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Ladder leaned against a wall; forces

  1. Oct 31, 2012 #1
    1. The problem statement, all variables and given/known data
    There is a 5m ladder leaned against a wall, making an angle of θ with the ground. The ground is rough (getting at friction), and the wall is frictionless. Here's an ASCII attempt:

    |\
    |..\
    |....\5
    |......\
    |......θ\

    We want to make a force diagram for the ladder, then find the value of the forces acting on it.

    Dealing purely with variables aside from the 5m length.

    2. Relevant equations
    Sum of F = ma
    Torque = F x l

    3. The attempt at a solution
    I'm fine with the force diagram, I think. You've got gravity downwards, the force of friction to the left, some normal force upwards, and some normal force by the wall directly to the right.

    As for what they are...

    Force of gravity = mg
    FNfloor = mg?
    Ff = ?
    FNwall = ?
    Ff = FNwall

    I think FNfloor = mg, and Ff = FNwall, because this is an instance of static equilibrium, so sum of F = 0. Is that correct? And how do I account for torque?
     
    Last edited: Oct 31, 2012
  2. jcsd
  3. Oct 31, 2012 #2

    haruspex

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    For the torque, pick any convenient point and calculate the moments about it.
     
  4. Oct 31, 2012 #3
    Ok, I'll pick the bottom of the ladder, where it meets the ground.

    For force of g: T = mg x 2.5cos θ
    For force of FNwall: T = ? x 5cos θ
    For the other two forces, T = 0

    I'm not sure where that brings me, though.
     
  5. Oct 31, 2012 #4

    haruspex

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    Not quite. Try that second one again. And pick a consistent direction for positive torque. And you already have a name for the unknown (the '?').
    Is there any rotational acceleration?
     
  6. Oct 31, 2012 #5
    Another go for FNwall: 5sin θ x Ff

    Here's a diagram to help explain what I'm not getting:

    ||\
    ||..\
    ||....\-->
    ||......\
    v|......θ\

    Vertical arrow is what I now think torque for FNwall should be, 5sin θ x Ff. Horizontal is torque for g. Is that correct? If so, is one positive and the other negative? I just don't get that, but that's what I'm led to think given your advice.

    EDIT: No rotational acceleration.
     
    Last edited: Nov 1, 2012
  7. Nov 1, 2012 #6

    haruspex

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    No, a torque is neither horizontal nor vertical. In the present case it is clockwise or counterclockwise.
    The force FNWall is clearly horizontal, but what is the distance from its line of action to the base of the ladder? Is it 5 cos θ, 5 sin θ, 2.5 cos θ or 2.5 sin θ? Does its torque act clockwise?
     
  8. Nov 1, 2012 #7
    Ok, good news is I knew that about torque, at least. That's what I didn't understand, how a "torque" could work if it were simply a line. (We have only discussed it as clockwise/counterclockwise.)

    |\-LOA->
    |..\.....|
    |....\...|
    |......\.|
    |......θ\

    I think that is the line of action, and the vertical is what I'm trying to find, so I'm going to say 5sinθ is the answer. (Whoops, that's actually what I meant last time, but I'm still not positive whether it's correct.) I don't think it's 2.5 anything because the force is not applied to the ladder's center of gravity.

    I think that moment (I think it's a moment) acts clockwise. If I made the LOA a tangent line to a circle, it would "point" in the clockwise direction, which is my rationale. It would follow that the moment caused by Fg is counterclockwise.
     
    Last edited: Nov 1, 2012
  9. Nov 1, 2012 #8

    ehild

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    It looks good, so what is the net torque then?

    ehild
     
  10. Nov 1, 2012 #9
    Well, net torque must be 0 if it's in equilibrium. So,
    mg 2.5cosθ - 5sinθ Ff = 0

    I used a minus just to indicate that one direction is negative. (I arbitrarily picked clockwise.)

    Going off of that:
    mg 2.5cosθ = 5sinθ Ff
    Ff = (mg 2.5cosθ)/(5sinθ)
    Ff = FNwall, so
    FNwall = (mg 2.5cosθ)/(5sinθ)

    Did I bring it home?
     
    Last edited: Nov 1, 2012
  11. Nov 1, 2012 #10

    ehild

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    Ok so far, but you can simplify it.

    ehild
     
  12. Nov 1, 2012 #11
    = 0.5 mg cotθ

    I think that's all. Thanks a lot, ehild and haruspex; I've been out of school sick recently, going through this problem with me really did help me catch up. Once I'm a bit more learned when it comes to physics, I'll do my diligence to help out some other people on here, too. :)
     
  13. Nov 1, 2012 #12

    ehild

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    Nice plans! Go ahead.:smile:

    ehild
     
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