# Mechanical Principles

Size: px
Start display at page:

Transcription

1 Unit 4: Mechanical Principles Unit code: F/601/1450 QCF level: 5 Credit value: 15 OUTCOME 4 POWER TRANSMISSION TUTORIAL 2 BALANCING 4. Dynamics of rotating systems Single and multi-link mechanisms: slider crank and three/four bar mechanisms; production of vector diagrams and determination of relationships between velocity, acceleration, power and efficiency Balancing: single plane and multi-plane rotating mass systems; Dalby s method for determination of out-of-balance forces and couples and the required balancing masses Flywheels: angular momentum; kinetic energy; coefficient of fluctuation of speed; coefficient of fluctuation of energy; calculation of flywheel mass/dimensions to give required operating conditions Effects of coupling: conservation of angular momentum; energy loss due to coupling; final common rotational speed On completion of this tutorial you should be able to do the following. Explain the importance of balancing. Explain static and dynamic balance. Solve problems involving coplanar balance. Solve problems involving balancing in four planes. Construct vector diagrams. Use vector resolution to solve problems. Extend the work to more complex problems. It is assumed that the student is already familiar with the following concepts. Angular motion. Centrifugal force. Basic vector diagram construction. Basic trigonometry. All these above may be found in the pre-requisite tutorials. D.J.Dunn freestudy.co.uk 1

2 CONTENTS 1. Introduction 2. Balancing in One Plane 3. Masses Not in Same Plane 4. Solution by Using Vector Resolution 5. More Advanced Problems D.J.Dunn freestudy.co.uk 2

3 1. INTRODUCTION The balancing of rotating bodies is important to avoid vibrations. In heavy industrial machines such as steam turbines and electric generators, vibration could cause catastrophic failure. Vibrations are noisy and uncomfortable and when a car wheel is out of balance, the ride is quite unpleasant. In the case of a simple wheel, balancing simply involves moving the centre of gravity to the centre of rotation but as we shall see, for longer and more complex bodies, there is more to it. For a body to be completely balanced it must have two things. 1. Static Balance. This occurs when there is no resultant centrifugal force and the centre of gravity is on the axis of rotation. 2. Dynamic Balance. This occurs when there is no resulting turning moment along the axis. 2. BALANCING IN ONE PLANE If the system is a simple disc then static balance is all that is needed. Consider a thin disc or wheel on which the centre of gravity is not the same as the centre of rotation. A simple test for static balance is to place the wheel in frictionless bearings. The centre of gravity will always come to rest below the centre of rotation (like a pendulum). If it is balanced it will remain stationary no matter which position it is turned to. Figure 1 If the centre of gravity is distance r from the centre of rotation then when it spins at rad/s, centrifugal force is produced. This has a formula C.F.= M 2 r where M is the mass of the disc. This is the out of balance force. In order to cancel it out an equal and opposite force is needed. This is simply done by adding a mass M 2 at a radius r 2 as shown. The two forces must have the same magnitudes. M 2 r = M 2 2 r 2 M r = M 2 r 2 Placing a suitable mass at a suitable radius moves the centre of gravity to the centre of rotation. This balance holds true at all speeds down to zero hence it is balanced so long as the products of M and r are equal and opposite. Figure 2 D.J.Dunn freestudy.co.uk 3

4 Now consider that our disc is out of balance because there are three masses attached to it as shown. The 3 masses are said to be coplanar and they rotate about a common centre. The centrifugal force acting on each mass is F = M r 2. The radius of rotation is r and the angular velocity is in radians/second. The force acting on each one is hence F 1 =M 1 r 1 2 F 2 =M 2 r 2 2 F 3 =M 3 r 3 2 Figure 3 These are vector quantities and we can add them up to find the resultant force as shown. Figure 4 If the system was balanced, there would be no resultant force so the force needed to balance the system must be equal and opposite of the resultant (the vector that closes the polygon). The balancing mass M 4 is then added at a suitable radius and angle such that the product M r is correct. Figure 5 The result obtained would be the same whatever the value of and when = 0 we have static balance. In order to make the solution easier, we may make = 1 and calculate M r for each vector. This is called the M r polygon or vector diagram. Note that angles will be given in normal mathematical terms with anticlockwise being positive from the x axis as shown. Figure 6 D.J.Dunn freestudy.co.uk 4

5 WORKED EXAMPLE No.1 Three masses A, B and C are placed on a balanced disc as shown at radii of 120 mm, 100 mm and 80 mm respectively. The masses are 1 kg, 0.5 kg and 0.7 kg respectively. Find the 4 th mass which should be added at a radius of 60 mm in order to statically balance the system. SOLUTION Figure 7 First draw up a table to calculate the value of M r for each mass. Mass radius M r A B C D M D M D Draw the M r polygon to find the value of M r for the 4 th mass. Figure 8 The resultant is kg mm and is equal to 60M D. The mass required is 144.3/60 = 2.4 kg 208 o anticlockwise of A as shown. D.J.Dunn freestudy.co.uk 5

6 SELF ASSESSMENT EXERCISE No. 1 Find the 4 th mass that should be added at a radius of 50 mm in order to statically balance the system shown. Mass A is 1 kg at 100 mm radius. Mass B is 1.5 kg at 75 mm radius Mass C is 2.0 kg at 90 mm radius. Answer 0.52 kg at 156 o clockwise from A. Figure 9 D.J.Dunn freestudy.co.uk 6

7 3. MASSES NOT IN THE SAME PLANE Consider 2 masses statically balanced as shown but acting at different places along the axis. For static balance M A r A = M B r B Figure 10 It is clear that even with static balance, centrifugal force will produce a turning moment about the centre of gravity for the system. In this simple case, the problem is solved by adding equal and opposite forces at the two points as shown. Figure 11 Consider the turning moment due to a single mass. Figure 12 The centrifugal force produced is F = Mr 2 The turning moment about the reference plane = T.M. = F x = Mr 2 x For dynamic and static balance we must work out the resultant turning moment and add masses at appropriate points to cancel it out. The appropriate points will be on two planes not coplanar with any of the original masses. This involves drawing two vector diagrams and since is common to all vectors we can again take =1 and draw vectors representing Mr and Mrx. This is best explained with a worked example. D.J.Dunn freestudy.co.uk 7

8 WORKED EXAMPLE No.2 Find the mass and the angle at which it should be positioned in planes A and D at a radius of 60 mm in order to produce complete balance of the system shown. SOLUTION Figure 13 Note that the diagram has been drawn with B vertical. It is a good idea to always start by making one of the known masses horizontal or vertical to make the construction of the vector diagrams easier. All angles should be expressed in absolute terms. Plane A is the reference plane. All values of x are measured from plane A thus making Mrx for A equal to zero. It follows that it does not appear in the vector diagram. Make up a table as follows leaving unknowns as symbols. M r Mr x Mrx A M A 60 60M A 0 0 B C D M D 60 60M D M D Now draw a polygon of Mrx vectors in order to find the value of Mrx at D. Start with B in this case because it is vertical. Figure 14 D.J.Dunn freestudy.co.uk 8

9 Scaling the vector D which closes the triangle we find Mrx for D = = M D Hence M D = 94000/22500 = 4.17 kg and it is positioned 254o Now we calculate Mr for D. Mr for D = x 60 = Next we draw the polygon for the Mr values as shown. Figure 15 The vector which closes the polygon represents Mr for point A Mr for A is M A = 195 M A = 195/60 = 3.25 kg at 6 o to the vertical. The answer is best shown with an end view. Figure 16 D.J.Dunn freestudy.co.uk 9

10 SELF ASSESSMENT EXERCISE No A shaft has 4 discs A, B, C and D along its length 100 mm apart. A mass of 0.8 kg is placed on B at a radius of 20 mm. A mass of 2 kg is placed on C at a radius of 30 mm and rotated 120 o from the mass on B. Finf the masses to be placed on A and D at a radius of 25 mm that wil produce total balance. (Answers kg and 1.52 kg) 2. The diagram below shows masses on two rotors in planes B and C. Determine the masses to be added on the rotors in planes A and D at radius 40 mm which will produce static and dynamic balance.( 1.9 kg at 177 o and 2.2 kg at 141 o ) Figure 17 D.J.Dunn freestudy.co.uk 10

11 4 SOLUTION BY USING VECTOR RESOLUTION A more accurate approach to solving the vector diagrams in the preceding work is to resolve each vector into vertical and horizontal components. The resultant vector is then found by adding these components. Consider worked example No.1 again. Figure 18 Each vector has a component in the x direction given by Mr cos and in the y direction it is MRsin Figure 19 Work out these for each vector and include it in the table. Mass radius M r Mr cos MRsin A o B o C o Totals The resultant vector has x and y components of and This can be solved with Pythagoras. Resultant Mr = ( ) ½ = kg mm as before. The mass required is 144.3/60 = 2.4 kg. The angle = atan (67.9/127.3) or tan -1 (67.9/127.3) = 28 o The balancing force is 180 o anticlockwise of this so the balancing mass must be placed at an angle of 208 o. D.J.Dunn freestudy.co.uk 11

12 Here is the solution to worked example No.2 using the same method. Mass radius M r x Mrx A M A M A 0 0 unknown B C D M D M D M D unknown MRx cos MRxsin If balanced, the totals must be zero A 0 0 so the unknown values may be B deduced. C D Totals 0 0 Resultant Mrx for D = ( ) ½ = This is the value for D so M D = M D =9360/22500 = 4.16 kg = tan -1 (90000/25980) = o so the true angle is 180 o o = o Now fill in the table for the Mr values. Again the totals must be zero so we can deduce the values for A. M r M r Mr cos MRsin A M A M A B C D o Totals 0 0 Resultant Mr vector is ( ) ½ = and this is equal to 60 MA. M A = 185.8/60 = 3.1 kg = = tan -1 (185/17.3) = 84.6 o so the true angle is 180 o o = o This gives a more accurate answer than that produced by drawing and scaling. D.J.Dunn freestudy.co.uk 12

13 5. MORE ADVANCED PROBLEMS Some other points to be considered are: The reference plane can be any of the existing planes. The moment about any point can be made zero by placing correction masses on one plane only if static balance is not important. WORKED EXAMPLE No.3 A rigid rotor caries three thin discs A, B and C mounted on a shaft 1 metre long running from O to Z. The distances from O are designated x in the table and the out of balance moment as MR in gramme metre. The out of balance force is at angle degrees. Figure 20 TABLE x (m) MR (g m) (degrees) A B C (a) (b) only does Calculate correcting forces needed on planes A and C to bring the system into balance. The rotor is suspended from a bearing at O with the other end unrestrained. It can be balanced on disc C. Calculate the corrective balance needed to ensure that end Z not move. Calculate the magnitude of the force resulting at O when the shaft revolves at 500 rad/s. SOLUTION Make A the reference plane and re-tabulate as follows. x (m) MR (g m) (degrees) MRx (g m 2 ) A B C D.J.Dunn freestudy.co.uk 13

14 Construct the MR polygon. Figure 21 The red vector produces balance and may be calculated or measured. The length is 26.5 g m and the angle is 19.1 o. This is placed on plane A. The table may be further completed. x (m) MR (g m) (degrees) MRx (g m 2 ) A B C A Next construct the MRx polygon. This is simply two vectors at opposite angles giving a closing vector of 10.5 at 120 o. Figure 22 This is to be placed on C so the value of MR is 10.5/0.8 = g m This now produces an imbalance of on the static balance (MR) so an equal and opposite force is added to A The complete table is x (m) MR (g m) (degrees) MRx (g m 2 ) A B C A A C A2 and A3 can be combined to a single MR value of 18.5 at 187 degrees. The final table is (b) The key here is that the moment about end O must be zero to stop the other end moving and static balance must be ignored. Make O the reference plane and re-tabulate as follows with x measured from O D.J.Dunn freestudy.co.uk 14

15 x (m) MR (g m) (degrees) MRx (g m 2 ) A B C Construct the MR polygon. The closing vector is 11.6 at o Figure 23 This produces zero moment about end O and the balance needed on C is 11.6/0.85 = g m at o There is now a static imbalance that may be found from the MR polygon. x (m) MR (g m) (degrees) MRx (g m 2 ) A B C C Figure 24 A static imbalance of 17.4 g m at o now exists and this will be countered by a force in the bearing at O. Centrifugal force = m 2 R so the force on the bearing is MR x 2 = 17.4 x 10-3 (kg m) x = 4350 N Note there are various methods of solving the vector diagrams such as drawing to scale or calculating the vertical and horizontal components of each vector and summing them to find the resultant. D.J.Dunn freestudy.co.uk 15

### SOLID MECHANICS BALANCING TUTORIAL BALANCING OF ROTATING BODIES

SOLID MECHANICS BALANCING TUTORIAL BALANCING OF ROTATING BODIES This work covers elements of the syllabus for the Edexcel module 21722P HNC/D Mechanical Principles OUTCOME 4. On completion of this tutorial

### EDEXCEL NATIONAL CERTIFICATE/DIPLOMA MECHANICAL PRINCIPLES AND APPLICATIONS NQF LEVEL 3 OUTCOME 1 - LOADING SYSTEMS

EDEXCEL NATIONAL CERTIFICATE/DIPLOMA MECHANICAL PRINCIPLES AND APPLICATIONS NQF LEVEL 3 OUTCOME 1 - LOADING SYSTEMS TUTORIAL 1 NON-CONCURRENT COPLANAR FORCE SYSTEMS 1. Be able to determine the effects

### ENGINEERING COUNCIL DYNAMICS OF MECHANICAL SYSTEMS D225 TUTORIAL 1 LINEAR AND ANGULAR DISPLACEMENT, VELOCITY AND ACCELERATION

ENGINEERING COUNCIL DYNAMICS OF MECHANICAL SYSTEMS D225 TUTORIAL 1 LINEAR AND ANGULAR DISPLACEMENT, VELOCITY AND ACCELERATION This tutorial covers pre-requisite material and should be skipped if you are

### 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

### ANALYTICAL METHODS FOR ENGINEERS

UNIT 1: Unit code: QCF Level: 4 Credit value: 15 ANALYTICAL METHODS FOR ENGINEERS A/601/1401 OUTCOME - TRIGONOMETRIC METHODS TUTORIAL 1 SINUSOIDAL FUNCTION Be able to analyse and model engineering situations

### 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

### Center of Gravity. We touched on this briefly in chapter 7! x 2

Center of Gravity We touched on this briefly in chapter 7! x 1 x 2 cm m 1 m 2 This was for what is known as discrete objects. Discrete refers to the fact that the two objects separated and individual.

### MECHANICAL PRINCIPLES OUTCOME 4 MECHANICAL POWER TRANSMISSION TUTORIAL 1 SIMPLE MACHINES

MECHANICAL PRINCIPLES OUTCOME 4 MECHANICAL POWER TRANSMISSION TUTORIAL 1 SIMPLE MACHINES Simple machines: lifting devices e.g. lever systems, inclined plane, screw jack, pulley blocks, Weston differential

### Rotation: Moment of Inertia and Torque

Rotation: Moment of Inertia and Torque Every time we push a door open or tighten a bolt using a wrench, we apply a force that results in a rotational motion about a fixed axis. Through experience we learn

### Tennessee State University

Tennessee State University Dept. of Physics & Mathematics PHYS 2010 CF SU 2009 Name 30% Time is 2 hours. Cheating will give you an F-grade. Other instructions will be given in the Hall. MULTIPLE CHOICE.

### 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

### 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

### Problem Set 1. Ans: a = 1.74 m/s 2, t = 4.80 s

Problem Set 1 1.1 A bicyclist starts from rest and after traveling along a straight path a distance of 20 m reaches a speed of 30 km/h. Determine her constant acceleration. How long does it take her to

### Lecture 16. Newton s Second Law for Rotation. Moment of Inertia. Angular momentum. Cutnell+Johnson: 9.4, 9.6

Lecture 16 Newton s Second Law for Rotation Moment of Inertia Angular momentum Cutnell+Johnson: 9.4, 9.6 Newton s Second Law for Rotation Newton s second law says how a net force causes an acceleration.

### 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

### 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

### 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

### 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

### AP Physics Circular Motion Practice Test B,B,B,A,D,D,C,B,D,B,E,E,E, 14. 6.6m/s, 0.4 N, 1.5 m, 6.3m/s, 15. 12.9 m/s, 22.9 m/s

AP Physics Circular Motion Practice Test B,B,B,A,D,D,C,B,D,B,E,E,E, 14. 6.6m/s, 0.4 N, 1.5 m, 6.3m/s, 15. 12.9 m/s, 22.9 m/s Answer the multiple choice questions (2 Points Each) on this sheet with capital

### 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

### 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

### Figure 1.1 Vector A and Vector F

CHAPTER I VECTOR QUANTITIES Quantities are anything which can be measured, and stated with number. Quantities in physics are divided into two types; scalar and vector quantities. Scalar quantities have

### 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.

### Practice Exam Three Solutions

MASSACHUSETTS INSTITUTE OF TECHNOLOGY Department of Physics Physics 8.01T Fall Term 2004 Practice Exam Three Solutions Problem 1a) (5 points) Collisions and Center of Mass Reference Frame In the lab frame,

### 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

### Universal Law of Gravitation

Universal Law of Gravitation Law: Every body exerts a force of attraction on every other body. This force called, gravity, is relatively weak and decreases rapidly with the distance separating the bodies

### Mechanical Principles

Unit 4: Mechanical Principles Unit code: F/60/450 QCF level: 5 Credit value: 5 OUTCOME 3 POWER TRANSMISSION TUTORIAL BELT DRIVES 3 Power Transmission Belt drives: flat and v-section belts; limiting coefficient

### Fric-3. force F k and the equation (4.2) may be used. The sense of F k is opposite

4. FRICTION 4.1 Laws of friction. We know from experience that when two bodies tend to slide on each other a resisting force appears at their surface of contact which opposes their relative motion. The

### PHYSICS 111 HOMEWORK SOLUTION #9. April 5, 2013

PHYSICS 111 HOMEWORK SOLUTION #9 April 5, 2013 0.1 A potter s wheel moves uniformly from rest to an angular speed of 0.16 rev/s in 33 s. Find its angular acceleration in radians per second per second.

### Dynamics of Rotational Motion

Chapter 10 Dynamics of Rotational Motion PowerPoint Lectures for University Physics, Twelfth Edition Hugh D. Young and Roger A. Freedman Lectures by James Pazun Modified by P. Lam 5_31_2012 Goals for Chapter

### 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,

### 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

### 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

### 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 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,

### Chapter 11. h = 5m. = mgh + 1 2 mv 2 + 1 2 Iω 2. E f. = E i. v = 4 3 g(h h) = 4 3 9.8m / s2 (8m 5m) = 6.26m / s. ω = v r = 6.

Chapter 11 11.7 A solid cylinder of radius 10cm and mass 1kg starts from rest and rolls without slipping a distance of 6m down a house roof that is inclined at 30 degrees (a) What is the angular speed

### 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.

### Examples of Scalar and Vector Quantities 1. Candidates should be able to : QUANTITY VECTOR SCALAR

Candidates should be able to : Examples of Scalar and Vector Quantities 1 QUANTITY VECTOR SCALAR Define scalar and vector quantities and give examples. Draw and use a vector triangle to determine the resultant

### STATICS. Introduction VECTOR MECHANICS FOR ENGINEERS: Eighth Edition CHAPTER. Ferdinand P. Beer E. Russell Johnston, Jr.

Eighth E CHAPTER VECTOR MECHANICS FOR ENGINEERS: STATICS Ferdinand P. Beer E. Russell Johnston, Jr. Introduction Lecture Notes: J. Walt Oler Texas Tech University Contents What is Mechanics? Fundamental

### Midterm Solutions. mvr = ω f (I wheel + I bullet ) = ω f 2 MR2 + mr 2 ) ω f = v R. 1 + M 2m

Midterm Solutions I) A bullet of mass m moving at horizontal velocity v strikes and sticks to the rim of a wheel a solid disc) of mass M, radius R, anchored at its center but free to rotate i) Which of

### SOLID MECHANICS DYNAMICS TUTORIAL MOMENT OF INERTIA. This work covers elements of the following syllabi.

SOLID MECHANICS DYNAMICS TUTOIAL MOMENT OF INETIA This work covers elements of the following syllabi. Parts of the Engineering Council Graduate Diploma Exam D5 Dynamics of Mechanical Systems Parts of the

### 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 11 Equilibrium

11.1 The First Condition of Equilibrium The first condition of equilibrium deals with the forces that cause possible translations of a body. The simplest way to define the translational equilibrium of

### Torque Analyses of a Sliding Ladder

Torque Analyses of a Sliding Ladder 1 Problem Kirk T. McDonald Joseph Henry Laboratories, Princeton University, Princeton, NJ 08544 (May 6, 2007) The problem of a ladder that slides without friction while

### Mechanics lecture 7 Moment of a force, torque, equilibrium of a body

G.1 EE1.el3 (EEE1023): Electronics III Mechanics lecture 7 Moment of a force, torque, equilibrium of a body Dr Philip Jackson http://www.ee.surrey.ac.uk/teaching/courses/ee1.el3/ G.2 Moments, torque and

### Sample Questions for the AP Physics 1 Exam

Sample Questions for the AP Physics 1 Exam Sample Questions for the AP Physics 1 Exam Multiple-choice Questions Note: To simplify calculations, you may use g 5 10 m/s 2 in all problems. Directions: Each

APPLIED MATHEMATICS ADVANCED LEVEL INTRODUCTION This syllabus serves to examine candidates knowledge and skills in introductory mathematical and statistical methods, and their applications. For applications

### Physical Quantities and Units

Physical Quantities and Units 1 Revision Objectives This chapter will explain the SI system of units used for measuring physical quantities and will distinguish between vector and scalar quantities. You

### Linear Motion vs. Rotational Motion

Linear Motion vs. Rotational Motion Linear motion involves an object moving from one point to another in a straight line. Rotational motion involves an object rotating about an axis. Examples include a

### Chapter 8: Rotational Motion of Solid Objects

Chapter 8: Rotational Motion of Solid Objects 1. An isolated object is initially spinning at a constant speed. Then, although no external forces act upon it, its rotational speed increases. This must be

### 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

### 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

### 11. Rotation Translational Motion: Rotational Motion:

11. Rotation Translational Motion: Motion of the center of mass of an object from one position to another. All the motion discussed so far belongs to this category, except uniform circular motion. Rotational

### Force on Moving Charges in a Magnetic Field

[ Assignment View ] [ Eðlisfræði 2, vor 2007 27. Magnetic Field and Magnetic Forces Assignment is due at 2:00am on Wednesday, February 28, 2007 Credit for problems submitted late will decrease to 0% after

### 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

### 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

### Physics 125 Practice Exam #3 Chapters 6-7 Professor Siegel

Physics 125 Practice Exam #3 Chapters 6-7 Professor Siegel Name: Lab Day: 1. A concrete block is pulled 7.0 m across a frictionless surface by means of a rope. The tension in the rope is 40 N; and the

### 3 Work, Power and Energy

3 Work, Power and Energy At the end of this section you should be able to: a. describe potential energy as energy due to position and derive potential energy as mgh b. describe kinetic energy as energy

### Copyright 2011 Casa Software Ltd. www.casaxps.com. Centre of Mass

Centre of Mass A central theme in mathematical modelling is that of reducing complex problems to simpler, and hopefully, equivalent problems for which mathematical analysis is possible. The concept of

### Work Energy & Power. September 2000 Number 05. 1. Work If a force acts on a body and causes it to move, then the force is doing work.

PhysicsFactsheet September 2000 Number 05 Work Energy & Power 1. Work If a force acts on a body and causes it to move, then the force is doing work. W = Fs W = work done (J) F = force applied (N) s = distance

### Physics 11 Assignment KEY Dynamics Chapters 4 & 5

Physics Assignment KEY Dynamics Chapters 4 & 5 ote: for all dynamics problem-solving questions, draw appropriate free body diagrams and use the aforementioned problem-solving method.. Define the following

### 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

### 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

### 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

### Lecture 17. Last time we saw that the rotational analog of Newton s 2nd Law is

Lecture 17 Rotational Dynamics Rotational Kinetic Energy Stress and Strain and Springs Cutnell+Johnson: 9.4-9.6, 10.1-10.2 Rotational Dynamics (some more) Last time we saw that the rotational analog of

### G U I D E T O A P P L I E D O R B I T A L M E C H A N I C S F O R K E R B A L S P A C E P R O G R A M

G U I D E T O A P P L I E D O R B I T A L M E C H A N I C S F O R K E R B A L S P A C E P R O G R A M CONTENTS Foreword... 2 Forces... 3 Circular Orbits... 8 Energy... 10 Angular Momentum... 13 FOREWORD

### 3. KINEMATICS IN TWO DIMENSIONS; VECTORS.

3. KINEMATICS IN TWO DIMENSIONS; VECTORS. Key words: Motion in Two Dimensions, Scalars, Vectors, Addition of Vectors by Graphical Methods, Tail to Tip Method, Parallelogram Method, Negative Vector, Vector

### 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

### SOLID MECHANICS DYNAMICS TUTORIAL NATURAL VIBRATIONS ONE DEGREE OF FREEDOM

SOLID MECHANICS DYNAMICS TUTORIAL NATURAL VIBRATIONS ONE DEGREE OF FREEDOM This work covers elements of the syllabus for the Engineering Council Exam D5 Dynamics of Mechanical Systems, C05 Mechanical and

### 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

### SOLID MECHANICS DYNAMICS TUTORIAL PULLEY DRIVE SYSTEMS. This work covers elements of the syllabus for the Edexcel module HNC/D Mechanical Principles.

SOLID MECHANICS DYNAMICS TUTORIAL PULLEY DRIVE SYSTEMS This work covers elements of the syllabus for the Edexcel module HNC/D Mechanical Principles. On completion of this tutorial you should be able to

### Vector Algebra II: Scalar and Vector Products

Chapter 2 Vector Algebra II: Scalar and Vector Products We saw in the previous chapter how vector quantities may be added and subtracted. In this chapter we consider the products of vectors and define

### MECHANICAL PRINCIPLES HNC/D MOMENTS OF AREA. Define and calculate 1st. moments of areas. Define and calculate 2nd moments of areas.

MECHANICAL PRINCIPLES HNC/D MOMENTS OF AREA The concepts of first and second moments of area fundamental to several areas of engineering including solid mechanics and fluid mechanics. Students who are

### 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

### Unit - 6 Vibrations of Two Degree of Freedom Systems

Unit - 6 Vibrations of Two Degree of Freedom Systems Dr. T. Jagadish. Professor for Post Graduation, Department of Mechanical Engineering, Bangalore Institute of Technology, Bangalore Introduction A two

### SOLID MECHANICS DYNAMICS TUTORIAL CENTRIPETAL FORCE

SOLID MECHANICS DYNAMICS TUTORIAL CENTRIPETAL FORCE This work coers elements of the syllabus for the Engineering Council Exam D5 Dynamics of Mechanical Systems C10 Engineering Science. This tutorial examines

### THEORETICAL MECHANICS

PROF. DR. ING. VASILE SZOLGA THEORETICAL MECHANICS LECTURE NOTES AND SAMPLE PROBLEMS PART ONE STATICS OF THE PARTICLE, OF THE RIGID BODY AND OF THE SYSTEMS OF BODIES KINEMATICS OF THE PARTICLE 2010 0 Contents

### PHYSICAL QUANTITIES AND UNITS

1 PHYSICAL QUANTITIES AND UNITS Introduction Physics is the study of matter, its motion and the interaction between matter. Physics involves analysis of physical quantities, the interaction between them

### Gravitational Potential Energy

Gravitational Potential Energy Consider a ball falling from a height of y 0 =h to the floor at height y=0. A net force of gravity has been acting on the ball as it drops. So the total work done on the

### CHAPTER 6 WORK AND ENERGY

CHAPTER 6 WORK AND ENERGY CONCEPTUAL QUESTIONS. REASONING AND SOLUTION The work done by F in moving the box through a displacement s is W = ( F cos 0 ) s= Fs. The work done by F is W = ( F cos θ). s From

### Slide 10.1. Basic system Models

Slide 10.1 Basic system Models Objectives: Devise Models from basic building blocks of mechanical, electrical, fluid and thermal systems Recognize analogies between mechanical, electrical, fluid and thermal

### 226 Chapter 15: OSCILLATIONS

Chapter 15: OSCILLATIONS 1. In simple harmonic motion, the restoring force must be proportional to the: A. amplitude B. frequency C. velocity D. displacement E. displacement squared 2. An oscillatory motion

### AP1 Oscillations. 1. Which of the following statements about a spring-block oscillator in simple harmonic motion about its equilibrium point is false?

1. Which of the following statements about a spring-block oscillator in simple harmonic motion about its equilibrium point is false? (A) The displacement is directly related to the acceleration. (B) The

### Mobile field balancing reduces vibrations in energy and power plants. Published in VGB PowerTech 07/2012

Mobile field balancing reduces vibrations in energy and power plants Published in VGB PowerTech 07/2012 Dr. Edwin Becker PRÜFTECHNIK Condition Monitoring PRÜFTECHNIK Condition Monitoring GmbH 85737 Ismaning

### 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-

### Rotational Inertia Demonstrator

WWW.ARBORSCI.COM Rotational Inertia Demonstrator P3-3545 BACKGROUND: The Rotational Inertia Demonstrator provides an engaging way to investigate many of the principles of angular motion and is intended

### 4.2 Free Body Diagrams

CE297-FA09-Ch4 Page 1 Friday, September 18, 2009 12:11 AM Chapter 4: Equilibrium of Rigid Bodies A (rigid) body is said to in equilibrium if the vector sum of ALL forces and all their moments taken about

### EXPERIMENT: MOMENT OF INERTIA

OBJECTIVES EXPERIMENT: MOMENT OF INERTIA to 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 as mass plays in

### Physics Midterm Review Packet January 2010

Physics Midterm Review Packet January 2010 This Packet is a Study Guide, not a replacement for studying from your notes, tests, quizzes, and textbook. Midterm Date: Thursday, January 28 th 8:15-10:15 Room:

### Physics 1120: Simple Harmonic Motion Solutions

Questions: 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 Physics 1120: Simple Harmonic Motion Solutions 1. A 1.75 kg particle moves as function of time as follows: x = 4cos(1.33t+π/5) where distance is measured

### VELOCITY, ACCELERATION, FORCE

VELOCITY, ACCELERATION, FORCE velocity Velocity v is a vector, with units of meters per second ( m s ). Velocity indicates the rate of change of the object s position ( r ); i.e., velocity tells you how

### CHAPTER 15 FORCE, MASS AND ACCELERATION

CHAPTER 5 FORCE, MASS AND ACCELERATION EXERCISE 83, Page 9. A car initially at rest accelerates uniformly to a speed of 55 km/h in 4 s. Determine the accelerating force required if the mass of the car

### Exam 2 is at 7 pm tomorrow Conflict is at 5:15 pm in 151 Loomis

* By request, but I m not vouching for these since I didn t write them Exam 2 is at 7 pm tomorrow Conflict is at 5:15 pm in 151 Loomis There are extra office hours today & tomorrow Lots of practice exams

### 6. Vectors. 1 2009-2016 Scott Surgent (surgent@asu.edu)

6. Vectors For purposes of applications in calculus and physics, a vector has both a direction and a magnitude (length), and is usually represented as an arrow. The start of the arrow is the vector s foot,

### Chapter 6 Work and Energy

Chapter 6 WORK AND ENERGY PREVIEW Work is the scalar product of the force acting on an object and the displacement through which it acts. When work is done on or by a system, the energy of that system

### 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,