5. is made of a different material than bar two

Transcription

1 Example Problems: Bar one has a Young s modulus that is bigger than that of bar Two. This means that bar one: 1. is longer than bar two 2. is shorter than bar two 3. has a greater cross-sectional area than bar two 4. has a smaller cross-sectional area than bar two 5. is made of a different material than bar two

2 Example Problems: Bar one has a Young s modulus that is bigger than that of bar Two. This means that bar one: 1. is longer than bar two 2. is shorter than bar two 3. has a greater cross-sectional area than bar two 4. has a smaller cross-sectional area than bar two 5. is made of a different material than bar two

3 Density and Pressure Density = Mass / Volume ρ = M V A material property (such as Young s Modulus)! Specific gravity: density / density of water ρ ρ water ρ water = 10 3 kg m 3 =1 kg dm 4 C

4 What has more mass: 1kg of water or 1kg of steel?

5 What has more mass: 1kg of water or 1kg of steel? What has a higher density: 1kg of water or 1kg of steel?

6 What has more mass: 1kg of water or 1kg of steel? What has a higher density: 1kg of water or 1kg of steel? What has a larger volume: 1kg of water of 1kg of steel?

7 Density and Pressure Pressure = Force / Area Examples: Bulk modulus Air pressure P = F A More precise: It is the force component perpendicular (normal) to the area. Can, but does not has to come from all sides A A F F

8 Example: Tire pressure: The pressure in your tires should be 35psi. Your car has a mass of 1500kg. How large is the area of the tires which make contact to the street?

9 Example: Tire pressure: The pressure in your tires should be 35psi. Your car has a mass of 1500kg. How large is the area of the tires which make contact to the street? Pressure in SI-units: 35psi = 35 lbs in N lbs 39.37in m 2 = 241 kpa Has to support the weight of the car: F = W = 1500kg g = 14.7 kn A = F P = 14.7 kn 241 kpa =0.06 m2 A tire = A 4 =0.015 m2 Typical Rim width: w =6.5 = 16.5 cm l 9 cm

10 Variations of Pressure with depth : Block of water in water

11 Variations of Pressure with depth : Block of water in water Fluid at rest: Pressure/Force must be same on both sides, otherwise would move. F F Fluid at rest: Pressure/Force from bottom must be higher than from top to compensate gravity, otherwise would move. Mg F F

12 F P B A P T A = Mg Mg F y B y T M = ρv = ρa(y T y B ) P B = P T + ρg(y T y B ) The pressure increases with depth! Normally written for pressure P at a depth h below the surface of a liquid: P = P 0 + ρgh where P 0 is the pressure at the surface (often air pressure)

13 Pascal s principle: A change in pressure applied to an enclosed fluid is transmitted undiminished to every point of the fluid and to the walls of the container.

14 Pascal s principle: A change in pressure applied to an enclosed fluid is transmitted undiminished to every point of the fluid and to the walls of the container. F 1 F 1 + F P 1 P 1 + P P 2 P 1 = ρgh P 2 P 2 + P P = F A is the same everywhere

15 Pascal s principle: A change in pressure applied to an enclosed fluid is transmitted undiminished to every point of the fluid and to the walls of the container. F 1 F 1 + F F 2 F 2 + P A 2 A 1 F 2 = F A 2 A 1 Hydraulics: Used to lift heavy stuff

16 Pressure Measurements: P 0 P h P = P 0 + ρgh A way to measure P-P0. Need to know at either P or P0.

17 Pressure Measurements: P 0 P h P = P 0 + ρgh A way to measure P-P0. Need to know at either P or P0. P =0 P 0 P 0 = ρgh Pick P = 0

18 Bouyant Force: F 1 Free Body Equation: Ma = Mg + F 1 F 2 h use: Mg F 2 F 2 F 1 = A (P 2 P 1 )=A P A P = Aρ fl gh = gρ fl V = gm fl Gives: Ma = Mg gm fl The pressure difference in the liquid due to its own weight gives rise to new force on the mass in the water: Bouyant Force B = gm fl pushes against gravity.

19 Bouyant Force: F 1 Bouyant Force: B = gm fl h Equal to the weight of the Mg displaced fluid F 2 Ma = Mg B = g (M M fl ) Three Situations: M>M fl M sinks M = M fl M stays at constant height M<M fl M rises until it reaches surface, then M floats

20 Bouyant Force: M<M fl M rises until it reaches surface, then M floats V displ. h Ma = Mg B = g (M M fl ) Mg F 2 In equilibrium: we have: a =0 M = M fl = ρ fl V displ. For a homogenous mass: ρ Mass = constant Floating: V mass >V displ. ρ Mass < ρ fl Stuff that is lighter than water floats!

21 Question: An aircraft carrier appears to be built mainly from iron and steel. It swims on water because: A. The density of iron and steel is smaller than the density of water B. Most of the material used to build it is actually light wood and plastic C. It swims because the speed of the carrier creates a lift similar to the lift that makes an airplane fly D. Most of the displaced volume is filled with air which is much lighter than water. E. The Navy sends a team of divers which pushes it up 24h/7d.

22 Question: An aircraft carrier appears to be built mainly from iron and steel. It swims on water because: A. The density of iron and steel is smaller than the density of water B. Most of the material used to build it is actually light wood and plastic C. It swims because the speed of the carrier creates a lift similar to the lift that makes an airplane fly D. Most of the displaced volume is filled with air which is much lighter than water. AND if you replace it with water, the ship sinks. E. The Navy sends a team of divers which pushes it up 24h/7d.

23 HITT: How large a force is necessary to stretch a 2.0mm diameter steel wire (Y = 200 GPa) by 1%? kN kN kN kN kN

24 HITT: How large a force is necessary to stretch a 2.0mm diameter steel wire (Y = 200 GPa) by 1%? kN F A = Y δl L kN kN F A = Y δl L = N m 2 A = π 4 d2 = π 10 6 m kN kN F = π N = 6300N

25 HITT: How large a force is necessary to stretch a 3.0mm diameter steel wire (Y = 200 GPa) by 1%? kN kN kN kN kN

26 HITT: How large a force is necessary to stretch a 3.0mm diameter steel wire (Y = 200 GPa) by 1%? kN F A = Y δl L kN kN kN kN F A = Y δl L = N m 2 A = π 4 d2 = m 2 F = N = 14.1kN