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    introduction:Density and Pressure in Solids and Fluids Key unit competence The learner should be able to define pressure and explain factors affecting it.

     My goals By the end of this unit, I will be able to:

     ࿤ define and explain the pressure as a relationship of force acting on a surface area.

     ࿤ identify force and area as factors affecting pressure in solids.

     ࿤ give the relationship between force, pressure and area.

     ࿤ explain how pressure varies with force and the area of contact.

     ࿤ describe liquid (mercury) in glass barometer.

     ࿤ explain floating and phenomena.

     ࿤ describe how to measure atmospheric pressure.

     ࿤ carry out calculations using the equation. Pressure = force/area and P = ρgh.

     ࿤ explain the variation in atmospheric pressure with altitude.

     ࿤ explain the change in pressure by reducing or increasing area of contact and vice versa.

     ࿤ measure atmospheric pressure using a barometer, liquid in glass barometer. 70 Physics for Rwanda Secondary Schools Learner’s Book Senior Two

     ࿤ explain the functioning of aneroid barometer.

     ࿤ describe and explain pressure transmission in hydraulic systems.

     ࿤ explain functioning of a hydraulic press and hydraulic brakes. Key concepts

     1. How experimentally can one determine the effect of force exerted on a solid?

     2. How can one define pressure in a fluid?

     3. How can pressure in liquids be measured?

     4. What instrument should be used in measuring pressure?

     5. Where can pressure in solids and liquids be applied in real life?

     Š Vocabulary Pressure, atmospheric pressure, fluids, hydrostatic pressure, barometer, manometer. y Reading strategy As you read this unit – mark the paragraphs that contain definitions of key terms.

     Use all the information to write a definition of each key term in your own words.

     Perform calculations related to pressure in solids and fluids.

     4.1 Force exerted by solids Activity 4

    .1: Investigation pressure of a solid Materials:

     ⚫ One concrete brick 

    ⚫ Balance

     ⚫ A pile of sand 

    ⚫ A ruler

     ⚫ A long beam of wood



    1 Unit 4: Density and Pressure in Solids and Fluids Procedures:

     Measure the mass (m) of the brick and calculate its weight (w = mg).

     Pour two bucketfuls of sand outside your laboratory such that it forms a pile as shown in (i). Use the long wooden beam to spread the sand such that you have a fairly large plain surface on top of the sand pile, as shown in (ii).

     ⚫ Take measurement of dimensions of one of the large surface side and calculate its area A1. 

    ⚫ Take measurement of the dimensions of the small side and calculate its area A2.\

     ⚫ Gently place the brick in the sand on its big side and let it rest on the sand for 15s. Carefully remove the brick from the sand. Note the depression formed on the sand and carefully measure its depth. Measure the depth at four different places and determine the average depth of the depression on the sand left by the brick. Calculate pressure exerted by the brick using P1 = W A2

     ⚫ Gently place the brick on the sand on its smaller side but at a point away from the first experiment. Calculate the pressure exerted by the brick P2 = W A2 Compare and discuss the result obtained. (i) Sand pile (ii) Levelled sand pile Fig. 4.1: Force exerted by a solid 72 Physics for Rwanda Secondary Schools Learner’s Book Senior Two If the force is concentrated on a small area, it will exert higher pressure than if the same force is distributed over a larger surface area. As an example of varying pressures,

     ⚫ A finger can be pressed against a wall without making any lasting impression; however, the same finger pushing a thumbtack can easily penetrate the wall. Although the force applied to the surface is the same, the thumbtack applies more pressure because the point concentrates that force into a smaller area. ⚫ If we try and cut a fruit with the flat side of the knife it obviously won’t cut. But if we take the sharp side, it will cut smoothly. The reason is, the flat side has a greater surface area(less pressure) and so it does not cut the fruit. When we take the thin side, the surface area is very small and so it cuts the fruit easily and quickly. ⚫ A bus or truck is heavy. It may have large tyres, so that its weight is spread over a large area. This means that the pressure on the ground is reduced; so it is less likely to sink in soft ground. This is one example of a practical application of pressure. 4.2 Definition and units of pressure Pressure (symbol “p”) is the force acting normally per unit area applied in a direction perpendicular to the surface of an object. Gauge pressure is the pressure relative to the local atmospheric or ambient pressure. The pressure is directly proportional of force and proportional of square area. In mathematical terms, pressure can be expressed as: p = F A The pressure within a fluid (gas or liquid) is a scalar quantity—that is, it has magnitude but no particular direction associated with it in space. Pressure arises from two fundamentally different kinds of sources: ambient and localised. 1. Ambient sources of pressure are usually a gas or a liquid in which an entity is immersed, such as a human being on the surface of the earth or a fish in a stream or lake. Life forms are generally insensitive to ambient pressures and become aware of the source of that pressure when currents become strong enough that the fluid exerts a non-uniform localised pressure on the life form, such as when the wind blows. 73 Unit 4: Density and Pressure in Solids and Fluids 2. Localised pressure sources are usually discrete objects, such as the finger pressing on the wall, or the tyres of a car pressed against the pavement. A liquid or gas can become the source of a localised pressure if either of them is forced through a narrow opening. 4.2.1 Unit of pressure The unit is the pascal and is named after Blaise Pascal, the eminent French mathematician, physicist, and philosopher noted for his experiments with a barometer, an instrument to measure air pressure. The name Pascal was adopted for the SI unit Newton per square meter by the 14th CGPM in 1971. The Pascal (symbol: Pa) is the SI derived unit of pressure. It is a measure of force per unit area, equivalent to l Pa ≡ 1 N/m2 ≡ 1 kg/(m·s2). The Pascal is perhaps best known from meteorological barometric pressure reports, where it occurs in the form of hectopascals or millibar (1hPa ≡ 100 Pa ≡ 1 mbar). barye ≡ 1 μbar = 1 dyn/cm2 The standard atmosphere (atm) of pressure is approximately equal to air pressure on earth above mean sea level and is defined as: In 1985, the International Union of Pure and Applied Chemistry (IUPAC) recommended that standard atmospheric pressure should be harmonised to 100,000 Pa = 1 bar = 750 Torr. Another unit for pressure measurement is millimeters of mercury (.1 mmH g = 9.80669Pa) We use a manometer to measure pressure in liquids and a barometer to measure air pressure. Activity 4.2: Using a Manometer ⚫ Manometer ⚫ Water ⚫ Beaker ⚫ A ruler (Optional) Procedures: Try to refer to the Fig. 4.2 and do the following: a) Pour water into the beaker. b) Note the level of the manometer liquid. c) Lower the manometer nozzle in water. d) Note the change in the level of the manometer liquid. e) Lower it deeper than before and note the new changes. 74 Physics for Rwanda Secondary Schools Learner’s Book Senior Two Questions: 1. What changes are you observing? 2. Discuss the meaning and the cause of that change. 3. Prove that liquids exert pressure on a body submerged in. Fig. 4.2: A Manometer The figure above is a simple pressure gauge and it measures differences in pressure exerted at the two ends of the apparatus. It is called a Manometer. The mouth of a thistle funnel is tightly covered with a thin plastic sheet. The thistle funnel is connected to a U-tube manometer containing water, by rubber tubing. Now, lower the mouth of the funnel into a glass vessel containing water. You will notice that the deeper it goes; greater is the difference in the levels of the water in the manometer. This indicates that the pressure in a liquid increases with depth. Repeat the experiment by turning the thistle funnel in different directions keeping the depth constant. You will observe that as long as the depth remains the same, there is no change in the level of the water in the manometer. Thus, the pressure exerted by a liquid at a given depth is the same in all directions. Now, lower the thistle funnel to the same depth in a number of liquids having different densities. You will notice that in liquids having greater density, the pressure at the same depth is greater. This indicates that the greater the density of the liquid, the greater the pressure at the same depth. 75 Unit 4: Density and Pressure in Solids and Fluids 4.3 Static fluid pressure The pressure exerted by a static fluid depends only upon the depth of the fluid, the density of the fluid, and the acceleration of gravity. The pressure in a static fluid arises from the weight of the fluid and is given by the expression: Pstaticfluid = ρf gh Where ρ is the density of fluid; g is acceleration of gravity; and h is depth of fluid. The pressure from the weight of a column of liquid of area A and height h is: V = hA = volume weight = mg Static fluid pressure does not depend on the shape, total mass, or surface area of the liquid. Pressure = weight area mg pVg pgh P = pgh h h A A = = = h h P1 P2 Fig. 4.3: Pressure in liquid The most remarkable thing about this expression is what it does not include. The fluid pressure at a given depth does not depend upon the total mass or total volume of the liquid. The above pressure expression is easy to understand for the straight, unobstructed column, but not obvious for the cases of different geometry which are shown. Because of the ease of visualising a column height of a known liquid, it has become common practice to state all kinds of pressures in column height units, like mmHg. Pressures are often measured by manometers in terms of a liquid column height. 76 Physics for Rwanda Secondary Schools Learner’s Book Senior Two 4.3.1 Fluid pressure calculation h P1 P2 A Fluid column height in the relationship with pressure difference is given by: Pressure difference ∆P = P2 – P1 = eh2g – eh1g = ehg – 0 = eng Note: The pressure P1 is zero because h1 = 0 Static fluid pressure is dependent on density and depth only and independent of total mass, weight, volume, etc. of the fluid. Fig. 4.4: Pressure at a depth h P = ρgh 4.4 Atmospheric pressure 4.4.1 Measuring atmospheric pressure 760 mm column of mercury needed to Pressure of mercury Pressure of air balance pressure of air at sea level h The atmospheric pressure is the weight exerted by the overhead mass of air on a unit area of surface. It can be measured with a mercury barometer, consisting of a long glass tube full of mercury inverted over a pool of mercury: Fig. 4.5: Mercury barometer When the tube is inverted over the pool, mercury flows out of the tube, creating a vacuum in the head space, and stabilises at an equilibrium height, h over the surface of the pool. This equilibrium requires that the pressure exerted on the mercury at two points on the horizontal surface of the pool, (inside the tube) and (outside the tube), be equal. The pressure 77 Unit 4: Density and Pressure in Solids and Fluids at the point inside the tube is that of the mercury column overhead, while the pressure at that point is that of the atmosphere overhead. We obtain, from measurement of h, P = eHg gh where ρHg = 13.6 gcm–3 is the density of mercury and g = 9.8ms–2 is the acceleration of gravity. The mean value of h measured at sea level is 76.0 cm, and the corresponding atmospheric pressure is 1.013 × 105 kg m–1 S–2 in SI units. The most commonly used pressure unit is the atmosphere (atm) (1atm = 1.013 × 105 Pa) the bar (b) (1 b = 1 × 105 Pa), the millibar (mb) (1 mb = 100 Pa) and the torr. (1 torr = 1 mm Hg = 134 Pa) The use of millibars is slowly giving way to the equivalent SI unit of hectoPascals. The mean atmospheric pressure at sea level is given equivalently as; P = 1.013 × 105 Pa = 1013hPa = 1013mb = 1 atm = 760torr. R is the mean height of the atmosphere from the surface of earth. 4.4.2 Mass of the atmosphere The global mean pressure at the surface of the Earth is slightly less than the mean sea-level pressure because of the elevation of land. We deduce the total mass of the atmosphere ma: ma = 4π R2 Ps g Centre of earth Earth Maximum height for Atmosphere R = R2 – R1 R2 R1 R Na = ma Ma = 1.8 × 1020 moles. 78 Physics for Rwanda Secondary Schools Learner’s Book Senior Two 4.4.3 Torricelli’s experiment Mercury Atmospheric pressure 760 mm (29.92 in) Vacuum Glass tube Fig. 4.6: Torricelli experiment of a simple mercury barometer Torricelli’s experiment was a curious project made in 1643 by the Italian physicist and mathematician Evangelista Torricelli (1608-1647) in a laboratory that attained to measure the atmospheric pressure for the first time. Process: Torricelli filled a 1 metre long tube with mercury, (closed at one end) and inverted it on a tray full of mercury. Immediately the column of mercury went down several centimetres, remaining static at some 76cm (760mm) of height. As it was observed that the pressure was directly proportional to the height of the mercury column (Hg), the millimetre of mercury was adopted as a measurement of pressure. That way, the pressure corresponded to a column of 760mm. Conclusion: The column of mercury did not fall due to the fact that the atmospheric pressure exerted on the surface of the mercury was able to balance the pressure exerted by his weight. 760 mmHg = 1 atm 1 atm = 1.013 mbar or hPa 1 mbar or hPa = 0,7502467 mmHg 79 Unit 4: Density and Pressure in Solids and Fluids 4.5 Gas pressure Activity 4.3: Investigating gas pressure Imagine the case of pumping air in the ball (e.g Football): Materials: ⚫ Bicycle tube ⚫ Brick ⚫ Plank of wood ⚫ Bicycle pump Procedures: ⚫ Take an empty bicycle tube. ⚫ Lay a plank of wood across it. ⚫ Place the brick at the centre of the plank. ⚫ Take the pump and start to pump the gas into the bicycle tube. ⚫ Note the changes in the position of the brick as one continues to pump. Questions: 1. Explain why pumping air into the bicycle tube results in rise of the brick placed on the wooden plank. 2. Explain what would happen, if one continues to pump in more and more gas. Pressure is determined by the flow of mass from a high pressure region to a low pressure region. Air exerts a pressure which we are so accustomed to that we ignore it. The pressure of water on a swimmer is more noticeable. You may be aware of pressure measurements in relation to your bicycle tyres. Atmospheric pressure varies with height just as water pressure varies with depth. As a swimmer dives deeper, the water pressure on his/ her increases. As a mountain climber ascends to higher altitudes, the atmospheric pressure on him/her decreases. His body is compressed by a smaller amount of air above it. The atmospheric pressure at 6100m is only one-half of that at level because about half of the entire atmosphere is below this elevation. 80 Physics for Rwanda Secondary Schools Learner’s Book Senior Two 4.6 Simple pressure related applications 4.6.1 Drinking straw Activity 4.4: Investigation of atmospheric pressure in using drinking straws Materials: ⚫ Drinking straws ⚫ Very clean bottles of mineral water ⚫ Safe drinking water ⚫ 50mm beakers Procedures: ⚫ Make a hole on the cover of the mineral water bottle that exactly fits the straw tube. ⚫ Insert the straw tube through the hole such that its bottom is about 5mm from the bottom of the plastic water bottle. ⚫ Fill the water bottle with clean and safe water and close. ⚫ Make sure that the contacts between the tube and cover are airtight. ⚫ Fill your beaker with safe drinking water and suck it using a straw. ⚫ Take water/juice in the bottle (closed such that no air can get inside), but with an opening of the straw only and suck. Questions: 1. What have you noticed when drinking from the glass? 2. What have you noticed when drinking from the bottle? 3. Discuss and explain the causes of your observations. Fig. 4.7: Drinking straw 81 Unit 4: Density and Pressure in Solids and Fluids A drinking straw is used to create suction with your mouth. This causes a decrease in air pressure inside the straw. Since the atmospheric pressure is greater on the outside of the straw, liquid is forced into and up the straw. 4.6.2 Siphon Activity 4.5: Investigation of the siphon Materials: ⚫ A jerrycan ⚫ Bucket ⚫ Water ⚫ A long flexible plastic pipe ⚫ Table Procedures: ⚫ Fill a jerrycan of water on the table. ⚫ Place the bucket down on the lower level of the table. ⚫ Lower one end of the plastic pipe in the jerrycan. ⚫ Let the other end of the plastic pipe be at a lower level than of the water in the jerrycan. ⚫ Suck water from the jerrycan, and release after the water has come to your mouth. ⚫ Let water flow from the jerrycan to the bucket freely. Questions: 1. What causes the water to flow from the jerrycan to the bucket? 2. Why does the water continue to flow without sucking again? 3. Discuss and explain where this can be applied. Siphoning gas using mouth Fig. 4.8: The Siphon 82 Physics for Rwanda Secondary Schools Learner’s Book Senior Two With a siphon water can be made to flow “uphill”. A siphon can be started by filling the tube with water (perhaps by suction). Once started, atmospheric pressure upon the surface of the upper container forces water up the short tube to replace water flowing out of the long tube. 4.7 Pressure with altitudes Pressure decreases with increasing altitude The pressure at any level in the atmosphere may be interpreted as the total weight of the air above a unit area at any elevation. At higher elevations, there are fewer air molecules above a given surface than a similar surface at lower levels. For example, there are fewer molecules above the 50km surface than are found above the 12km surface, that is why the pressure is less at 50km as shown in the Fig. 4.10 below. 1000 mb 100 mb 1 mb 50 km 12 km Unit Area What this implies is that atmospheric pressure decreases with increasing height. Since most of the atmosphere’s molecules are held close to the earth’s surface by the force of gravity, air pressure decreases rapidly at first, then more slowly at higher levels. Fig. 4.19: Pressure variation with altitude Pressure (mb) Mt. Everest Altitude (km) Altitude (miles) 50 40 30 30 Above 99.9% Above 99% Above 90% Above 50% 20 20 10 10 100 500 900 5.5 Fig. 4.10: Pressure variations with altitude 83 Unit 4: Density and Pressure in Solids and Fluids Since more than half of the atmosphere’s molecules are located below an altitude of 5.5km, atmospheric pressure decreases roughly 50% (to around 500mb) within the lowest 5.5km (Fig 4.10). Above 5.5km, the pressure continues to decrease but at an increasingly slower rate. 4.8 Common observations of pressure Activity 4.6: Investigation of application of pressure in real life Study the following questions. Discuss them and give other more examples in each case. 1. Why do heavy lorries have many tyres? 2. Why do tractors (excavators, loaders, etc) have wide tyres? 3. Discuss and explain why ducks, geese have webbed feet? 4. Why do the feet of camels and elephants have wide pads? 4.8.1 Ducks, geese, and swans all have webbed feet. The primary use for webbed feet is paddling through water Here’s how it works: as the bird pulls its foot backwards through the water, the toes spread apart, causing the webs to spread out. The webs push more water than just a bird foot with spread-out toes would push. (It would be like trying to swim with your fingers spread apart.) The webbed feet propel the bird through the water. Fig. 4.11: Duck webbed feet When the bird pulls its foot forward for the next push, the toes come together, folding up the webs. The foot is instantly less resistant, moving through the water easily to get into place for the next stroke without pushing the bird backwards. Webbed feet are useful on land as well as on water because they allow birds to walk more easily on mud. Most swimming or paddling birds have 84 Physics for Rwanda Secondary Schools Learner’s Book Senior Two their legs and feet located at the rear of their body. This adaptation is an advantage on the water as it helps to propel the birds along. But what’s good on the water isn’t necessarily good on land. Having their legs and feet located at the rear of their body makes walking more difficult for these birds. 4.8.2 Camel or elephant wide pads Fig. 4.12: Wide pads of a Camel and an Elephant Why do camels or elephants have wide and large feet? Those features help them walk across desert sands. So that they are able to walk across sand without sinking in. To walk on sand so they have a bigger surface area to handle their weight and the objects that are put on it to carry - so they don’t sink into the sand to spread their weight out over the sand, which helps to prevent them from sinking into it. Camels are adapted to walk long distances in deserts, hence, they have evolved to form large, broad, flat feet. More surface area means less pressure exerted on that surface, and vice-versa as the pressure is distributed on a large area, because it would give less pressure on the sand which prevents it from sinking. 4.8.3 Lorries with many tyres Why do trucks that carry heavy loads have so many wheels often eighteen? 85 Unit 4: Density and Pressure in Solids and Fluids To distribute the load over a greater area. Fig. 4.13: Lorries with many tyres Pavement strength is all about pressure (or stress more correctly). This is a force divided by area. If you increase the area (number of tyres) that the load is distributed over, there will be less pressure (stress) on the pavement. Think about an extreme example: Say a really heavy dump truck has only four tyres, then there are only four places for the load to go to. Further still, think about a dump truck with only one tyre! all the weight of the truck will be put onto that one tyre! This is the same principle by which knives work, you are applying a small force, but over a very small area, which makes the thing you are cutting, cut. If you have a dull knife, it will be harder to cut because the force is distributed over a larger a  


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