Statics, UniKL Malaysian Institute of Aviation Technology PDF

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UniKL Malaysian Institute of Aviation Technology

2016

Wan Nur Shaqella Bte Wan Abdul Razak

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statics forces mechanics engineering

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This document is a set of lecture notes on statics. It covers topics like forces, moments and couples, vector representation, center of gravity, elements of theories, and pressure/buoyancy in liquids, and is suitable for an undergraduate level course in engineering.

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2. STATICS   February  11,  2016   LEARNING  OUTCOMES   On  compleAon  of  this  topic  you  should  be  able  to:   Describe  about  staAcs.   1. Forces,  moments  and  couples,  representaAon   as  vectors.   2....

2. STATICS   February  11,  2016   LEARNING  OUTCOMES   On  compleAon  of  this  topic  you  should  be  able  to:   Describe  about  staAcs.   1. Forces,  moments  and  couples,  representaAon   as  vectors.   2. Centre  of  gravity.   3. Elements  of  theories.   4. Nature  of  properAes.   5. Pressure  and  buoyancy  in  liquids  (Barometer).   Prepared By: Wan Nur Shaqella Bte Wan Abdul Razak 2 StaAc   qIf   a   Force   is   applied   to   a   body   it   will   cause   that   body   to   move   in   the   direc8on   of   the   applied   force,   a   force   has   both   magnitude   (size)   and   direcAon.   qSome   forces   require   contact   between   the   two   objects:          -­‐  e.g.  the  force  of  fric8on  between  car  8res              and  the  road  as  the  car  corners.   qSome  forces  do  not  require  contact:            -­‐  e.g.  the  force  between  two  magnets.     qStaAcs  is  used  to  describe  study  of  bodies  at  rest   when  forces  are  balanced.   Prepared By: Wan Nur Shaqella Bte Wan Abdul Razak 3 2.1 FORCES   Prepared By: Wan Nur Shaqella Bte Wan Abdul Razak 4 Force   q Force   –   anything   that   tends   to   cause   mo8on,   change   of   mo8on,   stop  mo8on  or  prevent  mo8on.   q Work   is   the   product   of   a   force   applied   to   an   object   8mes   the   distance  the  object  moves.   q Force  has  a  unit  of  Newtons  (N).   q One   Newton   is   defined   as   the   force   which   gives   a   mass   of   1   kg   an   accelera8on  (or  decelera8on)  of  1  m/s2,  i.e.  1  N  =  1  kg  m/s2.     Prepared By: Wan Nur Shaqella Bte Wan Abdul Razak 5 Forces   q Normally  more  than   one  force  acts  on  an   object.     q An  object  res8ng  on  a   table  is  pulled  down   by  its  weight  W  and   pushed  back  upwards   by  a  force  R  due  to   the  table  suppor8ng   it.     Prepared By: Wan Nur Shaqella Bte Wan Abdul Razak 6 2.1 SCALAR & VECTOR QUANTITY (Cont.)   Prepared By: Wan Nur Shaqella Bte Wan Abdul Razak 7 SCALAR QUANTITY VECTOR QUANTITY v Quantity (by a single number) v Quantity (by a number / magnitude and a direction) v Number with units (+ve, -ve, 0) v Magnitude of vector: |F| = F always +ve v Example: length, time, v Example: Force, momentum, temperature, mass, density, velocity, displacement, volume acceleration v Acceptable symbol for vector is F Prepared By: Wan Nur Shaqella Bte Wan Abdul Razak 8 Vector  AddiAon   If a particles undergoes a displacement A, followed by a second displacement B. The final result is the same as if the particle had started at the same initial point and undergone a single displacement C. We call the displacement C as Vector Sum or Resultant. B     A   C   Prepared By: Wan Nur Shaqella Bte Wan Abdul Razak 9 Two  or  more  forces  may  act  upon  the  same  point  so  producing  a  resultant  force.     If  the  forces  act  in  the  same  straight  line  the  resultant  is  found  by  simple   subtracAon  or  addiAon.     If  the  forces  are  do  not  act  in  a  straight  line  then  they  can  be  added  together   using  the  ‘parallelogram  law’.     Prepared By: Wan Nur Shaqella Bte Wan Abdul Razak 10 Example   A UniKL Miat student walks 12 km east one day and 5km east the following day. Find the resultant vector for the journey of the student? First  day  12  km   Second  day  5  km   17  km  to  the  east   A UniKL Miat student walks 12km east one day and 5km west the next day. Find the resultant vector for the journey of the student? Prepared By: Wan Nur Shaqella Bte Wan Abdul Razak 11 EQUILIBRANT  Force   1. A single force that can hold the original system of forces in equilibrium is known as the EQUILIBRANT. 2. It is equal in magnitude to the resultant but it is opposite in sense. B   A   A   C   B   Equilibrant Force Prepared By: Wan Nur Shaqella Bte Wan Abdul Razak 12   Vectors  in  2  Dimension  form.  (axis  –  x  and  axis  –  y) A vector in two dimensions may be resolved into two component vectors acting along any two mutually perpendicular directions. +y   A = Ax + Ay Ax = Acos θ Ay = Asin θ Ay   A   Magnitude, |A| = √(Ax2 + Ay2) θ Direction, tan θ = Ay / Ax +x   Prepared By: Wan Nur Shaqella Bte Wan Abdul Razak 13 Component  vector  along  x  and  y  axis  depend  on  the  angle,  θ   Bx  –  NegaAve   Ax  –  PosiAve   By  -­‐  PosiAve   Ay  -­‐  PosiAve   B   A   Cx  –  NegaAve   Dx  –  PosiAve   C   D   Cy  -­‐  NegaAve   Dy  -­‐  NegaAve   Prepared By: Wan Nur Shaqella Bte Wan Abdul Razak 14 Obtain  the  Resultant  Force  of  the  Following  Vectors?   |B|  =  180  N   θ  =  25o   |A|  =  150  N   θ  =  20o   Prepared By: Wan Nur Shaqella Bte Wan Abdul Razak 15 Obtain  the  Resultant  Force  of  the  Following  Vectors?   |A|  =  60  N   θ  =  35o   θ  =  30o   |B|  =  80  N   Prepared By: Wan Nur Shaqella Bte Wan Abdul Razak 16 Obtain  the  Resultant  Force  of  the  Following  Vectors?   |C|  =  60  N   |A|  =  160  N   θ  =  40o   θ  =  35o   θ  =  30o   |B|  =  80  N   Prepared By: Wan Nur Shaqella Bte Wan Abdul Razak 17 2.1 MOMENTS AND COUPLES (Cont.)   Prepared By: Wan Nur Shaqella Bte Wan Abdul Razak 18 Moment  of  a  Force   Ø A   force   can   also   be   used   to   produce   rota8on,   as   occurs   when   opening   a   door   or   8ghtening   a   nut   with   a  spanner.   Ø This   turning   effect   of   the   force   is   known   as   “the   moment  of  the  force”.   Ø It   depends   on   the   magnitude   of   the   force   and   a   distance   called   the   lever   arm.   This   is   the   perpendicular  distance  from  the  force  to  the  axis  of   rota8on.     Prepared By: Wan Nur Shaqella Bte Wan Abdul Razak 19 Moment     Moment  (Nm)    =    Magnitude  of  the  force  (N)    x       Perpendicular   distance   (d)   Applying  the  force  in  such  a  way  that  its  line  of  acAon  passes   through  the  pivot  will  not  produce  a  turning  effect.   In  SI  units,    Newton  metres    =    Newton    x    metres   Prepared By: Wan Nur Shaqella Bte Wan Abdul Razak 20 Line  of  acAon  of  the   Applied  force   force   Pivot   Pivot   Line  of  acAon  of  the   force   Applied  force   Pivot     If   the   force   causes   the   lever   to   move   in   a   clockwise   direc8on,   the   moment   is   said   to   be   a   clockwise   moment,  and  vice  versa.   If   the   force   is   inclined,   the   turning   effect   is   reduced   i.e.   moment   is   reduced   because   the   perpendicular   distance  is  reduced.   Prepared By: Wan Nur Shaqella Bte Wan Abdul Razak 21 Example   In  the  diagram  above  a  force  of  5  N  is  applied  at  a   distance  of  3  m  from  the  fulcrum,  therefore:    Moment  =  5  N  x  3  m        =  15  N  m   Prepared By: Wan Nur Shaqella Bte Wan Abdul Razak 22 Moments  and  Equilibrium   v Equilibrium  is  where  all  the  forces  and  all  the  moments  ac8ng  on  a   body  cancel  each  other  and  the  net  effect  on  the  body  is  zero.   v In   other   words   it   will   not   move   if   it   is   in   a   state   of   rest,   and   if   in   mo8on  it  will  not  slow  down  or  accelerate  or  change  direc8on.   P1 P2 S1 S2 Anticlockwise Clockwise tendency tendency Ø The  product  P2  x  S2  produces  a  clockwise  moment  about  the  pivot   and   the   product   P1   x   S1   produces   an   anA-­‐clockwise   moment   about  the  pivot.   Ø For  equilibrium  of  rota8on,  these  two  moments  must  be  equal.   Prepared By: Wan Nur Shaqella Bte Wan Abdul Razak 23 Example  of  One  Unknown  Force   1. In  this  case  the  requirement  is  to  balance  the  arrangement  in   figure  15  by  determining  the  unknown  force  (P).   Prepared By: Wan Nur Shaqella Bte Wan Abdul Razak 24 Example  of  Pivot  LocaAon   2. A   uniform   bar   AB   (figure   16)   7m   long   has   forces   of;   25N   at   a   point  0.5m  from  A,  12N  at  a  point  3m  from  A,  and  12N  at  a   point  1m  from  B  applied  to  it.  Find  the  posi8on  of  the  pivot   which  will  allow  the  beam  to  balance,  i.e.  be  in  the  state  of   equilibrium.   Prepared By: Wan Nur Shaqella Bte Wan Abdul Razak 25 Example  of  Mass  &  Forces     Pivot  LocaAon   3. A  uniform  beam  AB,  4m  long  and  200N  weight  has  forces  of   125N  and  20N  applied  respec8vely  to  its  ends  A  and  B.  Find   the  point  about  which  the  beam  will  balance.   Prepared By: Wan Nur Shaqella Bte Wan Abdul Razak 26 Principle  of  Moments   ‘When  a  body  is  in  equilibrium  under  the  ac5on  of  a   number  of  forces,  the  sum  of  the  clockwise  moments   about  any  point  is  equal  to  the  sum  of  the  an5clockwise   moments  about  that  point.’     a) Type  1  –  Beam  balances  where  arms  are  of  equal  length.   b) Type   2   –   Lever   arrangement   can   best   be   seen   in   design   of   a   wheelbarrow.   c) Type   3   –   Large   effort   moves   through   small   distance   to   overcome  small  load,  which  moves  through  a  large  distance.   Prepared By: Wan Nur Shaqella Bte Wan Abdul Razak 27 Couple   Ø A   special   case   of   moments   is   a   couple.   A   couple   consists   of   two   parallel   forces   that   are   equal   in   magnitude,   opposite   in   sense   and   do   not   share   a   line   of  ac8on.   Ø It  does  not  produce  any  translaAon,  only  rotaAon.  The   resultant   force   of   a   couple   is   zero.   BUT,   the   resultant   of  a  couple  is  not  zero;  it  is  a  pure  moment.     Prepared By: Wan Nur Shaqella Bte Wan Abdul Razak 28 Example   q In  some  situa8ons,  for  example  the  winding  up  of  a  clockwork   mechanism  the  forces  that  are  applied  to  the  winding  key  are   equal  in  magnitude  but  opposite  in  sense.   Prepared By: Wan Nur Shaqella Bte Wan Abdul Razak 29 q In  this  case  the  resultant  force  on  the  pivot  is  zero  and  there   is  only  pure  rota8on  present  with  no  tendency  for  the  pivot  to   move   sideways.   The   value   of   the   resultant   moment   (   P   x   d   )   produces  rota8on.   q Such   arrangement   of   forces   is   called   a   ‘COUPLE’   and   the   resultant  moment  of  a  couple  is  called  a  TORQUE.   Prepared By: Wan Nur Shaqella Bte Wan Abdul Razak 30 Example  of  Resultant  Moment   Ø Calculate  the  resultant  moment  of  a  pivot  ac8ng  on  a  bell         crank  lever,  refer  to  diagram  below.            AO  =  100  mm            OC  =  20  mm            BC  =  20  mm   Prepared By: Wan Nur Shaqella Bte Wan Abdul Razak 31 2.2 CENTRE OF GRAVITY   Prepared By: Wan Nur Shaqella Bte Wan Abdul Razak 32 Centre  of  Gravity   Ø Gravity  is  a  force  which  is  always  present  and  is  a  pulling   force  in  the  direc8on  of  the  center  of  the  earth.     Ø The  centre  of  gravity  is  the  force  acts  on  every  body  through   an  imaginary  point  .  A  point  where  all  the  weight  of  a  body   appears  to  be  concentrated.  (total  weight  can  be  considered   to  act  through  that  datum  posi8on  )   Ø In  flight,  both  airplanes  and  rockets  rotate  about  their  centre   of  gravity.  Determining  the  centre  of  gravity  is  very   important  for  any  flying  object.     Prepared By: Wan Nur Shaqella Bte Wan Abdul Razak 33 Example  of  Centroid   Prepared By: Wan Nur Shaqella Bte Wan Abdul Razak 34 Stability  /  Balancing   The  lower  the  C  of  G,  the  stable  an  object  is.   The  wider  the  base,  the  more  stable  an  object   is  –  C  of  G  towards  the  base.     Prepared By: Wan Nur Shaqella Bte Wan Abdul Razak 35 Ø The  point  of  O  is  the  C  of  G  of  the  rod   Ø Only  at  the  par8cular  point  O,  the  rod  can  stay   in  a  horizontal  posi8on.     Prepared By: Wan Nur Shaqella Bte Wan Abdul Razak 36 Ø When  force  applied  to  C   of  G,  the  body  will  not   rotate.   Ø But  if  the  force  is   applied  offset  of  the  C  of   G,  the  body  will  rotate,   or  torque  will  produced.     Prepared By: Wan Nur Shaqella Bte Wan Abdul Razak 37 C  of  G  of  an  aircrai   Prepared By: Wan Nur Shaqella Bte Wan Abdul Razak 38 v Similar  to  aircrai,  force  applied  will  be  acted  through  the   C    of  G,  resul8ng  in  torque.   v The  aircrai  rotate  about  its  C  of  G.   Prepared By: Wan Nur Shaqella Bte Wan Abdul Razak 39 The  Importance  of  C  of  G   q To  ensure  the  aircrai  is  safe  to  fly,  the  center-­‐of-­‐gravity  must  fall   within  specified  limits  established  by  the  manufacturer.   q To  ensure  the  C  of  G  range  –  C  of  G  limits  are  specified  longitudinal   (forward  and  ai)  and/or  lateral  (lei  and  right)  limits  within  which  the   aircrai's  center  of  gravity  must  be  located  during  flight.   q To  evenly  load  the  aircrai  –  equipments,  passengers,  baggage,  cargo,   fuel,  etc.   q So  that  C  of  G  range  will  not  be  exceeded  –  prevent  aircrai  unstable   during  flight.     q Also  affects  C  of  G  in  flight  –  fuel  usage,  passengers’  movement,  etc.     Prepared By: Wan Nur Shaqella Bte Wan Abdul Razak 40 2.3 ELEMENTS OF THEORIES   Prepared By: Wan Nur Shaqella Bte Wan Abdul Razak 41 Stress   Ø If  force  is  exerted  on  a  body,  there  will  be  mechanical  pressure  ac8ng  on   the  body  which  is  called  the  stress.     Ø A  body  with  having  twice  the  size  of  other  body  subjected  to  a  force,  it  will   be  stronger  and  less  likely  to  fail  due  to  applied  the  applied  force.   Ø So,  stress  is  said  :    Stress  =                                                                                                            or          σ    =          *units  :  Newton  metre  -­‐2  ,  Nm-­‐2   Ø Components  will  fail  due  to  over-­‐stressed,  not  over-­‐loaded.   Prepared By: Wan Nur Shaqella Bte Wan Abdul Razak 42 Example   Eg.  A  tennis  ball  sealed  from  atmospheric  pressure.  So,  as   long  as  the  external  forces  ac8ng  on  it  does  not  exceed   the  internal  forces,  the  ball  will  maintain  its  shape.       Prepared By: Wan Nur Shaqella Bte Wan Abdul Razak 43 Forces  applied  to  the  body  will  cause  distor8on  of  the  body  and  change   to  the  material’s  cross-­‐sec8onal  area  ;    eg.  Tensile  Forces  will  cause  elonga8on  .                Compressive  Force  will  cause  reduc8on  in  dimension.   Most  material  have  elas8c  proper8es  (  it  will  to  return  to  its  original   shape  aier  the  force  is  removed  )  -­‐  provided  forces  does  not  exceed   limit  of  elas8city.   There  are  5  types  of  stress  in  mechanical  bodies  :   i. Tension     ii. Compression     iii. Torsion   iv. Bending   v. Shear   Prepared By: Wan Nur Shaqella Bte Wan Abdul Razak 44 Tensile         v The  force  that  tends  to  pull  an  object  apart   v Flexible   steel   cable   used   in   aircrai   control   systems   is   an   example   of   a   component  that  is  in  designed  to  withstand  tension  loads.       Compression     v The  resistance  to  an  external  force  that                tries  to  push  an  object  together.   v Example:  aircrai  rivets.     Prepared By: Wan Nur Shaqella Bte Wan Abdul Razak 45 Torsion   v Torsional  stress  is  applied  to  a  material  when  it  is  twisted.   v Torsion  is  actually  a  combina8on  of  both  tension  and  compression   v Example:  an  engine  crankshai.   Bending   v In  flight,  the  force  of  lii  tries  to  bend  an  aircrai's  wing  upward.     Prepared By: Wan Nur Shaqella Bte Wan Abdul Razak 46 Shear   v Combina8on   of     tension   and   compression   is   the   shear   stress,   which   tries   to   slide  an  object  apart.   v Shear   stress   exists   in   a   clevis   bolt   when   it   is   used   to   connect   a   cable   to   a   sta8onary  part  of  a  structure.     v A  fork  filng,  such  as  drawn  below,  is  fastened  onto  one  end  of  the  cable,   and   an   eye   is   fastened   to   the   structure.   The   fork   and   eye   are   held   together   by  a  clevis  bolt.     v When   the   cable   is   pulled   there   is   a   shearing   ac8on   that   tries   to   slide   the   bolt  apart.  This  is  a  special  form  of  tensile  stress  inside  the  bolt  caused  by   the  fork  pulling  in  one  direc8on  and  the  eye  pulling  in  the  other.         Prepared By: Wan Nur Shaqella Bte Wan Abdul Razak 47 Strain   u Stress  is  a  force  inside  the  object  caused  by  an  external  force.   u If  the  outside  force  is  great  enough  to  cause  the  object  to   change  its  shape  or  size,  the  object  is  not  only  under  stress,   but  is  also  strained.   u If  a  length  of  elas8c  is  pulled,  it  stretches.    If  the  pull  is   increases,  it  stretches  more;  if  the  pull  is  reduced,  it  contracts.       Hooke’s law states that the amount of stretch (elongation) is proportional to the applied force. Prepared By: Wan Nur Shaqella Bte Wan Abdul Razak 48 v The  graph  below  shows  how  stress  varies  with  stress  when  a   steel  wire  is  stretched  un8l  it  breaks.     v Strain  can  be  defined  as  the  degree  of  distor8on  then  has  to  be   the  actual  distor8on  divided  by  the  original  length  (in  other   words,  elonga8on  per  unit  length).     Strain,  ε  =  change  in  dimension  /  original  dimension  (No  units).   Prepared By: Wan Nur Shaqella Bte Wan Abdul Razak 49 Example   Tensile  strain   If  a  cable  of  10  m  length  is  loaded  with  a  100  kg  weight  so   that  it  is  stretched  to  11  m,  what  is  the  strain  placed  on  the   cable?       Prepared By: Wan Nur Shaqella Bte Wan Abdul Razak 50 Example   Compressive  strain   A  25  cm  rod  is  subjected  to  a  compressive  load  so  that  its  length   changes  by  5  mm.    How  much  strain  is  the  rod  under  when   loaded?       Prepared By: Wan Nur Shaqella Bte Wan Abdul Razak 51 Shear  strain   Torsion  strain       q When   the   applied   load   causes   q Form   of   shear   stress   resul8ng   one   'layer'   of   material   to   from  a  twis8ng  ac8on.   move  rela8ve  to  the  adjacent   layers.   q If  a  torque,  or  twis8ng  ac8on  is   applied   to   the   bar   shown,   one   end  will  twist,  or  deflect  rela8ve   to  the  other  end.   q Twist   will   be   propor8onal   to   the   applied  torque.     52 Prepared By: Wan Nur Shaqella Bte Wan Abdul Razak 2.4 NATURE OF PROPERTIES   Prepared By: Wan Nur Shaqella Bte Wan Abdul Razak 53 ProperAes  of  Solid   v Strength   – A  strong  material  requires  a  strong  force  to  break  it.    The  strength  of   some  materials  depends  on  how  the  force  is  applied.     – For   example,   concrete   is   strong   when   compressed   but   weak   when   stretched,  i.e.  in  tension.     v SAffness   – A  s8ff  material  resists  forces  which  try  to  change  its  shape  or  size.    It  is   not  flexible.     v ElasAcity   – When  the  force  distor8ng  a  substance  is  removed,  and  that  substance   has   a   strong   tendency   to   return   to   its   original   shape,   it   is   said   to   be   elas8c.       Prepared By: Wan Nur Shaqella Bte Wan Abdul Razak 54 ProperAes  of  Solid   v Toughness   – This  is  the  ability  of  a  substance  to  resist  breakage  when  deforming  or   impact   forces   are   applied   to   it.   Hard   substances   are   usually   tough,   many  soier  substances  are  tough  e.g.  hammer  heads.     v Hardness   – A  hard  substance  has  a  high  resistance  to  indenta8on,  or  to  any  ac8on   tending  to  penetrate  its  surface.  In  other  words,  hardness  is  the  ability   of   a   material   to   resist   scratching,   indenta8on   or   penetra8on.   The   harder  a  material  the  more  difficult  it  is  to  scratch  it,  dent  it  or  cut  it.       v Brilleness   – Briqle   substances   break   with   liqle   or   no   change   of   shape.   In   most   applica8ons,   especially   where   sudden   impact-­‐type   forces   are   applied,   briqleness  is  undesirable.  At  room  temperature  and  below,  glass,  cast  iron,   and  very  hard  steel  are  example  of  briqle  materials.         Prepared By: Wan Nur Shaqella Bte Wan Abdul Razak 55 ProperAes  of  Solid   v Malleability   – Malleable   materials   can   be   beaten,   rolled,   or   pressed   into   shape   without  fracture  e.g.  red  hot  steel.  Malleable  metal  can  be  shaped  into   a  design  by  hilng  it.  It  could  also  lose  that  design  easily  by  being  hit   against  countertops,  cash  register  drawers,  and  other  hard  surfaces.     v DucAlity   – Duc8le   materials   can   be   stretched   into   new   shapes   without   pulling   them   apart,   and   keep   their   new   shape   aier   stretching   force   is   removed.       v PlasAcity   – Plas8city   is   the   ability   of   a   material   to   have   its   shape   permanently   changed   without   fracturing   by   stretching,   squashing   or   twis8ng.   In   other   words,   plas8city   is   described   as   a   material   that   does   not   spring   back  to  its  original  shape  when  the  load  is  removed.       Prepared By: Wan Nur Shaqella Bte Wan Abdul Razak 56 ProperAes  of  Fluid   v Viscosity   – As  the  molecules  of  a  liquid  move  about  due  to  thermal  energy,  the   aqrac8ve  forces  between  them  try  to  slow  the  mo8on  down.  The   stronger  the  forces  are  the  more  impediments  there  is  to  flow.  Such   resistance  to  the  flow  of  a  liquid  is  called  viscosity.  Viscosity  is  defined,   as  the  amount  of  force  one  layer  of  liquid  of  unit  area  will  exert  on  an   adjacent  layer.     v Surface  Tension   – The  molecules  of  a  liquid  within  the  body  of  the  liquid  are  subjected  to   forces  from  all  direc8ons.  The  molecules  at  the  surface  are  subjected   to  aqrac8ve  forces  from  within  and  to  the  sides.     – However  there  are  no  forces  from  the  outer  side  of  the  surface  to   balance  the  others.  This  places  the  surface  molecules  under  a  kind  of   tension.  This  surface  tension  tends  to  cause  the  surface  molecules  to   move  together  and  make  the  surface  area  as  small  as  possible.     Prepared By: Wan Nur Shaqella Bte Wan Abdul Razak 57 Surface  Tension   This   suggests   that   the   surface   of   a   liquid   behaves   as   if   it   is   covered   with   an   elas8c  skin  that  is  trying  to  shrink.  The  surface  tension  can  be  reduced  if  the   liquid   is   ‘contaminated’,   adding   a   detergent   to   the   water   will   cause   our   needle   to   sink.   In   a   liquid,   the   molecules   sAll   parAally   bond   together   and   prevents  liquid  from  spreading  nag  expanding  out.   Prepared By: Wan Nur Shaqella Bte Wan Abdul Razak 58 Example  of  Surface  Tension   Prepared By: Wan Nur Shaqella Bte Wan Abdul Razak 59 ProperAes  of  Fluid   v EvaporaAon   – A  change  of  a  substance  from  one  state  of  maqer  to  another  is  called  a   phase  change.  The  phase  change  from  liquid  to  a  gas  is  called  in   general  vaporiza8on.     – Vaporiza8on  from  the  surface  of  a  liquid  at  any  temperature  is  called   evapora8on.     v Boiling  Point   – The  molecules  are  in  mo8on  with  a  range  of  energies.  Some  can  escape   when  they  reach  the  surface  if  they  have  enough  energy.  Any  liquid  will   have  within  its  body  however,  microscopic  bubbles.  These  may  be  due   to  dissolved  air  or  momentary  pockets  of  vapour.       Prepared By: Wan Nur Shaqella Bte Wan Abdul Razak 60 2.5 PRESSURE AND BUOYANCY IN LIQUIDS (BAROMETERS)   Prepared By: Wan Nur Shaqella Bte Wan Abdul Razak 61 Pressure   Ø The  equivalent  term  associated  with  fluids  is  pressure:    Pressure  (P)    =    Force  (F)/  Area  (A)   Ø Pressure  is  the  internal  reac8on  or  resistance  to  that  external   force.     Ø SI  system  for  pressure  is  1  Pa  =  1N/m2     Pascal’s Law : “Pressure acts equally and in all directions throughout that fluid.” Prepared By: Wan Nur Shaqella Bte Wan Abdul Razak 62 Atmospheric  Pressure   The  atmosphere  is  the  whole  mass  of  air   surrounding  the  earth.  The  surface  of  the   earth  is  at  the  boqom  of  an  atmospheric  sea.     The  standard  atmospheric  pressure  is   measured  in  various  units:            𝟏  𝒂𝒕𝒎𝒐𝒔𝒑𝒉𝒆𝒓𝒆  =  𝟕𝟔𝟎𝒎𝒎𝑯𝒈  =            𝟐𝟗.  𝟗𝟐𝒊𝒏𝑯𝒈  =  𝟏𝟒.  𝟕𝒍𝒃/𝒊𝒏𝟐  =  𝟏𝟎𝟏.  𝟑𝒌𝑷𝒂     Prepared By: Wan Nur Shaqella Bte Wan Abdul Razak 63 Measurement  of  Atmospheric   Pressure     Atmospheric  pressure   is  typically  measured   in  inches  of  mercury   (in.Hg.)  by  a  mercurial   barometer.     Prepared By: Wan Nur Shaqella Bte Wan Abdul Razak 64 Barometer   Prepared By: Wan Nur Shaqella Bte Wan Abdul Razak 65 Gauge Pressure Ø Gauge  Pressure  is  the  reading   taken  directly  from  the  gauge   devices   Ø It  is  a  pressure  rela8ve  to  the   ambient  pressure.   Ø Gauge  pressure  is  used  to   measure  engine  oil  pressure,   hydraulic  pressure  and  other   operaAonal  pressures  built  up   by  pumps.     Ø This  is  because  atmospheric   pressure  acts  on  the  fluid  as  it   enters  and  as  it  leaves  the  pump   –  only  the  pressure  above   atmospheric  is  of  interest.     Prepared By: Wan Nur Shaqella Bte Wan Abdul Razak 66 PRESSURE GAUGE Prepared By: Wan Nur Shaqella Bte Wan Abdul Razak 67 Absolute Pressure Absolute  Pressure  is  the  sum  of  the  available   atmospheric  pressure  and  the  gauge  pressure.   Absolute  Pressure  (PSIA)     =     Gauge  Pressure  +  Atmospheric  Pressure         Prepared By: Wan Nur Shaqella Bte Wan Abdul Razak 68 CalculaAon  Example  :  Given     (Gauge  Pressure)  =  150  psig     (Atmospheric  Pressure)  =  14.7psi     Absolute  Pressure  =  150  psig  +  14.7  psi                      =  164.7  PSIA   Prepared By: Wan Nur Shaqella Bte Wan Abdul Razak 69 Buoyancy   Archimedes’ Principle states that when an object is submerged in a liquid, the object displaces a volume of liquid equal to its volume and is supported by a force equal to the weight of the liquid displaced. THE  BUOYANCY  OF  A  SUBMERGED  BODY  =     WEIGHT  OF  DISPLACED  LIQUID  –  WEIGHT  OF  THE  BODY   1.  The  body  will  float-­‐-­‐if  the  buoyancy  is  posiAve   2.The  body  will  sink-­‐-­‐if  the  buoyancy  is  negaAve   3.The  body  will  be  stuck-­‐-­‐if  the  buoyancy  is  neutral   Prepared By: Wan Nur Shaqella Bte Wan Abdul Razak 70 Buoyancy   Archimedes  Principle   q When  an  object  is  submerged  in  a  liquid,   the  object  displaced  a  volume  of  liquid   equal  to  its  volume  and  is  supported  by   a  force  equal  to  the  weight  of  the  liquid   it  displaced.   q The  buoyant  force  of  an  object  which  is   submerged  in  a  fluid  is  equal  to  the   weight  of  the  fluid  displaced  by  the   object.   q A  net  upward  ver8cal  force  results   because  pressure  increases  with  depth   and  the  pressure  forces  ac8ng  from   below  are  larger  than  the  pressure   forces  ac8ng  from  above.   Buoyant Force, FB = ρgV Prepared By: Wan Nur Shaqella Bte Wan Abdul Razak 71 Archimedes  Principle   “Any  object  completely  or  parAally   submerged  in  a  fluid  experiences  an   upward  force  equal  in  magnitude  to   the  weight  of  the  fluid  displaced  by   the  object.”   Prepared By: Wan Nur Shaqella Bte Wan Abdul Razak 72 THIS  EXPLAINS  WHY  BIG  NAVAL  SHIP  CAN  FLOAT  !!!!!!   A steel ship can encompass a great deal of empty space and so have a large volume and a relatively small density. Weight of ship = weight of water displaced Prepared By: Wan Nur Shaqella Bte Wan Abdul Razak 73

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