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Ladder against a frictionless wall

  1. Aug 8, 2011 #1
    A uniform ladder leans safely against a vertical wall at an angle 45° to the horizontal
    ground. The wall is 100% slippery, while at the ground the coefficient of friction is µ .
    Prove that since the ladder is not falling, µ [itex]\geq[/itex] 0.5

    I've attempted to solve this, and think I have, granted that Friction=[itex]\mu[/itex]mg, but is this true? If so, is it always true? I was taught that Friction=[itex]\mu[/itex]*Normal Force, and normal force=mg, but there are 2 normal forces in this problem, that of the wall on the ladder and that of the ground on the ladder.
     
  2. jcsd
  3. Aug 8, 2011 #2
    But what if the wall is not vertical, but sloping away from the ladder at an angle to the ground of 2theta? Arrange the ladder so it makes the same angle with the wall as it makes to the ground


    Wouldn't 2theta just equal 90 degrees then? Because 2theta+theta+theta=180 degrees
    in which case the wall isn't actually sloping away from the ladder at all...
     
  4. Aug 8, 2011 #3

    Doc Al

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    Staff: Mentor

    Yes, for this problem, since you want the smallest value of µ.
    So? Only the ground surface has friction.
     
  5. Aug 8, 2011 #4

    PeterO

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    Homework Helper

    If the wall was not vertical, then there would be a vertical component to the Normal Reaction Force from the wall to take into account.
     
  6. Aug 8, 2011 #5
    Yes, but I don't understand how it could be sloping away from the ladder using the given values for the angles. If the wall makes an angle of 2theta with the ground, and the angles the ladder makes with the wall and the ground are equal, wouldn't 2theta=90 degrees?
     
  7. Aug 8, 2011 #6

    PeterO

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    Homework Helper

    In the first problem, the wall was vertical, the angle between wall and ground = 90
    The ladder was placed so that the ladder made two 45 degree angles, one with the wall, one with the ground. The Ground, wall and ladder form an isosceles triangle

    The sloping wall situation appears to be changing the situation between wall, ground and ladder, but retaining the isosceles triangle condition. Ther will not be 45 degree angles anymore.

    To physically arrange this: suppose you have have a smooth vertical wall at one end of a basketball court, and a smooth angled wall at the other end.
    First you place the ladder against the vertical wall, creating an isosceles triangle.
    Next you pick up the ladder, carry it to the other end of the court and place the ladder against that wall - again creating an isosceles triangle.
    The foot of the ladder will be different distances from the base of the wall in each case!
     
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