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Rotational Statics problem

  1. Jun 9, 2006 #1
    [​IMG]
    "One end of a uniform meter stick is placed against a vertical wall View Figure . The other end is held by a lightweight cord that makes an angle [itex]\theta[/itex] with the stick. The coefficient of static friction between the end of the meter stick and the wall is 0.400 ."

    I can't for the life of me get the equation for this.
    Part A:
    What is the maximum value the angle [itex]\theta[/itex] can have if the stick is to remain in equilibrium?
    So far I've tried using the last end as a pivot point.
    so that would give me
    Balancing torque:
    [tex]\mu T cos(\theta) = mg(1-x)[/tex]
    and balancing total forces
    [tex]\mu T cos(\theta) + T sin(\theta) = mg [/tex]
    combining the two
    produces
    [tex] T sin(\theta) = mgx[/tex]
    dividing the third by the first gives me
    [tex] tan(\theta) = \mu \frac{x}{1-x} [/tex]
    and there is no way I can get a "max theta". Naturally I can't get Part B and C but I figured if you guys can help me with A it will be easy to get the rest.

    Thanks in advanced.
     
    Last edited: Jun 9, 2006
  2. jcsd
  3. Jun 9, 2006 #2

    Hootenanny

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    What is the mass of the meter rule? Is it light?
     
  4. Jun 9, 2006 #3
    negligible I presume.
     
  5. Jun 9, 2006 #4

    Gokul43201

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    Can you draw the forces in that figure?
    This is the end of the ruler not touching the wall, right?

    I see a mistake here. What torque does the string exert about the point of attachment of the string? And are there no other forces on the ruler that exert a torque about this point?

    Again, never do a force problem without a free body diagram.
     
    Last edited: Jun 9, 2006
  6. Jun 9, 2006 #5
    Yeah that's the end of the ruler not touching the wall. I thought about the string itself exhibiting torque but since I chose it as a pivot point it wouldn't make much sense if it did. Can you suggest an alternate pivot point? I also considered using the end touching the wall as pivot point and that gave me the equation
    [tex]T sin(\theta) = mgx [/tex]
    which was already derived from the first two. I also considered using the ruler's center of mass a pivot point and that didn't give me a sensical answer either.
    [​IMG]
    Pretty sure that's right.
     
    Last edited: Jun 9, 2006
  7. Jun 9, 2006 #6

    Hootenanny

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    Perhaps taking moments about the centre of the rule since we don't know its mass? I haven't worked through this problem so don't take my word for it.
     
  8. Jun 9, 2006 #7
    [tex] 1/2 \mu T cos(\theta) + mg(x-1/2) = 1/2 T sin(\theta) [/tex]
    [tex] \mu T cos(\theta) + T sin(\theta) = mg [/tex]

    so...
    [tex] \mu T cos(\theta) + 2mgx -mg= T sin(\theta) [/tex]
    [tex] \mu T cos(\theta) + 2mgx -\mu T cos(\theta) - Tsin(\theta)= T sin(\theta) [/tex]
    an equation I already had
    [tex] mgx = T sin(\theta) [/tex]
    and reapplying the second equation
    you get
    [tex] \mu T cos(\theta) + T sin(\theta) = mg [/tex]
    [tex] \mu T cos(\theta) + mgx = mg [/tex]
    [tex]\mu T cos(\theta) = mg(1-x)[/tex]
    and we're back where we started. So so far, I've chosen the left end (ruler touching wall), the right end (string pulling on ruler), and midpoint and gotten the same answer for all of them. Argrhgrh what am I doing wrong?
     
    Last edited: Jun 9, 2006
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