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Homework Help: A Person Standing on a Leaning Ladder

  1. Oct 23, 2005 #1
    A uniform ladder with mass [tex]m_2[/tex] and length L rests against a smooth wall. A do-it-yourself enthusiast of mass [tex]m_1[/tex] stands on the ladder a distance d from the bottom (measured along the ladder). The ladder makes an angle [tex]\theta[/tex] with the ground. There is no friction between the wall and the ladder, but there is a frictional force of magnitude f between the floor and the ladder. [tex]N_1[/tex] is the magnitude of the normal force exerted by the wall on the ladder, and [tex]N_2[/tex] is the magnitude of the normal force exerted by the ground on the ladder. Throughout the problem, consider counterclockwise torques to be positive. None of your answers should involve [tex]\pi[/tex]. (i.e., simplify your trig functions).

    1. What is the minimum coeffecient of static friction [tex]mu_min[/tex] required between the ladder and the ground so that the ladder does not slip?

    I'm thinking [tex]f_s + (-n_1) = 0[/tex].
    [tex]n_2 + (-9.80m_1) + (-9.80m_2) = 0[/tex]
    [tex]n_2 = 9.80(m_1 + m_2)[/tex]
    [tex]\sum \tau_B = (n_1)(Lcos(\theta)) - (9.80)(m_1)(d)(cos(\theta)) - (9.80)(m_2)(1/2)(L)(cos(\theta)) = 0[/tex]
    [tex]n_1 = (9.80)(dm_1 + (1/2)(Lm_2)/L[/tex] and [tex]f_s=n_1[/tex].
    [tex]\mu_s=f_s/f_n[/tex], where [tex]f_s[/tex] is maximum force of static friction.
    [tex]\mu_s = (dm_1 + (1/2)(Lm_2))/(L(m_1 + m_2))[/tex].

    My answer is not correct. Apparently, I'm missing some trig.

    What have I excluded or considered incorrectly?

    Attached Files:

    Last edited: Oct 23, 2005
  2. jcsd
  3. Oct 23, 2005 #2
    Well, I figured out my mistake and I got the right answer.

    But, I fell into the "B-Trap" for the second part of this question.

    Suppose that the actual coefficent of friction is one and a half times as large as the value of mu_min. That is, [tex]\mu_{\rm s} = (3/2)\mu_{\rm min}.[/tex] Under these circumstances, what is the magnitude of the force of friction f that the floor applies to the ladder?

    The ladder is no longer on the verge of slipping. So, [tex]f_s = (\mu_s)(n_2)[/tex] is not correct.

    What do I reconsider?
  4. Jun 9, 2008 #3

    Check this out, it may not help you much, but its something.
  5. Jun 10, 2008 #4


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    Hi highcoughdrop! :smile:

    erm … erik-the-red's last activity on this forum was about ten months ago! :rolleyes:
  6. Jun 10, 2008 #5
    Yeah, I'm aware of that, but the only reason I posted anything on here was so that the people who use google to search the physics forums have something when they do find this.

    Besides... that's how I got here, lol
  7. Oct 27, 2008 #6
    Awesome, highcoughdrop! That's pretty much how I got here too. You were pretty helpful there, you know.
  8. Mar 22, 2009 #7
    is the answer for part b same as that for part a? i'm stuck at b. :)
  9. Mar 29, 2009 #8
    any one can tell me if the answer for part b is the same as part a?
    i managed to solve only part a..
  10. Jun 4, 2009 #9
    Hey if anyone is still having trouble with this, (m_1*d+m_2*L/2)*cos(theta)*g/(L*sin(theta))
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