civl%2131%)%stacs% - the university of memphis three.pdf · 2010-01-22 · the force f = 450 lb...

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1 CIVL 2131 Sta-cs Resolu-on of a vector along an axes system Cartesian Coordinates Madness takes its toll. Please have exact change. Objec-ves Resolve a resultant vector into its components given any pair of axes View and Review the Cartesian Coordinate/axis system as a basis for vector opera-ons January 22, 2010 Basic Vector Operations II 2

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CIVL  2131  -­‐  Sta-cs  Resolu-on  of  a  vector  along  an  axes  system  

Cartesian  Coordinates  

Madness  takes  its  toll.  Please  have  exact  change.    

Objec-ves  

  Resolve  a  resultant  vector  into  its  components  given  any  pair  of  axes  

  View  and  Review  the  Cartesian  Coordinate/axis  system  as  a  basis  for  vector  opera-ons  

January 22, 2010 Basic Vector Operations II 2

2

Tools  

  Law  of  Sines  

  Law  of  Cosines  

  Basic  Trigonometry  

  Pythagorean  Theorem  

January 22, 2010 Basic Vector Operations II 3

Review  

  Both  Forces  and  Moments  can  be  mathema-cally  represented  as  vectors  which  allows  us  to  have  both  a  consistent  representa-on  and  a  convenient  set  of  tools  for  manipula-ng  the  elements  for  analysis  

January 22, 2010 Basic Vector Operations II 4

3

Review  

  Vectors  are  mathema-cal  representa-ons  which  have  two  components  

 Magnitude  and  

 Direc-on  

January 22, 2010 Basic Vector Operations II 5

Resolving  a  Vector  along  Axes  

  This  -me  we  will  start  with  the  resultant  and  try  to  find  the  components  that  were  combined  to  get  the  resultant  

 We  will  need  to  be  given  the  direc-ons  of  the  axes  

  Remember,  they  do  not  need  to  look  like  our  typical  x  and  y  axes  

January 22, 2010 Basic Vector Operations II 6

4

Resolving  a  Vector  along  Axes  

  If  we  are  given  two  axes,  a’  and  b’  

January 22, 2010 Basic Vector Operations II 7

Resolving  a  Vector  along  Axes  

 We  would  like  to  find  the  components  of  the  vector  F  that  lie  along  the  a’  and  b’  axes  

January 22, 2010 Basic Vector Operations II 8

5

Resolving  a  Vector  along  Axes  

  The  a’  and  b’  axes  in  this  case  are  not  perpendicular  (though  them  may  appear  so  in  my  drawing)  

January 22, 2010 Basic Vector Operations II 9

Resolving  a  Vector  along  Axes  

 When  we  added  vectors,  we  chose  one  to  remain  sta-onary  and  one  to  move  

January 22, 2010 Basic Vector Operations II 10

6

Resolving  a  Vector  along  Axes  

  This  -me,  we  will  keep  one  axis  sta-onary  and  move  the  other  axis  so  that  it  intersects  with  the  head  of  the  resultant  

January 22, 2010 Basic Vector Operations II 11

Resolving  a  Vector  along  Axes  

  Keeping  the  a’  axis  fixed,  we  move  the  b’  axis  un-l  it  intersects  the  head  of  the  resultant  

January 22, 2010 Basic Vector Operations II 12

7

Resolving  a  Vector  along  Axes  

 Now,  one  of  the  components  will  go  from  the  tail  of  the  resultant  to  the  intersec-on  of  the  a’  axis  and  the  shiWed  b’  axis  

January 22, 2010 Basic Vector Operations II 13

Resolving  a  Vector  along  Axes  

  The  other  component  will  go  from  the  head  of  the  vector  just  constructed  to  the  head  of  the  resultant  

January 22, 2010 Basic Vector Operations II 14

8

Resolving  a  Vector  along  Axes  

  Knowing  some  magnitudes  and  angles,  you  can  then  find  the  components  using  trig  and  geometry  

January 22, 2010 Basic Vector Operations II 15

An  Example  Problem  

January 22, 2010 Basic Vector Operations II 16

F2-5. The force F = 450 lb acts on the frame. Resolve the force into components acting along members AB and AC, and determine the magnitude of each component.

9

17 January 22, 2010 Basic Vector Operations II 17

Cartesian  Coordinates  

  There  is  a  special  case  when  the  axes  are  perpendicular  

  The  coordinate  system  or  axis  system  in  this  case  is  known  as  a  Cartesian  system  

  This  is  the  case  that  we  typically  see  

  In  the  most  common  2D  case,  we  have  the  tradi-onal  x  and  y  axis  as  we  developed  in  the  previous  class  

18 January 22, 2010 Basic Vector Operations II 18

Cartesian  Coordinates  

  Combining  component  forces  in  this  type  of  system  allows  for  the  use  of  the  Pythagorean  formula  

10

19 January 22, 2010 Basic Vector Operations II 19

Cartesian  Coordinates  

  For  example,  consider  two  forces  F1  and  F2  whose  direc-ons  are  perpendicular  to  each  other  

20 January 22, 2010 Basic Vector Operations II 20

Cartesian  Coordinates  

 We  can  extend  their  lines  of  ac-on  to  the  point  where  they  intersect  

  Remember  that  force  vectors  are  sliding  vectors  and  so  they  can  be  moved  anywhere  along  their  line  of  ac-on.  

The  line  of  ac-on  of  a  vector  is  the  line  that  would  be  generated  if  we  extended  the  vector  infinitely  in  both  direc-ons.  

11

21 January 22, 2010 Basic Vector Operations II 21

Cartesian  Coordinates  

  So  we  can  move  (slide)  the  vectors  un-l  we  reach  a  point  where  the  tails  of  the  vectors  intersect  

22 January 22, 2010 Basic Vector Operations II 22

Cartesian  Coordinates  

 We  can  set  the  origin  of  our  coordinate  system  at  this  intersec-on  

12

23 January 22, 2010 Basic Vector Operations II 23

Cartesian  Coordinates  

  If  we  place  a  Cartesian  (right  angle)  set  of  axis  on  the  forces,  we  can  see  that  they  are  also  perpendicular  

24 January 22, 2010 Basic Vector Operations II 24

Cartesian  Coordinates  

  Again  for  convenience,  we  will  label  the  horizontal  axis  as  the  x-­‐axis,  and  the  ver-cal  axis  as  the  y-­‐axis  

13

25 January 22, 2010 Basic Vector Operations II 25

Cartesian  Coordinates  

 We  can  move  our  new  reference  axis  to  coincide  with  the  intersec-on  of  the  two  lines  of  ac-on  or  the  origin  we  specified  

26 January 22, 2010 Basic Vector Operations II 26

Cartesian  Coordinates  

 We  are  going  to  use  the  same  method  to  add  there  two  forces  (vectors)  together  that  we  used  before  

14

27 January 22, 2010 Basic Vector Operations II 27

Cartesian  Coordinates  

 We  move  the  F1  vector  so  that  its  tail  lies  on  the  head  of  the  F2  vector  

28 January 22, 2010 Basic Vector Operations II 28

Cartesian  Coordinates  

  The  resultant  of  these  two  vectors,  F,  is  formed  in  the  usual  manner  by  connec-ng  the  tail  of  the  sta-onary  vector  to  the  head  of  the  moved  vector  

15

29 January 22, 2010 Basic Vector Operations II 29

Cartesian  Coordinates  

  In  this  case,  since  F2  lies  along  the  x-­‐axis,  it  is  known  as  the  x-­‐component  of  F  

30 January 22, 2010 Basic Vector Operations II 30

Cartesian  Coordinates  

  In  this  case,  since  F1  lies  along  the  y-­‐axis,  it  is  known  as  the  y-­‐component  of  F  

16

31 January 22, 2010 Basic Vector Operations II 31

Cartesian  Coordinates  

 We  oWen  rewrite  this  as  

32 January 22, 2010 Basic Vector Operations II 32

Cartesian  Coordinates  

 No-ce  that  the  resultant  makes  an  angle  α  that  is  CCW  with  the  posi-ve  x-­‐axis.  

17

33 January 22, 2010 Basic Vector Operations II 33

Cartesian  Coordinates  

  From  the  drawing,  we  have  the  following  rela-onships  

34 January 22, 2010 Basic Vector Operations II 34

Cartesian  Coordinates  

  All  of  these  are  drawn  from  the  basic  trig  rela-onships  

18

35 January 22, 2010 Basic Vector Operations II 35

Cartesian  Coordinates  

  Knowing  the  magnitudes  of  Fx  and  Fy,  we  can  calculate  the  magnitude  of  the  resultant  F  using  the  Pythagorean  theorem.  

36 January 22, 2010 Basic Vector Operations II 36

Cartesian  Coordinates  

 Unfortunatly,  component  vectors  don’t  always  lie  along  the  +x  and  +  y  axis,  nor  do  resultants  always  sit  so  nicely  in  the  +x/+y  plane  

 We  need  to  look  at  other  cases  

19

37 January 22, 2010 Basic Vector Operations II 37

Cartesian  Coordinates  

  For  example,  if  we  had  a  system  like  this  one  

38 January 22, 2010 Basic Vector Operations II 38

Cartesian  Coordinates  

 Here  the  x  component  is  in  the  nega-ve  x-­‐direc-on  and  the  y  component  is  in  the  posi-ve  y-­‐direc-on  

20

39 January 22, 2010 Basic Vector Operations II 39

Cartesian  Coordinates  

  If  we  use  our  same  method  we  can  move  either  vector  so  that  the  tail  of  the  moved  vector  is  on  the  head  of  the  sta-onary  vector.  

40 January 22, 2010 Basic Vector Operations II 40

Cartesian  Coordinates  

 Here  the  Fy  vector  is  sta-onary  and  Fx  vector  is  moved  

21

41 January 22, 2010 Basic Vector Operations II 41

Cartesian  Coordinates  

 We  can  s-ll  use  our  trig  rela-onships  to  calculate  the  components  but  we  have  to  be  careful  of  just  what  angle  we  are  looking  at  

42 January 22, 2010 Basic Vector Operations II 42

Cartesian  Coordinates  

  In  this  case,  our  angle  is  the  CCW  angle  from  the  +y  axis  

  This  is  no  special  angle,  just  what  were  given  

22

43 January 22, 2010 Basic Vector Operations II 43

Cartesian  Coordinates  

  For  example,  if  we  had  a  system  like  this  one  

44 January 22, 2010 Basic Vector Operations II 44

Cartesian  Coordinates  

  Please  remember  that  this  way  will  only  give  you  the  magnitude  of  the  components,  not  their  direc-on.  

23

45 January 22, 2010 Basic Vector Operations II 45

Cartesian  Coordinates  

  The  direc-on  (sign)  of  the  components  will  be  based  on  the  direc-ons  of  the  +x  and  +y  axex  

46 January 22, 2010 Basic Vector Operations II 46

Cartesian  Coordinates  

  REMEMBER,  draw  the  picture  before  you  try  to  use  trig,  even  though  the  trig  is  always  consistent,  it  does  depend  on  where  you  draw  the  angle.  

24

Example  

January 22, 2010 Basic Vector Operations II 47

2-­‐42  (not  complete).  Determine  the  components  of  the  forces  FA  and  FB  along  the  x  and  y  axes  if  FB  =  600  N  and  θ  =  20°.