Oct. 29, 2001Dr. Larry Dennis,

FSU Department of Physics 1

Physics 2053C – Fall 2001

Chapter 10Fluids

2

Pressure and Density

Density = Mass/Volume A property of the material.

Pressure = Force/Area Depends on the height of the fluid. Same in all directions. Units are:

Force/Area = N/m2. Pascals 1 Pa = 1 N/m2. Atmosphere 1 atm = 1.013 X 105 N/m2.

3

Buoyant Forces Force exeted by a displaced liquid.

Ft-Fb = B

gAht - gAhb = B

B = gA(ht – hb) = Wt - Wb = B

B = A(ht – hb) * g

4

Equation of Continuity

Flow1 = Flow2

assuming 1 = 2 (same liquid)

1A1v1 = 2A2v2

A1v1 = A2v2

so v2 = x v1

A1

A2

5

Bernoulli’s Equation

P = Pressurev = velocity = density of fluidy = heightg = acceleration due to gravity

P1 + ½v12 + gy1 = P2 + ½v2

2 + gy2

1

2

6

CAPA #1

What is the absolute pressure on the bottom of a swimming pool 20.0 m by 11.60 m whose uniform depth is 1.92 m?

Pw = gh = (1.0x103 kg/m3)(9.8 m/s2)(1.92m)

= 1.89x104 N/m2

But, we need absolute pressure…P = Pw + Patm

P = 1.89x104 N/m2 + 1.013x105 N/m2

= 1.20x105 N/m2

7

CAPA #2-3

2. What is the total force on the bottom of that swimming pool?

Area = 20.0 m x 11.60 mF = P x A = (1.20x105 N/m2)(20.0 m)(11.60

m) = 2.79x107 N

3. What will be the pressure against the side of the pool near the bottom?The pressure near the bottom is the same as on the bottom

P = 1.20x105 N/m2

8

CAPA #4

4. A dentist’s chair of mass 236.0 kg is supported by a hydraulic lift having a large piston of cross-sectional area 1434.0 cm2. The dentist has a foot pedal attached to a small piston of cross-sectional area 76.0 cm2. What force must be applied to the small piston to raise the chair?

Fchair F?Fchair = mg

= (236.0 kg)(9.80 m/s2) = 2312.8 N

9

CAPA #4(cont)

4. A dentist’s chair of mass 236.0 kg is supported by a hydraulic lift having a large piston of cross-sectional area 1434.0 cm2. The dentist has a foot pedal attached to a small piston of cross-sectional area 76.0 cm2. What force must be applied to the small piston to raise the chair?

P1 = P2 P = F/A

F1 F2

A1 A2

F2 = (A2/A1)F1 = ((76.0 cm2)/(1434.0 cm2)) x (2312.8 N)

F2 = 123 N

Fchair F?

___ = ___

10

Floating Objects Buoyant force must equal the

weight if the object is to float.

Weight = Mg = Vg

Buoyant Force = wVg

Apparent Weight = W – B = ( - w ) Vg

11

Floating Objects How much of an

ice cube is above the water line?

Mg

B

F = 0 = B – Mg

B = Mg

water gVbelow = Mg

Vbelow = M/water = iceVtotal/water

In order to float we must have Vbelow < Vtotal

Only floats if ice < water.

12

Applications of Bernoulli’s Principle

Why does an airplane fly?

Air moving over the top of the wing is moving faster than the air moving below the wing.

The air pressure on the top top of the wing is lower than the pressure on the bottom of the wing.

F = Pressure * Wing Area

Ptop + ½V2top = Pbottom + ½V2

bottom

Pbottom– Ptop = ½(V2top - V2

bottom )

13

Application to CAPA Problems Airplane wing:

F = P*A Pb + ½vb

2 = Pt + ½vt2

P = Pb – Pt = ½(vt2 – vb

2)

Roof in a Hurricane F = P*A Pb + ½vb

2 = Pt + ½vt2

P = Pb – Pt = ½(vt2 – vb

2)

Vb = 0 m/s

14

Quiz 7 Density: = M/V Pressure: P = F/A Buoyant Forces: B = liquidgVdisplaced

Continuity Eqn: A1v1 = A2v2

Bernoulli’s Principle:P1 + ½v1

2 + gy1 = P2 + ½v22 +

gy2

15

Quiz 7 Sample Questions:

Chap. 10: 10, 20, 27

Sample Problems: Chap. 10: 23, 29, 43, 71

16

Next Time

Quiz 7. Begin Chapter 11. Please see me with any questions

or comments.

See you Wednesday.