# Chapter 15. m. The symbolic equation for mass density is: ρ= m V. Table of Densities

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1 Chapter 15 Density Often you will hear that fiberglass is used for racecars because it is lighter than steel. This is only true if we build two identical bodies, one made with steel and one with fiberglass. If I have a fiberglass car body and a small piece of steel, the steel will be lighter. Where did we go wrong? The mass per unit kg volume is known as the mass density. Therefore, the units of density are 3 m. The symbolic equation for mass density is: ρ= m V Table of Densities Material Density kg 3 m Gold 19,320 Lead 11,344 Iron 7874 Tin 7310 Aluminum 2699 Ice 917 Seawater 1025 Water 1000 Physics 210 Page 1

2 Example 1: A 1 kg ball of aluminum is sitting next to a 1 kg ball of tin. What is the ratio of the radius of the aluminum ball to that of the tin ball? Physics 210 Page 2

3 Example 2: You have two balls, one made of gold and the other made of lead. 3 Each ball has a volume of 0.01 m. What is the ratio of the mass of the gold ball to that of the lead one? Physics 210 Page 3

4 Pressure When you look at a glass of water, we know that the water is pushing on the sides of the glass. How much force is the water applying to the glass? This is a difficult question to answer because the force is distributed over a surface instead of being concentrated at a single point. Therefore, to simplify our analysis, we look at the force per unit area or pressure. So the pressure is: P = F (1) A The units of pressure are 1 N/m 2 = 1 Pa = 1 Pascal. Pressure at Depth in a Fluid When you go swimming underwater, you can feel the water pushing on your body. As you continue to go deeper, it feels like the water is pushing harder and harder. Since our surface are does not change, we know that the pressure increases as our depth increases. This pressure is due to the weight of the water pushing against our body. How is pressure related to depth numerically? F mg P = = A A Recall that the mass can be expressed in terms of the mass density and the volume: m ρ= m =ρ V V Substituting this mass equation into our pressure equation give us: ρ P = Vg A Physics 210 Page 4

5 The volume of water is equal to the surface area times the depth (V=Ah). Therefore, our pressure is: ρ( ) Ah g P = = ρgh A When computing the total pressure at a depth h, we must include the fact that the air above the liquid is pushing down as well. This is known as atmospheric pressure and has a numerical value of: 5 P = Pa Therefore, the total pressure at depth h is: 0 P = P0 + ρgh (2) Which lake (on average) exerts more pressure on the wall? You might have thought that the lake on the left exerts more pressure because it is larger, but we now know that it s the lake on the right because of extends to a greater depth. Since the lake on the right is twice as deep, the pressure at the bottom will be twice as large. Physics 210 Page 5

6 kg Example 3: The density of water is What is the pressure at the m 3 bottom of a 200 m deep lake? What depth of seawater kg ρ= would m have the same pressure? Physics 210 Page 6

7 Buoyancy Is it easier to lift an object when it is underwater? Yes and this is because the water is also pushing up on the object. The force exerted by the water is called the buoyant force. Since the pressure at the bottom of the block is greater that the pressure above (why?), the buoyant force acts up. How big is the buoyant force? The magnitude of the buoyant force is given to us by Archimedes Principle: An immersed object is buoyed up by a force equal to the weight of the fluid is displaces. Mathematically, the buoyant force is: B = ρgv (3) kg Let s imagine that we have an object that has a density of 900 m 3 and has a 3 volume of 1.0 m. Will this object float? An object will float if the maximum possible value for the buoyant force must be greater than (or equal to) the weight of the object. In our case, the object has a weight of: kg 3 m w = mg =ρ Vg = 900 ( 1 m ) 9.8 = 8820 N 3 2 m s Physics 210 Page 7

8 The maximum possible buoyant force occurs when the object is completely submerged. This means that the volume of fluid displaced is equal to the volume of the object. Therefore, the maximum buoyant force in this case is: kg m 3 Bmax =ρ gvobject = ( 1.0 m ) = 9800 N 3 2 m s Since the maximum buoyant force is greater than the weight of the object, we know that the object will float. How much of the object is below the surface? In order to find this out, we must compute the actual buoyant force. Since we know that the object floats, the buoyant force will be equal to the weight of the object. Therefore: B = w (if object floats) F = ρgv G displaced = kg m 8820 N V m 3 s 2 displaced V = m 3 = 0.9V displaced So, 90% of our objects volume is below the surface of the water. That means that 10% of it is above the surface. In general: If ρ object <ρfluid, the object will float with part of its volume above the surface. In this case the buoyant force is equal to the weight of the object. If ρ object >ρfluid, the object will sink. In this case, the buoyant force has its maximum possible value (the volume of fluid displaced equals the volume of the object) but that force is less than the weight of the object. If ρ object =ρfluid, the object is completely submerged but it does not sink. In this case the maximum buoyant force is equal to the weight of the object. object Physics 210 Page 8

9 Example 4: A 4.00 kg iron block is suspended from a spring scale and submerged in a fluid of unknown density. The spring scale reads 8.20 N. What is the density of the fluid? kg ρ = iron m Physics 210 Page 9

10 Example 5: When a 50 N stone is attached to a spring scale and is submerged in water, the spring scale reads 35 N. What is the density of the stone? Physics 210 Page 10

11 Example 6: Consider a boat loaded with scrap iron in a swimming pool. If the iron is thrown overboard, will the water level at the edge of the pool rise, fall, or remain unchanged? Physics 210 Page 11

12 Flow of Fluids In general, the flow of fluids can be very complex. If we consider steady-state flow of an ideal fluid, no dissipation of mechanical energy will occur. We will also assume that our fluid is incompressible (which is only true with liquids) which means that the density will be uniform throughout. For the pipe above, the volume in region 1 is: The volume of liquid in region 2 is: V1 = Av 1 1 t V2 = Av 2 2 t Since we assume fluids are incompressible, an equal amount of fluid must flow past any point in an equal time interval. Therefore: V = V 1 2 Av t= Av t Av 1 1 = Av (4) 2 2 This is known as the continuity equation. This equation tells us that if we reduce the radius of a pipe, the speed of the water flowing through it will increase. The quantity Av is known as the volume flow rate (or volume flux) Physics 210 Page 12

13 and is given the symbol I V. Therefore, we can express the continuity equation as: IV = constant Example 7: Water exits a circular tap moving straight down with a flow rate of 3 cm s a) If the diameter of the tap is 1.2 cm, what is the speed of the water? b) What would the speed be if the diameter of the tap were doubled? Physics 210 Page 13

14 Bernoulli s Equation The work done in region 1 (for the pipe above) is: The work done in region 2 is: W1 = F1 x1 = P1A1 x1 = PV 1 1 W2 = F2 x2 = P2A2 x2 = P2v 2 Since x1 = vt, 1 x2 = vt, and 2 V1 = V2 = V for an incompressible fluid, we know that: ( ) W = W + W = P P V net The change in the potential energy of the liquid is: The change in the kinetic energy: U = mgh2 mgh K = mv mv 2 2 Now we apply the work-energy theorem: W= U+ K Physics 210 Page 14

15 1 ( ) ( ) m 2 2 ( v 2 v 1) P1 P2 V mg h2 h1 = + 2 V V V 1 P P =ρg h h + ρ v v ( ) ( ) P +ρ gh + ρ v = P +ρ gh + ρv (5) This is known as Bernoulli s equation and it is usually written as: 1 2 P +ρ gy + ρ v = constant 2 (6) Example 8: The pressure in a section of horizontal pipe with a radius of 2.00 cm is 142 kpa. Water flows through the pipe at 2.80 L/s. If the pressure at a certain point is to be reduced to 101 kpa by constricting a section of the pipe, what should the diameter of the constricted section be? Physics 210 Page 15

16 Example 9: Water flows through the pipe and exits at C. The diameter of pipe is 2.0 m at A, 1.5 m at B and 1.0 m at C. The pressure in section A is 250 kpa and the flow rate is 8000 L/s. a) What is the fluid velocity in section A? b) What is the height h? A c) What is the fluid velocity in section B? d) What is the height h? B e) What is the pressure in section B? f) What is the fluid velocity in section C? Physics 210 Page 16

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