1. PF Contest - Win "Conquering the Physics GRE" book! Click Here to Enter
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Torque and ladder problem

  1. Mar 27, 2015 #1
    1. The problem statement, all variables and given/known data

    2. Relevant equations
    F = μ FN

    τ = r F

    3. The attempt at a solution
    how do I approach this problem? Mass isn't given. And If i set the top of the ladder to be the pivot point, the pivot point will move vertically downwards if the ladder slides - is that even allowed?
  2. jcsd
  3. Mar 27, 2015 #2


    User Avatar
    Homework Helper

    If the ladder does not slip, the pivot point is also in rest. Collect all the forces and all the torques. What is the condition that the ladder does not start to move?
  4. Mar 27, 2015 #3

    does this look correct?
  5. Mar 27, 2015 #4


    User Avatar
    Science Advisor
    Homework Helper
    Gold Member

    First, what is "Fgy"? The horizontal component of the gravitational force?!
    Secondly, what force stops the ladder from slipping (until it does)?
  6. Mar 27, 2015 #5
    Force of the wall should equal force of friction.
  7. Mar 27, 2015 #6


    User Avatar
    Homework Helper

    Is the x axis horizontal, and the y axis vertical? I think, you mean Fg the force of gravity. It acts at the centre of mass of the ladder, not at the ends. What is its direction? What are the x and y components?

    The normal force at the wall is not the same as the normal force from the ground.

    And you completely ignored the force of friction at the ground.
  8. Mar 27, 2015 #7
    mg = FN at the bottom of the ladder.

    so Force of wall = force of friction since ladder doesn't move.

    Fwall = Ffriction = μ FN = μ mg

    torques : Torque of wall pushing ladder should equal torque of ladder at the center of mass.
    let's pick the bottom of the ladder as the pivot point since the ladder doesn't move.
    Fwall⋅ d ⋅sin 55.2 = mg ⋅ d/2 ⋅ cos 55.2

    plugging in μ mg for Fwall

    μ mg ⋅ d ⋅sin 55.2 = mg ⋅ d/2 ⋅ cos 55.2

    canceling out mg and d from both sides, μ should come out to be 0.347
  9. Mar 27, 2015 #8


    User Avatar
    Homework Helper

    It looks correct.
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook

Have something to add?
Draft saved Draft deleted

Similar Threads - Torque ladder problem Date
Stability: Ladder Problem Dec 16, 2014
Ladder problem, sum of forces and sum of torque Mar 29, 2013
Ladder Problem using Torque Dec 3, 2012
Difficult Ladder Against Wall Torque Problem Sep 26, 2011
Torque Problem with Ladder Mar 3, 2011