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Ladder problem

  1. Dec 16, 2003 #1
    a ladder of mass M and length L leans at an angle theta against a frictionless wall. coefficient of static friction between ladder and ground is MU, what is the minimum angle at which the ladder will not slip? this problem sounds fairly simple, but ive been looking at it for the past 20 minutes and have no idea where to begin.
  2. jcsd
  3. Dec 16, 2003 #2
    The approach is the same as the boom problem.

    Begin by drawing a force diagram.

    Then figure out the horizontal & vertical components.

    Then pick an axis & figure out the torques.

    Hint: Even though it may not be obvious, you do have to consider rotation to solve this.

    Another hint: there are 4 forces acting on the ladder. Can you identify them all?
  4. Dec 16, 2003 #3
    the forces are gravity, friction, normal force from wall and normal force from ground right?
    Fn from ground = mg sin 90-theta correct?
    but how would i found the normal force from the wall?
  5. Dec 16, 2003 #4
    How many horizontal forces are there? What are they? What must their sum be, remembering that the ladder (hopefully) isn't moving?

    Before I answer that, what is the magnitude of the normal force? How many vertical forces are there? What must their sum be?
  6. Dec 16, 2003 #5
    2 horizontal forces and two vertical forces. each is a component of the normal force.

    normal force = m * g, sum of vertical forces must be zero.
  7. Dec 16, 2003 #6
    So do you want to change your statement about the magnitude of the friction force?

    I'm not sure I understand what you're saying there. There are TWO normal forces in this problem, aren't there?
  8. Dec 16, 2003 #7
    there is a normal force from the wall and a normal force from the ground. there is a horizontal force and a vertical force component vector of each.
    Friction = m*g*tan theta * mu right?

    what kind of problems are these anyways? ive never seen anything this complicated.
  9. Dec 16, 2003 #8
    REALLY think about this. What is the direction of the normal force from the ground? Does it really have a horizontal component? What is the direction of ANY Normal force relative to the surface that applies it?

    Now you're just guessing, right? Why tan theta? Why anything theta?

    It all depends on your point of view. Wait till you see problems on electrical and magnetic fields. Then these statics problems will seem simple.
  10. Dec 16, 2003 #9
    is this right? i did sum of torques to find normal force from wall, which ended up being d*m*g/2*h.
    than i figured theta = arctan mg/(dmg/2h)

    am i getting close?
  11. Dec 16, 2003 #10
    Where did d and h come from? Don't introduce new variables. You already have enough of them.
  12. Dec 16, 2003 #11
    d is distance from base of ladder to wall and h is height. dont i need these for the sum of torques?
  13. Dec 16, 2003 #12
    No. You should express any distance that you need in terms of L and theta.
    i.e. some fraction of L times some function of θ

    Also, I just realized, for this problem you can't use the point where the ladder touches the floor as the pivot for your torque equation,
    [[because then μ drops out, and you need μ for the solution.]]
    edit: the preceding statement should be:
    because then θ drops out, and you need θ for the solution.

    So figure the torques about a pivot either where the ladder touches the wall, or at the center of mass. Either way, there will be 3 torques in the equation.
    Last edited: Dec 17, 2003
  14. Dec 16, 2003 #13
    Unless I blew it, you should end up ultimately solving for θ as a function of μ.

    All the other factors -- m, g and L -- drop out.
  15. Dec 16, 2003 #14
    sum of torques = 1/2L * m + cos theta * L * m + sin theat * L * m = 0
    thats center of mass, point on wall and base of ladder.
    how do i figure in mu?
  16. Dec 16, 2003 #15
    Take the point where ladder meets wall as the pivot.

    Then (assuming the wall is on the left), the clockwise torques are
    due to gravity:

    and due to the friction:

    The counterclockwise torque due to the normal force from the floor is

    Now, that's really all you need. Make an equation from that & solve for θ.
  17. Dec 16, 2003 #16
    so theta = arcsin(sin theta/2 + mu cos theta)
  18. Dec 17, 2003 #17
    (Sigh) :frown:

    (mgL/2)sinθ + μmgLcosθ = mgLsinθ

    divide by mgL:

    (1/2)sinθ + μcosθ = sinθ

    μcosθ = (1/2)sinθ

    divide by cosθ, multiply by 2:

    tanθ = 2μ

    θ = arctan(2μ)

    If θ is any bigger than that, (1/2)sinθ will be greater than μcosθ, and the ladder falls.
  19. Dec 17, 2003 #18
    sorry dude, i dont mean to look like a total idiot.
    after i posted the answer last night i realized it was wrong. i figured it out to be the arctan statement you gave. thanks for confirmation.
  20. Dec 17, 2003 #19
    No problem. If I didn't enjoy helping I wouldn't be here.
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