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Homework Help: Ladder problem

  1. Nov 21, 2009 #1
    Problem: An 8 m, 200 N uniform ladder rests against a smooth wall. The coefficient of static friction between the ladder and the ground is .45, and the ladder makes a 60 degree angle with the ground. How far up can an 800 N person climb before the ladder begins to slip?

    a. 7.8 m
    b. 6.8 m
    c. 5.8 m
    d. 4.8 m

    I don't know where to start with this problem. Any help would be appreciated. Please go through each step with detail, because I need to do it myself so I understand what I am doing.
    Last edited: Nov 21, 2009
  2. jcsd
  3. Nov 21, 2009 #2
    Draw the free body diagram. You should have:
    -Normal force from the ground.
    -frictional force from the ground
    -force of gravity on the ladder
    -force of gravity on the person
    -normal force from the wall.

    Set up your equations so net force is 0. Set up your torque equation using the maximum torque the friction can provide and set it equal to 0. The distance of the person from your point of rotation is your unknown. Solve it and you'll have your answer.
  4. Nov 21, 2009 #3
    In my free body diagram, I have all of those except gravity on person because I do not know where that goes on the ladder.

    I also don't understand how to set up the equations. Am I using the right arm rule, where clockwise is negative, and counter-clockwise is positive? And which one does weight (mg) fall under?
  5. Nov 21, 2009 #4
    If you set your origin to the right end of the ladder then the weight of the person is counter-clockwise.
  6. Nov 21, 2009 #5
    Oh, okay. I didn't see that at first.

    When I set those equations up, does the force equation break into the total forces acting on the y-axis and the total forces acting on the x-axis?
  7. Nov 21, 2009 #6
  8. Nov 21, 2009 #7
    Okay, let me make sure this is right so far:

    [tex]\sum[/tex]Fx = 0
    fs - Nw = 0

    [tex]\sum[/tex]Fy = 0
    Ng - mg = 0

    (Nw is the normal force by the wall, Ng is normal force by the ground)

    Is this correct so far?
  9. Nov 21, 2009 #8
    Equation for x direction is correct. Equation for y direction needs the weight of the ladder and the weight of the person, but I only see one weight term.
  10. Nov 21, 2009 #9
    Well, I get lost at that because how do I know where to place the weight of the person? Or does position not matter for this part?
  11. Nov 21, 2009 #10
    It doesn't matter for the force equations because linear acceleration does not depend on where the force is located.

    Position does matter for your torque equations but in that equation, position is the unknown which you are solving for.
  12. Nov 21, 2009 #11
    Well I did solve for [tex]\sum[/tex]Fy, and the three forces in it. But for [tex]\sum[/tex]Fx, there is static friction and normal force of the wall, which I don't know either. All I know is the coefficient of static friction. So how do I find those?
  13. Nov 21, 2009 #12
    Just before the ladder starts to slip, static friction is maximized. The maximum force of static friction is equal to your y normal force times the coefficient of static friction.

    From your x equation you know the frictional force is equal to the x normal force.
  14. Nov 21, 2009 #13
    I know, but how do I find what the y normal force is?
  15. Nov 21, 2009 #14
    It's equal to the frictional force and you know the frictional force.
  16. Nov 21, 2009 #15
    I'm really confused. I don't see how I know the frictional force. To find the frictional force, I use the equation

    fs = [tex]\mu[/tex]k n

    and I know the coefficient of static friction, but what gets plugged in for n in that equation?
  17. Nov 21, 2009 #16
    Weight of the person plus weight of the ladder.
  18. Nov 21, 2009 #17
    so I combine those 2 and plug that number into n?
  19. Nov 21, 2009 #18
    Yes. It should be clear from your free body diagram that the y normal force minus the combined weight of the ladder and person gives the net force acting in the y-direction. Acceleration is zero so the y-normal force equals the combined weight of ladder and person.
  20. Nov 21, 2009 #19
    So this is what I got:

    [tex]\mu[/tex]kn - nw = 0
    (.45)(800 N + 200 N) - (450 N) = 0

    And then in the y direction, I have:

    Ng - mlg - mmg = 0
    1000 N - 200 N - 800 N = 0

    (mlg is weight of the ladder, mmg is weight of the man, Ng is normal force by the ground)
  21. Nov 21, 2009 #20
    Yes. Now get your torque equation set up so you can find the position of the person. It is most convenient to use the right end of the ladder as your origin.
  22. Nov 21, 2009 #21
    Okay, I think I got it.

    Nw + fs - mlg - mmg - Ng = 0

    Is that right? And does Nw cancel because that is my point of origin?
  23. Nov 21, 2009 #22
    Oops. For some reason I thought you did your free body diagram the same way I did. I have my free body diagram set up a bit different so the origin is at the bottom. It doesn't matter in any case. Where you set it won't affect your final answer.

    The force of friction and the weight will tend to rotate the ladder in the same direction. Their sign should be the same. Ng will rotate it in the opposite direction. The sign should be the opposite. You also need distances and angles in your equations.

    Origin at the bottom.

    Origin at the top.

    If you choose to go with the second equation, your x will be the distance from the top of the ladder. Subtract it from 8 to get the distance from the bottom of the ladder.
  24. Nov 22, 2009 #23
    I used your top formula, and it worked out to 6.79 m, which 6.8 m is one of the answers. So thank you very much for your help. I understood it afterwards.
  25. Nov 22, 2009 #24
    also if you wanna look at it again there are some MIT lectures where the lecturer goes through the equations. very helpful
  26. Nov 22, 2009 #25
    Thanks. I will keep that in mind. 8)
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