# Chapter (1) Fluids and their Properties

Save this PDF as:

Size: px
Start display at page:

## Transcription

1 Chapter (1) Fluids and their Properties

2 Fluids (Liquids or gases) which a substance deforms continuously, or flows, when subjected to shearing forces. If a fluid is at rest, there are no shearing forces acting. In fluids we usually deal with continuous streams of fluid without beginning or end. Shear Stress in Moving Particles If fluid is in motion, shear stress are developed this occur if the fluid particles move relative to each other with different velocities. However, if the fluid velocity is the same at every point (fluid particles are at rest relative to each other), no shear stress will be produced. The following figure exhibit the velocity profile in a circular pipe: y v=f(y) Note that fluid next to the pipe wall has zero velocity (fluid sticks to wall), But if the fluid moved away from the wall, velocity increases to maximum. Change in velocity (v) with distance(y) is (velocity gradient): Velocity gradient = dv dy This also called (rate of shear strain) v Page (1)

3 Newton s Law of Viscosity: τ = μ dv dy (Pa = N/m ) τ = shear stress. μ = dynamic viscosity (will be discussed later). dv dy = velocity gradient Newtonian & Non-Newtonian Fluid Newtonian Fluids: Fluids obey (تتبع) Newton s law of viscosity are Newtonian fluids. For this type of fluids, there is a linear relationship between shear stress and the velocity gradient. Dynamic viscosity (μ) is the slope of the line. Dynamic viscosity (μ) is constant for a fluid at the same temperature. As temperature increase (μ) decreases slope decreases. Most common fluids are Newtonian, for example: Air, Water, Oil, etc The following graph explains the linear variation of shear stress with rate of shear strain (velocity gradient) for common fluids: Slope = μ Page (2)

4 Non-Newtonian Fluids: Fluids don t obey Newton s law of viscosity are Non-Newtonian fluids. For this type of fluids, there is no linear relationship between shear stress and the velocity gradient and the slope of the curves is varies. There are several types of Non-Newtonian fluids based on the relationship between shear stress and the velocity gradient. The general relationship is τ = A + B, the values of A, B, and n depends on the type of Non-Newtonians fluid. For Newtonian fluids A = 0.0, B = μ, n = 1 Newtonian fluids is a special case from the above equation. The following graph explains the variation of shear stress with rate of shear strain (velocity gradient) for Different types of Non-Newtonian Fluids: Page (3)

5 Properties of Fluids Density: There are many ways of expressing density: 1. Mass Density (ρ): Mass per unit volume ρ = M V (Kg/m ) Where, M = mass of fluid (Kg), V = volume of fluid (m ) Typical values: Water= 1000 Kg/m, Mercury = Kg/m, Air = 1.23Kg/m 2. Specific Weight (γ): also called (unit weight): is the weight per unit volume. γ = ρ g = W V (N/m ) Where, W = weight of fluid(n) = M g, V = volume of fluid (m ) Typical values: Water= 9814 N/m, Mercury = N/m, Air = 12.07Kg/m 3. Relative Density (σ): also called (specific gravity SG ): σ = SG = ρ ρ (Unitless) Typical values: Water = 1, Mercury = Specific Volume (ν): is the reciprocal of mass density. ν = 1 ρ (m /Kg) Page (4)

6 Viscosity: it s the ability of fluid to resist shear deformation due to cohesion and interaction between molecules of fluid. There are two types of viscosity: 1. Coefficient of Dynamic Viscosity (μ): τ = μ dv dy μ = τ Force Force = Area = Area dv Velocity Distance dy Distance Distance Time = Force Time Area But Force = Units: = kg m s μ = = Another Unit: N = Kg = = =. S = Pa. S Note that (μ) is often expressed in Poise (P) where 10P = 1Pa. S 2. Coefficient of Kinematic Viscosity (ν): ν = μ ρ Units: = = Note that (ν) is often expressed in Stokes (St) where 10 St = 1 Surface Tension(σ): is a force per unit length (N/m) pπr σ Page (5)

7 R is the radius of the droplet, σ is the surface tension, p is the pressure difference between the inside and outside pressure. The force developed around the edge due to surface tension along the line: F σ = σ Perimeter = σ 2πR The force produced from the pressure difference p: F p = p Area = p πr 2 (Inside direction, because inside pressure largere thanoutside pressure) Now, these two forces are in equilibrium so we equating them: σ 2πR = p πr p = 2σ R = p p Capillarity Capillary effect: is the rise or fall of a liquid in a small-diameter tube caused by surface tension. h: is the height of water rises. R: is the radius of the tube (d=2r). θ: is the angle of contact between liquid and solid 2πRσ = πdσ θ The weight of the fluid is balanced with the vertical force caused by surface tension. W = m g = ρ V g πdσcos(θ) = ρ π 4 d h g h = 4σ cos(θ) (only for tube) ρgd if any thing else you should derive the equation with the same concept (see p. 12) Notes: If θ < 90 water is rise, If θ > 90 water is fall, If θ = 90 no rise or fall (h = 0.0). If d is too small θ is too small and may be neglected. For clean glass in contact with Water θ = 0.0 (rise). For clean glass in contact with Mercury θ = 130 (fall). Page (6)

8 Compressibility & Bulk Modulus All materials are compressible under the application of an external force. The compressibility of a fluid is expressed by its bulk modulus (K) such that: K = Change in Pressure = p = P P Volumetric Strain = V V V K = p V/V = V V The ve sign refer to the volume decrease as the pressure increase. K can be expressed by another form as following: K = ρ p ρ Large values of bulk modulus refer to incompressibility which indicates the large pressures are needed to compress volume slightly. Famous Unit Conversion 1m= ft 1ft=12 inch 1inch=2.54 cm 1Ib = 4.45 N γ = 9810 g = 9.81 (. ).. = 32.2ft/S To transform from to v = ω r 1slug = Kg. = 62.4 Ib/ft multiply by where, ω = velocity in (ثلث حجم الاسطوانة) Volume of cone (V ) = R h 1m = 10 Liter, 1Liter = 10 ml, 1m = 10 cm., r = radius of rotation (m) 1m = 10 ml. Page (7)

9 Problems 1. If 9 m3 of oil weighs 70.5 KN, Calculate its: A. Specific Weight: γ = Weight Volume = = KN/m = N/m. B. Density: γ(newton s) = ρ g ρ = γ = g 9.81 C. Specific Volume: ν = 1 ρ = = m /Kg. = Kg/m. 2. A cylindrical tube of volume 335 ml is filled with soda. If the Mass of tube when it s filled by soda is 0.37 Kg and the weight of tube when it s empty is N. Determine the specific weight, density, and specific gravity for soda. γ = W V V = V = 335ml = m M = 0.37Kg W = M g = = 3.63 N W = 0.153N W = W W = = 3.47 N 3.47 γ = = N/m. γ = ρ g ρ = γ g = S. G = ρ = = = Kg/m. Page (8)

10 3. A quantity of soda of mass M is filled in a container having volume V. The density of soda in this container is 1005 Kg/m. If the soda is distributed from this container to three similar containers having the same volume V. Calculate specific gravity and specific weight of the soda in each one of three containers after distribution occurs. The mass of soda (M) is the same before and after distribution M = M ( Key) ρ = M V M = ρv ρ V = ρ V (ρ = 1005, V = V, ρ =??, V = 3V) 1005V = ρ 3V 1005 = 3ρ ρ = 335 Kg/m ρ = 335 Kg/m (For soda in each container) S. G = ρ = 335 = (For soda in each container) γ = ρ g = = N/m. 4. A 10-kg block slides down a smooth inclined surface as shown in Figure below. Determine the terminal velocity of the block if the 0.1-mm gap between the block and the surface contains oil having dynamic viscosity of 0.38 Pa.s. Assume the velocity distribution in the gap is linear, and the area of the block in contact with the oil is 0.1 m 2. Page (9)

11 Important Notes: In all problems, the direction of shear force is in reverse direction of movement, because it s the resistance of fluid against moving object. So the shear force resists the movement of the object exactly like friction force (if the object moves above rough surface). The summation of forces equal zero in all directions, because the velocity is constant (we assume it constant) and so the acceleration is zero. Return to the above problem, the free body diagram of the block is: Page (10)

12 F = 0.0 Wsin(20) = τa τ = Wsin(20) A W = Mg = = 98.1 N, A = 0.1 m (given) 98.1 sin (20) τ = = Pa. 0.1 τ = μ v y v = τ y μ = 0.38 = m/s. 5. A cylinder of 40 cm length and 10 cm diameter rotates about a vertical axis inside a fixed cylindrical tube of 105 mm diameter, and 0.4 m length. If the space between the tube and the cylinder is filled with liquid of dynamic viscosity of 0.2 N.s/m 2. Determine the external torque that led the cylinder to rotate by speed of 700 rev. /min. Torque = Shear force radius of rotating cylinder Shear force (F) = τ Side Area of rotating cylinder (A ) τ = μ du dy μ = 0.2N. s/m du = velocity of rotating cylinder in (m/s). ω = 700rev/min = 700 2π 60 = 73.3 rad/s. v = ω radius of rotating cylinder = v = 3.66 m/s. dy = ( always is: distance perpendicular to the direction of shear force) dy = = 2.5mm = m 2 Page (11)

13 τ = = N/m mm A = πdl = π = m Shear force (F) = = 36.7 N. Torque = = N. m. 100 mm 6. A cylindrical body of 70 mm diameter and 150 mm length falls freely in a 80 mm diameter circular tube as shown in the figure below. If the space between the cylindrical body and the tube is filled with oil of viscosity 0.9 poise. Determine the weight of the body when it falls at a speed of 1.5 m/s. Page (12)

14 The weight of cylindrical body (downward) equals the resist shear force (upward) due to the equilibrium. F = 0.0 W = τ A The shear stress caused by cylindrical tube is: τ = μ du dy μ = 0.9 pois, but 10pois = 1pa. s 0.9poise =. = 0.09 pa. s. du = velocity of freely falling = 1.5m/s dy = = 5mm = 0.005m 2 τ = = 27 pa Shear force = side area of cylindrical body τ F = π d L τ = F = 0.89 N The weight of cylindrical body = F = 0.89 N. 7. A movable plate of 0.5m 2 area (for each face) is located between two large fixed plates as shown in the figure below. Two different Newtonian fluids having the viscosities indicated are contained between the plates. Determine the magnitude of the force acting on the movable plate when it moves at a speed of 4.0 m/s. Page (13)

15 The free body diagram of the plate is: τ A τ A A F F = 0.0 F = τ A + τ A F = A (τ + τ ) u = u = 4m/s. u τ = μ = = Pa. y u τ = μ = = Pa. y F = 0.5 ( ) = N. Page (14)

16 8. Fluid flow through a circular pipe is one-dimensional, and the velocity profile for laminar flow is given by u(r) = u 1 where R is the radius of the pipe, r is the radial distance from the center of the pipe, and u is the maximum flow velocity, which occurs at the center. Determine: (a). a relation for the drag force applied by the fluid on a section of the pipe of length L. (b). the value of the drag force for water flow at 20 C with R = 0.08m, L = 15m, u = 3m/s, μ = 0.001kg/m. s. a) The shear stress at the surface of the pipe (at r = R) is given by: τ = μ du (Negative sign is because u decresed with r increased) dr τ = μ d dr u 1 r R (But, u is constant) τ = μu d r 1 dr R = μu 0 2r R (because R is constant) τ = 2rμu R (at the surface of the pipe r = R) τ = 2μu R. Page (15)

17 Note: Always we calculate the shear stress at the fixed surface (like pipe surface in the above problem) because it gives maximum shear stress. The drag force that causes shear stress on the pipe surface is: F = 0.0 F = τ A (A = side area of tube = 2πRL) F = 2μu 2πRL R F = 4πLμu. b) F = 4πLμu F = 4π = N. 9. A layer of water flows down an inclined fixed surface with the velocity profile shown in Figure below. Determine the magnitude of shearing stress that the water exerts on the fixed surface for U= 2m/s and h=0.1m μ = pa. s The shear stress at the fixed surface (at y = 0.0) is given by: τ = +μ du (Positive sign is because u incresed with y increased) dy u U = 2 y h y h u = U 2 y h y h (But, U = 2m/s and h = 0.1m) Page (16)

18 u = 2 2y 0.1 y 0.1 u = 40y 200y du dy = y (But, at the fixed surface y = 0.0) du dy = 40 τ = = N/m. 10. When a 2-mm-diameter tube is inserted into a liquid in an open tank, the liquid is observed to rise 10 mm above the free surface of the liquid. The contact angle between the liquid and the tube is zero, and the specific weight of the liquid is 1.2 x 10 4 N/m 3. Determine the value of the surface tension for this liquid. For tube the capillary rise can be expressed by the relation tha we previously derived: 4σ cos(θ) h = σ = ρgd hρgd 4 cos(θ) h = 0.01m, ρg = γ = N/m, d = 0.002m, θ = 0 σ = cos(0) = 0.06 N/m. 11. Derive an expression for the capillary height change h for a fluid of surface tension σ and contact angle θ between two vertical parallel plates a distance W apart, as in figure. Page (17)

19 Assume the side length of each plate is (b). F = 0.0 F = Weight of water column F = 2 σ b cos(θ) (2 because there are two plates) Weight = M g = ρ V g (V = W h b) Weight = W h b ρ g 2 σ b cos(θ) = W h b ρ g h = 2σ cos(θ) ρgw. 12. Assume that the surface tension of 7.34x10-2 N/m act at an angle θ relative to the water surface as shown in Figure below. a. If the mass of the double-edge blade is 0.64 x 10-3 Kg, and the total length of its sides is 206 mm. Determine the value of θ required to maintain equilibrium between the blade weight and the resultant surface tension force. b. If the mass of the single-edge blade is 2.61 x 10-3 Kg, and the total length of its sides is 154 mm. Explain why this blade sinks. Support your answer with the necessary calculations. σ σ Page (18)

20 a. Note: The surface tension σ exists about all sides of the blade so to calculate the surface tension force F we should multiply the value of σ by the total length of blade sides as following: F = σ total length = = N F = N with inclination of θ with horizontal The weight of the blade is: W = Mg = = N (downward) By Equilibrium F = sin(θ) = θ = b. F = σ total length = = N F, = sin(θ) W = Mg = = N Now, the maximum value of sin(θ) = 1 so, the maximum value of F, = N. Note that the value of W = N (downward) is greater than the value of F, = N so, the blade will sink downward. 13. A rigid cylinder (15 mm inside diameter) contains column of water 500mm length. If the bulk modulus of water is K water = 2.05 x 10 9 N/m 2. What will be the column length if a 2 KN force is applied it s end by frictionless plunger? Assume no leakage. Page (19)

21 K = p V/V K = Pa. p = P P (P = 0.0 because no applied forces in first case) P = Force Area = 2 10 π = Pa. p = = Pa. V V = V V = L A 0.5 A = L 0.5 V 0.5 A 0.5 K = =... L = m. Page (20)

### 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

### 1. Introduction, fluid properties (1.1, and handouts)

1. Introduction, fluid properties (1.1, and handouts) Introduction, general information Course overview Fluids as a continuum Density Compressibility Viscosity Exercises: A1 Applications of fluid mechanics

### 1.4 Review. 1.5 Thermodynamic Properties. CEE 3310 Thermodynamic Properties, Aug. 26,

CEE 3310 Thermodynamic Properties, Aug. 26, 2011 11 1.4 Review A fluid is a substance that can not support a shear stress. Liquids differ from gasses in that liquids that do not completely fill a container

### Higher Technological Institute Civil Engineering Department. Lectures of. Fluid Mechanics. Dr. Amir M. Mobasher

Higher Technological Institute Civil Engineering Department Lectures of Fluid Mechanics Dr. Amir M. Mobasher 1/14/2013 Fluid Mechanics Dr. Amir Mobasher Department of Civil Engineering Faculty of Engineering

### These slides contain some notes, thoughts about what to study, and some practice problems. The answers to the problems are given in the last slide.

Fluid Mechanics FE Review Carrie (CJ) McClelland, P.E. cmcclell@mines.edu Fluid Mechanics FE Review These slides contain some notes, thoughts about what to study, and some practice problems. The answers

### 1. Fluids Mechanics and Fluid Properties. 1.1 Objectives of this section. 1.2 Fluids

1. Fluids Mechanics and Fluid Properties What is fluid mechanics? As its name suggests it is the branch of applied mechanics concerned with the statics and dynamics of fluids - both liquids and gases.

### 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

### 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)

### 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

### CE 3500 Fluid Mechanics / Fall 2014 / City College of New York

1 Drag Coefficient The force ( F ) of the wind blowing against a building is given by F=C D ρu 2 A/2, where U is the wind speed, ρ is density of the air, A the cross-sectional area of the building, and

### 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

### CH-205: Fluid Dynamics

CH-05: Fluid Dynamics nd Year, B.Tech. & Integrated Dual Degree (Chemical Engineering) Solutions of Mid Semester Examination Data Given: Density of water, ρ = 1000 kg/m 3, gravitational acceleration, g

### Fluids and Solids: Fundamentals

Fluids and Solids: Fundamentals We normally recognize three states of matter: solid; liquid and gas. However, liquid and gas are both fluids: in contrast to solids they lack the ability to resist deformation.

Week 8 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

### Solution: (a) For a positively charged particle, the direction of the force is that predicted by the right hand rule. These are:

Problem 1. (a) Find the direction of the force on a proton (a positively charged particle) moving through the magnetic fields as shown in the figure. (b) Repeat part (a), assuming the moving particle is

### Differential Relations for Fluid Flow. Acceleration field of a fluid. The differential equation of mass conservation

Differential Relations for Fluid Flow In this approach, we apply our four basic conservation laws to an infinitesimally small control volume. The differential approach provides point by point details of

### 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

### 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

### 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

### 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

### Physics 200A FINALS Shankar 180mins December 13, 2005 Formulas and figures at the end. Do problems in 4 books as indicated

1 Physics 200A FINALS Shankar 180mins December 13, 2005 Formulas and figures at the end. Do problems in 4 books as indicated I. Book I A camper is trying to boil water. The 55 g aluminum pan has specific

### Stress and Deformation Analysis. Representing Stresses on a Stress Element. Representing Stresses on a Stress Element con t

Stress and Deformation Analysis Material in this lecture was taken from chapter 3 of Representing Stresses on a Stress Element One main goals of stress analysis is to determine the point within a load-carrying

### CE 6303 MECHANICS OF FLUIDS L T P C QUESTION BANK PART - A

CE 6303 MECHANICS OF FLUIDS L T P C QUESTION BANK 3 0 0 3 UNIT I FLUID PROPERTIES AND FLUID STATICS PART - A 1. Define fluid and fluid mechanics. 2. Define real and ideal fluids. 3. Define mass density

### Applied Fluid Mechanics

Applied Fluid Mechanics 1. The Nature of Fluid and the Study of Fluid Mechanics 2. Viscosity of Fluid 3. Pressure Measurement 4. Forces Due to Static Fluid 5. Buoyancy and Stability 6. Flow of Fluid and

### FLUID DYNAMICS. Intrinsic properties of fluids. Fluids behavior under various conditions

FLUID DYNAMICS Intrinsic properties of fluids Fluids behavior under various conditions Methods by which we can manipulate and utilize the fluids to produce desired results TYPES OF FLUID FLOW Laminar or

### 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

### CBE 6333, R. Levicky 1 Review of Fluid Mechanics Terminology

CBE 6333, R. Levicky 1 Review of Fluid Mechanics Terminology The Continuum Hypothesis: We will regard macroscopic behavior of fluids as if the fluids are perfectly continuous in structure. In reality,

### SURFACE TENSION. Definition

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

### Announcements. Dry Friction

Announcements Dry Friction Today s Objectives Understand the characteristics of dry friction Draw a FBD including friction Solve problems involving friction Class Activities Applications Characteristics

### 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

### CE 204 FLUID MECHANICS

CE 204 FLUID MECHANICS Onur AKAY Assistant Professor Okan University Department of Civil Engineering Akfırat Campus 34959 Tuzla-Istanbul/TURKEY Phone: +90-216-677-1630 ext.1974 Fax: +90-216-677-1486 E-mail:

### Chapter 8 Fluid Flow

Chapter 8 Fluid Flow GOALS When you have mastered the contents of this chapter, you will be able to achieve the following goals: Definitions Define each of the following terms, and use it in an operational

### 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

### 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

### The Viscosity of Fluids

Experiment #11 The Viscosity of Fluids References: 1. Your first year physics textbook. 2. D. Tabor, Gases, Liquids and Solids: and Other States of Matter (Cambridge Press, 1991). 3. J.R. Van Wazer et

### Chapter 13, example problems: x (cm) 10.0

Chapter 13, example problems: (13.04) Reading Fig. 13-30 (reproduced on the right): (a) Frequency f = 1/ T = 1/ (16s) = 0.0625 Hz. (since the figure shows that T/2 is 8 s.) (b) The amplitude is 10 cm.

### Homework 4. problems: 5.61, 5.67, 6.63, 13.21

Homework 4 problems: 5.6, 5.67, 6.6,. Problem 5.6 An object of mass M is held in place by an applied force F. and a pulley system as shown in the figure. he pulleys are massless and frictionless. Find

### Physics 172H. Lecture 6 Ball-Spring Model of Solids, Friction. Read

Physics 172H Lecture 6 Ball-Spring Model of Solids, Friction Read 4.1-4.8 Model of solid: chemical bonds d radial force (N) 0 F linear If atoms don t move too far away from equilibrium, force looks like

### Notes on Polymer Rheology Outline

1 Why is rheology important? Examples of its importance Summary of important variables Description of the flow equations Flow regimes - laminar vs. turbulent - Reynolds number - definition of viscosity

### Fluid Mechanics Prof. T. I. Eldho Department of Civil Engineering Indian Institute of Technology, Bombay. Lecture No. # 36 Pipe Flow Systems

Fluid Mechanics Prof. T. I. Eldho Department of Civil Engineering Indian Institute of Technology, Bombay Lecture No. # 36 Pipe Flow Systems Welcome back to the video course on Fluid Mechanics. In today

### Practice Problems on Boundary Layers. Answer(s): D = 107 N D = 152 N. C. Wassgren, Purdue University Page 1 of 17 Last Updated: 2010 Nov 22

BL_01 A thin flat plate 55 by 110 cm is immersed in a 6 m/s stream of SAE 10 oil at 20 C. Compute the total skin friction drag if the stream is parallel to (a) the long side and (b) the short side. D =

### THIS IS A NEW SPECIFICATION

THIS IS A NEW SPECIFICATION ADVANCED SUBSIDIARY GCE PHYSICS A Mechanics G481 *CUP/T63897* Candidates answer on the question paper OCR Supplied Materials: Data, Formulae and Relationships Booklet Other

### Properties of Fluids

CHAPTER Properties of Fluids 1 1.1 INTRODUCTION A fluid can be defined as a substance which deforms or yields continuously when shear stress is applied to it, no matter how small it is. Fluids can be subdivided

### 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

### For Water to Move a driving force is needed

RECALL FIRST CLASS: Q K Head Difference Area Distance between Heads Q 0.01 cm 0.19 m 6cm 0.75cm 1 liter 86400sec 1.17 liter ~ 1 liter sec 0.63 m 1000cm 3 day day day constant head 0.4 m 0.1 m FINE SAND

### PHY231 Section 2, Form A March 22, 2012. 1. Which one of the following statements concerning kinetic energy is true?

1. Which one of the following statements concerning kinetic energy is true? A) Kinetic energy can be measured in watts. B) Kinetic energy is always equal to the potential energy. C) Kinetic energy is always

### 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

### PHY231 Section 1, Form B March 22, 2012

1. A car enters a horizontal, curved roadbed of radius 50 m. The coefficient of static friction between the tires and the roadbed is 0.20. What is the maximum speed with which the car can safely negotiate

### 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

### Lecture 5 Hemodynamics. Description of fluid flow. The equation of continuity

1 Lecture 5 Hemodynamics Description of fluid flow Hydrodynamics is the part of physics, which studies the motion of fluids. It is based on the laws of mechanics. Hemodynamics studies the motion of blood

### Ch 6 Forces. Question: 9 Problems: 3, 5, 13, 23, 29, 31, 37, 41, 45, 47, 55, 79

Ch 6 Forces Question: 9 Problems: 3, 5, 13, 23, 29, 31, 37, 41, 45, 47, 55, 79 Friction When is friction present in ordinary life? - car brakes - driving around a turn - walking - rubbing your hands together

### Viscosity: The Fluids Lab Teacher Version

Viscosity: The Fluids Lab Teacher Version California Science Content Standards: 1. Motion and Forces: Newton's laws predict the motion of most objects. 1b. Students know that when forces are balanced,

### Physics 126 Practice Exam #3 Professor Siegel

Physics 126 Practice Exam #3 Professor Siegel Name: Lab Day: 1. Which one of the following statements concerning the magnetic force on a charged particle in a magnetic field is true? A) The magnetic force

### Rotational inertia (moment of inertia)

Rotational inertia (moment of inertia) Define rotational inertia (moment of inertia) to be I = Σ m i r i 2 or r i : the perpendicular distance between m i and the given rotation axis m 1 m 2 x 1 x 2 Moment

### 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

### Experiment 3 Pipe Friction

EML 316L Experiment 3 Pipe Friction Laboratory Manual Mechanical and Materials Engineering Department College of Engineering FLORIDA INTERNATIONAL UNIVERSITY Nomenclature Symbol Description Unit A cross-sectional

### Unit 1 INTRODUCTION 1.1.Introduction 1.2.Objectives

Structure 1.1.Introduction 1.2.Objectives 1.3.Properties of Fluids 1.4.Viscosity 1.5.Types of Fluids. 1.6.Thermodynamic Properties 1.7.Compressibility 1.8.Surface Tension and Capillarity 1.9.Capillarity

### charge is detonated, causing the smaller glider with mass M, to move off to the right at 5 m/s. What is the

This test covers momentum, impulse, conservation of momentum, elastic collisions, inelastic collisions, perfectly inelastic collisions, 2-D collisions, and center-of-mass, with some problems requiring

### 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.

### CHAPTER 2: LIQUID VISCOSITY MEASUREMENT

CHAPTER 2: LIQUID VISCOSITY MEASUREMENT Objective Calculate viscosity (dynamic or absolute, and kinematic) and determine how this property varies with changes in temperature for a constant-composition

### du u U 0 U dy y b 0 b

BASIC CONCEPTS/DEFINITIONS OF FLUID MECHANICS (by Marios M. Fyrillas) 1. Density (πυκνότητα) Symbol: 3 Units of measure: kg / m Equation: m ( m mass, V volume) V. Pressure (πίεση) Alternative definition:

### 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

### 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

### The Viscosity of Fluids

Experiment #11 The Viscosity of Fluids References: 1. Your first year physics textbook. 2. D. Tabor, Gases, Liquids and Solids: and Other States of Matter (Cambridge Press, 1991). 3. J.R. Van Wazer et

### Physics 41 HW Set 1 Chapter 15

Physics 4 HW Set Chapter 5 Serway 8 th OC:, 4, 7 CQ: 4, 8 P: 4, 5, 8, 8, 0, 9,, 4, 9, 4, 5, 5 Discussion Problems:, 57, 59, 67, 74 OC CQ P: 4, 5, 8, 8, 0, 9,, 4, 9, 4, 5, 5 Discussion Problems:, 57, 59,

### B) 40.8 m C) 19.6 m D) None of the other choices is correct. Answer: B

Practice Test 1 1) Abby throws a ball straight up and times it. She sees that the ball goes by the top of a flagpole after 0.60 s and reaches the level of the top of the pole after a total elapsed time

### I. INTRODUCTION: Phenomenology. N / m. N / m (2) kg G. Let s consider a shear force experiment on a solid cube. (Fig. 1)

I. INTRODUCTION: Phenomenology Let s consider a shear force experiment on a solid cube. (Fig. 1) We can easily verify that for the solid not to move or rotate, all forces acting on the cube must be of

### 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

### HW Set VI page 1 of 9 PHYSICS 1401 (1) homework solutions

HW Set VI page 1 of 9 10-30 A 10 g bullet moving directly upward at 1000 m/s strikes and passes through the center of mass of a 5.0 kg block initially at rest (Fig. 10-33 ). The bullet emerges from the

### 5.2 Rotational Kinematics, Moment of Inertia

5 ANGULAR MOTION 5.2 Rotational Kinematics, Moment of Inertia Name: 5.2 Rotational Kinematics, Moment of Inertia 5.2.1 Rotational Kinematics In (translational) kinematics, we started out with the position

### F N A) 330 N 0.31 B) 310 N 0.33 C) 250 N 0.27 D) 290 N 0.30 E) 370 N 0.26

Physics 23 Exam 2 Spring 2010 Dr. Alward Page 1 1. A 250-N force is directed horizontally as shown to push a 29-kg box up an inclined plane at a constant speed. Determine the magnitude of the normal force,

### University Physics 226N/231N Old Dominion University. Newton s Laws and Forces Examples

University Physics 226N/231N Old Dominion University Newton s Laws and Forces Examples Dr. Todd Satogata (ODU/Jefferson Lab) satogata@jlab.org http://www.toddsatogata.net/2012-odu Wednesday, September

### PHYSICS 111 HOMEWORK SOLUTION #10. April 8, 2013

PHYSICS HOMEWORK SOLUTION #0 April 8, 203 0. Find the net torque on the wheel in the figure below about the axle through O, taking a = 6.0 cm and b = 30.0 cm. A torque that s produced by a force can 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

### VISUAL PHYSICS School of Physics University of Sydney Australia. Why do cars need different oils in hot and cold countries?

VISUAL PHYSICS School of Physics University of Sydney Australia FLUID FLOW VISCOSITY POISEUILLE'S LAW? Why do cars need different oils in hot and cold countries? Why does the engine runs more freely as

### PHYS 101-4M, Fall 2005 Exam #3. MULTIPLE CHOICE. Choose the one alternative that best completes the statement or answers the question.

PHYS 101-4M, Fall 2005 Exam #3 Name MULTIPLE CHOICE. Choose the one alternative that best completes the statement or answers the question. 1) A bicycle wheel rotates uniformly through 2.0 revolutions in

### SIZE OF A MOLECULE FROM A VISCOSITY MEASUREMENT

Experiment 8, page 1 Version of April 25, 216 Experiment 446.8 SIZE OF A MOLECULE FROM A VISCOSITY MEASUREMENT Theory Viscous Flow. Fluids attempt to minimize flow gradients by exerting a frictional force,

### 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

### Chapter 24 Physical Pendulum

Chapter 4 Physical Pendulum 4.1 Introduction... 1 4.1.1 Simple Pendulum: Torque Approach... 1 4. Physical Pendulum... 4.3 Worked Examples... 4 Example 4.1 Oscillating Rod... 4 Example 4.3 Torsional Oscillator...

### 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

### Physics Notes Class 11 CHAPTER 5 LAWS OF MOTION

1 P a g e Inertia Physics Notes Class 11 CHAPTER 5 LAWS OF MOTION The property of an object by virtue of which it cannot change its state of rest or of uniform motion along a straight line its own, is

### Chapter 5 Using Newton s Laws: Friction, Circular Motion, Drag Forces. Copyright 2009 Pearson Education, Inc.

Chapter 5 Using Newton s Laws: Friction, Circular Motion, Drag Forces Units of Chapter 5 Applications of Newton s Laws Involving Friction Uniform Circular Motion Kinematics Dynamics of Uniform Circular

### SHORT ANSWER. Write the word or phrase that best completes each statement or answers the question.

Exam Name SHORT ANSWER. Write the word or phrase that best completes each statement or answers the question. 1) A person on a sled coasts down a hill and then goes over a slight rise with speed 2.7 m/s.

### No Brain Too Small PHYSICS. 2 kg

MECHANICS: ANGULAR MECHANICS QUESTIONS ROTATIONAL MOTION (2014;1) Universal gravitational constant = 6.67 10 11 N m 2 kg 2 (a) The radius of the Sun is 6.96 10 8 m. The equator of the Sun rotates at a

### ME19b. SOLUTIONS. Feb. 11, 2010. Due Feb. 18

ME19b. SOLTIONS. Feb. 11, 21. Due Feb. 18 PROBLEM B14 Consider the long thin racing boats used in competitive rowing events. Assume that the major component of resistance to motion is the skin friction

### 2.1 Force and Motion Kinematics looks at velocity and acceleration without reference to the cause of the acceleration.

2.1 Force and Motion Kinematics looks at velocity and acceleration without reference to the cause of the acceleration. Dynamics looks at the cause of acceleration: an unbalanced force. Isaac Newton was

### Lecture L2 - Degrees of Freedom and Constraints, Rectilinear Motion

S. Widnall 6.07 Dynamics Fall 009 Version.0 Lecture L - Degrees of Freedom and Constraints, Rectilinear Motion Degrees of Freedom Degrees of freedom refers to the number of independent spatial coordinates

### 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

### 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

### Newton s Laws of Motion

Section 3.2 Newton s Laws of Motion Objectives Analyze relationships between forces and motion Calculate the effects of forces on objects Identify force pairs between objects New Vocabulary Newton s first

### Lecture 2 PROPERTIES OF FLUID

Lecture 2 PROPERTIES OF FLUID Learning Objectives Upon completion of this chapter, the student should be able to: Define three states of matter: Solid, liquid and gas. Define mass density, specific weight

### B) 286 m C) 325 m D) 367 m Answer: B

Practice Midterm 1 1) When a parachutist jumps from an airplane, he eventually reaches a constant speed, called the terminal velocity. This means that A) the acceleration is equal to g. B) the force of

### 7. Kinetic Energy and Work

Kinetic Energy: 7. Kinetic Energy and Work The kinetic energy of a moving object: k = 1 2 mv 2 Kinetic energy is proportional to the square of the velocity. If the velocity of an object doubles, the kinetic

### SOLUTIONS TO PROBLEM SET 4

MASSACHUSETTS INSTITUTE OF TECHNOLOGY Department of Physics Physics 8.01X Fall Term 2002 SOLUTIONS TO PROBLEM SET 4 1 Young & Friedman 5 26 A box of bananas weighing 40.0 N rests on a horizontal surface.

### 1) The gure below shows the position of a particle (moving along a straight line) as a function of time. Which of the following statements is true?

Physics 2A, Sec C00: Mechanics -- Winter 2011 Instructor: B. Grinstein Final Exam INSTRUCTIONS: Use a pencil #2 to ll your scantron. Write your code number and bubble it in under "EXAM NUMBER;" an entry

### Chapter 4: Buoyancy & Stability

Chapter 4: Buoyancy & Stability Learning outcomes By the end of this lesson students should be able to: Understand the concept of buoyancy hence determine the buoyant force exerted by a fluid to a body

### Angular acceleration α

Angular Acceleration Angular acceleration α measures how rapidly the angular velocity is changing: Slide 7-0 Linear and Circular Motion Compared Slide 7- Linear and Circular Kinematics Compared Slide 7-

### Bead moving along a thin, rigid, wire.

Bead moving along a thin, rigid, wire. odolfo. osales, Department of Mathematics, Massachusetts Inst. of Technology, Cambridge, Massachusetts, MA 02139 October 17, 2004 Abstract An equation describing