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Homework Help: Mechanics 2 - Statics of rigid bodies. Please help.

  1. Jun 23, 2004 #1
    I'm having difficulty with the question, perhaps mainly because it is consists completely of unknowns, with little or no numbers. Help, in the form of steps to reach the proof would be much appreciated:

    'A uniform ladder of mass M rests in limiting equilibrium with one end on a rough horizontal ground and the other end against a rough vertical wall. The coefficient of friction between the ladder and the ground is U. The coefficient of friction between the ladder and the wall is U'. Given that the ladder makes an angle a with the horizontal show that:
    tan a = (1- UU')/2U
  2. jcsd
  3. Jun 23, 2004 #2


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    Note that in order to have an equilibrium situation you need:
    The sum of forces acting upon the object must be zero.
    The torque of forces computed about a point must be zero.
    Since you have a 2-D problem, you have a total of 3 equations.

    What are your unknowns?
    The 2 magnitudes of the normal forces, and the angle a.

    Try and set up your equations with this information!
  4. Jun 23, 2004 #3
    After checking the notice topic i thought i'd make it clear this isn't homework, i'm just starting next years work early, as i have recently finished my AS exams.

    Here's where i've got so far:

    Resolving horizontally: UR = S
    Vertically: R +U'S = Mg

    Taking moments about mg:
    Clockwise = anticlockwise.
    Rxcos(theta) = XSin(theta) + XU'Scos(theta) + RxUsin(theta)
    Divide by x:
    Rcos(theta) = Sin(theta) + U'SCos(theta) + RUSin(theta)
    Divide by cos(theta)
    R=Tan(theta) + U'S + URTan(theta)
    Tan(theta) + URTan(theta) = R - U'S
    Tan(theta) + URTan(theta) = R - U'UR

    Here i'm stuck. I would really appreciate any help.
  5. Jun 23, 2004 #4
    Yes, i set that up. However, it seems i have more unknowns which include:
    Length of rod (2X), but this may later be cancelled out.
    R (upward reaction on the floor)
    UR (Friction on floor)
    S(reaction at wall)
    U'S (friction at wall)
  6. Jun 23, 2004 #5


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    The mass is given as M, so that is "known". U and U' are also givens.
    True, the length has not been specified, but that cancels out (Just give it a default value L, or something (2X?))

    Hence, in you notation, you have a,R,S as unknowns.
    I'll check up on post 3, and post some comments.
  7. Jun 23, 2004 #6


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    There's a nasty trap in this problem:
    You do not know at the outset which way (up or down) the friction along the vertical wall goes!!

    Hence, when computing moments about a point, you should use the contact point on the vertical wall!
    I have to think about this..
  8. Jun 23, 2004 #7
    Hmm, i assumed naturally that the ladder would want to slip down, so friction would act up, though this isn't given in the question.
  9. Jun 23, 2004 #8
    Cripes, just got it after trying for the fourth time or something. Here's the solution if anyone is interested:

    Let 2x = length (as it is a uniform rod the distance from the top or bottom to mg = x)
    R = Reaction at base.
    UR = Friction at base.
    mg = weight action.
    S = Reaction at wall.
    U'S = Friction at wall.

    Horizontally: UR = S
    Vertically: R + U'S = mg

    Taking moments at A (bottom)
    Clockwise = anticlockwise.
    xcos(theta)mg = 2xsin(theta)S + 2U'Sxcos(theta)
    Cos(theta)Mg = 2sin(theta)S + 2U'Scos(theta)

    Perform some substitution:
    Rcos(theta) + U'Scos(theta) = 2Ssin(theta) + 2U'Scos(theta)
    Rcos(theta)+U'URcos(theta) = 2URsin(theta) + 2U'Scos(theta)
    R+U'UR = 2URtan(theta) + 2U'S
    R+U'UR = 2URtan(theta) + 2U'UR
    1+U'U = 2Utan(theta)+2U'U
    2Utan(theta) = 1+U'U - 2U'U
    2Utan(theta) = 1-UU'
    Tan(theta) = (1-UU')/(2U)

    Sweet, pleased with myself now >_<
    Last edited: Jun 23, 2004
  10. Jun 23, 2004 #9


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    That's what I got as well, and while I agree with argument about the direction, you also get a solution with your minus swapped to a plus..
    Smart of you to use a contact point in calculating torques.
  11. Jun 23, 2004 #10
    To be honest i haven't done any work on torques yet. I just resolved at A because i saw i would be able to cancel two forces, and the unknown 'mg' could be translated into a more useful form via one of the other equations.
    Interesting question, getting it right always gives a good sense of accomplishment.
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