1 atm = 1.01x10 Pa = 760 Torr = 14.7 lb / in

Transcription

1 Last class we began discussion of ressure in fluids, with ressure defined as, F = ; units N 1 Pa = 1 m 2 There are a number of other ressure units in common use having the following equivalence, atm = 1.01x10 Pa = 760 Torr = 14.7 lb / in We also discussed a ressure gauge based on an evacuated cylinder & sring arrangement.

2 device for measuring atmosheric ressure is the barometer. This is a close bottomed tube, filled to overflowing with a fluid and then turned over in an oen bath of the same fluid. Done on the moon which has essentially no atmoshere, the fluid would just run out until h = 0. vacuum On earth where the ressure on the exterior surface is around 1 atm the fluid at oint x inside the tube has ressure equal to o, from the atmosheric ressure o outside the tube. o x h That means a force u given by solving o F = with the cross-section of the tube.

3 Since the fluid is stationary, there must be an equal but oosite force down at oint x having the same magnitude, but where we recognize the force as due to the weight of the fluid column, o F = = But since the density (rho) this can be rearranged to give, m = ρv = ρh, ρ= m V h vacuum which substituted above gives o F ρgh = = = o = ρgh o x

4 o = ρgh Taking h = 0 as zero ressure (absolute) the height of the fluid column is roortional to the ressure. Mercury is often used. The height of a column of mercury at sea level is on average 760 mm, so you will often hear atmosheric ressures quoted in mm (or equivalently Torr) around this value. o x vacuum h Since the density of mercury deends on the temerature, and the ressure reading deends on density, for accurate readings of atmosheric ressure this must be corrected for the local temerature.

5 HITT Suerman (of course) has suer suction, caable of ulling an infinite vacuum. Given an aroriate straw how high could he suck the fresh water from a lake? ) as high as he wants B) 32.0 m C) 10.3 m h o = 1.01x10 5 Pa ρ = 1000 kg/m 3

6 HITT The barometer also has a vacuum in the uer art of the tube. Making that vacuum more erfect changes the height of the water negligibly. There we found o = ρgh so, h ρg kg m m s 5 o Pa = = = 10.3 m h o = 1.01x10 5 Pa ρ = 1000 kg/m 3

7 The oen tube manometer works similarly to measure the gauge ressure of a gas inside an otherwise closed tank, that is oen to the manometer on one side. Here g = ρgh Where it measures gauge ressure (i.e. relative to the atmosheric ressure) because it is oen to the atmoshere on the end of the tube not connected to the tank.

8 Pascal s Princile Consider the circumstance to the right in which we have a cylinder of cross-section filled with fluid. d 1 The ressure at a deth d below the surface deends on the deth. Let s now add a iston and aly an additional force to it. The ressure at deth d increases by F F = but that s true at every oint of the fluid. d 2 The ressure change is transmitted throughout the entire fluid.

9 Such a change in ressure would be transmitted indeendent of the source of the ressure change. For examle, if the temerature rises and the fluid exands, the resulting change in the ressure would occur throughout the volume of fluid. This is Pascal s rincile: the change in ressure occurring in an incomressible fluid, in a closed container, is transmitted undiminished to every ortion of the fluid and to the walls of the container.

10 This allows for a hydraulic lever, which consists of different area istons/cylinders connected together as shown here. force is alied at the inut iston and the force occurs 1 at the outut iston. By Pascal s rincile, the change in ressure is the same everywhere so But then, F F 2 F = = F = F F So the outut force is the inut force, times the ratio of the iston areas.

11 For round cylinder/istons: πr 2 R 2 2 = 1 = 2 1 = 1 1 R 1 R 1 F F F F π So if the outut cylinder has a 10 cm radius and the inut cylinder a 1 cm radius the force multilier is (10/1) 2 = 100. This is how hydraulic lifts and the brakes in your car work. Does this scheme defy conservation of energy? (a wise thing to ask when we seem to be getting something seemingly extraordinary)

12 If the inut iston moves down a distance x 1 the volume of fluid it dislaces is V = x 1 1. Since the fluid is incomressible this must be the same volume dislaced by the oosing iston so, V= x 1 1 = x 2 2 so that, x = x This is the inverse ratio of the force multilier so the distance moved by the outut iston is roortionately smaller than the distance moved by the inut iston. Since work is force times distance the same work is done on both sides & energy is conserved.

13 U-tubes (determining the density of an immiscible fluid) Tube cross-sectional area. h H-h H The ressure on the two sides at the lowest dashed line must be equal (or the fluid would move). o left = m = right u + = o + u m k k Initially fluid of known density ρ k dd column h of fluid of unknown density ρ u ( ρ >ρ case) u k ρ h =ρ u ρ = u k H h H ρ k

14 If ρ <ρ u k The ressure on the two sides at the lowest dashed line must be equal (or the fluid would move). h-h left = right h H o m = u + = o + ρ h =ρ u u m k k H k Initially fluid of known density ρ k dd column h of fluid of unknown density ρ u So again, ρ = u H h ρ But now H < h. k

15 rchimedes' rincile Consider the ressure on the to and lower faces of the cube of water labeled C, somewhere in the tall stationary column of water. The ressure at the uer face, t, is due to the mass of the column of water above C. The ressure at the lower face, l, is due to the mass of the column above C lus the mass of C itself. t C l With m f the mass of the fluid contained in C, we can write that, l = + t f area

16 Or since, F = Fl Ft f = + Fl = Ft + f F l is the force downwards due to C and the column of water above it but since the water is stationary there must be an equal and oosite force, due to the surrounding water on the lower face of C uwards. F t C F l m f g Hence the net forces acting on C are Fl Ft = Ft + f Ft = f

17 Hence we ve found that the water that surrounds C rovides a buoyant force F b uwards given by, F b = f This net force from the surrounding water it is actually indeendent of what material occuies the volume. C m g Hence if we relace the cube of water with a different material having a different mass, m, Newton s 2 nd law gives, F b = m f g Fb = ma mfg = ma

18 a m m m g f = (ignoring hydrodynamic drag) If the material has a smaller mass than the equal volume of water that it dislaced (meaning that it has a lower density than water, e.g. Styrofoam) then the acceleration due to the buoyant force will be ositive (i.e. uwards). F b = m f g If the material has a greater mass than the equal volume of water that it dislaced (meaning that it has a higher density than water, e.g. metal) then the acceleration due to the buoyant force will be negative (i.e. downwards), but smaller than its free fall acceleration if only gravity were acting. Styro foam m g

19 nother way to look at this is to to consider the aarent weight of the object. Suose the object is laced on a sring scale (in the fluid). Then F + F = ma = 0 S b So the scale reads, F = F S b C Thus its aarent weight is the actual weight minus the buoyant force. F B F S The buoyant force deends only on the weight of the fluid dislaced, i.e. F b = m f g This is rchimedes' rincile.

20 If the buoyant force exactly balances the weight of the object the aarent weight F S = 0 and the object neither rises nor sinks. The object is then said to be neutrally buoyant. F = F = 0 S B This is what ermits a submarine to hover at articular deth. submarine has internal ballast tanks that are designed to be filled with seawater. To dive a neutrally buoyant submarine ums water into these chambers making the weight of the hull lus the water taken on greater than the weight of the water dislaced by the hull (i.e. greater than the buoyant force). neutral sinks

21 s the desired deth is aroached this water is blown out, making the shi neutrally buoyant again. neutral To rise again, still more water is blown out, making the submarine have an aarent weight that is negative (F b > ). rises When the shi gets to the surface, it continues to rise until its total weight equals the weight of the water it dislaces.

22 This sets the deth at which an object floats, i.e. an object will sink into a fluid to a level until the weight of the object equals the weight of fluid dislaced. Examle Floating in the very salt rich (dense) waters of the Dead Sea kees about 1/3 of your body above the water line. What is the density of the water there? ssume your density to be 1 g/cm 3. Let your mass be m, F B = 0 but F B = m f g, m f = dislaced water F B 1/3 2/3 so m f g = 0 m f = m ρ f V f = ρv

23 m f g = 0 m f = m ρ f V f = ρv ρ f = density of the seawater ρ = 1 g/cm 3 (your density) Now the volume of fluid dislaced is 2/3 of your volume so, V f = 2/3V then, ρ f 2/3V = ρv ρ f 2/3 = ρ F B 1/3 2/3 ρ f = 3/2ρ ρ f = 1.5 g/cm 3

24 HITT merry-go-round of radius angular acceleration α. R has a constant t the instant shown the net linear acceleration of the oint in red on the rim is: 1) long vector 2) long vector B 3) Between vectors & B R B

25 Since there is an angular acceleration α the red sot has a linear acceleration tangential to the rim along that has magnitude, at = Rα Since the sot is moving on a circular ath it has a centrietal acceleration toward the center of the circle along B of magnitude, 2 v a = c R + B B B with v the linear seed of the sot at that instant. The answer is the sum of these vectors so between & B.