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Static equilibrium - man on a ladder

  1. May 3, 2015 #1
    1. The problem statement, all variables and given/known data
    Here's a screen shot of the problem statement:
    http://i.imgur.com/IRI02Ne.png

    For the purpose of this post, I'll label the given information as:
    a = left side of ladder (2.5m)
    a' = distance from floor to the point where the man is standing (2.0m)
    b = right side of ladder (2.5m)
    c = distance between the feet of the ladder (1.8m)
    m = mass of man (78kg)
    T = force of tension on the horizontal support of the ladder
    T_A = torque on the left foot of the ladder
    T_B = torque on the right foot of the ladder
    T_C = torque on the hinge of the ladder
    N_l = normal force of the ground against the left foot of the ladder
    N_r = normal force of the ground against the right foot of the ladder
    N_man = normal force of the step against the man

    Using the law of cosines, I found:
    angles A and B = 68.9 degrees
    angle C = 42.2 degrees

    I know FBD must be drawn for each side of the ladder
    I know that N_man = mg
    I know that T = the horizontal component of mg
    I know that net torque = 0 and net forces are 0 also

    I'm having a really hard time mapping out where all the forces are and where equilibrium actually exists...even after looking at similar post like this one: https://www.physicsforums.com/threads/triangle-ladder-equilibrium-problem.796257/

    I'm really disappointed that I can't figure this out even though there's all this information.

    2. Relevant equations

    net torque = 0
    net forces = 0

    3. The attempt at a solution

    Here's a terrible FBD of the left hand side: https://sketch.io/render/sketch55467fe4bc86c.png

    T_A = r x F = a' * cos(68.9) x (78)(9.8) = 550.3 N
    T_C = r x F = 0.5 * sin(C/2) x (78)(9.8) = 137.6 N
    T_B = ... (this part doesn't make sense to me)

    I'm so frustrated and confused at this point.

    Where is the force `R` coming from in the similar post I linked to?! Why does N_l exist if mg is canceled out by N_man?!
     
  2. jcsd
  3. May 3, 2015 #2

    haruspex

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    Consider the forces between the two sides of the ladder at the top hinge. What direction is the force one exerts on the other? Add it to your FBD. Since the situation is not symmetric, you will need an FBD for each side.
     
  4. May 3, 2015 #3

    D H

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    I would attack this problem in a completely different way: Break it down into parts.

    First, look at the ladder as a whole. There are only three external forces acting on the ladder, all of them vertical. The ladder as a whole isn't moving, so the forces have to balance. The ladder as a whole isn't rotating, so the net torque about any chosen reference point has to be zero. There are three obvious choices of reference points, which are the points at which the various forces are applied (the torque at that point from the force applied at that point is necessarily zero). This gives you two equations in two unknowns.

    What about the tensioner in the middle, or the hinge at the top? Those are internal forces. They don't come into play at this point. So let's bring that into play. Look at one of the legs. It's not rotating. Once again, the torques have to balance. You want the tension, so pick a reference point where this tension results in a torque. Keeping it simple, pick the hinge. There are unknown forces at the hinge, but they don't matter (yet). Keeping it simpler yet, the right leg only has two forces that contribute a torque. One equation, one unknown, and done.

    Next, there's the hinge. I'll leave that up to you.

    Finally, there's a sanity test (multiple sanity tests are better). Do the forces balance, as a whole and on each leg? Do the torques balance, as a whole and on each leg, with different choices of reference points?


    Sometimes this breaking things down into parts doesn't work and you have to look at the whole mess. That's why we have computer programs that do complex static analysis for us. Intelligently breaking things down into simpler, easily solvable pieces: That's why we humans still run the computers.
     
  5. May 3, 2015 #4
    Ok. So we have three external forces and they are:
    N_left (normal force from the ground on the left foot of the ladder)
    N_right (normal force from the ground on the right foot of the ladder)
    mg (force of gravity on the man)

    So right off the bat, we know that N_left + N_right = mg = (78 * 9.8) = 764.4 N...so this is one of those two equations and it has two unknowns.
    The other equation is...not obvious to me.

    So I look to torque at this point to fill in some holes...splitting the ladder up into the two sides we have:

    Ladder - left:
    Torque on hinge = [1.25cos(21.1) * tension] + [0.5sin(21.1) * mg] - [2.5sin(21.1) * N_left] = 0
    Horizontal forces = T - mg_x - L_left_x = 0 (L_x is the horizontal component of the force the right side of the ladder exerts on the hinge, right?)
    Vertical forces = N_left + L_left_y - mg = 0 (again, L_y is the vertical component of the force the right side of the ladder exerts on the hinge?)

    Ladder - right:
    Torque on hinge = [2.5sin(21.1) * N_right] - [1.25cos(21.1) * tension] = 0
    Horizontal forces = L_right_x - T = 0
    Vertical forces = N_right - L_right_y = 0

    I'm sorry, I can't make sense of this. I can't find tension because I don't know N_left, and I can't find that because the only other equation that has that in it has another unknown, L_left_y...I have no idea how to get that.
     
  6. May 3, 2015 #5

    haruspex

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    How about torque on the whole system (about some point)?
     
  7. May 3, 2015 #6

    D H

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    Don't do that then! Instead, try doing what I suggested, which is to *not* split the ladder into two sides. At least not yet. Find your N_left and N_right by looking at the ladder as a whole. Then you can start splitting the ladder into parts.
     
  8. May 3, 2015 #7
    Woohoo! I got it...finally!

    I can't believe it took me literally all day to get this...

    I guess I just needed to take a few hours to do something else and then come back to it with a fresh outlook. Thanks for all your help, both of you.
     
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