1+½v12+gy1= P2+½v22+gy2
P+½v2+gy=constant
The sum of the pressure (P), the kinetic energy per unit
volume (½v2) and the potential energy per unit volume (gy)
is constant at all points along a path of flow.
Note that for an incompressible fluid:
A1v1=A2v2
This is called the equation of
continuity.
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Applications of Bernoulli’s law
The examples shown are with air, not with fluid.
Remember that we derived this law for an
incompressible fluid. Air is not incompressible,
so the situation is typically more complicated…
But easier to show!
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Applications of Bernoulli’s law: moving a cart
No spin, no movement
Vair
V2=Vair-v
P2
Spin and movement
P1 V1=Vair+v
Near the surface of the rotating cylinder: V1>V2
P1+½v12= P2+½v22
P1-P2= ½[(vair-v)2-(vair+ v) 2]
P2>P1 so move to the left
P1-P2= ½(v22- v12)
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Applications of Bernoulli’s law: the golf ball
Neglecting the small change in height between the
top and bottom of the golf ball:
P1+½v12= P2+½v22
P1-P2= ½(v22- v12)
P1
P2
P1-P2= ½(v22- v12)=0
v2=v1
No pressure difference, no lift
P2
P1-P2= ½(v2-v)2-(v1+ v) 2=0
P2>P1 so:
Upward force: the ball goes higher
and thus travels faster
P1
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Not the whole story: the dimples in the
golf ball reduce the drag
The drag is the force you feel when you are biking. The
pressure in front of you is higher than behind you, so you
feel a force against the direction of your motion.
A
B
P1
P2
P1
P3
P3>P2 : there is less drag in case B
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Demo
A floating globe
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Energy
Surface tension
Two liquid molecules like to
sit close to each other (energy
is gained)
0
-Emin
R
R
2 liquid molecules
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A bunch of liquid molecules
Inside the liquid
1
6
2 3
Near the surface of the liquid
1
5
4
4
2 3
The molecule near the
surface only gains 4 times
The molecule in the center Emin of energy. The summed
gains 6 times Emin of energy. energy is only reduced by
The summed energy is
4Emin.
reduced by 6Emin It is energetically favorable to keep
the surface of the liquid as small
as possible
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Why does water make droplets on a
surface and does not spread out?
The liquid surface is smallest:
energetically favorable.
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Surface tension
If you make the surface of the fluid larger, it tries to
‘push’ back. The force with which this is done: Fs=L
where L is the length over which the force acts and
 is the surface tension. The force works parallel to the
water surface. Example: a needle on water
Top view
Fs
Fs

L
Fg
Horizontal: Fscos-Fscos=0
Vertical:
Fssin+Fssin-Fg=0 Units of : N/m=J/m2
Energy per unit surface
=m
g/(2Lsin)
needle
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Walking on water
The insect uses surface tension!
Surface tension depends on the type of liquid.
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Forces between molecules
Cohesive forces: forces between like molecules
Adhesive forces: forces between unlike molecules
Adhesive
Cohesive
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More on surfaces
If cohesive forces are stronger than the adhesive ones
like molecules in the drop try to stay together to reduce
the total energy of the system; if adhesive forces are
stronger the drop will spread out to reduce the total
energy of the system. The spreading will stop when the
surface tension becomes too strong.
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Same thing
Adhesive > Cohesive
The water wants to
cover as much of
the glass as its
surface tension allows
Adhesive < Cohesive
The mercury wants to
cover as little of the glass
as its surface tension allows
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Capillary action
Fsurface tesion=L= 2r
Vertical (upwards) component:
FSTvertical=2rcos
The weight of the liquid in the
tube: w=Mg=r2gh
The liquid stops going up when:
FSTvertical=w
h=2cos/(gr)
If r very large: h very small!
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Viscosity
Viscosity: stickiness of a fluid
One layer of fluid feels a large
resistive force when sliding
along another one or along a
surface of for example a tube.
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Viscosity
Contact surface A
moving
fixed
F=Av/d
=coefficient of viscosity
unit: Ns/m2
or poise=0.1 Ns/m2
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Poiseuille’s Law
How fast does a fluid flow
through a tube?
Rate of flow Q= v/t=
R4(P1-P2)
8L
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(unit: m3/s)
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Example
Flow rate Q=0.5 m3/s
Tube length: 3 m
=1500E-03 Ns/m2
P=105 Pa
PP=106 Pa
What should the radius of the tube be?
Rate of flow Q=
R4(P1-P2)
8L
R=[8QL/((P1-P2))]1/4=0.05 m
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