lecture 33 solving rigid-body equilibrium problems

Lecture  33:  Solving  rigid-­‐body  equilibrium  problems

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Lecture  Objective:  

Apply  conditions  of  equilibrium  to  rigid  bodies

1st  condition  for  equilibrium:  Balanced  Forces

2nd  condition  for  equilibrium:  Balanced  Torques

Equilibrium  of  a  rigid  body

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1.) The rigid body must not be rotating and accelerating.2.) Draw the free body diagram. Rigid bodies cannot be represented as a point.3.) Choose the coordinate axis and a positive sense of rotation. 4.) Convert the forces into its component forms.5.) Note that for forces acting at one point, the torque at that point is zero.6.) Write equations expressing equilibrium position.

7.) Count the number of unknowns and equations.8.) Do the math.

Example  12.1  (Tipler)

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Given:length, 3 mboard mass, 2 kgbox mass, 6 kgx1, 2.5 mx2, 0.5 mtarget variable, F1 and F2

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Apply the two conditions of equilibrium

F1 + F2 - mboardg - mboxg = 0 First condition-F1L + mboardgL/2 + mboxgx2 + F2 x(0) = 0 Second condition

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Apply the two conditions of equilibrium

F1 + F2 - mboardg - mboxg = 0 First condition-F1L + mboardgL/2 + mboxgx2 + F2 x(0) = 0 Second condition

can also be written as:F1 = 1/2mboardg + x2mbox/L

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Apply the two conditions of equilibrium

F1 + F2 - mboardg - mboxg = 0 First condition-F1L + mboardgL/2 + mboxgx2 + F2 x(0) = 0 Second condition

can also be written as:F1 = 1/2mboardg + x2mbox/L

Two equations with two unknowns so just substitute F1 to getF2 = mboard + mboxg - F1 = 1/2mboardg + mboxg -x2mboxg/L

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Apply the two conditions of equilibrium

F1 + F2 - mboardg - mboxg = 0 First condition-F1L + mboardgL/2 + mboxgx2 + F2 x(0) = 0 Second condition

can also be written as:F1 = 1/2mboardg + x2mbox/L

Two equations with two unknowns so just substitute F1 to getF2 = mboard + mboxg - F1 = 1/2mboardg + mboxg -x2mboxg/L

The scale readings will be:F1 = 19.6 NF2 = 58.9 N

Sample  Problem  Sir  Lancelot  is  trying  to  rescue  Lady  Elayne  from  Castle  Von  Doom  by  climbing  a  uniform  ladder  that  is  5.0m  long,  weighing  180N.    Lancelot  weighs  800N,  stops  a  third  of  the  way  up  the  ladder.  The  bottom  of  the  ladder  rests  on  a  horizontal  stone  ledge  and  leans  across  the  moat  in  equilibrium  against  a  vertical  wall  that  is  frictionless  because  of  a  thick  layer  of  moss.  The  ladder  make  an  angle  of  53.1o  with  horizontal,  conveniently  forming  a  3-­‐4-­‐5  right  triangle.

(a) Find  the  normal  force  and  frictional  force  on  the  ladder  at  its  base.  (b) Find  the  minimum  μs  needed  to  prevent  slipping  at  the  base.  (c) Find  the  magnitude  and  direction  of  the  contact  force  on  the  ladder  at  the  

base.

FBD  of  Sir  Lancelot  and  ladder

(a)  Use  the  1st  condition  for  equilibrium: FBD  of  Lancelot  and  ladder  

Use  the  2nd    condition  for  equilibrium:  

The  contact  force  at  B  is  the  superposition  of  the  normal  force  and  the  static  friction  force

Seatwork

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SW  1:  Clea’s  weight  SW2:  x  (cg)

Seatwork  3,  4  and  5:

A B

SW  3:  Draw  the  FBD  of  the  situation  SW  4:  Calculate  the  maximum  weight  you  can  place  without  breaking  the  cable.  (use  1st  condition)  SW  5:  Where  should  you  put  the  weight?  (use  2nd    condition,  calculate  for  CG)

Weight,  w

Seatwork  answers

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Apply  the  first  and  second  conditions  of  equilibrium  to  Clea

The  normal  force  at  her  feet  is  greater  than  at  her  rear  feet,  so  her  center  of  gravity  is  closer  to  her  front  feet  ☺

Seatwork  1  and  2

A B

SW  3:  Draw  the  FBD

wbar  =  350N

TA  =  500N TB  =  400N

w  =  ???

Weight,  w

SW4:  What  is  the  heaviest  weight  the  you  can  put  in  the  bar  without  breaking  the  cable?  (add  another  weight)

ΣFy  =0  -­‐w  -­‐wbar  +  TA  +  TB=  0  -­‐w  –  350N  +  500N  +  400N  =  0  w  =  550N

SW  3:  Draw  the  FBD

wbar  =  350N

TA  =  500N TB  =  400N

w  =  ???

SW  3:  Draw  the  FBD

wbar  =  350N

TA  =  500N TB  =  400N

w  =  ???

SW5:  where  should  you  put  it?  We  must  put  it  at  cg,  solve  for  cg  using  the  2nd  condition:Στ  =  0

-­‐TALcg  +  TB(1.5m-­‐Lcg)–wbar(0.75m-­‐Lcg)  =  0  (-­‐500N-­‐400N+350N)Lcg  =  (0.75m)(350N)-­‐(1.5m)(400N)  Lcg  =  0.614m