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Static equilibrium & force of friction

  1. Aug 30, 2009 #1
    1. The problem statement, all variables and given/known data
    A window cleaner of mass 95 kg places a 22-kg ladder against a frictionless wall at an angle of 65 degrees with the horizontal. The ladder is 10m long and rests on a wet floor with a coefficient of static friction equal to 0.40. What is the maximum length that the window cleaner can climb before the ladder slips?


    2. Relevant equations
    T = rF


    3. The attempt at a solution
    Sin65 = opp/10, therefore opposite = 9.06
    Cos65 = adj/10, therefore adjacent = 4.23
    m = 22; M=95
    Max Ffr = uFN = u(mg + Mg) = Fw
    Fy=FN-mg-Mg
    T=Fw(9.06)-mg(4.23/2)

    Any help is appreciated!
     
  2. jcsd
  3. Aug 31, 2009 #2

    Hootenanny

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    How do you know that the normal force (at the base of the ladder) will be equal to the combined weight of the ladder and man?
     
  4. Aug 31, 2009 #3
    I originally thought it would be equal to that because (mg + Mg) is the force exerted onto the ground ... but I see that my logic there is wrong.
    Should it be an unknown?
    Sorry, I am very lost with this question!
     
  5. Aug 31, 2009 #4

    Hootenanny

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    I would start by examining the turning moments about the top of the ladder, i.e. assume that the point where the ladder touches the wall is the pivot.
     
  6. Aug 31, 2009 #5
    By turning moments, do you mean torque?
    I thought this would be the correct eq'n for torque:
    T=Fw(9.06)-mg(4.23/2)

    And I know that T must equal 0 for the system to be at equilibrium.
     
  7. Aug 31, 2009 #6

    Hootenanny

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    Yes, I mean torques. What do each of those terms represent?
    Indeed, that is correct.
     
  8. Aug 31, 2009 #7
    Because T=rF
    T = Fw(9.06) - mg(4.23/2)
    The first force is Fw (= force of wall)
    9.06 is the height of the wall (i.e. where the ladder touches the wall)

    The negative force, is given by F=mg
    And r=4.23/2

    I'm sorry that I am so confused by this question, but I appreciate your help a lot!
     
  9. Aug 31, 2009 #8

    Hootenanny

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    Ahh, I see, you are taking moments about the base of the ladder, rather than the top. No matter, since you are going to have to do both eventually. However, I think that you're forgetting one small thing ...
    ... what about the man?
    No need to apologise, we wouldn't be here if students weren't asking for help! :smile:
     
  10. Aug 31, 2009 #9
    Okay - so does the torque equation have to be re-written to include the mass of the man? And is that then the unknown?
     
  11. Sep 1, 2009 #10

    Hootenanny

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    Indeed it does.
    Yes, that will be an unknown, but your also going to have another unknown: the reaction force from the wall.
     
  12. Sep 4, 2009 #11
    Okay....thanks for your help...and sorry for my delay in responding, but I still haven't had much success with this question.
    I have the torque equation as:
    T=FN(10)sin65-22(2.8)x1/2cos65=0
    as well as: T=FNr=(95)(9.8)r

    But I don't know how to incorporate the mass of the man into the normal force of the ladder (to calculate force of static friction).
     
  13. Sep 5, 2009 #12

    Hootenanny

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    I'm not quite sure what your doing there, your torque equation should have three terms in it: one term for the weight of the ladder, one term of the weight of the man and one term for the normal reaction force of the ladder.

    Try drawing a diagram of the situation. I'll help you out and write the torque equation if we take moments about the base of the ladder (i.e. we take the base of the ladder as the pivot).

    [tex]F_N\ell\sin\theta - W_\ell\frac{\ell}{2}\cos\theta - W_mx\cos\theta = 0[/tex]

    The first term is the normal reaction force of the wall, with [itex]\ell[/itex] being the length of the ladder. The second term is the weight if the ladder and the final term is the weight of the man.

    Do you follow? If so, can you do the same for the torques about the upper point of the ladder?
     
    Last edited: Sep 5, 2009
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