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Moments of a Ladder Propped Up Against A Wall

  1. Aug 22, 2015 #1
    1. The problem statement, all variables and given/known data
    A uniform ladder rests against a vertical wall where there is negligible friction. The bottom of the ladder rests on rough ground where there is friction. The top of the ladder is at a height h above the ground and the foot of the ladder is at a distance 2a from the wall. The diagram shows the forces that act on the ladder.
    upload_2015-8-22_10-50-55.png
    Which equation is formed by taking moments?
    (A) Wa + Fh = 2Wa
    (B) Fa + Wa = F h
    (C) Wa + 2Wa = Fh
    (D) Wa – 2Wa = 2Fh


    2. Relevant equations
    moment=F x perpendicular distance

    3. The attempt at a solution
    Should the answer be (C)?
    The anticlockwise moments are Wa and 2Wa and the clockwise moment is Fh, so they should be equated.
    But which is the pivot? Is the pivot one of the ends of the ladder or some other point on the ladder. Shouldn't the forces have to be split into components because the force and distance must be perpendicular to each other. And shouldn't the distance be taken along the length of the ladder and not the ground?
     
  2. jcsd
  3. Aug 22, 2015 #2
    You're right,C can be an option if moments are taken about the topmost point of ladder.

    About the pivot,I'm not sure what you mean.

    The force should be split into components and we should then take the component which is perpendicular to the distance.But,in this situation,the forces and the distances given are perpendicular to each other already.Hence,we don't need to break up the force into components.

    And if you take distance along the ladder you will also have to take the components of the forces perpendicular to this distance.By a bit of trigonometry you'll see that this will give you the same result as the one you got earlier.
     
  4. Aug 22, 2015 #3
    The answer given is (A)
    Around which point is the ladder going to rotate? Around the centre of gravity or does it rotate around one of the ends of the ladder? Where is the hinge of the turning effect located?
     
  5. Aug 22, 2015 #4

    haruspex

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    No. Those two moments are only in the same sense if you pick an axis between the two forces. But then the distances a and 2a will not both be right.
    Taking an axis at the base of the wall, the ladder's weight has a clockwise moment and the reaction from the ground has an anticlockwise moment.
     
  6. Aug 22, 2015 #5
    But why are they adding the clockwise and anticlockwise moments? How did they get 2Wa on the right hand side of the equation?
     
  7. Aug 22, 2015 #6
    Shouldn't you equate the two moments as the ladder does not turn and is in equilibrium?
     
  8. Aug 22, 2015 #7

    haruspex

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    In equation C? To produce a deliberately incorrect equation.
    You can either write ##\Sigma## clockwise moments = ##\Sigma## anticlockwise moments, or write ##\Sigma## moments = 0, where each moment is given a sign according to its sense.
     
  9. Aug 22, 2015 #8
    No, in option (a), the clockwise moments have been added to the anticlockwise moments to get 2Wa. Wa is the clockwise moment and Fh is the anticlockwise moment, but what is 2Wa?
     
  10. Aug 22, 2015 #9

    haruspex

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    No, they haven't done that.
    They have not stated what axis they are taking moments about in each case. You have to try to find an axis for which the equation is correct.
    How might a force W produce a moment 2Wa?
     
  11. Aug 22, 2015 #10
    Is the axis here then along the W force that acts upwards?
    But how do we figure out where the axis is if there are no options and if it was a written question?
     
  12. Aug 22, 2015 #11

    haruspex

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    In the diagram, there are two forces magnitude F and two magnitude W.
    In equation A, there are moments Wa, 2Wa and Fh.
    Clearly the Fh moment must be a result of one of the two F forces and the other F force has no moment. This leaves only two possibilities for the Y coordinate of the axis. What are they?
    Similarly, to get those two moments from the two W forces, there are only two possible values for the X coordinate of the axis.
    Finally, see if any of those four X, Y combinations give the right signs.

    Edit: That said, it's probably quicker just to pick a fairly obvious axis, work out the balance of moments equation for yourself, and see if it can be manipulated to match one of the given equations.
     
  13. Aug 22, 2015 #12
    The F that acts along the horizontal at the base of the ladder has no moments. So the two possibilities for the Y coordinate of the axis is along the point on the ladder where its weight acts (W) or through the foot of the ladder where there is a force of W.
    And the X coordinate acts either through F (the one at the upper tip of the ladder) or through the centre where its weight acts.
    So the axis of rotation is through the W in the middle of the ladder?
    But then where does 2Wa come from?
     
  14. Aug 22, 2015 #13

    haruspex

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    That is only true if you pick an axis somewhere along that horizontal line.
    If you pick an axis at the middle of the ladder the weight will have no moment, so there will only be one W term in the equation.
    Seems like you do not know how to take moments. Let's try a specific example. Suppose you pick the point where the floor meets the wall. For each of the four forces, how far is its line of action from that point? What four moments do they result in?
     
  15. Aug 23, 2015 #14
    For the ladder, I think I understand how it works.
    For the example, the forces acting is the weight of the ladder acting downwards and the normal reaction force that acts upwards, friction acts at the point and the other force is opposing this, right?
     
  16. Aug 23, 2015 #15
    How about considering the origin (say, vertical is y axis and horizontal is x axis) as the point of rotation of the rod?

    Clearly :
    1. The force acting horizontally at the ground will not produce any moment.
    2. The force at the height h will produce moment Fh in clockwise direction.
    3. The weight acting at the center of the ladder will produce moment Wa in clockwise direction.
    4. The weight acting at the end of the rod will produce moment W(2a) in anti-clockwise direction.

    You can deduce the result from here, right?

    And for how to solve such questions - Always prefer to imagine the situation and taking the origin as the pivot point and try to deduce equations accordingly. You can also try to work out by taking other points as the point of rotation and see the difference in the result.
     
  17. Aug 23, 2015 #16

    haruspex

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    Yes, those are the forces, but what moments do they exert about the point where the floor meets the wall?
    Let's just start with one: what moment does the normal reaction from the floor have about that point, and in what direction?
     
  18. Aug 28, 2015 #17
    The moment is zero as the normal force acts at the point, so the perpendicular distance is zero.
     
  19. Aug 28, 2015 #18
    Thank you!
     
  20. Aug 28, 2015 #19

    haruspex

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    No, the normal force from the floor acts where the ladder meets the floor. The point I specified is where the wall meets the floor. Did you misread my question?
     
  21. Aug 29, 2015 #20
    Are the moments Fh, Wa and 2Wa?
     
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