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Ladder equilibrium proof

  1. Nov 9, 2004 #1
    If a ladder makes an angle with the wall, has a mass m and is uniform in design, show that the minimum force of friction which must exist between the ladder and floor to keep the ladder from slipping is given by

    [tex]
    F_f = \frac{1}{2}mg\tan\theta
    [/tex]

    Now, [tex]\mu = \frac{F_g_x}{F_g_y}[/tex], and in a previous calculation I got [tex]F_g_x = \frac{\frac{L\cos\theta}{2}mg}{L\sin\theta}[/tex] and [tex]F_g_y = mg[/tex], with [tex]L[/tex] being the length of the ladder.

    When I plug all this together, I get [tex]\frac{\cot}{2}[/tex], not even close to what I'm intended to get. Can somebody tell me what I'm doing wrong?
     
  2. jcsd
  3. Nov 10, 2004 #2
  4. Nov 10, 2004 #3
    Thanks, that site looks great, but I'm already familiar with the basics behind the ladder problem. The equations that I'm using worked on previous problems, and I'm not sure why I'm not getting the same answer in this particular problem.
     
  5. Nov 10, 2004 #4

    Doc Al

    User Avatar

    Staff: Mentor

    I'm not sure what you are doing here, but the problem is simple. No need to find [itex]\mu[/itex], since all you need is the frictional force. Assuming there is no friction at the wall, the frictional force must equal the normal force exerted by the wall on the ladder. To find that normal force, apply equilibrium conditions for torques on the ladder about the ground contact point. Note that the angle is with respect to the wall, not the floor.
     
  6. Nov 10, 2004 #5
    Coldie,

    Make sure that the calculations done / formula's found in previous exercises do apply for this problem. If you describe for what exercises the formula's you found apply, we could give you better hint to where, and why your result is not 100%

    This is the point where physics requires you to understand what you are doing, in stead of just being a mathematical game.

    Greetz,
    Leo
     
  7. Nov 10, 2004 #6
    Gah, thank you, that's a big screw-up on my part. Also, since [tex]\theta[/tex] is at the wall pointing downwards, the [tex]\sin[/tex] and [tex]\cos[/tex] functions I'm using would be reversed, since I had been using the angle with respect to the floor! I think this solves my problem! Thanks!
     
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