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A ladder against a frictionless wall

  1. Apr 19, 2006 #1
    A uniform ladder of length 9 m leans against a frictionless vertical wall making an angle of 47° with the ground. The coefficient of static friction between the ladder and the ground is 0.41. If your mass is 74 kg and the ladder's mass is 33 kg, how far up the ladder can you climb before it begins to slip?

    [tex]\Sigma F_x=0[/tex]
    [tex]f_s-F_1 =0[/tex]

    [tex]\Sigma F_y=0[/tex]
    [tex]w-F_n =0[/tex]

    Where [tex]F_1[/tex] is the natural force of the frictionless wall on the ladder, [tex]F_n[/tex] is the natural force of the floor on the ladder, and [tex]w[/tex] is the weight of the ladder.

    [tex]\Sigma \tau =0[/tex]
    [tex]F_1(sin\theta L)-w(\frac{cos\theta L}{2})=0[/tex]

    Solving for [tex]F_1[/tex] gives you [tex]f_s[/tex]

    I just don't know how to put this together and answer the question. The torque part is also probably wrong. Any advice would be appreciated.
  2. jcsd
  3. Apr 19, 2006 #2
    Just wondering, do you have the numerical answer to this question?

    Here are some of my thoughts about your work...

    1) You said that [tex]f_s=F_1[/tex]. What is [tex]f_s[/tex]?

    2) You also wrote that [tex]w-F_n =0[/tex]. Well, what about the downward force you exert on the ladder when you stand on it?

    3) When taking moments, you need to specify a pivot. I think you have chosen the pivot correctly, but again, you need to consider the moment exerted by you standing on the ladder.

    Hope this helps! All the best!
  4. Apr 20, 2006 #3
    I don't have the answer.

    I've set the point where the ladder hits the floor as the pivot.

    I've thought about this a little more, but I'm still having trouble.

    This is when nobody is on the ladder:
    [tex]F_1(sin\theta L)-w(\frac{cos\theta L}{2})=0[/tex]

    When the person gets on the ladder, the first part stays the same, but the distance for the weight will change because the center of mass will change. [tex]F_1(sin\theta L)-w(x)=0[/tex] where x is the x-value of the center of mass of the ladder with the person on it. But then I have 2 unknowns and I have no idea how to find either of them.
    Last edited: Apr 20, 2006
  5. Apr 20, 2006 #4
    I agree that it is difficult to find the centre of mass of the ladder when the person is on it. So, why don't you consider the forces separately?

    Here's a general formula for you:

    By Principle of Moments, setting the point where the ladder hits the floor as the pivot,

    Sum of anticlockwise moments on ladder = Sum of clockwise moments on ladder


    Moment due to [tex]F_1[/tex] = Moment due to weight of ladder + Moment due to contact force on ladder by person standing on it

    By the way, what is [tex]f_s[/tex]?
  6. Apr 20, 2006 #5
    [tex]f_s=\mu mg[/tex]

    Is that what you're talking about? Or is there a way to actually solve for [tex]f_s[/tex]?

    Sorry I don't understand what you're talking about with the moments.
    Last edited: Apr 20, 2006
  7. Apr 20, 2006 #6


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    Moments it another word for torque

    [tex]\tau = F\cdot d[/tex]

    Where [itex]d[/itex], is the perpendicular distance from the point where the force acts to the pivot. This is what causes the rotation of a body. So for example if we take moments about the base of your ladder, the torque due to the weight of the ladder is given by;

    [tex]\tau_{L} = (33\times 9.81 \times 4.5)\sin\theta[/tex]

    The [itex]\sin\theta[/itex] is used because the force is not acting perpendicular to the ladder, therefore to find the perpendicular distance we have use trigonometry.

    For a body to remain at equilibrium (i.e. not move) the sum of all torques about any point must be zero. You can use this fact to solve this problem. You can take torques about any point on the ladder, however I would recommed taking them about the base of the ladder. Any forces acting through the pivot do not produce any moments/torques.

    You were correct in saying that [itex]F_{1} = F_{s}[/itex]. In this case I would obtain an equation of [itex]F_{s} = ...[/itex], which you can then equate with [itex]\mu mg[/itex]. Note: you will have an uknown in your equation for [itex]F_{1}[/itex], this will be the distance you are from the pivot and you should solve for this once you have equated [itex]\mu mg[/itex] with [itex]F_{1}[/itex].

    Hope this helps
  8. Apr 20, 2006 #7

    [tex]f_s=\mu mg[/tex]

    What's the m? Mass of ladder? Mass of person? Or mass of both ladder and person?

    I think it's best to define the mass of the ladder as [tex] m_{1} [/tex], the mass of the person as [tex] m_{2} [/tex] and the mass of the ladder and the person as [tex] m_{1}\ +\ m_{2} [/tex]. This will help when calculating moments, as the weights of the person and the ladder may not be of the same distance from the pivot. It also reduces confusion, I hope.:tongue2:

    Oh, and I guess [tex]f_s[/tex] stands for the frictional force between the ground and the ladder.
    Last edited: Apr 20, 2006
  9. Apr 20, 2006 #8
    Just to give you a different option of solving...or checking (those given by Pizzasky and Hootenanny are perfectly adequate)

    May I assume that you have done work on three force equilibrium? There are 3 forces in this set up that can help you find the centre of gravity of the ladder and the person on it.
    (actually there are more than 3 forces here but there are ways of reducing the number...for example: two perpendicular forces acting at a single point can be considered as components of one vector)

    Force one is the normal reaction exerted by the wall on the ladder...as it is smooth can you guess which way this force acts?

    Force two is the resultant between the normal and frictional force at the floor? (what value of the friction force is important to solving this problem)...

    If you find where these two forces intersect then the third force (ie the centre of gravity of (ladder + person on ladder)) hits the intersection of these 2 forces (if it didn't can you guess what would happen?) ...and from this combined centre of gravity you can then find where the man is

    (one more hint...I would use this centre of mass as a pivot...you know the magnitude, direction, and (position (barring the mans location)) of all the other forces, :wink: )

    *edit* I assume you are allowed to model the man as a particle :)
    Last edited: Apr 20, 2006
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