# SURFACE TENSION. Definition

Size: px
Start display at page:

Transcription

1 SURFACE TENSION Definition In the fall a fisherman s boat is often surrounded by fallen leaves that are lying on the water. The boat floats, because it is partially immersed in the water and the resulting buoyant force balances its weight, as Section 11.6 discusses. The leaves, however, float for a different reason. They are not immersed in the water, so the weight of a leaf is not balanced by a buoyant force. Instead, the force balancing a leaf s weight arises because of the surface tension of the water. Surface tension is a property that allows the surface of a liquid to behave somewhat as a trampoline does. When a person stands on a trampoline, the trampoline stretches downward a bit and, in so doing, exerts an upward elastic force on the person. This upward force balances the person s weight. The surface of the water behaves in a similar way. In Figure 1, for instance, you can see the indentations in the water surface made by the feet of an insect known as a water strider, because it can stride or walk on the surface just as a person can walk on a trampoline. Figure 1 The indentations in the water surface made by a water strider are readily seen in this photograph. The surface tension of the water allows the insect to walk on the water without sinking. ( Hermann Eisenbess/Photo Researchers) Figure illustrates the molecular basis for surface tension by considering the attractive forces that molecules in a liquid exert on one another. Part a shows a molecule within the bulk liquid, so that it is surrounded on all sides by other molecules. The 1 of 9

2 surrounding molecules attract the central molecule equally in all directions, leading to a zero net force. In contrast, part b shows a molecule in the surface. Since there are no molecules of the liquid above the surface, this molecule experiences a net attractive force pointing toward the liquid interior. This net attractive force causes the liquid surface to contract toward the interior until repulsive collisional forces from the other molecules halt the contraction at the point when the surface area is a minimum. If the liquid is not acted upon by external forces, a liquid sample forms a sphere, which has the minimum surface area for a given volume. Nearly spherical drops of water are a familiar sight, for example, when the external forces are negligible. Figure (a) A molecule within the bulk liquid is surrounded on all sides by other molecules, which attract it equally in all directions, leading to a zero net force. (b) A molecule in the surface experiences a net attractive force pointing toward the liquid interior, because there are no molecules of the liquid above the surface. To help us define the surface tension we use the apparatus shown in Figure 3. It consists of a C-shaped wire frame, on which is mounted a wire that can slide with negligible friction. The frame and sliding wire contain a thin film of liquid. Because surface tension causes the liquid surface to contract, a force F is needed to move the slider to the right and extend the surface. The surface tension is denoted by the Greek letter gamma (γ) and, as indicated by Equation 1, is the magnitude F of the force per unit length over which it acts. Table 1 gives the value of the surface tension for some typical materials. of 9

3 DEFINITION OF SURFACE TENSION The surface tension γ is the magnitude F of the force exerted parallel to the surface of a liquid divided by the length L of the line over which the force acts: γ = F L (1) SI Unit of Surface Tension: N/m For the specific case illustrated in Figure 3, there is an upper surface and a lower surface, as the blow-up drawing indicates. Thus, the force F acts along a total length of L = ", where " is the length of the slider. Example 1 deals with a demonstration of the effects of surface tension that you can try yourself. Upper film surface Lower film surface Figure 3 This apparatus, consisting of a C-shaped wire frame and a wire slider, can be used to measure the surface tension of a liquid. Table 1 Surface Tensions of Common Liquids Liquid Surface Tension γ (N/m) Benzene (0 C) 0.09 Blood (37 C) Glycerin (0 C) Mercury (0 C) 0.47 Water (0 C) Water (100 C) EXAMPLE 1 Floating a Needle on the Surface of Water A needle has a length of 3. cm. When placed gently on the surface of the water (γ = N/m) in a glass, this needle will float if it is not too heavy. What is the weight of the heaviest needle that can be used in this demonstration? 3 of 9

4 Reasoning As the end view in Figure 4 shows, three forces act on the needle, its weight W and the two forces F 1 and F due to the surface tension of the water. The forces F 1 and F result from the surface tension acting along the length of the needle on either side. According to Equation 1, they have the same magnitude F 1 = F = γ L, where γ = N/m is the surface tension of water and L is the length of the needle. F 1 and F are each tangent to the indented water surface that is formed when the needle presses on the surface, with the result that each acts at an angle θ with respect to the vertical. The needle floats in equilibrium. Therefore, the net force ΣF acting on the needle is zero. In the vertical direction this means that the sum of the vertical components of F 1, F, and W equals zero. Solution Applying the fact that the net force acting on the needle is zero we have Vertical component of F ΣF = 0 ( 49. ) W + γ L cosθ + γ L cosθ = W = γ L cosθ Vertical component of F In other words, the sum of the vertical components of F 1 and F balances the weight of the needle. The forces due to the surface tension will balance the largest weight when they point completely vertically and θ = 0. Therefore, the weight of the heaviest needle that can be used in this demonstration is W = γ L cos θ = N/m m cos 0 = N End view of needle F 1 F 1 cos θ θ θ F F cos θ W Figure 4 A needle can float on a water surface, because the surface tension of the water can lead to forces strong enough to support the needle s weight. 4 of 9

5 Capillary Action We have seen that surface tension arises because of the intermolecular forces of attraction that molecules in a liquid exert on one another. These forces, which are between like molecules, are called cohesive forces. A liquid, however, is often in contact with a solid surface, such as glass. Then additional forces of attraction come into play. They occur between molecules of the liquid and molecules of the solid surface and, being between unlike molecules, are called adhesive forces. Consider a tube with a very small diameter, which is called a capillary. When a capillary, open at both ends, is inserted into a liquid, the result of the competition between cohesive and adhesive forces can be observed. For instance, Figure 5 shows a glass capillary inserted into water. In this case, the adhesive forces are stronger than the cohesive forces, so that the water molecules are attracted to the glass more strongly than to each other. The result is that the water surface curves upward against the glass. It is said that the water wets the glass. The surface tension leads to a force F acting on the circular boundary between the water and the glass. This force is oriented at an angle φ, which is determined by the competition between the cohesive and adhesive forces. The vertical component of F pulls the water up into the tube to a height h. At this height the vertical component of F balances the weight of the column of water of length h. F φ φ F h Figure 5 Water rises in a glass capillary due to the surface tension of the water and the fact that the water wets the glass surface. Figure 6 shows a glass capillary inserted into mercury, a situation in which the adhesive forces are weaker than the cohesive forces. The mercury atoms are attracted to each other more strongly than they are to the glass. As a result, the mercury surface curves downward against the glass and the mercury does not wet the glass. Now, in contrast to the situation illustrated in Figure 5, the surface tension leads to a force F, the vertical component of which pulls the mercury down a distance h in the tube. The behavior of the liquids in both Figures 5 and 6 is called capillary action. 5 of 9

6 h F F Figure 6 Mercury falls in a glass capillary due to the surface tension of the mercury and the fact that the mercury does not wet the glass surface. Pressure Inside a Soap Bubble and a Liquid Drop Anyone who s blown up a balloon has probably noticed that the air pressure inside the balloon is greater than on the outside. For instance, if the balloon is suddenly released, the greater inner pressure forces the air out, propelling the balloon much like a rocket. The reason for the greater pressure is that the tension in the stretched rubber tends to contract the balloon. To counteract this tendency, the balloon has a greater interior air pressure acting to expand the balloon. A soap bubble (see Figure 7a) has two spherical surfaces (inside and outside) with a thin layer of liquid in-between. Like a balloon, the pressure inside a soap bubble is greater than that on the outside. As we will see shortly, this difference in pressure depends on the surface tension γ of the liquid and the radius R of the bubble. For the sake of simplicity, let s assume that there is no pressure on the outside of the bubble (P o = 0). Now, imagine that the stationary soap bubble is cut into two halves. Being at rest, each half has no acceleration and so is in equilibrium. According to Newton s second law of motion (see Section 4.11), a zero acceleration implies that the net force acting on each half must be zero (ΣF = 0). We will now use this equilibrium relation to obtain an expression relating the interior pressure to the surface tension and the radius of the bubble. Forces due to inside air pressure Projected area = π R P o P i R (a) Spherical soap bubble Forces due to surface tension (b) Free-body diagram for right-half of soap bubble Figure 7 (a) The inner and outer pressures on the spherical soap bubble are P i and P o, respectively. (b) The forces pointing to the left are due to the surface tension. The forces pointing perpendicular to the hemispherical surface are due to the air pressure inside the bubble. 6 of 9

7 Figure 7b shows a free-body diagram for the right half of the bubble, on which two forces act. First, there is the force due to the surface tension in the film. This force is exerted on the right half of the bubble by the left half. The surface tension force points to the left and acts all along the circular edge of the hemispherical film. The magnitude of the force due to each surface of the film is the product of the surface tension γ and the circumference (π R) of the circular edge, or γ (π R). The total force due to the inner and outer surfaces is twice this amount or γ (π R). We have included the minus sign to denote that this force points to the left in the drawing. We have also assumed the film to be sufficiently thin enough that its inner and outer radii are nearly the same. Second, there is a force caused by the air pressure inside the bubble. At each point on the surface of the bubble, the force due to the air pressure is perpendicular to the surface and is directed outward. Figure 7b shows this force at six points on the surface. When these forces are added to obtain the total force due to the air pressure, all the components cancel, except those pointing to the right. The total force due to all the components pointing to the right is equal to the product of the pressure P i inside the bubble times the circular cross-sectional area of the hemisphere, or P i (π R ). Using these expressions for the forces due to the surface tension and air pressure, we can write Newton s second law of motion as: ΣF = 0 ( 49. ) γ( πr) + Pi ( πr ) = 0 Force due to surface tension Force due to pressure inside bubble Solving this equation for the pressure inside the bubble gives P i = 4γ /R. In general, the pressure P o outside the bubble is not zero. However, this result still gives the difference between the inside and outside pressures, so that we have Pi Po = 4γ (Spherical soap bubble) () R This result tells us that the difference in pressure depends on the surface tension and the radius of the sphere. What is surprising is that a greater pressure exists inside a smaller soap bubble (smaller value of R) than inside a larger one. A spherical drop of liquid, like a drop of water, has only one surface, rather than two surfaces, for there is no air within it. Thus, the force due to the surface tension is only one-half as large as that in a bubble. Consequently, the difference in pressure between the inside and outside of a liquid drop is one-half of that for a soap bubble: Pi Po = γ (Spherical liquid drop) (3) R 7 of 9

8 Equation 3 is known as Laplace s law for a spherical liquid drop, after the French physicist and mathematician Marquis Pierre Simon delaplace ( ). This result also holds for a spherical bubble in a liquid, such as a gas bubble inside a glass of beer. However, the surface tension γ is that of the surrounding liquid in which the trapped bubble resides. Example illustrates the pressure difference for a soap bubble and a liquid drop. EXAMPLE A Soap Bubble and a Liquid Drop (a) A student, using a circular loop of wire and a pan of soapy water, produces a soap bubble whose radius is 1.0 mm. The surface tension of the soapy water is γ =.5 10 N/m. Determine the pressure difference between the inside and outside of the bubble. (b) The same soapy water is used to produce a spherical droplet whose radius is one-half that of the bubble, or 0.50 mm. Find the pressure difference between the inside and outside of the droplet. Reasoning If the bubble and drop had the same radius, we would expect that the pressure difference between the inside and outside of the bubble to be twice as large as that for the drop. The reason is that the bubble has two surfaces, whereas the drop has only one. Thus, the bubble would have twice the force due to surface tension, and so the pressure inside the bubble would have to be twice as large to counteract this larger force. In fact, however, the bubble has twice the radius compared to the drop. The doubled radius means that the bubble has one-half the pressure difference. Consequently, we expect the larger bubble and smaller drop to have the same pressure difference. Solution (a) The pressure difference, P i P o, between the inside and outside of the soap bubble is given by Equation as 3 8 4γ N/m Pi Po = = 3 R m = N/m (b) The pressure difference, P i P o, between the inside and outside of the drop is given by Equation 3 as 3 8 γ N/m Pi Po = = 3 R m = N/m 8 of 9

9 PROBLEMS 1. A sliding wire of length 3.5 cm is pulling a liquid film, as Figure 3 shows. The pulling force exerted by the wire is N. From the table of surface tensions, determine the film material.. A circular ring (radius = 5.0 cm) is used to determine the surface tension of a liquid. The plane of the ring is positioned so that it is parallel to the surface of the liquid. The ring is immersed in the liquid and then pulled upward, so a film is formed between the ring and the liquid. In addition to the ring s weight, an upward force of N is required to lift the ring to the point where it just breaks free of the surface. What is the surface tension of the liquid? 3. Suppose that the C-shaped wire frame in Figure 3 is rotated clockwise by 90, so that the sliding wire would fall down if it were free to do so. The sliding wire has a length of 3.5 cm and a mass of 0.0 g. There is no friction between the sliding wire and the vertical sides of the C-shaped wire. If the sliding wire is in equilibrium, what is the surface tension of the film? 4. Two soap bubbles (γ = 0.05 N/m) have radii of.0 and 4.5 mm. For each bubble, determine the difference between the inside and outside pressures. 5. Α small bubble of air in water (γ = N/m) has a radius of 0.10 mm. Find the difference in pressures between the inside and outside of the bubble. *6. A drop of oil (γ = N/m) has a radius of mm. The drop is located a distance of.55 m below the surface of fresh water. The atmospheric pressure above the water is Pa. (a) What is the absolute pressure in the water at this depth? (b) Determine the absolute pressure inside the oil drop. **7. Suppose that a bubble has the shape of a long cylinder, rather than that of a sphere. Determine an expression for the difference between the inside and outside pressures; express your answer in terms of the surface tension γ and the radius R of the cylinder. (Hint: Review the reasoning that was used to obtain the difference in pressures for a spherical soap bubble. For the cylindrical bubble, cut the cylinder into two halves by slicing along a line that is parallel to the axis of the cylinder.) 9 of 9

### Fluid Mechanics: Static s Kinematics Dynamics Fluid

Fluid Mechanics: Fluid mechanics may be defined as that branch of engineering science that deals with the behavior of fluid under the condition of rest and motion Fluid mechanics may be divided into three

### Physics 1114: Unit 6 Homework: Answers

Physics 1114: Unit 6 Homework: Answers Problem set 1 1. A rod 4.2 m long and 0.50 cm 2 in cross-sectional area is stretched 0.20 cm under a tension of 12,000 N. a) The stress is the Force (1.2 10 4 N)

### Lecture 24 - Surface tension, viscous flow, thermodynamics

Lecture 24 - Surface tension, viscous flow, thermodynamics Surface tension, surface energy The atoms at the surface of a solid or liquid are not happy. Their bonding is less ideal than the bonding of atoms

### Ch 2 Properties of Fluids - II. Ideal Fluids. Real Fluids. Viscosity (1) Viscosity (3) Viscosity (2)

Ch 2 Properties of Fluids - II Ideal Fluids 1 Prepared for CEE 3500 CEE Fluid Mechanics by Gilberto E. Urroz, August 2005 2 Ideal fluid: a fluid with no friction Also referred to as an inviscid (zero viscosity)

### FLUID FORCES ON CURVED SURFACES; BUOYANCY

FLUID FORCES ON CURVED SURFCES; BUOYNCY The principles applicable to analysis of pressure-induced forces on planar surfaces are directly applicable to curved surfaces. s before, the total force on the

### Chapter 4. Forces and Newton s Laws of Motion. continued

Chapter 4 Forces and Newton s Laws of Motion continued 4.9 Static and Kinetic Frictional Forces When an object is in contact with a surface forces can act on the objects. The component of this force acting

### oil liquid water water liquid Answer, Key Homework 2 David McIntyre 1

Answer, Key Homework 2 David McIntyre 1 This print-out should have 14 questions, check that it is complete. Multiple-choice questions may continue on the next column or page: find all choices before making

### Stack Contents. Pressure Vessels: 1. A Vertical Cut Plane. Pressure Filled Cylinder

Pressure Vessels: 1 Stack Contents Longitudinal Stress in Cylinders Hoop Stress in Cylinders Hoop Stress in Spheres Vanishingly Small Element Radial Stress End Conditions 1 2 Pressure Filled Cylinder A

### Physics 181- Summer 2011 - Experiment #8 1 Experiment #8, Measurement of Density and Archimedes' Principle

Physics 181- Summer 2011 - Experiment #8 1 Experiment #8, Measurement of Density and Archimedes' Principle 1 Purpose 1. To determine the density of a fluid, such as water, by measurement of its mass when

### Surface Tension. the surface tension of a liquid is the energy required to increase the surface area a given amount

Tro, Chemistry: A Molecular Approach 1 Surface Tension surface tension is a property of liquids that results from the tendency of liquids to minimize their surface area in order to minimize their surface

### Mercury is poured into a U-tube as in Figure (14.18a). The left arm of the tube has crosssectional

Chapter 14 Fluid Mechanics. Solutions of Selected Problems 14.1 Problem 14.18 (In the text book) Mercury is poured into a U-tube as in Figure (14.18a). The left arm of the tube has crosssectional area

### Solution Derivations for Capa #11

Solution Derivations for Capa #11 1) A horizontal circular platform (M = 128.1 kg, r = 3.11 m) rotates about a frictionless vertical axle. A student (m = 68.3 kg) walks slowly from the rim of the platform

### Weight The weight of an object is defined as the gravitational force acting on the object. Unit: Newton (N)

Gravitational Field A gravitational field as a region in which an object experiences a force due to gravitational attraction Gravitational Field Strength The gravitational field strength at a point in

### Accelerometers: Theory and Operation

12-3776C Accelerometers: Theory and Operation The Vertical Accelerometer Accelerometers measure accelerations by measuring forces. The vertical accelerometer in this kit consists of a lead sinker hung

### Chapter 3.8 & 6 Solutions

Chapter 3.8 & 6 Solutions P3.37. Prepare: We are asked to find period, speed and acceleration. Period and frequency are inverses according to Equation 3.26. To find speed we need to know the distance traveled

### MECHANICS OF SOLIDS - BEAMS TUTORIAL 2 SHEAR FORCE AND BENDING MOMENTS IN BEAMS

MECHANICS OF SOLIDS - BEAMS TUTORIAL 2 SHEAR FORCE AND BENDING MOMENTS IN BEAMS This is the second tutorial on bending of beams. You should judge your progress by completing the self assessment exercises.

### Physics 112 Homework 5 (solutions) (2004 Fall) Solutions to Homework Questions 5

Solutions to Homework Questions 5 Chapt19, Problem-2: (a) Find the direction of the force on a proton (a positively charged particle) moving through the magnetic fields in Figure P19.2, as shown. (b) Repeat

### Chapter 10 Rotational Motion. Copyright 2009 Pearson Education, Inc.

Chapter 10 Rotational Motion Angular Quantities Units of Chapter 10 Vector Nature of Angular Quantities Constant Angular Acceleration Torque Rotational Dynamics; Torque and Rotational Inertia Solving Problems

### Chapter 13 - Solutions

= Chapter 13 - Solutions Description: Find the weight of a cylindrical iron rod given its area and length and the density of iron. Part A On a part-time job you are asked to bring a cylindrical iron rod

### Physics 2A, Sec B00: Mechanics -- Winter 2011 Instructor: B. Grinstein Final Exam

Physics 2A, Sec B00: Mechanics -- Winter 2011 Instructor: B. Grinstein Final Exam INSTRUCTIONS: Use a pencil #2 to fill your scantron. Write your code number and bubble it in under "EXAM NUMBER;" an entry

### p atmospheric Statics : Pressure Hydrostatic Pressure: linear change in pressure with depth Measure depth, h, from free surface Pressure Head p gh

IVE1400: n Introduction to Fluid Mechanics Statics : Pressure : Statics r P Sleigh: P..Sleigh@leeds.ac.uk r J Noakes:.J.Noakes@leeds.ac.uk January 008 Module web site: www.efm.leeds.ac.uk/ive/fluidslevel1

### PHY121 #8 Midterm I 3.06.2013

PHY11 #8 Midterm I 3.06.013 AP Physics- Newton s Laws AP Exam Multiple Choice Questions #1 #4 1. When the frictionless system shown above is accelerated by an applied force of magnitude F, the tension

### CHAPTER 3: FORCES AND PRESSURE

CHAPTER 3: FORCES AND PRESSURE 3.1 UNDERSTANDING PRESSURE 1. The pressure acting on a surface is defined as.. force per unit. area on the surface. 2. Pressure, P = F A 3. Unit for pressure is. Nm -2 or

### If you put the same book on a tilted surface the normal force will be less. The magnitude of the normal force will equal: N = W cos θ

Experiment 4 ormal and Frictional Forces Preparation Prepare for this week's quiz by reviewing last week's experiment Read this week's experiment and the section in your textbook dealing with normal forces

### Physics 201 Homework 8

Physics 201 Homework 8 Feb 27, 2013 1. A ceiling fan is turned on and a net torque of 1.8 N-m is applied to the blades. 8.2 rad/s 2 The blades have a total moment of inertia of 0.22 kg-m 2. What is the

### Experiment 9. The Pendulum

Experiment 9 The Pendulum 9.1 Objectives Investigate the functional dependence of the period (τ) 1 of a pendulum on its length (L), the mass of its bob (m), and the starting angle (θ 0 ). Use a pendulum

### Chapter 22: Electric Flux and Gauss s Law

22.1 ntroduction We have seen in chapter 21 that determining the electric field of a continuous charge distribution can become very complicated for some charge distributions. t would be desirable if we

### E/M Experiment: Electrons in a Magnetic Field.

E/M Experiment: Electrons in a Magnetic Field. PRE-LAB You will be doing this experiment before we cover the relevant material in class. But there are only two fundamental concepts that you need to understand.

### Centripetal Force. This result is independent of the size of r. A full circle has 2π rad, and 360 deg = 2π rad.

Centripetal Force 1 Introduction In classical mechanics, the dynamics of a point particle are described by Newton s 2nd law, F = m a, where F is the net force, m is the mass, and a is the acceleration.

### Unit 4 Practice Test: Rotational Motion

Unit 4 Practice Test: Rotational Motion Multiple Guess Identify the letter of the choice that best completes the statement or answers the question. 1. How would an angle in radians be converted to an angle

### 1. A wire carries 15 A. You form the wire into a single-turn circular loop with magnetic field 80 µ T at the loop center. What is the loop radius?

CHAPTER 3 SOURCES O THE MAGNETC ELD 1. A wire carries 15 A. You form the wire into a single-turn circular loop with magnetic field 8 µ T at the loop center. What is the loop radius? Equation 3-3, with

### Acceleration due to Gravity

Acceleration due to Gravity 1 Object To determine the acceleration due to gravity by different methods. 2 Apparatus Balance, ball bearing, clamps, electric timers, meter stick, paper strips, precision

### A drop forms when liquid is forced out of a small tube. The shape of the drop is determined by a balance of pressure, gravity, and surface tension

A drop forms when liquid is forced out of a small tube. The shape of the drop is determined by a balance of pressure, gravity, and surface tension forces. 2 Objectives Have a working knowledge of the basic

### AP Physics - Chapter 8 Practice Test

AP Physics - Chapter 8 Practice Test Multiple Choice Identify the choice that best completes the statement or answers the question. 1. A single conservative force F x = (6.0x 12) N (x is in m) acts on

### XI / PHYSICS FLUIDS IN MOTION 11/PA

Viscosity It is the property of a liquid due to which it flows in the form of layers and each layer opposes the motion of its adjacent layer. Cause of viscosity Consider two neighboring liquid layers A

### Student Outcomes. Lesson Notes. Classwork. Exercises 1 3 (4 minutes)

Student Outcomes Students give an informal derivation of the relationship between the circumference and area of a circle. Students know the formula for the area of a circle and use it to solve problems.

### PHYS 211 FINAL FALL 2004 Form A

1. Two boys with masses of 40 kg and 60 kg are holding onto either end of a 10 m long massless pole which is initially at rest and floating in still water. They pull themselves along the pole toward each

### Rotational Motion: Moment of Inertia

Experiment 8 Rotational Motion: Moment of Inertia 8.1 Objectives Familiarize yourself with the concept of moment of inertia, I, which plays the same role in the description of the rotation of a rigid body

### Eðlisfræði 2, vor 2007

[ Assignment View ] [ Pri Eðlisfræði 2, vor 2007 28. Sources of Magnetic Field Assignment is due at 2:00am on Wednesday, March 7, 2007 Credit for problems submitted late will decrease to 0% after the deadline

### FRICTION, WORK, AND THE INCLINED PLANE

FRICTION, WORK, AND THE INCLINED PLANE Objective: To measure the coefficient of static and inetic friction between a bloc and an inclined plane and to examine the relationship between the plane s angle

### Chapter 19. Mensuration of Sphere

8 Chapter 19 19.1 Sphere: A sphere is a solid bounded by a closed surface every point of which is equidistant from a fixed point called the centre. Most familiar examples of a sphere are baseball, tennis

### Math 1B, lecture 5: area and volume

Math B, lecture 5: area and volume Nathan Pflueger 6 September 2 Introduction This lecture and the next will be concerned with the computation of areas of regions in the plane, and volumes of regions in

### Serway_ISM_V1 1 Chapter 4

Serway_ISM_V1 1 Chapter 4 ANSWERS TO MULTIPLE CHOICE QUESTIONS 1. Newton s second law gives the net force acting on the crate as This gives the kinetic friction force as, so choice (a) is correct. 2. As

### Physics 25 Exam 3 November 3, 2009

1. A long, straight wire carries a current I. If the magnetic field at a distance d from the wire has magnitude B, what would be the the magnitude of the magnetic field at a distance d/3 from the wire,

### When the fluid velocity is zero, called the hydrostatic condition, the pressure variation is due only to the weight of the fluid.

Fluid Statics When the fluid velocity is zero, called the hydrostatic condition, the pressure variation is due only to the weight of the fluid. Consider a small wedge of fluid at rest of size Δx, Δz, Δs

### Lecture 12: Heterogeneous Nucleation: a surface catalyzed process

Lecture 1: Heterogeneous Nucleation: a surface catalyzed process Today s topics What is heterogeneous nucleation? What implied in real practice of materials processing and phase transformation? Heterogeneous

### Awell-known lecture demonstration1

Acceleration of a Pulled Spool Carl E. Mungan, Physics Department, U.S. Naval Academy, Annapolis, MD 40-506; mungan@usna.edu Awell-known lecture demonstration consists of pulling a spool by the free end

### Geometric Optics Converging Lenses and Mirrors Physics Lab IV

Objective Geometric Optics Converging Lenses and Mirrors Physics Lab IV In this set of lab exercises, the basic properties geometric optics concerning converging lenses and mirrors will be explored. The

### Temperature Measure of KE At the same temperature, heavier molecules have less speed Absolute Zero -273 o C 0 K

Temperature Measure of KE At the same temperature, heavier molecules have less speed Absolute Zero -273 o C 0 K Kinetic Molecular Theory of Gases 1. Large number of atoms/molecules in random motion 2.

### Grade 7 & 8 Math Circles Circles, Circles, Circles March 19/20, 2013

Faculty of Mathematics Waterloo, Ontario N2L 3G Introduction Grade 7 & 8 Math Circles Circles, Circles, Circles March 9/20, 203 The circle is a very important shape. In fact of all shapes, the circle is

### Chapter 9 Circular Motion Dynamics

Chapter 9 Circular Motion Dynamics 9. Introduction Newton s Second Law and Circular Motion... 9. Universal Law of Gravitation and the Circular Orbit of the Moon... 9.. Universal Law of Gravitation... 3

### 4. How many integers between 2004 and 4002 are perfect squares?

5 is 0% of what number? What is the value of + 3 4 + 99 00? (alternating signs) 3 A frog is at the bottom of a well 0 feet deep It climbs up 3 feet every day, but slides back feet each night If it started

### AS COMPETITION PAPER 2008

AS COMPETITION PAPER 28 Name School Town & County Total Mark/5 Time Allowed: One hour Attempt as many questions as you can. Write your answers on this question paper. Marks allocated for each question

### v v ax v a x a v a v = = = Since F = ma, it follows that a = F/m. The mass of the arrow is unchanged, and ( )

Week 3 homework IMPORTANT NOTE ABOUT WEBASSIGN: In the WebAssign versions of these problems, various details have been changed, so that the answers will come out differently. The method to find the solution

### Basic Principles in Microfluidics

Basic Principles in Microfluidics 1 Newton s Second Law for Fluidics Newton s 2 nd Law (F= ma) : Time rate of change of momentum of a system equal to net force acting on system!f = dp dt Sum of forces

### ENGINEERING SCIENCE H1 OUTCOME 1 - TUTORIAL 3 BENDING MOMENTS EDEXCEL HNC/D ENGINEERING SCIENCE LEVEL 4 H1 FORMERLY UNIT 21718P

ENGINEERING SCIENCE H1 OUTCOME 1 - TUTORIAL 3 BENDING MOMENTS EDEXCEL HNC/D ENGINEERING SCIENCE LEVEL 4 H1 FORMERLY UNIT 21718P This material is duplicated in the Mechanical Principles module H2 and those

### Conceptual Questions: Forces and Newton s Laws

Conceptual Questions: Forces and Newton s Laws 1. An object can have motion only if a net force acts on it. his statement is a. true b. false 2. And the reason for this (refer to previous question) is

### Vatten(byggnad) VVR145 Vatten. 2. Vätskors egenskaper (1.1, 4.1 och 2.8) (Föreläsningsanteckningar)

Vatten(byggnad) Vätskors egenskaper (1) Hydrostatik (3) Grundläggande ekvationer (5) Rörströmning (4) 2. Vätskors egenskaper (1.1, 4.1 och 2.8) (Föreläsningsanteckningar) Vätska som kontinuerligt medium

### 9. The kinetic energy of the moving object is (1) 5 J (3) 15 J (2) 10 J (4) 50 J

1. If the kinetic energy of an object is 16 joules when its speed is 4.0 meters per second, then the mass of the objects is (1) 0.5 kg (3) 8.0 kg (2) 2.0 kg (4) 19.6 kg Base your answers to questions 9

### Physics 210 Q1 2012 ( PHYSICS210BRIDGE ) My Courses Course Settings

1 of 11 9/7/2012 1:06 PM Logged in as Julie Alexander, Instructor Help Log Out Physics 210 Q1 2012 ( PHYSICS210BRIDGE ) My Courses Course Settings Course Home Assignments Roster Gradebook Item Library

### Chapter 18 Static Equilibrium

Chapter 8 Static Equilibrium 8. Introduction Static Equilibrium... 8. Lever Law... Example 8. Lever Law... 4 8.3 Generalized Lever Law... 5 8.4 Worked Examples... 7 Example 8. Suspended Rod... 7 Example

### SURFACE AREAS AND VOLUMES

CHAPTER 1 SURFACE AREAS AND VOLUMES (A) Main Concepts and Results Cuboid whose length l, breadth b and height h (a) Volume of cuboid lbh (b) Total surface area of cuboid 2 ( lb + bh + hl ) (c) Lateral

### 01 The Nature of Fluids

01 The Nature of Fluids WRI 1/17 01 The Nature of Fluids (Water Resources I) Dave Morgan Prepared using Lyx, and the Beamer class in L A TEX 2ε, on September 12, 2007 Recommended Text 01 The Nature of

### When showing forces on diagrams, it is important to show the directions in which they act as well as their magnitudes.

When showing forces on diagrams, it is important to show the directions in which they act as well as their magnitudes. mass M, the force of attraction exerted by the Earth on an object, acts downwards.

### Physics 111: Lecture 4: Chapter 4 - Forces and Newton s Laws of Motion. Physics is about forces and how the world around us reacts to these forces.

Physics 111: Lecture 4: Chapter 4 - Forces and Newton s Laws of Motion Physics is about forces and how the world around us reacts to these forces. Whats a force? Contact and non-contact forces. Whats a

### Magnetism. d. gives the direction of the force on a charge moving in a magnetic field. b. results in negative charges moving. clockwise.

Magnetism 1. An electron which moves with a speed of 3.0 10 4 m/s parallel to a uniform magnetic field of 0.40 T experiences a force of what magnitude? (e = 1.6 10 19 C) a. 4.8 10 14 N c. 2.2 10 24 N b.

### 8.012 Physics I: Classical Mechanics Fall 2008

MIT OpenCourseWare http://ocw.mit.edu 8.012 Physics I: Classical Mechanics Fall 2008 For information about citing these materials or our Terms of Use, visit: http://ocw.mit.edu/terms. MASSACHUSETTS INSTITUTE

### Physics 9e/Cutnell. correlated to the. College Board AP Physics 1 Course Objectives

Physics 9e/Cutnell correlated to the College Board AP Physics 1 Course Objectives Big Idea 1: Objects and systems have properties such as mass and charge. Systems may have internal structure. Enduring

### Objective: Equilibrium Applications of Newton s Laws of Motion I

Type: Single Date: Objective: Equilibrium Applications of Newton s Laws of Motion I Homework: Assignment (1-11) Read (4.1-4.5, 4.8, 4.11); Do PROB # s (46, 47, 52, 58) Ch. 4 AP Physics B Mr. Mirro Equilibrium,

### Newton s Law of Motion

chapter 5 Newton s Law of Motion Static system 1. Hanging two identical masses Context in the textbook: Section 5.3, combination of forces, Example 4. Vertical motion without friction 2. Elevator: Decelerating

### C B A T 3 T 2 T 1. 1. What is the magnitude of the force T 1? A) 37.5 N B) 75.0 N C) 113 N D) 157 N E) 192 N

Three boxes are connected by massless strings and are resting on a frictionless table. Each box has a mass of 15 kg, and the tension T 1 in the right string is accelerating the boxes to the right at a

### Chapter 21. Magnetic Forces and Magnetic Fields

Chapter 21 Magnetic Forces and Magnetic Fields 21.1 Magnetic Fields The needle of a compass is permanent magnet that has a north magnetic pole (N) at one end and a south magnetic pole (S) at the other.

### PERIMETER AND AREA. In this unit, we will develop and apply the formulas for the perimeter and area of various two-dimensional figures.

PERIMETER AND AREA In this unit, we will develop and apply the formulas for the perimeter and area of various two-dimensional figures. Perimeter Perimeter The perimeter of a polygon, denoted by P, is the

### SOLID MECHANICS TUTORIAL MECHANISMS KINEMATICS - VELOCITY AND ACCELERATION DIAGRAMS

SOLID MECHANICS TUTORIAL MECHANISMS KINEMATICS - VELOCITY AND ACCELERATION DIAGRAMS This work covers elements of the syllabus for the Engineering Council exams C105 Mechanical and Structural Engineering

### Chapter 4. Electrostatic Fields in Matter

Chapter 4. Electrostatic Fields in Matter 4.1. Polarization A neutral atom, placed in an external electric field, will experience no net force. However, even though the atom as a whole is neutral, the

### Review Questions PHYS 2426 Exam 2

Review Questions PHYS 2426 Exam 2 1. If 4.7 x 10 16 electrons pass a particular point in a wire every second, what is the current in the wire? A) 4.7 ma B) 7.5 A C) 2.9 A D) 7.5 ma E) 0.29 A Ans: D 2.

### CHAPTER 29 VOLUMES AND SURFACE AREAS OF COMMON SOLIDS

CHAPTER 9 VOLUMES AND SURFACE AREAS OF COMMON EXERCISE 14 Page 9 SOLIDS 1. Change a volume of 1 00 000 cm to cubic metres. 1m = 10 cm or 1cm = 10 6m 6 Hence, 1 00 000 cm = 1 00 000 10 6m = 1. m. Change

### OUTCOME 1 STATIC FLUID SYSTEMS TUTORIAL 1 - HYDROSTATICS

Unit 41: Fluid Mechanics Unit code: T/601/1445 QCF Level: 4 Credit value: 15 OUTCOME 1 STATIC FLUID SYSTEMS TUTORIAL 1 - HYDROSTATICS 1. Be able to determine the behavioural characteristics and parameters

### www.mathsbox.org.uk Displacement (x) Velocity (v) Acceleration (a) x = f(t) differentiate v = dx Acceleration Velocity (v) Displacement x

Mechanics 2 : Revision Notes 1. Kinematics and variable acceleration Displacement (x) Velocity (v) Acceleration (a) x = f(t) differentiate v = dx differentiate a = dv = d2 x dt dt dt 2 Acceleration Velocity

### HW6 Solutions Notice numbers may change randomly in your assignments and you may have to recalculate solutions for your specific case.

HW6 Solutions Notice numbers may change randomly in your assignments and you may have to recalculate solutions for your specific case. Tipler 22.P.053 The figure below shows a portion of an infinitely

### FLUID MECHANICS. TUTORIAL No.7 FLUID FORCES. When you have completed this tutorial you should be able to. Solve forces due to pressure difference.

FLUID MECHANICS TUTORIAL No.7 FLUID FORCES When you have completed this tutorial you should be able to Solve forces due to pressure difference. Solve problems due to momentum changes. Solve problems involving

### Structural Axial, Shear and Bending Moments

Structural Axial, Shear and Bending Moments Positive Internal Forces Acting Recall from mechanics of materials that the internal forces P (generic axial), V (shear) and M (moment) represent resultants

### Practice Test SHM with Answers

Practice Test SHM with Answers MPC 1) If we double the frequency of a system undergoing simple harmonic motion, which of the following statements about that system are true? (There could be more than one

### 1. Units of a magnetic field might be: A. C m/s B. C s/m C. C/kg D. kg/c s E. N/C m ans: D

Chapter 28: MAGNETIC FIELDS 1 Units of a magnetic field might be: A C m/s B C s/m C C/kg D kg/c s E N/C m 2 In the formula F = q v B: A F must be perpendicular to v but not necessarily to B B F must be

### Lab 7: Rotational Motion

Lab 7: Rotational Motion Equipment: DataStudio, rotary motion sensor mounted on 80 cm rod and heavy duty bench clamp (PASCO ME-9472), string with loop at one end and small white bead at the other end (125

### PHYS 101 Lecture 10 - Work and kinetic energy 10-1

PHYS 101 Lecture 10 - Work and kinetic energy 10-1 Lecture 10 - Work and Kinetic Energy What s important: impulse, work, kinetic energy, potential energy Demonstrations: block on plane balloon with propellor

### Pre-lab Quiz/PHYS 224 Magnetic Force and Current Balance. Your name Lab section

Pre-lab Quiz/PHYS 224 Magnetic Force and Current Balance Your name Lab section 1. What do you investigate in this lab? 2. Two straight wires are in parallel and carry electric currents in opposite directions

### Force on a square loop of current in a uniform B-field.

Force on a square loop of current in a uniform B-field. F top = 0 θ = 0; sinθ = 0; so F B = 0 F bottom = 0 F left = I a B (out of page) F right = I a B (into page) Assume loop is on a frictionless axis

### Aids needed for demonstrations: viscous fluid (water), tubes (pipes), injections, paper, stopwatches, vessels,, weights

1 Viscous and turbulent flow Level: high school (16-17 years) hours (2 hours class teaching, 2 hours practical excercises) Content: 1. Viscous flow 2. Poiseuille s law 3. Passing from laminar to turbulent

### CHAPTER 7: CAPILLARY PRESSURE

CHAPTER 7: CAPILLARY PRESSURE Objective To measure capillary pressure of unconsolidated sand packs. Introduction Capillary pressure is important in reservoir engineering because it is a major factor controlling

### Lecture 6. Weight. Tension. Normal Force. Static Friction. Cutnell+Johnson: 4.8-4.12, second half of section 4.7

Lecture 6 Weight Tension Normal Force Static Friction Cutnell+Johnson: 4.8-4.12, second half of section 4.7 In this lecture, I m going to discuss four different kinds of forces: weight, tension, the normal

### Buoyant Force and Archimedes' Principle

Buoyant Force and Archimedes' Principle Introduction: Buoyant forces keep Supertankers from sinking and party balloons floating. An object that is more dense than a liquid will sink in that liquid. If

### Surface Tension: Liquids Stick Together Teacher Version

Surface Tension: Liquids Stick Together Teacher Version In this lab you will learn about properties of liquids, specifically cohesion, adhesion, and surface tension. These principles will be demonstrated

### Chapter 28 Fluid Dynamics

Chapter 28 Fluid Dynamics 28.1 Ideal Fluids... 1 28.2 Velocity Vector Field... 1 28.3 Mass Continuity Equation... 3 28.4 Bernoulli s Principle... 4 28.5 Worked Examples: Bernoulli s Equation... 7 Example

### Physics 1A Lecture 10C

Physics 1A Lecture 10C "If you neglect to recharge a battery, it dies. And if you run full speed ahead without stopping for water, you lose momentum to finish the race. --Oprah Winfrey Static Equilibrium

### MENSURATION. Definition

MENSURATION Definition 1. Mensuration : It is a branch of mathematics which deals with the lengths of lines, areas of surfaces and volumes of solids. 2. Plane Mensuration : It deals with the sides, perimeters

### PHYSICS 111 HOMEWORK SOLUTION, week 4, chapter 5, sec 1-7. February 13, 2013

PHYSICS 111 HOMEWORK SOLUTION, week 4, chapter 5, sec 1-7 February 13, 2013 0.1 A 2.00-kg object undergoes an acceleration given by a = (6.00î + 4.00ĵ)m/s 2 a) Find the resultatnt force acting on the object