Question

I will give a medal to the best response:

I would love it if someone could please give me an explanation as to what I am doing wrong.

Please wait while I type my full question.

Answer 1

I already solved the first part of the question (please see attachment for picture) which reads:

"What is the minimum coefficient of static friction μmin required between the ladder and the ground so that the ladder does not slip?"

I got:
\[\frac{ \cos \theta(\frac{ 1 }{ 2 }m_2L+m_1d) }{ L(m_1+m_2)\sin \theta}\]

This is supposedly correct

Answer 2

What bothers me is the second part of the problem, which states:

"Suppose that the actual coefficient of friction is one and a half times as large as the value of μmin. That is, μs=(3/2)μmin. Under these circumstances, what is the magnitude of the force of friction f that the floor applies to the ladder?"

Answer 3

Now, I already found μmin (which is what I have above).

I also know(from previous calculations), that \[f=\mu(m_1+m_2)g\]

Answer 4

So, in order to find the answer I multiplied
\[\frac{3 }{ 2 }\times \mu \times(m_1+m_2)g\]

I got:
\[\frac{ 3gcos \theta(\frac{ 1 }{ 2 }m_2L+m_1d) }{ 2Lsin \theta}\]

But the correct answer is supposed to be:
\[\frac{ gcos \theta(\frac{ 1 }{ 2 }m_2L+m_1d) }{ Lsin \theta}\]

Answer 5

What am I doing wrong?

It looks almost as if the 3/2 is ignored. Where did I make a mistake? Could it actually be the program I am using for my homework that made the mistake?

Answer 6

Thank you to everyone who is willing to help!

Answer 7

I know that @AllTehMaffs and @douglaswinslowcooper are very good at solving this kind of problem.

Answer 8

I get a message that says:

"The ladder is no longer on the verge of slipping, so f≠μs∗N_2."

Answer 9

Maybe someone could really just help me solve the question:

"Suppose that the actual coefficient of friction is one and a half times as large as the value of μmin. That is, μs=(3/2)μmin. Under these circumstances, what is the magnitude of the force of friction f that the floor applies to the ladder?"

Answer 10

@ybarrap Please. I have math final on next Monday and Physics on next Thursday. Now I have to study for math and don't have time for this problem. However, I really, really want to know what happen. Please, please , please

Answer 11

I need to go out for a while, if no one has helped by the time I get back we'll work on this one

Answer 12

Sounds good!

Answer 13

Before I go, take a look at this lecture by my favorite Professor - Walter Lewin in Lec 25, where he goes through an extensive analysis of the forces and torques about a ladder with respect to static equilibrium (including a nice discussion on friction). Start it at about 2:44 where he begins talking about the ladder analysis: http://www.youtube.com/watch?v=lJfuU7D1DOA. He is lots of fun to watch and the best I've ever seen!

Answer 14

Sure, actually, I watched some of that very lecture yesterday--without it, I probably would not have been able to solve the first part of the problem.

I'll try finishing the lecture though!

Answer 15

This might be of better quality- http://teachingexcellence.mit.edu/inspiring-teachers/walter-lewin8-01-lecture-25-static-equilibrium-stability-rope-walker

Answer 16

Thanks !

Answer 17

I think the point is that if it is not slipping, the minimum amount of friction is sufficient and is all that actually applies.

Answer 18

When you solve all your equations (summing all forces and torques) with the person located on the ladder at a distance \(d\) from the bottom, you get that the frictional force is
$$
\large
F_s=g\cot\theta \left [\cfrac{m_2}{2}+\cfrac{m_1d}{L}\right ]
$$
The upper bound on the frictional force will always be
$$
F_s\le\mu_s\cdot \left [ m_1+m_2\right ]g\\
\text{So, }\\
g\cot\theta \left [\cfrac{m_2}{2}+\cfrac{m_1d}{L}\right ]\le\mu_s\cdot \left [ m_1+m_2\right ]g
$$
If you now \(u_{snew}=\cfrac{3}{4}\mu_s\cdot \left [ m_1+m_2\right ]g\). The new upper bound becomes
$$
g\cot\theta \left [\cfrac{m_2}{2}+\cfrac{m_1d}{L}\right ]\le\cfrac{3}{4}\mu_s\cdot \left [ m_1+m_2\right ]g
$$
But the lower bound doesn't change, it's still,
$$
\bf{\color{red}{g\cfrac{\cos\theta}{\sin \theta } \left [\cfrac{m_2}{2}+\cfrac{m_1d}{L}\right ]}}\\
=g\cot\theta \left [\cfrac{m_2}{2}+\cfrac{m_1d}{L}\right ]
$$
(red highlight above is the answer you showed above)

What the increase in static friction does is improve the safety of the ladder by increasing the \(upper\) bound, but it does not affect the lower bound, which is the actual force that friction will apply to keep the ladder up. If the frictional force was lower, then the ladder would not be in static equilibrium and would thus be moving.

Answer 19

@ybarrap sorry for taking so long.

Any way, when you say "lower bound" are you speaking of kinetic friction?

Answer 20

Well, I gave you the medal because what you said really makes sense.

I would still like to know what you meant by "lower bound" though.

Overall, thank you so much for your help @ybarrap!

Answer 21

The magnitude of the force of friction is what I'm calling the lower bound. This is the force that friction must provide to keep the ladder is static equilibrium. The range that the frictional force is allowed to take is between the value I call the lower bound and the upper bound determined by the coefficient of static friction. As the man moves up the ladder, his distance \(d\) will increase relative to the bottom. The lower bound will get closer to the upper bound, which is \(3\mu_s/2\times (m_1+m_2)g\), afterwhich the ladder begins to slip. Throughout this process, the the ladder remains static and is dependent on the static coefficient, not the kinetic.

Answer 22

Thank you so much @ybarrap, this really makes sense!

Answer 23

I'm going to leave this question opened a little longer: just to make it easier for @Loser66 to find.

Answer 24

@ybarrap Thanks for the explanation
@HomeshoolGrad Thanks for leaving it opened.