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Static friction-ladder

  1. Sep 15, 2010 #1
    1. The problem statement, all variables and given/known data
    A window cleaner uses a ladder of length L and weighs 200N. He places it agaisnt a wall where, at the point of contact the coefficient of static friction, [tex]\mu[/tex], is 0.4. Upon the ground is 0.6.

    The window cleaner has a mass of 80kg and is 0.8 of the way up the ladder when it begins to slip. At what angle to the ground did he set the ladder up?



    2. Relevant equations
    so i know this is a limiting friction question so i need to use F=[text]\mu[/tex]R and the forces should balance finding the normal using some trig i should get the angle.


    3. The attempt at a solution
    ok so i know the cleaner is 0.8L up the ladder and the weight of the ladder should act downwards 0.5L of the way up, so i have two downward forces of weight in different places on the ladder. Im having trouble putting the free body diagram together. i have a normal force resulting from the contact of the cleaner, two frictional forces one on the wall and one on the ground, (do i need further normal contact forces on the ground and against the wall?). Im not not sure how many total forces i have in order to resolve. could someone please help with the method desperate.
     
  2. jcsd
  3. Sep 15, 2010 #2

    Doc Al

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    Good.

    Where the ladder touches a surface you'll have both a normal force and a friction force.

    The contact forces that I just described plus the weights of ladder and cleaner.

    Hint: Don't forget to balance torques.
     
  4. Sep 15, 2010 #3
    Thanks allot, should i combine the total weights of the ladder and the cleaner or treat them separable as they are at different locations on the ladder? balance the torques? at the points of contacts when slipping?
     
  5. Sep 15, 2010 #4

    Doc Al

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    If you are just balancing forces, location doesn't matter; but location matters when computing torques.
    That would be a good choice.

    Hint: Get two force balance equations, for vertical and horizontal forces. Add a torque equation and you'll have all you need.
     
  6. Sep 16, 2010 #5
    thanks again, i have my two force balance equations, but for a torque equation will i not need information about the moment of inertia and or angular accelerations? inertia can be calculated but im confused how to build a torque equation with the information given.
     
    Last edited by a moderator: Sep 16, 2010
  7. Sep 16, 2010 #6

    Doc Al

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    You won't need to worry about moments of inertia or acceleration--the ladder's in equilibrium. (You're solving for the point just before it starts to slip.) You have all the forces identified (or at least you should); pick an axis and set the net torque equal to zero.
     
  8. Sep 16, 2010 #7
    ah yes thats what i thought initially, i just a little confused when you brought torques up thanks allot this has been really helpful.
     
  9. Sep 17, 2010 #8
    ok so ive resolved horizontally and vertically,

    Vertically, 200+80g=Rb+Ra+(80gcos(theta))cos(theta)

    Horizontally 0.6Ra=80gcos(theta)sin(theta)+0.4Rb

    where Ra and Rb are the contact forces at each end of the ladder ground to wall respectively, the 80gcos(theta) term is the result of the contact force of the climber on the ladder. 0.6 and 0.4 are the coefficients of friction ground to wall respectively.

    so i have these two equations (with some simplifying)

    200+80g-80g(cos(theta)^2=Rb+Ra
    which becomes
    200+80g(1-cos(theta)^2)=Rb+Ra
    which is

    200+80gsin^2(theta)=Rb+Ra vertical resolved
    i have simultaneous equations

    80gcos(theta)sin(theta)=0.6Ra-0.4Rb
    200+80gsin^2(theta)=Rb+Ra

    subtracting i get this
    200+80g[sin^2(theta)+cos(theta)sin(theta)]=0.4Ra+Rb

    not sure where to go next from here or if ive taken a wrong turn.
     
  10. Sep 17, 2010 #9

    Doc Al

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    I don't understand this.

    Downward forces: The weight of climber and ladder
    Upward forces: Normal force at ground (Ra) and friction force at wall.

    Horizontal forces: Normal force at wall (Rb); Friction force at ground.

    I assume Ra and Rb are the normal components of the contact force.
    The contact force of the climber on the ladder is just the weight of the climber. Even better: Just treat the 'climber + ladder' as a single composite system.

    Redo your force equations. And you still need to add a torque equation into the mix.
     
  11. Sep 17, 2010 #10
    yes i was referring to the normal forces Ra and Rb, i think i got the directions of the frictional and normal forces facing the wrong way in my diagram. treating the climber and ladder as "one" is much easier it appeared i could do so but i was unsure because of the different positions on the ladder. ok this has been really helpful ill get cracking thanks agian.
     
  12. Sep 17, 2010 #11
    wont the normal force from the cleaner be perpendicular to the incline/ladder and therefore wont i need to find its horizontal and vertical components in order to resolve correctly.
     
  13. Sep 17, 2010 #12
    also isnt 80gcos(theta) needed for the normal component of the climber since the weight acts downwards and the normal is perpendicular to the incline.
     
  14. Sep 17, 2010 #13

    Doc Al

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    Again, I suggest treating the cleaner as being just an 80 kg mass attached to the ladder. No point in treating the cleaner as a separate object. But if you want to treat the cleaner separately, the normal force between cleaner and ladder must exactly equal the cleaner's weight--after all, the ladder is supporting him. (You'll get the same answer either way.)

    If you treat the cleaner/ladder as one object, then the only normal forces you need to worry about are those against the floor and the wall. Much easier that way.
     
  15. Sep 17, 2010 #14
    ok perfect, will that work even though the cleaner isnt positioned at the centre of the ladder?

    ok i have these force equations,
    (80g+200)[i haven't added them yet as im using whole number]=Ra+0.4Rb
    0.6Ra=Rb
    added them i get
    (80g+200)=0.4Ra+1.4Rb
    for the torques
    L(length of ladder)Rbsin(theta)=LRa0.6cos(theta)
    which becomes tan(theta)=Ra0.6/Rb

    solve simultaneously from here?
     
  16. Sep 17, 2010 #15

    Doc Al

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    The position of the cleaner only matters when computing torques.

    Good.
    Rather than add them, solve them simultaneously for Ra and Rb.
    What are you using as your axis? Be sure to include the torque from every force.
     
  17. Sep 17, 2010 #16
    i was using each end of the ladder as an axis for the torques, should i work out the centre of mass instead and use that as my axis and work around there? A torque from every force? ok including the downward weights?
     
  18. Sep 17, 2010 #17

    Doc Al

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    You can't use both ends at the same time! Pick one axis--either end is fine--and write your torque equation.
    Any axis will do. I'd stick with one of the ends.
    Yes, every force. Including the weights!
     
  19. Sep 17, 2010 #18
    ah i get taking moments around a point, im with you now.
     
  20. Sep 17, 2010 #19
    like taking moments i should say
     
  21. Sep 18, 2010 #20
    ok i have my Rb and Ra done, for the torques i have

    0.5L*200sin(theta)+(80gsin(theta)*0.8L)=Rb*Lsin(theta)+(0.4Rbsin(theta)*L)

    but after simplifying this i divide and lose my angle so im sure something is wrong.
     
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