Physics 2053C – Fall 2001

Physics 2053C – Fall 2001 Chapter 10 Fluids Dr. Larry Dennis, FSU Department of Physics

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.

Buoyant Forces • Force exeted by a displaced liquid. Ft-Fb = B gAht - gAhb = B B = gA(ht – hb) = Wt - Wb = B B = A(ht – hb) * g

Equation of Continuity Flow1 = Flow2 1A1v1 = 2A2v2 assuming 1 = 2 (same liquid) A1v1 = A2v2 A1 so v2 = x v1 A2

Bernoulli’s Equation P = Pressure v = velocity  = density of fluid y = height g = acceleration due to gravity P1 + ½v12 + gy1 = P2 + ½v22 + gy2 2 1

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

CAPA #2-3 2. What is the total force on the bottom of that swimming pool? Area = 20.0 m x 11.60 m F = 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

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

Fchair F? 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 ___ = ___

Floating Objects • Buoyant force must equal the weight if the object is to float. Buoyant Force = wVg Apparent Weight = W – B = ( - w )Vg Weight = Mg = Vg

Floating Objects B • How much of an ice cube is above the water line? Mg • F = 0 = B – Mg • B = Mg • water gVbelow = Mg • Vbelow = M/water = iceVtotal/water In order to float we must haveVbelow <Vtotal Only floats ifice < water.

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 + ½V2top = Pbottom + ½V2bottom Pbottom– Ptop = ½(V2top - V2bottom )

Application to CAPA Problems • Airplane wing: • F = P*A • Pb +½vb2 = Pt +½vt2 • P = Pb – Pt = ½(vt2 – vb2) • Roof in a Hurricane • F = P*A • Pb +½vb2 = Pt +½vt2 • P = Pb – Pt = ½(vt2 – vb2) Vb = 0 m/s

Quiz 7 • Density:  = M/V • Pressure: P = F/A • Buoyant Forces: B = liquidgVdisplaced • Continuity Eqn: A1v1 = A2v2 • Bernoulli’s Principle: P1 + ½v12 + gy1 = P2 + ½v22 + gy2

Quiz 7 • Sample Questions: • Chap. 10: 10, 20, 27 • Sample Problems: • Chap. 10: 23, 29, 43, 71

Next Time • Quiz 7. • Begin Chapter 11. • Please see me with any questions or comments. See you Wednesday.

Physics 2053C – Fall 2001