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A Person Standing on a Leaning Ladder

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?


Last edited:
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?

Check this out, it may not help you much, but its something.


Science Advisor
Homework Helper
Check this out, it may not help you much, but its something.
Hi highcoughdrop! :smile:

erm … erik-the-red's last activity on this forum was about ten months ago! :rolleyes:
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
Awesome, highcoughdrop! That's pretty much how I got here too. You were pretty helpful there, you know.
is the answer for part b same as that for part a? i'm stuck at b. :)
any one can tell me if the answer for part b is the same as part a?
i managed to solve only part a..
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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