Physics 2 - College Physics 2

Physics2 Lesson 1 Chapter Review II

Chapter Review

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15-4 - 15-5 Archimedes' Principle and Buoyancy

When an object is submerged in a fluid, the volume taken up by the object displaces an equal volume of the fluid. The pressure applied by the fluid onto the object results in an upward force on the object; this phenomenon is known as buoyancy. This phenomenon is governed by Archimedes' principle:

The weight of the fluid displaced by the object equals the mass of this fluid times the acceleration due to gravity, mg. When dealing with buoyancy it is usually more convenient to express the mass in terms of the density, m =

V. Therefore, for an object submerged in a fluid, the buoyant force on it is

Archimedes' principle explains the phenomenon of floatation which occurs when the buoyant force acting on an object equals it's weight. Often, floating objects are not completely submerged in the fluid. The amount of volume submerged V_sub for a solid object of volume V_s floating in a fluid of density

_f is given by

where

_sis the density of the solid object.

Physlet Illustration: Archimedes' Principle

A metallic object hangs from a digital balance, which reads the object's mass in grams. The object is suspended above a graduated cylinder containing water. The grid is such that each grid square represents a volume increment of 1 cm³. Lower the apparatus into the water, and verify Archimedes' Principle. Start

Hints:

What are the balance readings before and after the object is immersed in the water?
From the difference in these readings, what is the buoyant force on the immersed object?
What are the water levels in the cylinder before and after the object is immersed in the water?
From the difference in these readings, what is the volume of water displaced by the object?
Since the density of water is 1 gm/cm³, what is the weight of the water displaced by the object?
Can you determine the density of the metallic object?

Exercise 15.5 The Secret of Magic: Many magic tricks are based on physical principles. In order to fool her audience a magician uses an object that sinks in the fresh water made available to the audience, but floats in the seawater that she uses on stage. What is the maximum percentage of the object's volume that will float above the seawater?

Picture the Problem The picture shows a floating object partially submerged in seawater.

Strategy According to the above discussion, more of an object will be submerged if its density approaches that of the fluid. So, you get the maximum above-surface float for the smallest possible object density. Since it must sink in fresh water, the smallest object density is 1000 kg/m³.

Solution

1. The minimum fraction of volume submerged is:
2. The maximum amount of volume floating above the surface is:
3. So the percentage is:

Insight What could the magician do to the seawater to make a larger percentage of the object float?

Practice Quiz


		An object of density 750 kg/m³ is half submerged in a fluid. What is the density of this fluid? 1500 kg/m³ 188 kg/m³ 375 kg/m³ 2250 kg/m³ 750 kg/m³


		If the volume of the object in question #5 above is 0.33 m³, what is the buoyant force on this object? 9810 N 1210 N 3240 N 1000 N 2430 N

Physlet Illustration: Archimedes' Principle and Melting Ice

An ice cube melts in a glass of water as shown in the animation (position is measured in centimeters and time is shown in minutes). Which animation correctly shows what the final water level will be?
Start

Start Animation 1

Start Animation 2

Start Animation 3

Hints:

Since the bouyant force is the weight of the water displaced, you can determine the mass of the ice cube.
How does the volume of the ice cube that is below the water level change with time?

15-6 Fluid Flow and Continuity

In this section we begin to discuss properties of fluid flow. During the smooth flow of a constrained fluid (e.g., through a pipe) we can assume that the same amount of mass passes through each cross section of pipe in a given amount of time. This smooth flow condition leads to what is known as the equation of continuity which says that mass m₁ flowing through an area A₁ in a given time equals the mass m₂ flowing through area A₂ in that same amount of time. The amount of mass per unit time of a fluid of density

flowing through area A at speed v is

Av. Therefore, the equation of continuity is

Usually, liquids are considered to be incompressible because the density of the liquid hardly changes as it flows from one place to another. In such cases the densities in the equation of continuity are equal,

₁ =

₂, and we can write it as

The quantity Av equals the volume flow rate of the fluid; the above equation then says that this volume flow rate is constant for an incompressible fluid.

Example 15.6 Continuity: Plastic bottles that are used to hold water for athletes often have a long slender nozzle out of which the water emerges. If the end of the nozzle has a diameter of 1.0 cm and you determine the water to emerge at 25 cm/s for a typical squeeze of the bottle, what is the initial speed of the water in the neck of the bottle if its diameter is 6.0 cm?

Picture the Problem The picture shows a squeeze water bottle with a thin nozzle on the end.

Strategy Since we don't expect the density of the water to change when flowing from inside the bottle to outside, we only need to use the fact that the volume flow rate is constant.

Solution

1. Using the equation of continuity gives:
2. Solving for v₁ gives:
3. Obtain the numerical result:

Insight The fact that a fluid flows more rapidly when squeezed is used in many different applications. Can you think of some others?

Practice Quiz


		If the area through which an incompressible fluid flows decreases, the speed of flow will... decrease increase stay the same


		If the area through which an incompressible fluid flows is cut in half, the speed of flow will... also be cut in half. double. decrease to its speed. triple. [None of the above.]

Physics2 Lesson 1 Chapter Review I

Fluids and Pressure

Chapter Review

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In this chapter we study fluids.

--> Fluids are characterized by their ability to flow; both liquids and gases are considered to be fluids. An understanding of fluid behavior is essential to life and applications of this understanding are essential to many of the conveniences of modern living.

15-1 - 15-2 Density and Pressure

One of the more convenient properties used to describe a fluid is its density

. The density of a substance is a measure of how compact the substance is; that is, how much mass is packed into a volume of the substance. On average the density of a substance is the amount of mass M divided by the volume V taken up by that mass

Another important quantity in the study of fluids is pressure P. On the average, the pressure that is applied to an object is the amount of force F (normal to the surface) divided by the area A over which the force spreads

As the above expressions indicates, the units of pressure are those of force divided by area; in SI units this is called the pascal (Pa): 1 Pa = 1 N/m². An important property of the pressure in a fluid is that it is equally applied in all directions (at a given depth) and applies forces that are perpendicular to any surface in the fluid.

When considering the pressure on or within a fluid it is important to recognize that for many applications we must account for a constant atmospheric pressure (P_at = 1.01 x 10⁵ Pa). Because of the constant presence of the atmosphere we are often only interested in the pressure above and beyond that applied by the atmosphere. This additional pressure is called the gauge pressureP_g

where P is the total pressure applied (sometimes called the absolute pressure).

Exercise 15.1 The Density of a Fluid: If 1.00 gallons of a certain fluid weighs 3.22 lb, what is its density?

Solution: We are given the following information:

Given: V = 1.00 gal, W = 3.22 lb Find:

Since the density is given by

= m/V we need to find the mass from the weight and convert everything to SI units. To get the mass we have

The volume can be immediately converted to give

We are now ready to calculate the density as

Example 15.2 Gauge Pressure of Water: A uniform cylindrical container has a radius of 7.8 cm and a height of 13.2 cm. If this container is completely filled with water, what gauge pressure does the water apply to the bottom of the container?

Picture the Problem The picture shows a cylindrical container filled with water.

Strategy The pressure applied by the water will be the force that the water applies (equal to its weight) divided by the bottom area of the cylinder.

Solution

1. The volume of water equals the volume of the cylinder:
2. Determine mass of the water
3. Determine the weight of the water:
4. Obtain the gauge pressure:

Insight The above answer is the gauge pressure because we have ignored atmospheric pressure.

Practice Quiz


		If it takes as much volume of fluid 1 to weigh the same as fluid 2, how do their densities compare? fluid 1 is twice as dense as fluid 2 fluid 1 is half as dense as fluid 2 fluid 1 and fluid 2 have equal densities fluid 1 is four times less dense as fluid 2 [None of the above]


		If a cylinder of height 0.850 m and radius 0.250 m is filled with water, what is the total pressure on the bottom of the cylinder? 8.34 kPa 133 kPa 2.08 kPa 109 kPa 46.2 kPa

15-3 Static Equilibrium in Fluids: Pressure and Depth

In this and the next several sections we will consider the properties of static fluids, that is, fluids that do not flow. In the case of static fluids, every part of the fluid and every object within the fluid is in static equilibrium. One of the basic properties of fluids that is very important to our ability to understand fluid behavior is known as Pascal's principle:

Pascal's principle is important in determining the dependence of pressure on the depth within a fluid. Without any external pressure applied to the outer surface of a fluid, the pressure measured at a depth h beneath the surface arises from the weight of the fluid above the given level. The amount of this increase in pressure is given by

gh. If there is external pressue on the fluid, such as from the atmosphere and/or any other source, this pressure is transmitted undiminished to every point in the fluid and it must be added to the pressure due to the weight of the fluid. Thus, for the dependence of pressure on depth we have

where P₂ is the pressure at a given level within the fluid and P₁ is the pressure at a height h above that level.

Example 15.3 Blood Pressure with Depth: Human blood has a density of approximately 1.05 x 10³ kg/m³. Use this information to estimate the difference in blood pressure between the brain and the feet in a person who is approximately six feet tall.

Picture the Problem The picture shows a person approximately six feet in height.

Strategy We attempt this approximation to two significant figures by applying the above result for the dependence of pressure on depth using blood as the fluid.

Solution

1. Convert the height to meters:
2. The difference in pressure is given by:
3. Obtain the numerical result:

Insight This is only an estimate for many reasons. The blood in the body is not a static fluid, but it flows; and between the head and the feet there is a pump (the heart) which will affect the result. If you plan to study medicine, see if you can find out the difference in blood pressure between the head and the feet.

Pascal's principle is also key to understanding the hydraulic lift. This device uses fluid pressure to convert a small input force into a large output force. The input force F₁ is applied to a fluid over a comparatively small area A₁ giving rise to a pressure change of F₁/A₁ = F₂/A₂, we must get a larger output force F₂ > F₁ if it is spread over a larger area A₂ > A₁. Therefore,

The consequence of getting this larger output force is that the distance through which this force can move an object at the output, d₂, is smaller than the distance at the input d₁. Since the same volume of fluid moves at the input and output, A₁d₁ = A₂d₂, the output distance is given by

Exercise 15.4 Hydraulic Lift: Your job at Dave's Manufacturing Company is to design a hydraulic lift for a client. This client typically needs to raise objects through a height of 0.500 meters. The system should be easily used by people of average height so the input distance should not exceed 5.00 ft. What would be a good ratio of output area to input area to consider?

Solution: We are given the following information.

Given: d₁ = 5.00 ft, d₂ = 0.500 m Find: A₂/A₁

From the information given, we know that the ratio of the input distance to the output distance is directly proportional to the ratio we seek

Therefore, a ratio of

The output area should be at least 3.05 times greater than the input area. Of course, in a more realistic situation you'd want more information from the client. What additional information might you want?

Practice Quiz


		What is the gauge pressure 3.4 m below the surface of a container filled with a fluid of density 550 kg/m³? 33 kPa 1900 Pa 18 kPa 100 kPa 550 Pa


		The gauge pressure at a particular location in a fluid of density 870 kg/m³ is 120 kPa. What is the gauge pressure in the fluid 5.9 m above this location? 70 kPa 50 kPa 62 kPa 58 kPa 20 kPa

Physics2 Lesson 1- Fluids

Fluids

(source: Cliffs Notes)

A fluid is a substance that cannot maintain its own shape but takes the shape of its container. Fluid laws assume idealized fluids that cannot be compressed.

Density and pressure

The density (ρ) of a substance of uniform composition is its mass per unit volume: ρ = m/ V. In the SI system, density is measured in units of kilograms per cubic meter.

Imagine an upright cylindrical beaker filled with a fluid. The fluid exerts a force on the bottom of the container due to its weight. Pressure is defined as the force per unit area: P = F/ A , or in terms of magnitude, P = mg/A, where mg is the weight of the fluid. The SI unit of pressure is N/m², called a pascal. The pressure at the bottom of a fluid can be expressed in terms of the density (ρ) and height (h) of the fluid:

or P = ρ hg. The pressure at any point in a fluid acts equally in all directions. This concept is sometimes called the basic law of fluid pressure.

Pascal's principle

Pascal's principle may be stated thus: The pressure applied at one point in an enclosed fluid under equilibrium conditions is transmitted equally to all parts of the fluid. This rule is utilized in hydraulic systems. In Figure 1 , a push on a cylindrical piston at point a lifts an object at point b.

Figure 1

Pascal's principle is used to easily lift a car.

Let the subscripts a and b denote the quantities at each piston. The pressures are equal; therefore, P_a= P_b. Substitute the expression for pressure in terms of force and area to obtain f_a/ A_a= ( F_b/ A_b). Substitute π r² for the area of a circle, simplify, and solve for F_b: F_b=( F_a)( r_b²/ r_a²). Because the force exerted at point a is multiplied by the square of the ratio of the radii and r_b> r_a, a modest force on the small piston a can lift a relatively larger weight on piston b.

Archimedes' principle

Water commonly provides partial support for any object placed in it. The upward force on an object placed in a fluid is called the buoyant force. According toArchimedes' principle, the magnitude of a buoyant force on a completely or partially submerged object always equals the weight of the fluid displaced by the object.

Archimedes' principle can be verified by a nonmathematical argument. Consider the cubic volume of water in the container of water shown in Figure 2 . This volume is in equilibrium with the forces acting on it, which are the weight and the buoyant force; therefore, the downward force of the weight ( W) must be balanced by the upward buoyant force ( B), which is provided by the rest of the water in the container.