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Ladder against a wall how simple

  1. Apr 18, 2005 #1
    OK, so this appears to be a very simple problem... but I can't seem to find the way to do it... Anyway the problem is:

    A uniform 6.74 m long ladder of mass 15.6 kg leans against a smooth wall (so the force exerted by the wall, FW, is perpendicular to the wall). The ladder makes an angle = 71.4° with respect to horizontal (see figure below), and the ground is rough.


    Calculate the x component of the force exerted by the ground on
    the ladder at its base (+x is to the right).

    Any tips would be greatly appreciated,
    Thanks...
     
  2. jcsd
  3. Apr 18, 2005 #2

    Doc Al

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    Staff: Mentor

    As usual with these kinds of problems, begin by identifying (draw a diagram) all the forces acting on the ladder. Then apply the conditions for equilibrium.
     
  4. Apr 18, 2005 #3
    Al since your here maybe you can help us both out. I was looking at this a minute ago and was explaining what you said, but then the idea that torques might be the better method came into my head and I wasn't sure how to explain/include that. What's your take on that?
     
  5. Apr 18, 2005 #4

    Doc Al

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    Yes, you will need to consider torques! (The "conditions for equilibrium" include torques: The net torque about any point must be zero. That's rotational equilibrium.)
     
  6. Apr 19, 2005 #5
    How can I find an expression involving torque involving the force in the x-dir? I just cant seem to find it....
     
  7. Apr 19, 2005 #6
    Its the perpendicular distance of every force from the chosen axis multipled by that force. So you probably mean the contribution to torque from the force acting in the x-direction. Well chose an axis first and examine the direction of the torque of that force about this axis. Fix signs accordingly and sum all the torques so obtained (for every force).
     
  8. Apr 20, 2005 #7

    Doc Al

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    Staff: Mentor

    For your torque equation, pick either the top or bottom of the ladder as your pivot point. Whichever you pick, you'll also need to use the fact that the net force (horizontal and vertical) on the ladder is zero.

    Did you indentify and label all the forces acting on that ladder?
     
  9. Apr 20, 2005 #8

    minger

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    I think an important step into this when doing this type of problem, especially when summing moments is making sure you have the correct components. When you sum your torque, say at the top, your weight is acting in the -y direction, however you must remember that you only factor in the component of the weight that is acting perpendicular to the ladder.

    Likewise, you will find a force at the bottom of the ladder, but that force will again, be perpendicular to the ladder, so you must find the correct component that you're looking for in the x-direction.
     
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