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I Ladder facing a wall

  1. Jul 7, 2017 #1
    Hey everybody. I just took a test where a problem described, "a ladder is leaning against a wall with an angle of A. It has a length L and it weighs mg. Assume no friction against the vertical wall and a frictional coefficient of B, find Ffriction"

    My dad, a quantum physicist, explained how the problem was done, saying that the normal force was mg because the forces in the Y direction added up to 0, the coefficient wasn't needed, and he used Torque (clockwise) = Torque (counterclockwise) to solve for Ffriction in a systems of equations. However, first of all, it wouldn't make sense that there was extra info in the problem, and second of all, I don't understand how the Fn could be mg if it was applied to the end (it just makes logical sense that if it is applied further from the weight center, it affects the translation force less.) I think he may be out of shape in terms of classical physics and forgot the exact way to do this problem and thus I'd like somebody to explain how to ACTUALLY do the problem.
  2. jcsd
  3. Jul 7, 2017 #2


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    What is the friction force that is needed to act on the foot of the ladder to ensure that the ladder doesn't fall down ?
  4. Jul 7, 2017 #3
    It sounds to me like your dad's analysis is right-on. Have you drawn a free body diagram of the ladder?
  5. Jul 7, 2017 #4
    yeah. the normal force is at the bottom of the ladder while the MG is at the middle (center of gravity) so how do those 2 cancel out? if you put a pencil on a table, pushing it in the center to make it translate if way easier than pushin on the edge because it rotates at the same time, so part of the force becomes rotational acceleration, right?
  6. Jul 7, 2017 #5
    In your problem, there is no rotational acceleration. It is in static equilibrium. Do you believe that, for a rigid body in static equilibrium, you must have a balance of forces and a balance of moment? Yes or no?
  7. Jul 7, 2017 #6
    I do. However, here what I've considered. If there were no side wall and the floor had no friction, what would happen is that the object would rotate & accelerate CCW but the center of mass would go downwards at the same time. That means the Fnormal is not as much as mg. Once you add the leaning wall, the F(leaning-wall-normal) will push the ladder sideways, but it still rotates and translates downwards. Only when the friction is added, another sideways force, then does the object reach equilibrium. So if the normal force wasn't enough to counteract the mg when there was no ground friction, why would it suddenly increase if you add a ground friction?

    essentially what happens is that if there is no friction on the ground, the normal force causes the ladder to rotate and it also keeps falling. If there is a leaning wall, it does the same thing, except the wall will apply a sideways force on the ladder and make it translate sideways while rotating and falling at the same time (the sideways translation keeps the end of the ladder touching the wall in the same X-coordinate). But once the ground friction is added, the leader stops both rotating and translating.
  8. Jul 7, 2017 #7
    that's what i'm trying to find out. the Ffriction on the ground. If you mean the leaning wall, it has no friction.
  9. Jul 7, 2017 #8
    It doesn't suddenly increase. You are talking about two different problems.
    This is all correct. So?

    My recommendation is that you solve both problems so you can compare the results and get a better understanding of their relationship. It should be pretty interesting.
  10. Jul 7, 2017 #9
    so the normal force is bigger if I add a frictional force sideways?
  11. Jul 7, 2017 #10


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    It is, and it is critical that you get used to this. FrI'm now on, in school and in real life, you may be given extra (or not enough!) information and have to figure out what you need and what you dont.
  12. Jul 8, 2017 #11
    We can continue speculating about this forever, of we can get down to business and actually solve both versions of the problem to see how this all plays out. I can help you solve the frictionless sliding version of the problem if you are game to try. Are you?
  13. Jul 8, 2017 #12
    Yep. Would love to. Let me get my pencil and paper
  14. Jul 8, 2017 #13
    not in the questions from the book's tests though. I mean it would make sense in real life if there was extra info, but not from that book's problems
  15. Jul 8, 2017 #14


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    It is good design to provide extra information in a problem. It helps train the student for real life. And it works to prevent the pattern matching "what formulas do I have that take a distance, a time and a force as inputs" approach to problem solving.
  16. Jul 8, 2017 #15
    The first step in the analysis is to quantify the kinematics of the ladder motion. We will use the figure below to address that:
    LADDER 2.png

    From the geometry of this figure, what are the x and y coordinates of the center of mass of the ladder (in terms of L and ##\theta##)?

    Attached Files:

  17. Jul 8, 2017 #16
    Well I've found some example problems of the exact same question online, and I know HOW to do it, but I don't know why. How come the normal force is larger when it is in equilibrium but less when there is no friction on the ground if it's the same ladder, same gravity, etc.?
  18. Jul 8, 2017 #17

    but for the sliding object: the center of mass is at .5lsin(theta) and .5lcos(theta).
  19. Jul 8, 2017 #18
    Like I said (several times), this will all reveal itself when we actually analyze the problem. Until then, we are just waving our hands.
  20. Jul 8, 2017 #19
    alright, lets go on then. Sorry for the delay by the way, I'm in a summer camp so I'm busy a lot of the time
  21. Jul 8, 2017 #20
    Excellent. Now, using these results, in terms of L, ##\theta##, and ##d\theta /dt##, what are the x and y components of the velocity of the center of mass?

    Please do me a favor. Please use LaTex to do the equations. There is a LaTex tutorial in the Physics Forums help.
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