1206 Concepts in Physics

Monday, October 19th

Notes• Problems with webpage over the weekend

• If this is not fixed by tonight, I will find another way to display the material

• Either webCT or printouts

• The assignments (#1 to #4) and solution (#1,#2) are displayed just outside my office. The solutions for #3 will be there after today’s tutorial as well

Notes• Today’s tutorial: formula sheet for midterm

(posted after)

• Some “Ansatz” training

• Question to assignment #4

• Graham will then write the solutions for assignment #3 onto the board - and more time for questions

FluidsFluids are materials that can flow, therefore they include both gases and liquids. Water is the most common liquid we know from experience and air the gas we are most familiar with.

Mass densityThe mass density of a liquid or gas is an important factor that determines its behavior as a fluid.

The mass density ρ is the mass m of a substance divided by its volume V: ρ = m/VSI unit of mass density: kg/m3

Note! Equal volumes of different substances generally have different masses, so the density depends on the nature of the material. Gases have the smallest densities because gas molecules are relatively far apart and gas contains a large fraction of empty space. In liquids and solids the

molecules are much more tightly packed and have therefore larger densities.

substancemass density ρ

[kg/m3]substance

mass density ρ [kg/m3]

substancemass density ρ

[kg/m3]

gold 19300 water (4°) 1000 air 1.29

ice 917 blood (37°) 1060 helium 0.179

wood (pine) 550 mercury 13600 hydrogen 0.0899

aluminum 2200 oil 800 nitrogen 1.25

Example:The body of a person whose weight is 690 N contains about 5.2 x 10-3 m3 of blood. Find the

blood’s weight and express it as a percentage of the body weight.

To find the weight W of the blood, we need the mass m, since W = mg, where g the magnitude of the acceleration due to gravity is. The density of blood is 1060 kg/m3 (from

table), so the mass of the blood can be found by using the given volume.

The mass and the weight of the blood are:

mass density: ρ = m/V therefore m = ρV = (1060 kg/m3)(0.0052 m3) = 5.5 kg

W = mg = (5.5 kg)(9.80 m/s2) = 54 N

Percentage:

(54 N)/(690 N) x 100 = 7.8%

Note! Make sure, that you don’t mix mass and weight. Be aware which one to use where.

Side noteA convenient way to compare densities is to use the concept of specific gravity. The specific

gravity of a substance is its density divided by the density of a standard reference material, usually chosen to be water at 4°C.

Specific gravity =Density of the substanceDensity of water at 4°C

=Density of the substance

1.000 x 103 kg/m3

Being the ratio of two densities, specific gravity has no units. The specific gravity of water is per definition 1.000.

Your turn:

The density of a diamond is 3520. What is it’s specific gravity?

PressureMost of you will have some idea what pressure is. Imagine a flat tire (on a car of bicycle) - in any case, once you have changed or fixed it, you will have to re-inflate it by pumping air into the tire. The under-inflated tire is soft because it contains an insufficient number of air molecules to push outward against the rubber and give the tire that solid feel. When air is added from a pump, the number of molecules and the collective force they exert are increased. The air molecules within a tire are free to wander throughout its entire volume and in doing so, they collide with on another and with the walls of the tire. The collisions with the wall allow the air to exert a force against every part of the wall surface. The pressure P exerted by a fluid is defined as the magnitude F of the force acting perpendicular to a surface divided by the area A over which the force acts:

P = F/A

The SI unit for pressure is Pascal Pa = N/m2

At the earth's surface, the pressure felt by an object due to the weight of the atmosphere above it is:

1 atmosphere = 1.01 x 105   Pa = 1.01325 bar

Note! This is a so-called statistic process. We have a huge number of molecules in a tire. It is not significant what one specific particle does, but all of them have macroscopic effects.

There are many different common units for pressure: torr, psi, mbar, ... 1 mbar = 1 Pa (weather)

1 psi = 6894.75 Pa (gas bottles)1 torr = 133.32 Pa

We can picture pressure to cause a force to act perpendicularly on each face of a cube for example. Both air (gases) and water (liquids) exert pressure on objects such as a swimmer in water who will fee the water pushing perpendicularly inward everywhere on his/her

body. In general a fluid cannot produce a force parallel to a surface. If it would do so, it would apply a reaction force to the fluid (Newton’s

third law) and in response the fluid would flow and not be static.

While fluid pressure can generate a force, pressure itself is not a vector quantity. In the definition for pressure p = F/A, the symbol F refers only to the magnitude of the force, so that pressure has

no directional characteristic. The force generated by the pressure of a static fluid is always perpendicular to the surface that the fluid contacts.

Example: Suppose the pressure acting on the back of a swimmer’s hand is 1.2 x 105 Pa.This is a realistic value near the bottom of a diving end of a pool. The surface area of the back of the hand is 8.4 x 10-3 m2. Determine the magnitude of the force that acts on it and discuss the direction of the force.

From the definition of pressure, we can see that the magnitude of the force is the pressure times the area. We assume that the back of the hand of the swimmer is parallel to the surface at the moment of our calculation. The direction o f the force is always perpendicular to the surface that the water contacts.

P = F/A, therefore F = P*A = 1.2 x 105 Pa * 8.4 x 10-3 m2 = 1.0 x 103 N

Atmospheric pressure Like all fluids, the air exerts a pressure on everything within and around it, although we are not aware of it. Pressure is a force, or weight, exerted on a surface per unit area, and is measured in Pascals (Pa). The pressure exerted by a kilogram mass on a surface equals 9.8 Pa. The pressure exerted by the whole atmosphere on the Earth’s surface is approximately 100,000 Pa. Usually, atmospheric pressure is quoted in millibars (mbar). 1 mbar is equal to 100 Pa, so standard atmospheric pressure is about 1000 mbar. In fact, actual values of atmospheric pressure vary from place to place and from hour to hour. At sea level, commonly observed values range between 970 mbar and 1040 mbar. Because pressure decreases with altitude, pressure observed at various stations must be adjusted to the same level, usually sea level.

Note! 1 atm is enough pressure to crumple a can, it the inside air is pumped out ...

Atmospheric pressure is measured by a barometer. A mercury barometer measures the pressure by noting the length of mercury which is supported by the weight of the atmosphere. One centimeter of mercury is equal to 13.33 mbar, so normal atmospheric pressure can support a column of mercury about 75 cm (or 30 inches) high. An aneroid barometer is a more compact instrument for measuring pressure. It consists of a box of partially exhausted air which expands and contracts as the pressure falls and rises. The box is connected through a system of levers to a pointer which, in conjunction with a dial, indicates the pressure.

Map for weather predictions - locale pressures

Pressure and Depth in a Static FluidOne might guess that the deeper you go into a liquid or gas, the greater the pressure on you from the surrounding fluid will be. The reason for the increased pressure is that the deeper into a fluid you go, the more fluid, and thus the more weight, you have over top of you.

We can calculate the variation of pressure with depth by considering a volume of fluid of height h and cross-sectional area A

  Variation of Pressure with Depth

If this volume of fluid is to be in equilibrium, the net force acting on the volume must be zero. There are three external forces acting on this volume of fluid. These forces are:

1.) The force PTA due to the pressure on top of the volume of fluid. If the fluid is open to the air, PT = PO = 1.01 x 105 Pa, which is atmospheric pressure.2.) The weight of the volume of fluid, W = mg . Remembering the definition of density, ρ= m/V , and that the volume of the fluid may be calculated as V = Ah , we can write the weight of the fluid as W = ghA .3.) The force pushing up on the bottom of the volume of fluid, PBA , due to the fluid below the volume under consideration.

If we take the up direction to be positive and add the forces we getPBA - ghA - PTA = 0,

which givesPB = PT + gh.

This provides the general formula relating the pressures at two different points in a fluid separated by a depth h .Note: Only the density of the fluid and the difference in depth affects the pressure. The shape and size of the container are irrelevant. Thus the water pressure 6 inches below the surface of the ocean is the same as it is 6 inches below the the surface of a glass of salt water.

Idea: Pascal's Principle states that any pressure applied to an enclosed fluid is transmitted undiminished to every point of the fluid.

Several differently shaped vases are all connected at the bottom, and fluid is put into all of them. Since atmospheric pressure is the same in each vase, the fluids will seek their own level, no matter what the shape of the container.

Example: The Hoover Dam. Lake Mead is the largest wholly artificial reservoir in the United States and was formed after the completion of the Hoover Dam in 1936. The water in the reservoir back up

behind the dam for about 200 km. Suppose that tall the water were removed, except for a relatively marrow vertical column in contact with the dam. The water in this situation has the same depth as the water in the dam. Would the hoover dam still be needed to contain the water in this hypothetical reservoir, or could a much less massive structure do the job?

land

dam water YOUR TURN

land

dam water

Since the hypothetical reservoir contains much less water than Lake Mead, it is tempting to say that a much less massive structure would be fine. However, the hoover dam would still be needed for our hypothetical

reservoir. The magnitude or the force on any square is the product of its area and the height. The horizontal distance doesn’t matter, it has no effect

on the pressure.

Example: Blood pressureBlood in the arteries is flowing, but as a first approximation, the effect of this flow can be

ignored and the blood can be treated as a static fluid. Estimate the amount by which the blood pressure P2 in the anterior tibial artery at the foot exceeds the blood pressure P1 in the aorta

at the heart when a person is (a) reclining horizontally

(b) standing up

(a) When a person is lying down (body horizontally), there is little or no vertical separation between the feet and the heart. Therefore h = 0.

P2 - P1 = ρgh = 0 Pa

(b) When an adult is standing up, the vertical separation between the feet and the heart is approximately 1.35 m. The density of the blood is given as 1060 kg/m3, so that:

P2 - P1 = ρgh = (1060 kg/m3)(9.80 m/s2)(1.35 m) = 1.40 x 104 Pa

Example: Sometimes fluid pressure places limits on how a job can be done. Water should be pumped out of a well. There are two possible locations for the pump - it can be submerged and placed at the bottom or it can stand on ground level. For a shallow well, both methods (locations) can be used, but for a deep well only one of them works. Which one?

To answer this question, we need to examine the nature of the job done by each pump. The pump at the bottom of the well literally pushes water up the pipe. For a very deep well, the column of water becomes very tall, and the pressure at the bottom of the pump becomes large. However as long as the pump can push with sufficient strength to overcome the large pressure, it can shove the next increment of water into the pipe, so the method can be used for very deep wells. In contrast, the pump at ground level does not push water at all. It removes air from the pipe.As the pump reduces the air pressure within the pipe, the greater air pressure outside the pipe pushes the water up the pipe. But even with the strongest pump you can only remove all the air. Once the air is completely removed, an increase in pump strength does not increase the height to which the water is pushed by external air pressure. Thus, a ground-level pump can only cause water to rise to a certain maximum height, so it cannot be used for very deep wells. Note! The pump at ground level is the same principle as drinking with a straw.

Measuring PressureIt is possible to measure pressure by relating pressure differences to the variation in the height of a column of liquid. Two devices for measuring pressure are: 1. Open-tube manometer

  

The pressure P of the gas in the closed cylinder on the left in the figure is related to atmospheric pressure P0 (exerted on the top of the column of liquid on the right) by:

P - P0 = ρgh,

where is the density of the fluid in the closed tube. The pressure P of the gas is called the Absolute Pressure, to distinguish it from what is directly measured from the height of the column, namely the Gauge Pressure. Thus:

P absolute = P gauge + P0.

2. Mercury Barometer

  

Atmospheric pressure is measured by ensuring that the pressure at the closed end of the barometer tube in the figure is zero (i.e. it is vacuum). Thus

P0 = ρgh

where ρ is the density of the mercury (13600 kg/m3).It turns out that the pressure of 1 atmosphere (1 atm), is equivalent to a column of mercury exactly 0.76 m high at a temperature of 0°C with g = 9.80665 m/s 2 and = 13.595 x 103 kg/m 3. From this information, we conclude that

1  atm = 1.013 x 105  Pa = 101.3  kPa.

3. A sphygmomanometer for measuring blood pressure.

A squeeze bulb is used to inflate the cuff with air, which cuts off the flow of blood through the artery below the cuff. When the release valve is opened, the cuff pressure drops. Blood begins to flow again when the pressure created by the heart at the peak of its beating cycle exceeds the cuff pressure.

Using a stethoscope to listen for the initial flow, the operator can measure the corresponding cuff gauge pressure with, for example, an open-tube manometer. This cuff gauge pressure is call the systolic pressure. Eventually, there comes a point when even the pressure created by the heart at the the low point of its beating cycle is sufficient to cause blood to flow. Identifying this point with the stethoscope, the operator can measure the corresponding cuff gauge pressure, which is referred to as the diastolic pressure. The systolic and diastolic pressure are reported in millimeters of mercury, and values of 120 and 80 , respectively, are typical of a young healthy heart.

Pascal’s PrincipleWe have seen, the pressure in a fluid increases with depth, due to the weight of the fluid above the point of interest. A completely enclosed fluid may be subjected to an additional pressure by the application of an external force. For example two interconnected cylindrical chambers. The chambers have different diameters and, together with the connecting tube, are completely filled with a liquid. The larger chamber is sealed with a cap, while the smaller one is fitted with a movable piston. Such devices are used to lift heavy loads, for example a so-called hydraulic car lift.

We now begin by asking what determines the pressure P1 at a point immediately beneath the piston. According to the definition of pressure, it is the magnitude F1 of the

external force divided by the area A1 of the piston: P1 = F1/A1

If it is necessary to know the pressure P2 at andy deeper place in the liquid, we just add to the value of P1 the

increment ρgh. This important feature here is: The pressure P1 adds to the pressure ρgh which takes

into account the depth h below the piston: P2 = P1 + ρghThis behavior is described by Pascal’s principle.

PASCAL’S PRINCIPLEAny change in the pressure applied to a completely enclosed fluid is transmitted undiminished to all

parts of the fluid and the enclosing walls.

This arrangement is very useful - as shown in the car lift picture. Let’s calculate the force F2 applied by the liquid to the cap on the right side.

The area of the cap is A2 and the pressure there is P2. As long as the tops of the left and right chambers are at

the same level, the pressure increment ρgh is zero. Therefore P2 = P1. Consequently F2/A2 = F1/A1 and

F2 = F1(A2/A1) The force F2 is determined by the ratio of the area’s. If area A2 is larger than area A1, a large force can be applied to the cap starting with a smaller force F1 on

the left.

Example in numbers:In a hydraulic car lift, the input piston has a radius of r1 = 0.0120 m and a negligible weight. The output plunger has a radius of r2 - 0.150 m. The combined weight of the car and the plunger is F2 = 20500 N. The lift uses hydraulic oil that has a density of 8.00 x 102 kg/m3. What input force F1 is needed to support the car and the output plunger when the bottom surfaces of the piston

and plunger at at (a) the same level and (b) the levels have a height different of 1.10 m.

You try ...

When the bottom surfaces of the piston and plunger are at the same levels, as in part we can just use the

ratio of the areas to calculate F1. However, for part (b) we have the output plunger at h = 1.10 m below the piston. Therefore, we have to take into account the

pressure increment ρgh. We still expect the input force to be smaller then the combined weight.

(a) Using A = πr2 for the circular areas of the piston and plunger, we rearrange F2 = F1(A2/A1) for F1:

F1 = F2(A1/A2) = F2 (πr12/πr22) = (20500 N)((0.0120 m)2/(0.150 m)2) = 131 N

(b) We need to apply P2 = P1 + ρgh with P2 = F2/A2 and P1 = F1/A1:

F2/(πr22) = F1/(πr12) + ρgh

F1 = F2(r22/r12) - ρgh(πr12) = (20500 N)((0.0120 m) 2/(0.150 m)2) + - (8.00 x 102 kg/m3) x (9.80 m/s2)(1.10 m) π (0.0120 m)2 = 127 N

The result for (b) is smaller than for (a) because the weight of the 1.10 m column of hydraulic oil provides some of the input force on the car.

YOUR TURN: Pumping Water into a Tank ProblemA tall tank has a cross-sectional area of 340 cm2 and is open to the air (Pa = 101 kPa). The top portion of the tank contains 156 liters of oil with a density of 890 kg/m3. The bottom portion of the tank contains 173 liters of water. A horizontal tube connected to the tank at the bottom has a cross-sectional area of 11.0 mm2. Water is forced into the tank through the tube which is connected to a pump.   (A) What is the pressure in the tube before the pump is turned on when there is no fluid flow in or out of the tank ?

   (B) If the pump pumps 3.80 liter of water through the tube every minute, what is the velocity of the water in the tube ?   (C) What is the pressure of the water in the tube when the water is flowing ?   (D) What is the rate at which the surface of the oil rises per minute ?

(B), (C) and (D) are for next lecture !!!

(A) What is the pressure in the tube before the pump is turned on when there is no fluid flow in or out of the tank ?Find the pressure Pb at the bottom of the tank before the pump is turned on.

The total pressure at the bottom of the tank will be the sum of the pressure of the air on the surface of the oil Pa plus the pressure of the oil alone on the water at their interface Po plus the pressure that the water alone exerts on the bottom of the tank Pw.

Note that if the problem had asked for the gauge pressure at the bottom of the tank we would only find the pressure of the oil and water.The densities and volumes of the liquids are given, but not their heights. We can find their heights since we were given the area of the bottom of the tank.