Leaning pole with friction problem

[shinobi20]

This is a problem from Mechanics by Kleppner and Kolenkow1. Homework Statement
Two identical masses M are pivoted at each end of a massless pole of length L. The pole is held leaning against frictionless surfaces at angle θ, as shown, and then released. Find the initial acceleration of each mass.

 Note:
|
o
|  \
|    \     
|      \       
|        \        
|_____o____

Homework Equations

y^2+x^2=L^2
yy''+(y')^2+xx''+(x')^2=0, if the masses are at initially at rest then yy''+xx''=0, x'' = (-y/x)y'' = -tan(θ) y''.

The forces that act on the upper mass is the weight (Mg), normal force (Ny) , and force of the pole (F) so the equations are:
My'' = Mg-Fsin(θ) and Ny = Fcos(θ) (No use, can be omitted)

The forces that act on the lower mass is the weight (Mg), normal force (Nx), force of the pole (F) and friction (f) so the equations are:
Mx'' = Fcos(θ)-f and Nx-Fsin(θ)-Mg=0

The Attempt at a Solution

By manipulation, I ended up getting y'' = (-g)/(1+µtan(θ)+(tan(θ))^2) and so x'' can easily be derived.
[/B]
My question here is that K&K didn't include Fsin(θ) in the second equation of the lower mass. This will result to Nx=Mg which then leads to a different answer. Why did K&K exclude Fsin(θ) knowing that it is a force from the pole that contributes to the downward force for the lower mass?

 

[haruspex]

The answer looks right (after setting mu to 0), but there's an easier way.
Let the accelerations be a_x and a_y. Resolving along the ladder they must have the same component: $a_x\sin(\theta)=a_y\cos(\theta)$.
Take moments about the point where the two normals intersect (so they don't figure in the equations). $MgL\sin(\theta)=Ma_xL\cos(\theta)+Ma_yL\sin(\theta)$.

 

[rivendell]

I'm working on this problem now in my book, and I'm not really sure if y^2+x^2=L^2 is really the way to go on this problem. I'd just like to know why it isn't (b-y)^2 + x^2 = L^2 since y is really the distance the top of the pole moves. If that's the case, then the answer the first guy who posted got might not be right. Without making sure of this, I'm a bit stuck.
Can someone please explain / clear this up? Thanks very much!

 

[haruspex]

I'm working on this problem now in my book, and I'm not really sure if y^2+x^2=L^2 is really the way to go on this problem. I'd just like to know why it isn't (b-y)^2 + x^2 = L^2 since y is really the distance the top of the pole moves. If that's the case, then the answer the first guy who posted got might not be right. Without making sure of this, I'm a bit stuck.
Can someone please explain / clear this up? Thanks very much!

The OP did not define y explicitly, but appears to have taken it as the height of the top of the pole. If you define it differently you will get a different equation.

 

[rivendell]

ok. just wondering, aren't we supposed to use b since it was given? and isn't this problem exactly the same as 2.7 in kleppner and kolenkow (the other 'leaning pole' one, I know this one came from 3.1) or am I just missing something here? Is it supposed to be done in a different way in this problem?

well since I defined it differently I got (y')^2 + y(y'') = (x')^2 + x(x''). Then I evaluated the FBDs and ended up simplifying everything down to 2W = M(y'' + x''). (or basically 2g = y'' + x'') So now I have a constraint equation and another from my FBDs, but I can't think how to solve for y'' and x'' from these two equations and what I have. Is there a way to do this that I'm not thinking of?

I'd like to be able to work my approach out to the answer. (I have no clue what that answer might look like). and sorry I'm don't exactly follow the 'easier way' you mentioned in the 2nd post as you explained there. I'd like to know the exact steps to that, it's probably a better way than I'm pursuing. I've been trying this problem in whatever way I know for a while and I think I need to stop. Thank you!

 

[haruspex]

aren't we supposed to use b

b? You mean L? By dimensional analysis, it cannot feature in the answer. Only g and theta can be involved.

isn't this problem exactly the same as 2.7 in kleppner and kolenkow

No idea, never seen that book.

I'm don't exactly follow the 'easier way' you mentioned in the 2nd post

I may have taken theta to be the complement of that intended. Looks like I took it as the angle to the vertical, and y measured down from the top, as you have.
What is the component of a_y along the pole? What is the component of a_x along the pole? How must they be related?

2g = y'' + x''

That looks wrong. If you want me to check your method you will need to post all your working.

 

[rivendell]

lol, I do mean b. It's given in this problem's diagram, where the setup is just like another problem in the book shown here (though I don't know if I can assume that motion exerted by the rod is in line with it as was stated there): https://www.physicsforums.com/attachments/kk-2-7-jpg.72173/

and here are my steps:
(here F is the force the rope exerts)
from FBD for top M
my" = W - Fsin(theta) = 0
N - Fcos(theta) = 0 --> N = Fcos(theta) (1)

from FBD for bottom M
mx" = Fcos(theta)
N - W - Fsin(theta) = 0 --> N = W + Fsin(theta) (2)

substituting 1 into 2: Fcos(theta) = W + Fsin(theta) => F(cos(theta) - sin(theta)) = W
so my" = Fcos(theta) - 2Fsin(theta)
and mx" = Fcos(theta)
so: my" = mx" - 2(W-my") = mx" - 2W + my" ==> 2W = m(y" + x") ==> 2g = y" + x"

maybe there's something here I'm not catching, it's not a great way, but it's the only one I could think of.
though I thought I might have come close, I'm not sure how to proceed from here to get y" and x",

 

[haruspex]

I do mean b. It's given in this problem's diagram

Ok, that was not shown in this thread. Anyway, same story - it cannot feature in the answer, other than replacing functions of theta by functions of b/L.

if I can assume that motion exerted by the rod is in line with it as was stated there

You mean force, yes? I don't understand the book's remark about the force only being in line with the rod at the start. If the rod has no mass, the forces it exerts must always be along it. Consider moments about one end of the rod.

my" = W - Fsin(theta) = 0

I assume the "=0" there is a typo.

substituting 1 into 2

Eh? You seem to be assuming the two normal forces are equal. Always try to avoid using the same symbol for two different quantities.

 

[rivendell]

oh. thanks for catching that mistake! :)

 

[MARX]

I am working out this problem now
why is there no friction on top mass upwards
problem says frictionless surfaces so why friction on bottom one then? "VERY CONFUSED"

 

[haruspex]

I am working out this problem now
why is there no friction on top mass upwards
problem says frictionless surfaces so why friction on bottom one then? "VERY CONFUSED"

I do not know why @shinobi20 included friction on the lower mass. As I wrote in post #2, the answer found in post#1 appears correct if that friction is set to zero.

 

[MARX]

Right thanks so much haruspex that's what I got. The solution manual (which I believe was done by the author am I right?!) has one friction force on the bottom mass which really confused me..

 

[haruspex]

Right thanks so much haruspex that's what I got. The solution manual (which I believe was done by the author am I right?!) has one friction force on the bottom mass which really confused me..

I take it you are also using K&K, which I have never seen.
The question asked in post #1 implies some other strangeness in the book solution. It sounds like the solution is to a different problem. I suspect this happens in textbooks mostly by the question being edited and the answer not being updated to match.

 

[MARX]

I take it you are also using K&K, which I have never seen.
The question asked in post #1 implies some other strangeness in the book solution. It sounds like the solution is to a different problem. I suspect this happens in textbooks mostly by the question being edited and the answer not being updated to match.

Yes you're absolutely correct I thought the same thing; the solution is for a different question especially that at the end of the solution (K & K Yes) it says that the nonfriction case was treated in the previous chapter I did the problem accounting for friction but then I used 2 frictions one for the top mass and one for the bottom mass of course I got a different answer that doesn't match theirs. Am I right in the sense that if we consider rough surfaces there would be two friction forces because either way they're solution would not be correct! Unless the question that they are solving explicitly states friction on one mass but not the other..

 

[MARX]

I take it you are also using K&K, which I have never seen.
The question asked in post #1 implies some other strangeness in the book solution. It sounds like the solution is to a different problem. I suspect this happens in textbooks mostly by the question being edited and the answer not being updated to match.

Overall K& K is a great book! The problems and examples are rich and tricky! If someone (as crazy and obsessed as myself lol) works out all the examples and problems it goes a long way to mastering this level of Mechanics! Best book I have seen as of yet in doing so! (so far at least, still on chapter 5! but I am optimistic it will be the same throughout).
you know of any such good books.

 

[fm_sahaf]

Yes you're absolutely correct I thought the same thing; the solution is for a different question especially that at the end of the solution (K & K Yes) it says that the nonfriction case was treated in the previous chapter I did the problem accounting for friction but then I used 2 frictions one for the top mass and one for the bottom mass of course I got a different answer that doesn't match theirs. Am I right in the sense that if we consider rough surfaces there would be two friction forces because either way they're solution would not be correct! Unless the question that they are solving explicitly states friction on one mass but not the other..

Um... hi Marx! I am another guy just passing through KKs book because of the enthusiasm. I'd be glad if you could reply soon with the answers you got from using two frictions. I solved it myself but I am not sure if I did it right; mainly because of the same confusion you had. An advanced thanks for helping. :D

 

[haruspex]

Um... hi Marx! I am another guy just passing through KKs book because of the enthusiasm. I'd be glad if you could reply soon with the answers you got from using two frictions. I solved it myself but I am not sure if I did it right; mainly because of the same confusion you had. An advanced thanks for helping. :D

MARX seems no longer to be a user of this forum.
Please post your solution.

 

[fm_sahaf]

MARX seems no longer to be a user of this forum.
Please post your solution.

Thanks for the information. Um- The thing is my calculation is like all jotted up. Nth gets cancelled(almost). The first 3 are the equation of mine and the 4th one a little below is the value of b" or y". Lemme know if that's okay. Please feel free to recalculate because I couldn't put much effort into it as I have exams going on.