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Torque - ladder against a wall

  1. Apr 8, 2010 #1
    1. The problem statement, all variables and given/known data

    A uniform ladder 5.1 m long rests against a frictionless, vertical wall with its lower end 3.1 m from the wall. The ladder weighs 165 N. The coefficient of static friction between the foot of the ladder and the ground is 0.39 . A painter weighing 739 N climbs slowly up the ladder.

    1) What is the actual size of the frictional force when the painter has climbed 0.9393 m along the ladder?

    2) How far along the ladder can the painter climb before the ladder starts to slip?


    2. Relevant equations

    [itex]\tau=r*Fsin(\theta[/itex])

    3. The attempt at a solution

    [itex] \theta = cos^{-1}\frac{3.1}{5.1} = 52.57\degree [/itex]

    This is the angle where the ladder meets with the floor

    For the first question i found that

    [itex] \tau_{man} + \tau_{ladder} - \tau_{friction} = 0 [/itex]

    I used my pivot point to be at the bottom of the ladder where it meets with the floor

    [itex]739N(0.9393m)(cos(52.57)) + 165N(\frac{5.1m}{2})(cos(52.57)) - f_s(5.1m)(cos(37.43)) = 0[/itex]

    [itex] f_s = 167.3 N [/itex]

    I know this is the right answer but i am a little confused on when to use sine or cosine and what angle to use with it. Could anyone explain this to me? I have a test in a few days and i really need to know this :)

    I Could not figure out part 2 so after i figure out how to use sine and cosine for part 1 i will give part 2 a shot.

    Thank you for any explanations on this :)
     
  2. jcsd
  3. Apr 8, 2010 #2

    PhanthomJay

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    That last term should be [tex]\tau_{normal force (N) between wall and ladder}[/tex]
    Again that last term should note N instead of f_s
    From Newton 1 in the x direction, f_s = N
    Sounds like you were taking a stab at sin vs. cos, and you didn't use the proper forces when calculating torques. There are 2 ways to calculate the magnitude of a torque: one is per your relevant equation, where theta is the included angle between the force and position vector. The other is Torque = force times perpendicular distance from line of action of the force to the pivot point. Either way gives you the same result, using geometry and trig. Be consistent in your approach. Watch plus and minus signs.
     
  4. Apr 8, 2010 #3
    Thank you Jay. I will go through the problem again tonight or tomorrow and if i get stuck somewhere i will post here again.

    Thanks again :)
     
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