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Need some help starting this problem.

  1. Apr 19, 2005 #1
    hello. im just a little stuck knowing how to start this problem. i dont know i just guess i keep looking at it and drawing a diagram is as far as i got so far.

    A straight ladder of length L has a mass M, which is uniformly distributed along its length. The ladder has one end on the horizontal ground and its other end leaning against a vertical wall. Assume that the ladder makes a flat contact at both the wall and the ground, and that the coefficients of static friction are Uw at the surface between the ladder and the wall, and Ug at the surface between the ladder and the ground. Let (theta) be the angle between the ladder and the ground. Find an expression for the value of (theta) at which the ladder will just begin to slip.

    I am just confused by the question. i know that static friction is at a maximum value at both the ground and wall when the ladder begins to slip.
     
  2. jcsd
  3. Apr 19, 2005 #2
    what school do you attend Hoppa? This is a serious question because i've had to solve this exact same problem last week at my school. BTW i know the answer.
     
  4. Apr 19, 2005 #3
    go to CSU Australia. what about you?
     
  5. Apr 19, 2005 #4
    Think about the situation where the ladder will begin to slip... When there is enough force to overcome the coeficient of static friction. Find an expression involving theta when there is enough force to overcome the static friction (IE: when the force due to gravity is equal to the force resultant of the static friction)

    Hope this helps a little...
     
  6. Apr 19, 2005 #5
    I go to college in florida and if i'm not mistaken the answer is 55 degrees.
     
  7. Apr 19, 2005 #6
    so would i use force due to gravity to be 9.8ms and then find an equation so that static friction equals that? but what does the static friction equal? do the length or mass of the ladder come into it at all?
     
  8. Apr 20, 2005 #7
    ok thanks hopefully i get the same answer then.
     
  9. Apr 20, 2005 #8

    OlderDan

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    Setting it up:

    The wall must push against the ladder in a horizontal direction. It also provides a vertical force by friction. Those two forces are proportional via the coefficient of friction. The gound must push up on the ladder in a vertical direction. It also provides a horizontal force by friction. Those two forces are proportional via the coefficient of friction. The ladder has weight, which may be treated as concentrated at its center of mass. Since the ladder is in equilibrium, the sum of the vertical forces is zero, the sum of the horizontal forces is zero, and the sum of the torques calculated around any axis of your choosing must be zero.
     
  10. Apr 20, 2005 #9
    i'm sorry about the answer i gave you of 55 degrees, that depends on the coefficient of static friction, which in my problem was 0.35.
     
  11. Apr 22, 2005 #10
    how did u find the coefficeint of static friction?
     
  12. Apr 22, 2005 #11

    OlderDan

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    The answer to your problem will be an expression that contains coefficients of friction that have unspecified values, along with unspecified values for L and M. It will then be useful for any specific values you plug in.
     
  13. Apr 22, 2005 #12
    it was given to us.
     
  14. Apr 24, 2005 #13

    how do i find that expression then?
     
  15. Apr 24, 2005 #14

    OlderDan

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    At the point of just slipping, the diagram shows the forces acting on the ladder in equlibrium. Summing vertical and horizontal forces gives

    [tex] \mu_wN_w + N_g = Mg [/tex]
    [tex] N_w = \mu_gN_g [/tex]

    By substitution

    [tex] \mu_w\mu_gN_g + N_g = Mg[/tex]
    [tex] (\mu_w\mu_g+1) N_g = Mg [/tex]
    [tex] N_g = \frac{ Mg }{ \mu_w\mu_g+1}[/tex]
    [tex] N_w = \frac{ \mu_gMg }{ \mu_w\mu_g+1}[/tex]
    [tex] \frac{Mg }{N_w} = \frac{\mu_w\mu_g+1}{ \mu_g}[/tex]

    Summing the torques about the point of contact with the ground gives

    [tex] N_wLsin\theta + \mu_wN_wLcos\theta = \Left[\frac{MgL}{2}\Right]cos\theta[/tex]
    [tex] N_wsin\theta = \Left[\frac{Mg}{2}\Right]cos\theta - \mu_wN_wcos\theta[/tex]
    [tex] tan\theta = \Left[\frac{Mg}{2N_w}\Right] - \mu_w[/tex]
    [tex] tan\theta = \Left[\frac{\mu_w\mu_g+1}{2 \mu_g}}\Right] - \mu_w[/tex]
    [tex] tan\theta = \Left[\frac{\mu_w\mu_g+1 - 2\mu_w\mu_g}{2 \mu_g}}\Right][/tex]
    [tex] tan\theta = \Left[\frac{1 - \mu_w\mu_g}{2 \mu_g}}\Right][/tex]
    [tex] \theta = tan^-1\Left[\frac{1 - \mu_w\mu_g}{2 \mu_g}}\Right][/tex]

    See if this is reasonable. If the ground has no friction [itex] \theta [/itex] will be 90 degrees at equilibrium. (Not zero. That would be a different problem with a force from the ground acting on both ends of the ladder.) If the wall has no friction, the ground must support the entire weight of the ladder.

    [tex] N_g = Mg [/tex]
    [tex] N_w = \mu_gMg [/tex]
    [tex] \mu_gMgLsin\theta = \Left[\frac{MgL}{2}\Right]cos\theta[/tex]
    [tex] tan\theta = \Left[\frac{1}{2\mu_g}\Right][/tex]

    Looks good
     

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  16. Apr 25, 2005 #15
    thanks for that, but where does the torques come into the problem? how come u had to use that?
     
  17. Apr 25, 2005 #16

    OlderDan

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    Without the torque, you know that forces have to be balanced, but you do not know the angle of the ladder that will cause both the frictional force at the wall and the frictional force at the ground to be at their maximum possible values. The equations that precede the torque calculation assume that critical configuration has been reached. Without setting the ladder at that critical angle, the earlier equations would not be valid.
     
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