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Ladder leaning on a wall problem -- Please help me

  1. Dec 3, 2014 #1
    1. The problem statement, all variables and given/known data
    A ladder 8m long weighing 200N is resting against a rough vertical wall. A man of 720N climbs the ladder at point M. Take the coefficient of friction for both surfaces to be 0.25.

    The top of the ladder is point A and the bottom is B. The ladder makes an angle of 30 degress to the wall.

    Here's the link to the image in case the picture isn't showing...
    2. Relevant equations
    How far should the man stand, from B, such that the ladder is at the point of slipping?

    3. The attempt at a solution
    Taking upwards as positive...
    ΣForcesx = RA - FfA = 0
    ∴ RA = μRB ⇒ 0.25RB

    Taking the rightward direction as positive...
    ΣForcesy = FfB - 200N - 720N + RB = 0
    ∴ μRA + RB = 920N
    ∴ 0.25(0.25RB) + RB = 920N
    ∴ 0.0625RB + RB = 920N
    ∴ RB(0.0625 + 1) = 920N
    ∴ RB = 865.8823529N

    ∴RA = 0.25RB
    ∴RA = 0.25 x 865.8823529N
    ∴RA = 216.4705882N

    Taking clockwise as positive...
    ΣMomentsB = -(720sin30)(x) - (200sin30)(4m) + (RAcos30)(8m) + (FfAcos60)(8m) = 0
    ∴ 360x - 400 + (216.4705882cos30)(8m) + (μRAcos60)(8m) = 0
    ∴ 360x - 400 + 1499.752228 + ((0.25)(216.4705882cos60))(8m) = 0
    ∴ 360x - 400 + 1499.752228 + 216.4705882 = 0
    ∴ 360x = -1316.222816
    ∴ x = -3.656174489m

    ∴x = 3.66m?

    The official final answer wasn't given and I'm quite certain that I made a mistake somewhere. On the internet, I've only seen ladder questions where only one of the surfaces is rough. Please help me! Many thanks in advanced.
  2. jcsd
  3. Dec 3, 2014 #2


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    I'm not so sure forces are what I'd be looking at. You are almost doing it towards the end. The ladder will not fall, it will slide and spin. What's newtons 2nd law equivalent for rotational motion?

    Also, it will be useful to keep in mind that an object which is not locked down (like a pivot point) will rotate about it's center of mass.
  4. Dec 3, 2014 #3


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    I would start by solving the problem completely algebraically, i.e., not inserting the actual values until the very end but instead represent the forces by variables. This will help you in discovering unphysical errors and it will help us to understand your solution better without gritting out every single number.
  5. Dec 4, 2014 #4
    Thank you for your response! Unfortunately I'm going to have to tell you that my lecturer hasn't introduced me to the concept of Newton's 2nd Law of Rotation yet, so this problem has to be attempted solely by forces and moments. I will however do research into the 2nd Law of rotation and apply it to this question! Thank you so much!
  6. Dec 4, 2014 #5
    I shall do this! Where I'm from, we have a tendency to not use variables when answering questions. Thanks so much for your input!
  7. Dec 4, 2014 #6
    Please explain what RA and RB are. Why do you think you made a mistake?
    Last edited: Dec 4, 2014
  8. Dec 4, 2014 #7


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    N2L for rotation uses moments. It's ##T = I \alpha##.

    This is what needs to be applied in this case.
  9. Dec 4, 2014 #8
    Reaction Force at A and Reaction Force at B.
    RA and RB are the reaction forces at point A and point B. :)
  10. Dec 5, 2014 #9
    To be exact T here is the net torque or ΣT=Iα. I think he already have applied that by setting ΣMomentsB = 0
    Last edited: Dec 5, 2014
  11. Dec 5, 2014 #10
    Yes, but they are called normal forces because they're always perpendicular to the surface, right? You didn't answer my question though. Why do you think you made a mistake?
  12. Dec 6, 2014 #11
    It was stated that the ladder is "at the point of slipping" which is another way of saying that the system is in equilibrium. This is why I let the sum of the moments equal to zero.
  13. Dec 6, 2014 #12
    I think I made a mistake because the original answer was a negative value. In a second post I typed out why I thought I was wrong, but I suppose it didn't go through. I apologise for this. As for the Reaction forces, "Reaction Force" is another name for normal force.
  14. Dec 6, 2014 #13
    That probably gave you negative answer. I mean, torques are positive counter-clockwise and negative clockwise. (right-hand-rule)
    Last edited: Dec 6, 2014
  15. Dec 6, 2014 #14
    It a
    It actually doesn't matter if I take clockwise as positive or not, because I let the sum of all the moments about point B equal to zero, thus implying that the sum of the clockwise moments are equal to the sum of the anti-clockwise moments about B. The right hand rule is relative to finding the direction of magnetic force. This, however, is a mechanics question.
    Last edited: Dec 6, 2014
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