Fluids Physics 116 2015 Fluid Sta4cs •  Density is the iner%al quan%ty associated with ﬂuids. It is an index that describes how much mass there is in each unit of volume of the ﬂuid. We assume that this quan4ty is a constant in our mathema4cal descrip4ons that follow. •  The pressure of a sta4c ﬂuid increases linearly with depth. Let the origin of y be at the surface of the ﬂuid, and the posi4ve direc4on be down: P(y)=Po + ρgy •  The diﬀerence in pressure on the top and boPom of an object submerged in a sta4c ﬂuid results in a buoyant force, which is a net force of the ﬂuid on the object, and always points upward. •  The force that the sta4c ﬂuid exerts on the object is equal to the weight of the water that has been displaced by the object. Fwater=ρﬂuidVsubmergedy (Archimedes principle) Density of object > density of ﬂuid Weight of the object Weight of the water displaced Density of object < density of ﬂuid Weight of the water displaced Weight of the object Density of object = density of ﬂuid nT11-CK1 Aluminum and steel The cylinders have the same mass,
but the aluminum cylinder is larger
than the steel cylinder. Let TS and
TA be the tensions in the strings
attached to the steel and aluminum
cylinders respectively. Also let BS
and BA be the buoyant forces on the
steel and aluminum cylinders.
Steel
Aluminum
a) TA < TS and BA < BS.
b) TA < TS and BA > BS.
c) TA < TS and BA= BS.
d)  TA > TS and BA < BS.
e)  TA > TS and BA > BS.
f) TA > TS and BA = BS.
nT11-CK2 Immersed cubes Two cubes with identical shapes and
volumes but different masses are immersed
in oil. Both cubes are completely below the
surface of the oil, but the lighter cube is
farther below the surface.
3 kg
1 kg
a) The buoyant force on the 3-kg cube is triple the buoyant force
on the 1-kg cube. The tensions are the same.
b) The buoyant force is greater for the 1-kg cube because it is
deeper in the water. The tension force is less on the 1-kg
cube than on the 3-kg cube.
c) We cannot compare the tensions unless we know the depths
of the two cubes.
d) The buoyant force is the same for the two cubes. The tension
is greater on the 3-kg mass than on the 1-kg mass.
nT11-CK3 Floating cubes A
B
Two cubes are floating at rest on the
surface of a fluid as shown. Which of the
following statements is true?
a)
b)
c)
d)
Cube A must be less dense than cube B.
Cube A must have less mass than cube B.
Cube A must have less volume than cube B.
The net force on cube A is less than the net force on cube B.
nT11-CK4 Buoyant Force Three cubes, each one 10 centimeters on a side, are
submerged in an incompressible fluid. Cubes A and B
are made of steel (ρ = 7 g/cm3) and cube C is made of
aluminum
(ρ = 2.7 g/cm3). Cubes A and C are at the same depth.
Which cube has the greatest buoyant force acting on
it?
C
A
B
(a)  Cube A.
(b)  Cube B.
(c)  Cube C.
(d)  A & B are equal and are greater than C.
(e)  All buoyant forces are the same.
nT11-CK5 Tension Three cubes, each one 10 centimeters on a side, are
submerged in a fluid. Cubes A and B are made of
steel (ρ = 7 g/cm3) and cube C is made of aluminum
(ρ = 2.7 g/cm3). Cubes A and C are at the same depth.
Which string has the greatest tension?
C
A
B
(a)  Cube A.
(b)  Cube B.
(c)  Cube C.
(d)  A & B are equal and are greater than C.
(e)  All buoyant forces are the same.
FLUID DYNAMICS Consider an IDEAL FLUID Fluid mo4on is very complicated. However, by making some assump4ons, we can develop a useful model of ﬂuid behavior. An ideal ﬂuid is: Incompressible – the density is constant Irrota:onal – the ﬂow is smooth, no turbulence Nonviscous – ﬂuid has no internal fric4on Steady ﬂow – the velocity of the ﬂuid at each point is constant in 4me. EQUATION OF CONTINUITY (conserva4on of mass) mass ﬂowing into a pipe = mass ﬂowing out of a pipe m1 = m2 ρV1 =ρV2 Since V=AΔx and v=Δx/Δt then V=A vΔt (where V represents volume and v represents speed) ρA1v1Δt =ρA2v2Δt A1 v1 = A2 v2 =volume ﬂow rate=Volume/unit of 4me =constant Conserva4on of energy – Bernoulli’s Equa4on There must be energy in a system that includes a pressurized ﬂuid and the earth by the virtue that work must be have done to establish the pressure.A ﬂuid that undergoes a pressure change undergoes an energy change. Assume: uniform density and no internal fric3on or turbulence Change in kine4c energy ΔK=1/2mv22 -‐ 1/2mv12 =1/2(ρΔV)v22 -‐1/2(ρΔV)v12 Change in poten4al energy ΔU = mgy2 –mgy1 = (ρ ΔV)gy2 – (ρΔV)gy1 Work done (Recall that work=force*displacement and force = pressure *area) Wnet =F1 Δx1 –F2 Δx2 =(p1 A1) Δx1 –(p2 A2) Δx2 Work done = change in kine4c + poten4al energy Wnet = P1 ΔV – P2 ΔV = ΔK + ΔU P1 ΔV–P2 ΔV ={ 1/2(ρΔV)v22 – 1/2(ρΔV)v12} + {(ρ ΔV)gy2 – (ρΔV)gy1} Dividing both sides by ΔV we deﬁne “energy density”, or the energy per unit of volume, which represents the eﬀect of a pressure diﬀerence on the total energy in the system for a ﬂuid with a uniform density P1 –P2 ={ 1/2ρv22 – 1/2ρv12} + {ρgy2 – ρgy1} called “Bernoulli’s Equa3on” Venturi Eﬀect P1 –P2 ={ 1/2ρv22 – 1/2ρv12} + {ρgy2 – ρgy1} Start with Bernoulli’s Eq and look at two points in a ﬂowing ﬂuid that are at the same eleva4on, y1=y2 P1 –P2 ={ 1/2ρv22 – 1/2ρv12} + {ρgy2 – ρgy1} Since the loca4on 1 has a wider diameter than the loca4on 2, then by the con%nuity equa%on v1 <v2. This means that the pressure at 2 has to be SMALLER than the pressure at 1, and we get the counter-‐
intui4ve result that in regions where a ﬂuid ﬂows faster, the pressure is LOWER. This is known as the Venturi Eﬀect, and is a founda4onal idea in air ﬂight. Below is a cross sec4on of an airplane wing. The air moves faster across the top of the wing. LiD on an airplane wing is due to the diﬀerent air speeds (and therefore pressures) on the upper and lower surfaces of the wing.