PHYSICS 231 Lecture 21: Buoyancy and Fluid Motion. Remco Zegers Walkin hour: Tue 45 pm Helproom. Previously. Young’s modulus. Solids:. Shear modulus. Bulk modulus Also fluids. General:. P=F/A (N/m 2 =Pa) F pressuredifference = PA =M/V (kg/m 3 ).
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PHYSICS 231Lecture 21: Buoyancy and Fluid Motion
Remco Zegers
Walkin hour: Tue 45 pm
Helproom
PHY 231
Young’s modulus
Solids:
Shear modulus
Bulk modulus
Also fluids
General:
P=F/A (N/m2=Pa) Fpressuredifference=PA
=M/V (kg/m3)
Pascal’s principle: a change in pressure applied
to a fluid that is enclosed is transmitted to the whole
fluid and all the walls of the container that hold the fluid.
PHY 231
Pascal’s principle: a change in pressure applied
to a fluid that is enclosed is transmitted to the whole
fluid and all the walls of the container that hold the fluid.
HOLDS FOR A FLUID FULLY ENCLOSED ONLY
PHY 231
In other words then before: a change in pressure applied
to a fluid that is enclosed in transmitted to the whole
fluid and all the walls of the container that hold the fluid.
P=F1/A1=F2/A2
If A2>>A1 then
F2>>F1.
So, if we apply a small
force F1, we can exert
a very large Force F2.
Hydraulic press demo
PHY 231
Horizontal direction:
P1=F1/A P2=F2/A
F1=F2 (no net force)
So, P1=P2
Vertical direction:
Ftop=PatmA
Fbottom=PbottomAMg=PbottomAgAh
Since the column of water is not moving:
FtopFbottom=0
PatmA=PbottomAgAh
Pbottom=Patm+ gh
PHY 231
Pdepth=h =Pdepth=0+ gh
Where:
Pdepth=h: the pressure at depth h
Pdepth=0: the pressure at depth 0
=density of the liquid
g=9.81 m/s2
h=depth
Pdepth=0=Patmospheric=1.013x105 Pa = 1 atm =760 Torr
From Pascal’s principle: If P0 changes then the pressures
at all depths changes with the same value.
PHY 231
A submarine is built in such a way that it can stand pressures
of up to 3x106 Pa (approx 30 times the atmospheric
pressure). How deep can it go?
PHY 231
NO!!
PHY 231
The opentube manometer.
The pressure at A and B is
the same:
P=P0+gh
so h=(PP0)/(g)
If the pressure P=1.01 atm, what
is h? (the liquid is water)
h=(1.011)*(1.0E+05)/(1.0E+03*9.81)=
=0.1 m
PHY 231
P0= mercurygh
mercury=13.6E+03 kg/m3
mercury,specific=13.6
PHY 231
P0
P0
h
h
h
P=P0+gh
P=P0+gh
P=P0+gh
PHY 231
P0
Ptop=P0+ wghtop
Pbottom=P0+ wghbottom
p= wg(htophbottom)
F/A= wgh
F= wghA=gV
B=wgV=Mwaterg
Fg=w=Mobjg
If the object is not moving:
B=Fg so: wgV=Mobjg

htop
hbottom
Archimedes (287 BC) principle: the magnitude of the buoyant
force is equal to the weight of the fluid displaced by the object
PHY 231
B=fluidgV Buoyant force
w=Mobjectg=objectgV
Stationary: B=w
object= fluid
If object> fluid the object goes down!
If object< fluid the object goes up!
PHY 231
A
w=Mobjectg=objectVobjectg
B=weight of the fluid displaced by
the object
=Mwater,displacedg
= waterVdisplacedg
= waterhAg
h: height of the object under water!
h
B
w
The object is floating, so there is no net force (B=w):
objectVobject= waterVdisplaced
h= objectVobject/(waterA) only useable if part of the object
is above the water!!
PHY 231
?? N
question
A hollow sphere with negligible weight
if not filled, is filled with water and
hung from a scale. The weight is 10 N.
It is then submerged. What is the
weight read from the scale?
10N of water inside
thin hollow sphere
PHY 231
?? N
A)
?? N
B)
7 kg iron sphere of
the same dimension
as in A)
1 kg of water inside
thin hollow sphere
Two weights of equal size and shape, but different mass are
submerged in water. What are the weights read out?
PHY 231
An air mattress 2m long 0.5m wide and 0.08m thick and has
a mass of 2.0 kg. A) How deep will it sink in water? B) How
much weight can you put on top of the mattress before it
sinks? water=1.0E+03 kg/m3
PHY 231
x2
x1
2
v2
v1
1
A1,1
A2,2
the mass flowing into area 1 (M1) must be the same as the
mass flowing into area 2 (M2), else mass would accumulate
in the pipe).
M1= M2
1A1x1= 2A2x2 (M=V=Ax)
1A1v1t=2A2v2t (x=vt)
1A1v1 =2A2v2
if is constant (liquid is incompressible) A1v1 =A2v2
same
W1=F1x1=P1A1 x1=P1V
W2=F2x2=P2A2 x2=P2V
Net Work=P1VP2V
m: transported fluid mass
KE=½mv22½mv12 & PE=mgy2mgy1
Wfluid= KE+ PE
P1VP2V=½mv22½mv12+ mgy2mgy1 use =M/V and div. By V
P1P2=½v22½v12+ gy2 gy1
P1+½v12+gy1= P2+½v22+gy2
P+½v2+gy=constant
P: pressure ½v2:kinetic Energy per unit volume
gy: potential energy per unit volume
Another conservation Law
PHY 231
When air is blown in between the
cans, the velocity is not equal to 0.
P2+½v22 (ignore y)
P0
Before air is blown in between
the cans, P0=P1; the cans remain
at rest and the air in between
the cans is at rest (0 velocity)
P1+½v12+gy1= Po
Top view
P1
case:
1: no blowing
2: blowing
P0
PHY 231
No spin, no movement
Vair
PHY 231