Verification Analysis of the Slope Stability

Size: px
Start display at page:

Download "Verification Analysis of the Slope Stability"

Transcription

1 Verifiction nul no. 3 Updte 04/016 Verifiction Anlysis of the Slope Stbility Progr: File: Slope Stbility Deo_v_en_03.gst In this verifiction nul you will find hnd-de verifiction nlysis of the stbility of slope nd nchored slope in pernent design sitution. The results of the hnd-de clcultions re copred with the results fro the GEO5 Slope Stbility progr. Ters of Reference: In Figure 1, n exple of slope is shown. The slope hs height H nd is djusted in 1:1.5 inclintion. At the top of the slope is lod f 0 kn/. The erth body is fored of sndy cly (CS). The properties of soil (effective vlues) re shown in Tble 1. The clcultion is divided into two stges. In the 1 st stge the stbility of the slope is clculted nd in the nd stge the stbility of n nchored slope is clculted. The slope stbility is clculted using Fellenius/Petterson ethod nd Bishop s siplified ethod (the circulr slip surfce). The verifiction ethodology of the slope stbility is done ccording to sfety fctors. Figure 1 Slope - diensions Soil Unit weight kn/ 3 Sturted unit weight st kn/ 3 Angle of internl friction ef Cohesion of soil kp c ef CS Tble 1 Soil properties effective vlues 1

2 1. Fellenius/Petterson ethod Verifiction of the Stbility of the Slope The slip surfce ws deterined. In this cse the slip surfce is deterined by circle with its centre t point O x, z ; nd rdius R Points Z sp nd K sp indicte the beginning nd end of the slip surfce. The slope ws divided into verticl blocks of width b i In Figure, slope divided into 0 blocks is shown. Figure Slope verticl blocks Figure 3 Sttic schee of the block Clcultion of the weight of the individul blocks of the slope. The weight of the blocks of the erth body bounded by the slip surfce re clculted. The overll clcultion is shown in Tble. An exple of the clcultion for block nuber 13 is done.

3 Deterintion of the re bove the ground wter tble (the re A ) nd under the ground wter tble (the re B ): A B Weight of the individul prts of the block: A W A kn/, BW, 13 B13 st kn/ Weight force of the block: W AW,13 BW, kn/ 13 Clcultion for ll blocks: Are of the prt A i [ ] B i [ ] Width of the block b i [ ] Weight of one prt A W, i [ kn/ ] B W, i [ kn/ ] Weight of the block W i [ kn/ ] Lod f i [ kn/ ] , Tble Weight nd forces of the lod 3

4 Deterintion of the inclintion of the slip surfce of the individul blocks nd clcultion of the pore pressure. To siplify the hnd-de clcultion the circle slip surfces of the individul blocks hve been replced by lines. The inclintion of the slip surfce is deterined by the ngle between the slip surfce nd the horizontl plne. The height of the ground wter tble ust be deterined for the clcultion of the pore pressure. The height of the ground wter tble h i is considered to the xis of the block. The unit weight of wter is w kn /. The heights of the ground wter tble on the left nd right side of the block ust be deterined for the clcultion of the horizontl forces of the pore pressure. The overll clcultion is shown in Tble 3. An exple of the clcultion for block nuber 13 is done. Inclintion of the slip surfce: Length of the slip surfce: b l cos( 13 ) cos(7.719) Inclintion of the ground wter tble: w, Height of the ground wter tble: h Clcultion of the reduced height of the ground wter tble: hr, 13 h13 cos( w,13) cos(5.0169) Clcultion of the pore pressure: u w hr, kp 13 Clcultion of the horizontl forces of the pore pressure: U U h L,13 w,13) w cos(5.0169) cos( 10 HL, 13 h P,13 w,13) w cos(5.0169) cos( 10 HP, kn/ 7.7 kn/ - left side - right side Clcultion for ll blocks: Inclintion of the slip surfce i [ ] Length of the slip surfce l i [ ] Inclintion of the ground wter tble w,i [ ] Ground wter tble Height of the ground wter tble h i [ ] Reduced height of the ground wter tble h r, i [ ] Pore pressure i u [ kp ] 4

5 Tble 3 Inclintions nd lengths of the slip surfces nd pore pressures Left side of the block h L, i [ ] U HL, i [ kn/ ] Right side of the block h R, i [ ] U HR, i [ kn/ ]

6 Tble 4 Horizontl forces of the pore pressure Clcultion of the sliding oent. The weight of the individul blocks including forces of the lod ct on the horizontl r fro xis of the block to the centre of the circulr slip surfce (to the point O). The rs of the forces re clculted fro the beginning of the slip surfce x, z 8.00; 5.00 ). The overll clcultion is in Tble 5. An exple of the clcultion for block ( Z sp nuber 13 is done. Clcultion of the oent r: b 1.0 r, 13 X zsp X O i b Clcultion of the sliding oent: W f r ( ) kn, ,13 / Clcultion for ll blocks: r, i [ ] Sliding oent, i [ kn/ ] r, i [ ] Sliding oent, i [ kn/ ] Tble 5 Sliding oents Resultnt sliding oent: 0 i1, i kn / Result fro the GEO5 Slope Stbility progr: kn / 6

7 Resultnt ctive force: 0, i i F kn/ R Result fro the GEO5 Slope Stbility progr: F kn/ Clcultion of the resisting oent. Norl forces N i of the individul blocks ust be clculted. The norl force cts upright to the slip surfce. The overll clcultion is shown in Tble 6. An exple of the clcultion for block 13 is done. Clcultion of the sfety fctor : p Clcultion of the norl force: N ( W13 f13) cos( 13) u13 l13 ( U HL,13 U HR,13) sin( 13) 13 N13 ( ) cos(7.719) ( ) sin( 7.719) Clcultion of the resisting oent: c l N tn( ) R tn(7.00) kn / p, kn/ Clcultion for ll blocks: Norl force N i [ kn/ ] Resisting oent p, i [ kn/ ] Norl force N i [ kn/ ] Resisting oent p, i [ kn/ ] Tble 6 Norl forces nd resisting oents 7

8 Resultnt resisting oent: 0 p p, i kn / i1 Result fro the GEO5 Slope Stbility progr: p kn / Resultnt pssive force: 0 p, i i Fp kn/ R Result fro the GEO5 Slope Stbility progr: F p kn/ Clcultion of the sfety fctor: p , NOT OK Result fro the GEO5 Slope Stbility progr: 1. 43, NOT OK Verifiction of the Stbility of Anchored Slope In Figure 4, n exple of nchored slope in the nd stge is shown. The nchor force is F A knnd the spcing is b A. 00. The position of the nchor hed is x, z 16.00; 9.00 H nchor. The nchor hed is on block nuber 9. Figure 4 Anchored slope - diensions 8

9 Clcultion of the sliding oent. The nchor cts s pssive eleent, which ens tht ctive oents will be the se s in the 1 st stge. Resultnt sliding oent: 0 i1, i kn / Result fro the GEO5 Slope Stbility progr: kn / Resultnt ctive force: 0, i i F kn/ R Result fro the GEO5 Slope Stbility progr: F kn/ Clcultion of the resisting oent. The norl forces N i of the individul blocks ust be clculted. The norl force cts perpendiculr to the slip surfce. Norl force on block nuber 9 is influenced by the nchor force. The overll clcultion is shown in Tble 7. An exple of the clcultion for block 13 is done. Anchor force t 1 : ' FA F kn b.00 A / A Clcultion of the r of the nchor force: ra Z O Z nchor Resisting oent of the nchor: ' p, A FA ra kn/ Clcultion of the sfety fctor : p Clcultion of the norl force: N 13 ( W13 f13) cos( 13) u13 l13 ( U HL,13 U HR,13) sin( 13) N13 ( ) cos(7.719) ( ) sin( 7.719) Clcultion of the effect of the nchor force (block nuber 9): N A, 9 F' Asin( 9 ) sin( ) kn/ kn/ Clcultion of the resisting oent: 9

10 c l N tn( ) R tn(7.00) kn / p, Clcultion for ll blocks: Norl force N i [ kn/ ] Resisting oent p, i [ kn/ ] Norl force N i [ kn/ ] Resisting oent p, i [ kn/ ] Tble 7 Norl forces nd resisting oents Resultnt resisting oent: 0 p p, i p, A kn / i1 Result fro the GEO5 Slope Stbility progr: p kn / Resultnt pssive force: 0 p, i p, A i Fp kn/ R Result fro the GEO5 Slope Stbility progr: F p kn/ Clcultion of the sfety fctor: p , SATISFACTORY Result fro the GEO5 Slope Stbility progr: 1. 54, SATISFACTORY 10

11 . Bishop s Siplified ethod Verifiction of the Stbility of the Slope The slip surfce is the se s in the first clcultion using the Fellenius/Petterson ethod (Figure ). The clcultion of the weight of the individul blocks is shown in Tble. Deterintion of the inclintion of the slip surfce of the individul blocks nd clcultion of the pore pressure. To siplify the hnd-de clcultion the circulr slip surfces of the individul blocks hve been replced by lines. The inclintion of the slip surfce is deterined by the ngle between the slip surfce nd the horizontl plne. The height of the ground wter tble ust be deterined for the clcultion of the pore pressure. The height of the ground wter tble h i is considered to the xis of the block. The unit weight of wter is w kn /. The resultnt effect of the horizontl forces of the pore pressure is not significnt nd hd been neglected. The overll clcultion is in Tble 8. An exple of the clcultion for block 13 is done. Inclintion of the slip surfce: Inclintion of the ground wter tble: w, Height of the ground wter tble: h Clcultion of the reduced height of the ground wter tble: hr, 13 h13 cos( w,13) cos(5.0169) Clcultion of the pore pressure: u w hr, kp 13 Clcultion for ll blocks: Inclintion of the slip surfce i Inclintion of the ground wter tble w,i Ground wter tble Height of the ground wter tble h i [ ] Reduced height of the ground wter tble h r, i Pore pressure u i [ kp ] [ ] [ ] [ ]

12 Tble 8 Inclintions of the slip surfces nd pore pressures Clcultion of the sliding oent. The weight of the individul blocks including forces of the lod ct on the horizontl r fro the xis of the block to the centre of the circle slip surfce (to the point O). The rs of the forces re clculted fro the edge of the slip surfce x, z 8.00; 5.00 ). The overll clcultion is shown in Tble 9. An exple of the clcultion for ( Z sp block 13 is done. Clcultion of the r of the force: b 1.0 r, 13 X zsp X O i b Clcultion of the sliding oent: W f r ( ) kn, ,13 / Clcultion for ll blocks: r, i [ ] Sliding oent, i [ kn/ ] r, i [ ] Sliding oent, i [ kn/ ] Tble 9 Sliding oents 1

13 Resultnt sliding oent: 0 i1, i kn / Result fro the GEO5 Slope Stbility progr: kn / Resultnt ctive force: 0, i i F kn/ R Result fro the GEO5 Slope Stbility progr: F kn/ Clcultion of the resisting oent. The clcultion of the resisting oents is itertive becuse the clcultion of the resisting oents ccording to Bishop s ethod depends on the sfety fctor. In the 1 st itertion the sfety fctor is considered. Five itertions re done in the hnd-de clcultion. The overll clcultion is shown in Tble 10. An exple of the clcultion for block 13 is done. Clcultion of the sfety fctor in the individul itertions: p Clcultion of the resisting oent, : c b13 ( W13 f13 u13 b13 ) tn( ) p,13 R tn( ) sin( 13 ) cos( 13 ) ( ) tn(7.00) tn(7.00) sin( ) cos(7.719 ) p, result of the 1 st itertion Clcultion of the resisting oent, : ( ) tn(7.00) tn(7.00) sin( ) cos(7.719 ) p, result of the nd itertion Clcultion of the resisting oent, : ( ) tn(7.00) tn(7.00) sin( ) cos(7.719 ) p, kn / kn / kn / 13

14 result of the 3 rd itertion Clcultion of the resisting oent, : ( ) tn(7.00) tn(7.00) sin( ) cos(7.719 ) p, result of the 4 th itertion Clcultion of the resisting oent, : ( ) tn(7.00) tn(7.00) sin( ) cos(7.719 ) p, result of the 5 th itertion Clcultion for ll blocks: kn / kn / 1 st itertion nd itertion 3 rd itertion 4 th itertion 5 th itertion p, i 1.546, p i p i p i p i [ kn/ ] [ kn/ ] [ kn/ ] [ kn/ ] [ kn/ ] TOTAL , Tble 10 Resisting oents nd sfety fctors, 1.554,

15 Resultnt resisting oent in 5 th itertion: 0 p p, i kn / i1 Result fro the GEO5 Slope Stbility progr: p kn / Resultnt pssive force: 0 p, i i Fp kn/ R Result fro the GEO5 Slope Stbility progr: F p kn/ Clcultion of the sfety fctor in 5 th itertion: p , SATISFACTORY Result fro the GEO5 Slope Stbility progr: 1. 56, SATISFACTORY Verifiction of the Stbility of Anchored Slope In Figure 4, n exple of nchored slope in nd stge is shown. The nchor force is F A knnd the spcing is b A. 00. The position of the nchor hed is x, z 16.00; 9.00 H nchor. Clcultion of the sliding oent. The nchor cts s pssive eleent, which ens tht ctive oents will be the se s in the 1 st stge. Resultnt sliding oent: 0 i1, i kn / Result fro the GEO5 Slope Stbility progr: kn / Resultnt ctive force: 0, i i F kn/ R Result fro the GEO5 Slope Stbility progr: F kn/ 15

16 Clcultion of the resisting oent. The nchor force enters the clcultion of the resisting oents. The clcultion of the resisting oents is itertive becuse the clcultion of the resisting oents using the Bishop s ethod depends on the sfety fctor. In the 1 st itertion the sfety fctor is Five itertions re done in the hnd-de clcultion. The overll clcultion is shown in Tble 11. An exple of the clcultion for block 13 is done. Anchor force t 1 : ' FA F kn b.00 A / A Clcultion of the r of the nchor force: ra Z O Z nchor Resisting oent of the nchor: ' p, A FA ra kn/ Clcultion of the sfety fctor in the individul itertions: p Clcultion of the resisting oent, : p,13 c b13 ( W cos( f13 u13 b13 ) tn( ) R tn( ) sin( 13 ) ) ( ,00) tn( 7.00) tn( 7.00) sin( ) cos(7.719 ) p, result of the 1 st itertion Clcultion of the resisting oent, : ( ) tn(7.00) tn(7.00) sin( ) cos(7.719 ) result of the nd itertion p, Clcultion of the resisting oent, : ( ) tn(7.00) tn(7.00) sin( ) cos(7.719 ) 1.66 p, result of the 3 rd itertion kn / kn / kn / 16

17 Clcultion of the resisting oent, : ( ) tn(7.00) tn(7.00) sin( ) cos(7.719 ) p, result of the 4 th itertion Clcultion of the resisting oent, : ( ) tn(7.00) tn(7.00) sin( ) cos(7.719 ) p, result of the 5 th itertion Clcultion for ll blocks: kn / kn / 1 st itertion nd itertion 3 rd itertion 4 th itertion 5 th itertion p, i 1.641, p i p i p i p i [ kn/ ] [ kn/ ] [ kn/ ] [ kn/ ] [ kn/ ] Anchor TOTAL , Tble 11 Resisting oents nd sfety fctors, 1.665,

18 Resultnt resisting oent in 5 th itertion: 0 p p, i p, A kn / i1 Result fro the GEO5 Slope Stbility progr: p kn / Resultnt pssive force: 0 p, i p, A i Fp kn/ R Result fro the GEO5 Slope Stbility progr: F p kn/ Clcultion of the sfety fctor in 5 th itertion: p , SATISFACTORY Result fro the GEO5 Slope Stbility progr: 1. 67, SATISFACTORY 18

Verification Analysis of the Redi Rock Wall

Verification Analysis of the Redi Rock Wall Verifiction Mnul no. Updte 06/06 Verifiction Anlysis of the Redi Rock Wll Progr File Redi Rock Wll Deo_v_etric_en_0.grr In this verifiction nul you will find hnd-de verifiction nlysis of the Redi Rock

More information

Solutions to Supplementary Problems

Solutions to Supplementary Problems Solutions to Supplementry Problems Chpter 8 Solution 8.1 Step 1: Clculte the line of ction ( x ) of the totl weight ( W ).67 m W = 5 kn W 1 = 16 kn 3.5 m m W 3 = 144 kn Q 4m Figure 8.10 Tking moments bout

More information

EFFECTIVE BUCKLING LENGTH OF COLUMNS IN SWAY FRAMEWORKS: COMPARISONS

EFFECTIVE BUCKLING LENGTH OF COLUMNS IN SWAY FRAMEWORKS: COMPARISONS IV EFFETIVE BUING ENGTH OF OUMN IN WAY FRAMEWOR: OMARION Ojectives In the present context, two different pproches re eployed to deterine the vlue the effective uckling length eff n c of colun n c out the

More information

A wire. 100 kg. Fig. 1.1

A wire. 100 kg. Fig. 1.1 1 Fig. 1.1 shows circulr cylinder of mss 100 kg being rised by light, inextensible verticl wire. There is negligible ir resistnce. wire 100 kg Fig. 1.1 (i) lculte the ccelertion of the cylinder when the

More information

Chapter 5 Bending Moments and Shear Force Diagrams for Beams

Chapter 5 Bending Moments and Shear Force Diagrams for Beams Chpter 5 ending Moments nd Sher Force Digrms for ems n ddition to illy loded brs/rods (e.g. truss) nd torsionl shfts, the structurl members my eperience some lods perpendiculr to the is of the bem nd will

More information

JUST THE MATHS SLIDES NUMBER INTEGRATION APPLICATIONS 12 (Second moments of an area (B)) A.J.Hobson

JUST THE MATHS SLIDES NUMBER INTEGRATION APPLICATIONS 12 (Second moments of an area (B)) A.J.Hobson JUST THE MATHS SLIDES NUMBER 13.12 INTEGRATION APPLICATIONS 12 (Second moments of n re (B)) b A.J.Hobson 13.12.1 The prllel xis theorem 13.12.2 The perpendiculr xis theorem 13.12.3 The rdius of grtion

More information

OXFORD H i g h e r E d u c a t i o n Oxford University Press, All rights reserved.

OXFORD H i g h e r E d u c a t i o n Oxford University Press, All rights reserved. Renshw: Mths for Econoics nswers to dditionl exercises Exercise.. Given: nd B 5 Find: () + B + B 7 8 (b) (c) (d) (e) B B B + B T B (where 8 B 6 B 6 8 B + B T denotes the trnspose of ) T 8 B 5 (f) (g) B

More information

C D o F. 30 o F. Wall String. 53 o. F y A B C D E. m 2. m 1. m a. v Merry-go round. Phy 231 Sp 03 Homework #8 Page 1 of 4

C D o F. 30 o F. Wall String. 53 o. F y A B C D E. m 2. m 1. m a. v Merry-go round. Phy 231 Sp 03 Homework #8 Page 1 of 4 Phy 231 Sp 3 Hoework #8 Pge 1 of 4 8-1) rigid squre object of negligible weight is cted upon by the forces 1 nd 2 shown t the right, which pull on its corners The forces re drwn to scle in ters of the

More information

Applications of Bernoulli s theorem. Lecture - 7

Applications of Bernoulli s theorem. Lecture - 7 Applictions of Bernoulli s theorem Lecture - 7 Prcticl Applictions of Bernoulli s Theorem The Bernoulli eqution cn be pplied to gret mny situtions not just the pipe flow we hve been considering up to now.

More information

Form 5 HKCEE 1990 Mathematics II (a 2n ) 3 = A. f(1) B. f(n) A. a 6n B. a 8n C. D. E. 2 D. 1 E. n. 1 in. If 2 = 10 p, 3 = 10 q, express log 6

Form 5 HKCEE 1990 Mathematics II (a 2n ) 3 = A. f(1) B. f(n) A. a 6n B. a 8n C. D. E. 2 D. 1 E. n. 1 in. If 2 = 10 p, 3 = 10 q, express log 6 Form HK 9 Mthemtics II.. ( n ) =. 6n. 8n. n 6n 8n... +. 6.. f(). f(n). n n If = 0 p, = 0 q, epress log 6 in terms of p nd q.. p q. pq. p q pq p + q Let > b > 0. If nd b re respectivel the st nd nd terms

More information

Contact Analysis on Large Negative Clearance Four-point Contact Ball Bearing

Contact Analysis on Large Negative Clearance Four-point Contact Ball Bearing Avilble online t www.sciencedirect.co rocedi ngineering 7 0 74 78 The Second SR Conference on ngineering Modelling nd Siultion CMS 0 Contct Anlysis on Lrge Negtive Clernce Four-point Contct Bll Bering

More information

Discussion Question 1A P212, Week 1 P211 Review: 2-D Motion with Uniform Force

Discussion Question 1A P212, Week 1 P211 Review: 2-D Motion with Uniform Force Discussion Question 1A P1, Week 1 P11 Review: -D otion with Unifor Force The thetics nd phsics of the proble below re siilr to probles ou will encounter in P1, where the force is due to the ction of n

More information

SOLUTIONS TO CONCEPTS CHAPTER 6

SOLUTIONS TO CONCEPTS CHAPTER 6 SOLUIONS O CONCEPS CHAPE 6 1. Let ss of the block ro the freebody digr, 0...(1) velocity Agin 0 (fro (1)) g 4 g 4/g 4/10 0.4 he co-efficient of kinetic friction between the block nd the plne is 0.4. Due

More information

Design Data 1M. Highway Live Loads on Concrete Pipe

Design Data 1M. Highway Live Loads on Concrete Pipe Design Dt 1M Highwy Live Lods on Concrete Pipe Foreword Thick, high-strength pvements designed for hevy truck trffic substntilly reduce the pressure trnsmitted through wheel to the subgrde nd consequently,

More information

Lecture 8. Newton s Laws. Applications of the Newton s Laws Problem-Solving Tactics. Physics 105; Fall Inertial Frames: T = mg

Lecture 8. Newton s Laws. Applications of the Newton s Laws Problem-Solving Tactics. Physics 105; Fall Inertial Frames: T = mg Lecture 8 Applictions of the ewton s Lws Problem-Solving ctics http://web.njit.edu/~sireno/ ewton s Lws I. If no net force ocects on body, then the body s velocity cnnot chnge. II. he net force on body

More information

Discussion Introduction P212, Week 1 The Scientist s Sixth Sense. Knowing what the answer will look like before you start.

Discussion Introduction P212, Week 1 The Scientist s Sixth Sense. Knowing what the answer will look like before you start. Discussion Introduction P1, Week 1 The Scientist s Sith Sense As scientist or engineer, uch of your job will be perforing clcultions, nd using clcultions perfored by others. You ll be doing plenty of tht

More information

Correct answer: 0 m/s 2. Explanation: 8 N

Correct answer: 0 m/s 2. Explanation: 8 N Version 001 HW#3 - orces rts (00223) 1 his print-out should hve 15 questions. Multiple-choice questions my continue on the next column or pge find ll choices before nswering. Angled orce on Block 01 001

More information

Physics. Friction.

Physics. Friction. hysics riction www.testprepkrt.co Tble of Content. Introduction.. Types of friction. 3. Grph of friction. 4. riction is cuse of otion. 5. dvntges nd disdvntges of friction. 6. Methods of chnging friction.

More information

Jackson 2.26 Homework Problem Solution Dr. Christopher S. Baird University of Massachusetts Lowell

Jackson 2.26 Homework Problem Solution Dr. Christopher S. Baird University of Massachusetts Lowell Jckson 2.26 Homework Problem Solution Dr. Christopher S. Bird University of Msschusetts Lowell PROBLEM: The two-dimensionl region, ρ, φ β, is bounded by conducting surfces t φ =, ρ =, nd φ = β held t zero

More information

First Semester Review Calculus BC

First Semester Review Calculus BC First Semester Review lculus. Wht is the coordinte of the point of inflection on the grph of Multiple hoice: No lcultor y 3 3 5 4? 5 0 0 3 5 0. The grph of piecewise-liner function f, for 4, is shown below.

More information

Phys101 Lecture 4,5 Dynamics: Newton s Laws of Motion

Phys101 Lecture 4,5 Dynamics: Newton s Laws of Motion Phys101 Lecture 4,5 Dynics: ewton s Lws of Motion Key points: ewton s second lw is vector eqution ction nd rection re cting on different objects ree-ody Digrs riction Inclines Ref: 4-1,2,3,4,5,6,7,8,9.

More information

SOLUTIONS TO CONCEPTS CHAPTER 10

SOLUTIONS TO CONCEPTS CHAPTER 10 SOLUTIONS TO CONCEPTS CHPTE 0. 0 0 ; 00 rev/s ; ; 00 rd/s 0 t t (00 )/4 50 rd /s or 5 rev/s 0 t + / t 8 50 400 rd 50 rd/s or 5 rev/s s 400 rd.. 00 ; t 5 sec / t 00 / 5 8 5 40 rd/s 0 rev/s 8 rd/s 4 rev/s

More information

ME 141. Lecture 10: Kinetics of particles: Newton s 2 nd Law

ME 141. Lecture 10: Kinetics of particles: Newton s 2 nd Law ME 141 Engineering Mechnics Lecture 10: Kinetics of prticles: Newton s nd Lw Ahmd Shhedi Shkil Lecturer, Dept. of Mechnicl Engg, BUET E-mil: sshkil@me.buet.c.bd, shkil6791@gmil.com Website: techer.buet.c.bd/sshkil

More information

We know that if f is a continuous nonnegative function on the interval [a, b], then b

We know that if f is a continuous nonnegative function on the interval [a, b], then b 1 Ares Between Curves c 22 Donld Kreider nd Dwight Lhr We know tht if f is continuous nonnegtive function on the intervl [, b], then f(x) dx is the re under the grph of f nd bove the intervl. We re going

More information

Homework: 5, 9, 19, 25, 31, 34, 39 (p )

Homework: 5, 9, 19, 25, 31, 34, 39 (p ) Hoework: 5, 9, 19, 5, 31, 34, 39 (p 130-134) 5. A 3.0 kg block is initilly t rest on horizontl surfce. A force of gnitude 6.0 nd erticl force P re then pplied to the block. The coefficients of friction

More information

PHYS 601 HW3 Solution

PHYS 601 HW3 Solution 3.1 Norl force using Lgrnge ultiplier Using the center of the hoop s origin, we will describe the position of the prticle with conventionl polr coordintes. The Lgrngin is therefore L = 1 2 ṙ2 + 1 2 r2

More information

Job No. Sheet 1 of 8 Rev B. Made by IR Date Aug Checked by FH/NB Date Oct Revised by MEB Date April 2006

Job No. Sheet 1 of 8 Rev B. Made by IR Date Aug Checked by FH/NB Date Oct Revised by MEB Date April 2006 Job o. Sheet 1 of 8 Rev B 10, Route de Limours -78471 St Rémy Lès Chevreuse Cedex rnce Tel : 33 (0)1 30 85 5 00 x : 33 (0)1 30 5 75 38 CLCULTO SHEET Stinless Steel Vloristion Project Design Exmple 5 Welded

More information

Purpose of the experiment

Purpose of the experiment Newton s Lws II PES 6 Advnced Physics Lb I Purpose of the experiment Exmine two cses using Newton s Lws. Sttic ( = 0) Dynmic ( 0) fyi fyi Did you know tht the longest recorded flight of chicken is thirteen

More information

E S dition event Vector Mechanics for Engineers: Dynamics h Due, next Wednesday, 07/19/2006! 1-30

E S dition event Vector Mechanics for Engineers: Dynamics h Due, next Wednesday, 07/19/2006! 1-30 Vector Mechnics for Engineers: Dynmics nnouncement Reminders Wednesdy s clss will strt t 1:00PM. Summry of the chpter 11 ws posted on website nd ws sent you by emil. For the students, who needs hrdcopy,

More information

HW Solutions # MIT - Prof. Kowalski. Friction, circular dynamics, and Work-Kinetic Energy.

HW Solutions # MIT - Prof. Kowalski. Friction, circular dynamics, and Work-Kinetic Energy. HW Solutions # 5-8.01 MIT - Prof. Kowlski Friction, circulr dynmics, nd Work-Kinetic Energy. 1) 5.80 If the block were to remin t rest reltive to the truck, the friction force would need to cuse n ccelertion

More information

Year 12 Mathematics Extension 2 HSC Trial Examination 2014

Year 12 Mathematics Extension 2 HSC Trial Examination 2014 Yer Mthemtics Etension HSC Tril Emintion 04 Generl Instructions. Reding time 5 minutes Working time hours Write using blck or blue pen. Blck pen is preferred. Bord-pproved clcultors my be used A tble of

More information

k ) and directrix x = h p is A focal chord is a line segment which passes through the focus of a parabola and has endpoints on the parabola.

k ) and directrix x = h p is A focal chord is a line segment which passes through the focus of a parabola and has endpoints on the parabola. Stndrd Eqution of Prol with vertex ( h, k ) nd directrix y = k p is ( x h) p ( y k ) = 4. Verticl xis of symmetry Stndrd Eqution of Prol with vertex ( h, k ) nd directrix x = h p is ( y k ) p( x h) = 4.

More information

UCSD Phys 4A Intro Mechanics Winter 2016 Ch 4 Solutions

UCSD Phys 4A Intro Mechanics Winter 2016 Ch 4 Solutions USD Phys 4 Intro Mechnics Winter 06 h 4 Solutions 0. () he 0.0 k box restin on the tble hs the free-body dir shown. Its weiht 0.0 k 9.80 s 96 N. Since the box is t rest, the net force on is the box ust

More information

Electromagnetism Answers to Problem Set 10 Spring 2006

Electromagnetism Answers to Problem Set 10 Spring 2006 Electromgnetism 76 Answers to Problem Set 1 Spring 6 1. Jckson Prob. 5.15: Shielded Bifilr Circuit: Two wires crrying oppositely directed currents re surrounded by cylindricl shell of inner rdius, outer

More information

Advanced Computational Analysis

Advanced Computational Analysis Advnced Computtionl Anlysis REPORT REPORT NO: S2149-2 Revision A Title: Closed-Form Anlysis Of Forces And Moments In Bungee Trmpoline Structure Client: Mr Jmes Okey Author: Dr M Lcey BSc PhD CEng F I Mech

More information

JUST THE MATHS UNIT NUMBER INTEGRATION APPLICATIONS 8 (First moments of a volume) A.J.Hobson

JUST THE MATHS UNIT NUMBER INTEGRATION APPLICATIONS 8 (First moments of a volume) A.J.Hobson JUST THE MATHS UNIT NUMBER 3.8 INTEGRATIN APPLICATINS 8 (First moments of volume) b A.J.Hobson 3.8. Introduction 3.8. First moment of volume of revolution bout plne through the origin, perpendiculr to

More information

ELE B7 Power Systems Engineering. Power System Components Modeling

ELE B7 Power Systems Engineering. Power System Components Modeling Power Systems Engineering Power System Components Modeling Section III : Trnsformer Model Power Trnsformers- CONSTRUCTION Primry windings, connected to the lternting voltge source; Secondry windings, connected

More information

DYNAMIC EARTH PRESSURE SIMULATION BY SINGLE DEGREE OF FREEDOM SYSTEM

DYNAMIC EARTH PRESSURE SIMULATION BY SINGLE DEGREE OF FREEDOM SYSTEM 13 th World Conference on Erthque Engineering Vncouver, B.C., Cnd August 1-6, 2004 per No. 2663 DYNAMIC EARTH RESSURE SIMULATION BY SINGLE DEGREE OF FREEDOM SYSTEM Arsln GHAHRAMANI 1, Seyyed Ahmd ANVAR

More information

Summer Work Packet for MPH Math Classes

Summer Work Packet for MPH Math Classes Summer Work Pcket for MPH Mth Clsses Students going into Pre-clculus AC Sept. 018 Nme: This pcket is designed to help students sty current with their mth skills. Ech mth clss expects certin level of number

More information

BME 207 Introduction to Biomechanics Spring 2018

BME 207 Introduction to Biomechanics Spring 2018 April 6, 28 UNIVERSITY O RHODE ISAND Deprtment of Electricl, Computer nd Biomedicl Engineering BME 27 Introduction to Biomechnics Spring 28 Homework 8 Prolem 14.6 in the textook. In ddition to prts -e,

More information

Mathematics of Motion II Projectiles

Mathematics of Motion II Projectiles Chmp+ Fll 2001 Dn Stump 1 Mthemtics of Motion II Projectiles Tble of vribles t time v velocity, v 0 initil velocity ccelertion D distnce x position coordinte, x 0 initil position x horizontl coordinte

More information

(b) Let S 1 : f(x, y, z) = (x a) 2 + (y b) 2 + (z c) 2 = 1, this is a level set in 3D, hence

(b) Let S 1 : f(x, y, z) = (x a) 2 + (y b) 2 + (z c) 2 = 1, this is a level set in 3D, hence Problem ( points) Find the vector eqution of the line tht joins points on the two lines L : r ( + t) i t j ( + t) k L : r t i + (t ) j ( + t) k nd is perpendiculr to both those lines. Find the set of ll

More information

JUST THE MATHS UNIT NUMBER INTEGRATION APPLICATIONS 12 (Second moments of an area (B)) A.J.Hobson

JUST THE MATHS UNIT NUMBER INTEGRATION APPLICATIONS 12 (Second moments of an area (B)) A.J.Hobson JUST THE MATHS UNIT NUMBE 13.1 INTEGATION APPLICATIONS 1 (Second moments of n re (B)) b A.J.Hobson 13.1.1 The prllel xis theorem 13.1. The perpendiculr xis theorem 13.1.3 The rdius of grtion of n re 13.1.4

More information

Problems (Motion Relative to Rotating Axes)

Problems (Motion Relative to Rotating Axes) 1. The disk rolls without slipping on the roblems (Motion Reltie to Rotting xes) horizontl surfce, nd t the instnt represented, the center O hs the elocity nd ccelertion shown in the figure. For this instnt,

More information

Individual Events I3 a 10 I4. d 90 angle 57 d Group Events. d 220 Probability

Individual Events I3 a 10 I4. d 90 angle 57 d Group Events. d 220 Probability Answers: (98-8 HKMO Finl Events) Creted by: Mr. Frncis Hung Lst updted: 8 Jnury 08 I 800 I Individul Events I 0 I4 no. of routes 6 I5 + + b b 0 b b c *8 missing c 0 c c See the remrk 600 d d 90 ngle 57

More information

MORE FUNCTION GRAPHING; OPTIMIZATION. (Last edited October 28, 2013 at 11:09pm.)

MORE FUNCTION GRAPHING; OPTIMIZATION. (Last edited October 28, 2013 at 11:09pm.) MORE FUNCTION GRAPHING; OPTIMIZATION FRI, OCT 25, 203 (Lst edited October 28, 203 t :09pm.) Exercise. Let n be n rbitrry positive integer. Give n exmple of function with exctly n verticl symptotes. Give

More information

50. Use symmetry to evaluate xx D is the region bounded by the square with vertices 5, Prove Property 11. y y CAS

50. Use symmetry to evaluate xx D is the region bounded by the square with vertices 5, Prove Property 11. y y CAS 68 CHAPTE MULTIPLE INTEGALS 46. e da, 49. Evlute tn 3 4 da, where,. [Hint: Eploit the fct tht is the disk with center the origin nd rdius is smmetric with respect to both es.] 5. Use smmetr to evlute 3

More information

Computer Graphics (CS 543) Lecture 3 (Part 1): Linear Algebra for Graphics (Points, Scalars, Vectors)

Computer Graphics (CS 543) Lecture 3 (Part 1): Linear Algebra for Graphics (Points, Scalars, Vectors) Coputer Grphics (CS 543) Lecture 3 (Prt ): Liner Alger for Grphics (Points, Sclrs, Vectors) Prof Enuel Agu Coputer Science Dept. Worcester Poltechnic Institute (WPI) Points, Sclrs nd Vectors Points, vectors

More information

CHAPTER 5 Newton s Laws of Motion

CHAPTER 5 Newton s Laws of Motion CHAPTER 5 Newton s Lws of Motion We ve been lerning kinetics; describing otion without understnding wht the cuse of the otion ws. Now we re going to lern dynics!! Nno otor 103 PHYS - 1 Isc Newton (1642-1727)

More information

Mathematics Extension 2

Mathematics Extension 2 S Y D N E Y B O Y S H I G H S C H O O L M O O R E P A R K, S U R R Y H I L L S 005 HIGHER SCHOOL CERTIFICATE TRIAL PAPER Mthemtics Extension Generl Instructions Totl Mrks 0 Reding Time 5 Minutes Attempt

More information

Solutions to Physics: Principles with Applications, 5/E, Giancoli Chapter 16 CHAPTER 16

Solutions to Physics: Principles with Applications, 5/E, Giancoli Chapter 16 CHAPTER 16 CHAPTER 16 1. The number of electrons is N = Q/e = ( 30.0 10 6 C)/( 1.60 10 19 C/electrons) = 1.88 10 14 electrons.. The mgnitude of the Coulomb force is Q /r. If we divide the epressions for the two forces,

More information

MASTER CLASS PROGRAM UNIT 4 SPECIALIST MATHEMATICS WEEK 11 WRITTEN EXAMINATION 2 SOLUTIONS SECTION 1 MULTIPLE CHOICE QUESTIONS

MASTER CLASS PROGRAM UNIT 4 SPECIALIST MATHEMATICS WEEK 11 WRITTEN EXAMINATION 2 SOLUTIONS SECTION 1 MULTIPLE CHOICE QUESTIONS MASTER CLASS PROGRAM UNIT 4 SPECIALIST MATHEMATICS WEEK WRITTEN EXAMINATION SOLUTIONS FOR ERRORS AND UPDATES, PLEASE VISIT WWW.TSFX.COM.AU/MC-UPDATES SECTION MULTIPLE CHOICE QUESTIONS QUESTION QUESTION

More information

Multiple Integrals. Review of Single Integrals. Planar Area. Volume of Solid of Revolution

Multiple Integrals. Review of Single Integrals. Planar Area. Volume of Solid of Revolution Multiple Integrls eview of Single Integrls eding Trim 7.1 eview Appliction of Integrls: Are 7. eview Appliction of Integrls: olumes 7.3 eview Appliction of Integrls: Lengths of Curves Assignment web pge

More information

The Atwood Machine OBJECTIVE INTRODUCTION APPARATUS THEORY

The Atwood Machine OBJECTIVE INTRODUCTION APPARATUS THEORY The Atwood Mchine OBJECTIVE To derive the ening of Newton's second lw of otion s it pplies to the Atwood chine. To explin how ss iblnce cn led to the ccelertion of the syste. To deterine the ccelertion

More information

In Mathematics for Construction, we learnt that

In Mathematics for Construction, we learnt that III DOUBLE INTEGATION THE ANTIDEIVATIVE OF FUNCTIONS OF VAIABLES In Mthemtics or Construction, we lernt tht the indeinite integrl is the ntiderivtive o ( d ( Double Integrtion Pge Hence d d ( d ( The ntiderivtive

More information

Level I MAML Olympiad 2001 Page 1 of 6 (A) 90 (B) 92 (C) 94 (D) 96 (E) 98 (A) 48 (B) 54 (C) 60 (D) 66 (E) 72 (A) 9 (B) 13 (C) 17 (D) 25 (E) 38

Level I MAML Olympiad 2001 Page 1 of 6 (A) 90 (B) 92 (C) 94 (D) 96 (E) 98 (A) 48 (B) 54 (C) 60 (D) 66 (E) 72 (A) 9 (B) 13 (C) 17 (D) 25 (E) 38 Level I MAML Olympid 00 Pge of 6. Si students in smll clss took n em on the scheduled dte. The verge of their grdes ws 75. The seventh student in the clss ws ill tht dy nd took the em lte. When her score

More information

Multiple Integrals. Review of Single Integrals. Planar Area. Volume of Solid of Revolution

Multiple Integrals. Review of Single Integrals. Planar Area. Volume of Solid of Revolution Multiple Integrls eview of Single Integrls eding Trim 7.1 eview Appliction of Integrls: Are 7. eview Appliction of Integrls: Volumes 7.3 eview Appliction of Integrls: Lengths of Curves Assignment web pge

More information

Mathematics Extension 2

Mathematics Extension 2 00 HIGHER SCHOOL CERTIFICATE EXAMINATION Mthemtics Etension Generl Instructions Reding time 5 minutes Working time hours Write using blck or blue pen Bord-pproved clcultors my be used A tble of stndrd

More information

Jackson 2.7 Homework Problem Solution Dr. Christopher S. Baird University of Massachusetts Lowell

Jackson 2.7 Homework Problem Solution Dr. Christopher S. Baird University of Massachusetts Lowell Jckson.7 Homework Problem Solution Dr. Christopher S. Bird University of Msschusetts Lowell PROBLEM: Consider potentil problem in the hlf-spce defined by, with Dirichlet boundry conditions on the plne

More information

PART 1 MULTIPLE CHOICE Circle the appropriate response to each of the questions below. Each question has a value of 1 point.

PART 1 MULTIPLE CHOICE Circle the appropriate response to each of the questions below. Each question has a value of 1 point. PART MULTIPLE CHOICE Circle the pproprite response to ech of the questions below. Ech question hs vlue of point.. If in sequence the second level difference is constnt, thn the sequence is:. rithmetic

More information

13.4 Work done by Constant Forces

13.4 Work done by Constant Forces 13.4 Work done by Constnt Forces We will begin our discussion of the concept of work by nlyzing the motion of n object in one dimension cted on by constnt forces. Let s consider the following exmple: push

More information

PROPERTIES OF AREAS In general, and for an irregular shape, the definition of the centroid at position ( x, y) is given by

PROPERTIES OF AREAS In general, and for an irregular shape, the definition of the centroid at position ( x, y) is given by PROPERTES OF RES Centroid The concept of the centroid is prol lred fmilir to ou For plne shpe with n ovious geometric centre, (rectngle, circle) the centroid is t the centre f n re hs n is of smmetr, the

More information

1 (=0.5) I3 a 7 I4 a 15 I5 a (=0.5) c 4 N 10 1 (=0.5) N 6 A 52 S 2

1 (=0.5) I3 a 7 I4 a 15 I5 a (=0.5) c 4 N 10 1 (=0.5) N 6 A 52 S 2 Answers: (98-84 HKMO Finl Events) Creted by Mr. Frncis Hung Lst updted: December 05 Individul Events SI 900 I 0 I (=0.5) I 7 I4 5 I5 80 b 7 b b 5 b 6 b 8 b 4 c c 4 c 0 x (=0.5) c 4 N 0 d 9 d 5 d 5 y d

More information

1. a) Describe the principle characteristics and uses of the following types of PV cell: Single Crystal Silicon Poly Crystal Silicon

1. a) Describe the principle characteristics and uses of the following types of PV cell: Single Crystal Silicon Poly Crystal Silicon 2001 1. ) Describe the principle chrcteristics nd uses of the following types of PV cell: Single Crystl Silicon Poly Crystl Silicon Amorphous Silicon CIS/CIGS Gllium Arsenide Multijunction (12 mrks) b)

More information

State space systems analysis (continued) Stability. A. Definitions A system is said to be Asymptotically Stable (AS) when it satisfies

State space systems analysis (continued) Stability. A. Definitions A system is said to be Asymptotically Stable (AS) when it satisfies Stte spce systems nlysis (continued) Stbility A. Definitions A system is sid to be Asymptoticlly Stble (AS) when it stisfies ut () = 0, t > 0 lim xt () 0. t A system is AS if nd only if the impulse response

More information

Linear Inequalities: Each of the following carries five marks each: 1. Solve the system of equations graphically.

Linear Inequalities: Each of the following carries five marks each: 1. Solve the system of equations graphically. Liner Inequlities: Ech of the following crries five mrks ech:. Solve the system of equtions grphiclly. x + 2y 8, 2x + y 8, x 0, y 0 Solution: Considerx + 2y 8.. () Drw the grph for x + 2y = 8 by line.it

More information

PhysicsAndMathsTutor.com

PhysicsAndMathsTutor.com 1. A uniform circulr disc hs mss m, centre O nd rdius. It is free to rotte bout fixed smooth horizontl xis L which lies in the sme plne s the disc nd which is tngentil to the disc t the point A. The disc

More information

APPLICATIONS OF DEFINITE INTEGRALS

APPLICATIONS OF DEFINITE INTEGRALS Chpter 6 APPICATIONS OF DEFINITE INTEGRAS OVERVIEW In Chpter 5 we discovered the connection etween Riemnn sums ssocited with prtition P of the finite closed intervl [, ] nd the process of integrtion. We

More information

DEFINITION OF ASSOCIATIVE OR DIRECT PRODUCT AND ROTATION OF VECTORS

DEFINITION OF ASSOCIATIVE OR DIRECT PRODUCT AND ROTATION OF VECTORS 3 DEFINITION OF ASSOCIATIVE OR DIRECT PRODUCT AND ROTATION OF VECTORS This chpter summrizes few properties of Cli ord Algebr nd describe its usefulness in e ecting vector rottions. 3.1 De nition of Associtive

More information

Vadose Zone Hydrology

Vadose Zone Hydrology Objectives Vdose Zone Hydrology 1. Review bsic concepts nd terminology of soil physics. 2. Understnd the role of wter-tble dynmics in GW-SW interction. Drcy s lw is useful in region A. Some knowledge of

More information

west (mrw3223) HW 24 lyle (16001) 1

west (mrw3223) HW 24 lyle (16001) 1 west (mrw3223) HW 24 lyle (16001) 1 This print-out should hve 30 questions. Multiple-choice questions my continue on the next column or pge find ll choices before nswering. Reding ssignment: Hecht, sections

More information

- 5 - TEST 2. This test is on the final sections of this session's syllabus and. should be attempted by all students.

- 5 - TEST 2. This test is on the final sections of this session's syllabus and. should be attempted by all students. - 5 - TEST 2 This test is on the finl sections of this session's syllbus nd should be ttempted by ll students. Anything written here will not be mrked. - 6 - QUESTION 1 [Mrks 22] A thin non-conducting

More information

16 Newton s Laws #3: Components, Friction, Ramps, Pulleys, and Strings

16 Newton s Laws #3: Components, Friction, Ramps, Pulleys, and Strings Chpter 16 Newton s Lws #3: Components, riction, Rmps, Pulleys, nd Strings 16 Newton s Lws #3: Components, riction, Rmps, Pulleys, nd Strings When, in the cse of tilted coordinte system, you brek up the

More information

ragsdale (zdr82) HW2 ditmire (58335) 1

ragsdale (zdr82) HW2 ditmire (58335) 1 rgsdle (zdr82) HW2 ditmire (58335) This print-out should hve 22 questions. Multiple-choice questions my continue on the next column or pge find ll choices before nswering. 00 0.0 points A chrge of 8. µc

More information

Physics Honors. Final Exam Review Free Response Problems

Physics Honors. Final Exam Review Free Response Problems Physics Honors inl Exm Review ree Response Problems m t m h 1. A 40 kg mss is pulled cross frictionless tble by string which goes over the pulley nd is connected to 20 kg mss.. Drw free body digrm, indicting

More information

Log1 Contest Round 3 Theta Individual. 4 points each 1 What is the sum of the first 5 Fibonacci numbers if the first two are 1, 1?

Log1 Contest Round 3 Theta Individual. 4 points each 1 What is the sum of the first 5 Fibonacci numbers if the first two are 1, 1? 008 009 Log1 Contest Round Thet Individul Nme: points ech 1 Wht is the sum of the first Fiboncci numbers if the first two re 1, 1? If two crds re drwn from stndrd crd deck, wht is the probbility of drwing

More information

AP Physics 1. Slide 1 / 71. Slide 2 / 71. Slide 3 / 71. Circular Motion. Topics of Uniform Circular Motion (UCM)

AP Physics 1. Slide 1 / 71. Slide 2 / 71. Slide 3 / 71. Circular Motion. Topics of Uniform Circular Motion (UCM) Slide 1 / 71 Slide 2 / 71 P Physics 1 irculr Motion 2015-12-02 www.njctl.org Topics of Uniform irculr Motion (UM) Slide 3 / 71 Kinemtics of UM lick on the topic to go to tht section Period, Frequency,

More information

PARABOLA EXERCISE 3(B)

PARABOLA EXERCISE 3(B) PARABOLA EXERCISE (B). Find eqution of the tngent nd norml to the prbol y = 6x t the positive end of the ltus rectum. Eqution of prbol y = 6x 4 = 6 = / Positive end of the Ltus rectum is(, ) =, Eqution

More information

An Introduction to Trigonometry

An Introduction to Trigonometry n Introduction to Trigonoetry First of ll, let s check out the right ngled tringle below. The LETTERS, B & C indicte the ngles nd the letters, b & c indicte the sides. c b It is iportnt to note tht side

More information

Space Curves. Recall the parametric equations of a curve in xy-plane and compare them with parametric equations of a curve in space.

Space Curves. Recall the parametric equations of a curve in xy-plane and compare them with parametric equations of a curve in space. Clculus 3 Li Vs Spce Curves Recll the prmetric equtions of curve in xy-plne nd compre them with prmetric equtions of curve in spce. Prmetric curve in plne x = x(t) y = y(t) Prmetric curve in spce x = x(t)

More information

MTH 4-16a Trigonometry

MTH 4-16a Trigonometry MTH 4-16 Trigonometry Level 4 [UNIT 5 REVISION SECTION ] I cn identify the opposite, djcent nd hypotenuse sides on right-ngled tringle. Identify the opposite, djcent nd hypotenuse in the following right-ngled

More information

Homework Assignment 6 Solution Set

Homework Assignment 6 Solution Set Homework Assignment 6 Solution Set PHYCS 440 Mrch, 004 Prolem (Griffiths 4.6 One wy to find the energy is to find the E nd D fields everywhere nd then integrte the energy density for those fields. We know

More information

Explain shortly the meaning of the following eight words in relation to shells structures.

Explain shortly the meaning of the following eight words in relation to shells structures. Delft University of Technology Fculty of Civil Engineering nd Geosciences Structurl Mechnics Section Write your nme nd study number t the top right-hnd of your work. Exm CIE4143 Shell Anlysis Tuesdy 15

More information

Name Class Date. Line AB is parallel to line CD. skew. ABDC } plane EFHG. In Exercises 4 7, use the diagram to name each of the following.

Name Class Date. Line AB is parallel to line CD. skew. ABDC } plane EFHG. In Exercises 4 7, use the diagram to name each of the following. Reteching Lines nd Angles Not ll lines nd plnes intersect. prllel plnes. prllel. } shows tht lines or plnes re prllel: < > < > A } ens Line A is prllel to line. skew. A } plne EFHG A plne FH } plne AEG

More information

SOLUTIONS TO CONCEPTS CHAPTER

SOLUTIONS TO CONCEPTS CHAPTER 1. m = kg S = 10m Let, ccelertion =, Initil velocity u = 0. S= ut + 1/ t 10 = ½ ( ) 10 = = 5 m/s orce: = = 5 = 10N (ns) SOLUIONS O CONCEPS CHPE 5 40000. u = 40 km/hr = = 11.11 m/s. 3600 m = 000 kg ; v

More information

99/105 Comparison of OrcaFlex with standard theoretical results

99/105 Comparison of OrcaFlex with standard theoretical results 99/105 Comprison of OrcFlex ith stndrd theoreticl results 1. Introduction A number of stndrd theoreticl results from literture cn be modelled in OrcFlex. Such cses re, by virtue of being theoreticlly solvble,

More information

ME 354 Tutorial, Week#11 Non-Reacting Mixtures Psychrometrics Applied to a Cooling Tower

ME 354 Tutorial, Week#11 Non-Reacting Mixtures Psychrometrics Applied to a Cooling Tower ME 5 Tutoril, Week# Non-Recting Mixtures Psychroetrics Applied to Cooling Toer Wter exiting the condenser of poer plnt t 5 C enters cooling toer ith ss flo rte of 5000 kg/s. A stre of cooled ter is returned

More information

ES.182A Topic 32 Notes Jeremy Orloff

ES.182A Topic 32 Notes Jeremy Orloff ES.8A Topic 3 Notes Jerem Orloff 3 Polr coordintes nd double integrls 3. Polr Coordintes (, ) = (r cos(θ), r sin(θ)) r θ Stndrd,, r, θ tringle Polr coordintes re just stndrd trigonometric reltions. In

More information

Date Lesson Text TOPIC Homework. Solving for Obtuse Angles QUIZ ( ) More Trig Word Problems QUIZ ( )

Date Lesson Text TOPIC Homework. Solving for Obtuse Angles QUIZ ( ) More Trig Word Problems QUIZ ( ) UNIT 5 TRIGONOMETRI RTIOS Dte Lesson Text TOPI Homework pr. 4 5.1 (48) Trigonometry Review WS 5.1 # 3 5, 9 11, (1, 13)doso pr. 6 5. (49) Relted ngles omplete lesson shell & WS 5. pr. 30 5.3 (50) 5.3 5.4

More information

Types of forces. Types of Forces

Types of forces. Types of Forces pes of orces pes of forces. orce of Grvit: his is often referred to s the weiht of n object. It is the ttrctive force of the erth. And is lws directed towrd the center of the erth. It hs nitude equl to

More information

Time : 3 hours 03 - Mathematics - March 2007 Marks : 100 Pg - 1 S E CT I O N - A

Time : 3 hours 03 - Mathematics - March 2007 Marks : 100 Pg - 1 S E CT I O N - A Time : hours 0 - Mthemtics - Mrch 007 Mrks : 100 Pg - 1 Instructions : 1. Answer ll questions.. Write your nswers ccording to the instructions given below with the questions.. Begin ech section on new

More information

Page 1. Motion in a Circle... Dynamics of Circular Motion. Motion in a Circle... Motion in a Circle... Discussion Problem 21: Motion in a Circle

Page 1. Motion in a Circle... Dynamics of Circular Motion. Motion in a Circle... Motion in a Circle... Discussion Problem 21: Motion in a Circle Dynics of Circulr Motion A boy ties rock of ss to the end of strin nd twirls it in the erticl plne. he distnce fro his hnd to the rock is. he speed of the rock t the top of its trectory is. Wht is the

More information

19 th INTERNATIONAL CONGRESS ON ACOUSTICS MADRID, 2-7 SEPTEMBER 2007 A NON-CONTACT SYSTEM FOR TRANSPORTING OBJECTS USING ULTRASONIC LEVITATION

19 th INTERNATIONAL CONGRESS ON ACOUSTICS MADRID, 2-7 SEPTEMBER 2007 A NON-CONTACT SYSTEM FOR TRANSPORTING OBJECTS USING ULTRASONIC LEVITATION 19 th INTERNATIONAL CONGRESS ON ACOUSTICS MADRID, -7 SEPTEMBER 007 A NON-CONTACT SYSTEM FOR TRANSPORTING OBJECTS USING ULTRASONIC LEVITATION PACS: 3.5.Uv Gudr, Tdeusz 1 ; Perkowski, Dniel ; Opielinski,

More information

Trigonometric Functions

Trigonometric Functions Exercise. Degrees nd Rdins Chpter Trigonometric Functions EXERCISE. Degrees nd Rdins 4. Since 45 corresponds to rdin mesure of π/4 rd, we hve: 90 = 45 corresponds to π/4 or π/ rd. 5 = 7 45 corresponds

More information

The Spring. Consider a spring, which we apply a force F A to either stretch it or compress it

The Spring. Consider a spring, which we apply a force F A to either stretch it or compress it The Spring Consider spring, which we pply force F A to either stretch it or copress it F A - unstretched -F A 0 F A k k is the spring constnt, units of N/, different for different terils, nuber of coils

More information

r = cos θ + 1. dt ) dt. (1)

r = cos θ + 1. dt ) dt. (1) MTHE 7 Proble Set 5 Solutions (A Crdioid). Let C be the closed curve in R whose polr coordintes (r, θ) stisfy () Sketch the curve C. r = cos θ +. (b) Find pretriztion t (r(t), θ(t)), t [, b], of C in polr

More information

p(t) dt + i 1 re it ireit dt =

p(t) dt + i 1 re it ireit dt = Note: This mteril is contined in Kreyszig, Chpter 13. Complex integrtion We will define integrls of complex functions long curves in C. (This is bit similr to [relvlued] line integrls P dx + Q dy in R2.)

More information

Problem Solving 7: Faraday s Law Solution

Problem Solving 7: Faraday s Law Solution MASSACHUSETTS NSTTUTE OF TECHNOLOGY Deprtment of Physics: 8.02 Prolem Solving 7: Frdy s Lw Solution Ojectives 1. To explore prticulr sitution tht cn led to chnging mgnetic flux through the open surfce

More information

INTRODUCTION. The three general approaches to the solution of kinetics problems are:

INTRODUCTION. The three general approaches to the solution of kinetics problems are: INTRODUCTION According to Newton s lw, prticle will ccelerte when it is subjected to unblnced forces. Kinetics is the study of the reltions between unblnced forces nd the resulting chnges in motion. The

More information

NOT TO SCALE. We can make use of the small angle approximations: if θ á 1 (and is expressed in RADIANS), then

NOT TO SCALE. We can make use of the small angle approximations: if θ á 1 (and is expressed in RADIANS), then 3. Stellr Prllx y terrestril stndrds, the strs re extremely distnt: the nerest, Proxim Centuri, is 4.24 light yers (~ 10 13 km) wy. This mens tht their prllx is extremely smll. Prllx is the pprent shifting

More information