# A Person Standing on a Leaning Ladder

#### erik-the-red

Question:
A uniform ladder with mass $$m_2$$ and length L rests against a smooth wall. A do-it-yourself enthusiast of mass $$m_1$$ stands on the ladder a distance d from the bottom (measured along the ladder). The ladder makes an angle $$\theta$$ 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. $$N_1$$ is the magnitude of the normal force exerted by the wall on the ladder, and $$N_2$$ 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 $$\pi$$. (i.e., simplify your trig functions).

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

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

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

What have I excluded or considered incorrectly?

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#### erik-the-red

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, $$\mu_{\rm s} = (3/2)\mu_{\rm min}.$$ 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, $$f_s = (\mu_s)(n_2)$$ is not correct.

What do I reconsider?

#### highcoughdrop

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

#### tiny-tim

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

erm … erik-the-red's last activity on this forum was about ten months ago!

#### highcoughdrop

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

#### shurane

Awesome, highcoughdrop! That's pretty much how I got here too. You were pretty helpful there, you know.

#### yandao87

is the answer for part b same as that for part a? i'm stuck at b. :)

#### yandao87

any one can tell me if the answer for part b is the same as part a?
i managed to solve only part a..

#### gorignak

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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