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Statics, ladder leaning against wall finding friction mass etc. really simple q.

  1. Feb 11, 2010 #1
    This Is really simple I just can't remember what way Sin and Cosine go

    1. The problem statement, all variables and given/known data
    Here is the problem:
    7 (b) A uniform ladder rests on rough
    horizontal ground and leans against
    a smooth vertical wall.
    The length of the ladder is 5 m and
    its weight is 80 N.
    The angle between the ladder and the
    ground is 60
    The ladder is on the point of slipping.
    (i) Show on a diagram all the forces acting on the ladder.
    (ii) Calculate the value of the coefficient of friction

    right so I realize the weight acts through the center of gravity in this case half way up. but it also acts towards the wall. is it the sin or cos of the angle that acts down. which goes horizontaly? I know how to do the rest just what way do the components of the weight break up? Is it 80 cos(60 or 80 Sin(60 to find the force acting to the right ?

    3. The attempt at a solution
  2. jcsd
  3. Feb 11, 2010 #2


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    Homework Helper

    Welcome to PF!

    Hi Gogarty! Welcome to PF! :smile:

    If you're asking about components of force, it's always cos of the angle between the direction of the force and the direction in which you're taking components.

    The only time you use sin is when that angle is already called (90º - θ) … so you use cos(90º - θ), which is sinθ. :wink:

    If you're asking about moments, use the distance from the point to the line (of force).
    Sorry, I'm not understanding this :redface:

    how can weight act towards the wall? …

    and why would you want to break the weight into components? :confused:

    Just find the normal force, then find the friction force …

    what do you get? :smile:
  4. Feb 12, 2010 #3
    Look for a point about which you can take moments to reveal an equation which gives you the answer you want. In my experience, people often get their sins and coses the wrong way round. and I try to avoid them by using geometrical ratios and similar triangles. But try the moment equation first.
  5. Feb 18, 2010 #4
    the force you're talking about acting against the wall is the normal force, and it is perpendicular to the point of contact.

    as for the weight, in the X direction it should be mgsin(60) and the Y -mgcos(60)
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