9 Solids Fluids elasticity of solids pressure and

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9 Solids & Fluids • elasticity of solids • pressure and pascal’s principle •

9 Solids & Fluids • elasticity of solids • pressure and pascal’s principle • buoyancy and archimedes’ principle • Homework: • 1, 3, 5, 37, 39, 56, 69, 70.

Solid Deformation • stress = force/area = pressure • strain = fractional change in

Solid Deformation • stress = force/area = pressure • strain = fractional change in length • Solid Moduli ~ stress/strain ~ stiffness

Solid Deformation • • • Length Change: stress = F/A, strain = DL/Lo, Young’s

Solid Deformation • • • Length Change: stress = F/A, strain = DL/Lo, Young’s Modulus, Y = stress/strain. Volume Change: stress = F/A = p, strain = -DV/Vo, Bulk Modulus, B = -Dp/(DV/Vo)

Pressure, p • p = force/area • Unit: pascal, Pa • 1 Pa =

Pressure, p • p = force/area • Unit: pascal, Pa • 1 Pa = 1 N/m 2. • Example: 100 N applied to 0. 25 m 2. Pressure = 100/0. 25 = 400 N/m 2.

Density • r = mass/volume • SI Unit: kg/m 3 • common unit: g/cm

Density • r = mass/volume • SI Unit: kg/m 3 • common unit: g/cm 3 • 1, 000 kg/m 3 = 1 g/cm 3

Gauge Pressure is Differential Pressure

Gauge Pressure is Differential Pressure

Depth and Fluid Pressure • • • P = Po + rgd P is

Depth and Fluid Pressure • • • P = Po + rgd P is the pressure at depth d. Po is the pressure at the fluid surface r is the fluid density Example: increase in pressure at a depth of 1 m in water = (1, 000)(9. 8)(1. 0) Pa

Barometer

Barometer

Pascal’s Principle Example: A 2/A 1 = 100 F 2/F 1 = 100

Pascal’s Principle Example: A 2/A 1 = 100 F 2/F 1 = 100

Buoyancy and Archimedes’ Principle

Buoyancy and Archimedes’ Principle

L=0. 5 m cube of steel in water • • • Vo = L

L=0. 5 m cube of steel in water • • • Vo = L 3 = (0. 5 m)3 = 0. 125 m 3 FB = (fl. density)(Vo)(g) = (1, 000 kg/m 3 )(0. 125 m 3 )(9. 8 N/kg) = 1225 N Compare FB to W of object: W = mg = (ob. density)(ob. volume)(g) = (7, 800)(0. 125)(9. 8) = 9555 N What is its weight when under water? W’ = W – FB = 8330 N

1 kg object is weighed in water • • W’ = 9. 2 N.

1 kg object is weighed in water • • W’ = 9. 2 N. What is its density? FB = W – W’ = 9. 8 N – 9. 2 N = 0. 6 N FB = (fl. den. )(ob. vol. )(g) = 0. 6 N FB = (1000)(ob. vol. )(g) = 0. 6 N (ob. vol. ) = 6. 12 x 10 -5 cubic meter (ob. den. ) = mass/vol. = 1 kg/(6. 12 x 10 -5) 16, 333 kg/m 3. = 16. 333 g/cm 3. This is not pure gold.

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Example: Object mass 0. 150 kg density 2 g/cm 3 is weighed in water.

Example: Object mass 0. 150 kg density 2 g/cm 3 is weighed in water.

Example: calculate speed water exits the hole in terms of the given parameters.

Example: calculate speed water exits the hole in terms of the given parameters.