Need help with force calculation involving a ball bearing 4

[tfry200]

Hi all,

I'm trying to solve the attached problem in order to understand the system so I can design a related tool. I have to admit I'm a little frustrated because I know at one time I could have solved this no problem, but I guess I've been away from it for too long!

I'm trying to follow the 600N force through the first wedge, through the single ball bearing, to determine the force on the wedge at the other end of the system. I'm pretty sure the normal force acting on the ball should be Fb=600/cos(75)=2318.22, but I'm not sure how to deal with the reactions on the 2nd wedge. Can anyone assist me in getting started?

Thanks,
Tim

 

[tfry200]

Okay,

Many many thanks for all the responses and assistance. Here's what I ended up with from summing the forces with and without friction. My numbers came out close to Zekeman's, and they also are similar to the graphical method mentioned. My results are attached. Think I've got everything covered?

Tim

 

[desertfox]

hi tfry200

Well I am glad you got where you wanted to be and if I made a error in my post then thats is that, I can see now working out as been posted as opposed to just answers where our figures differ and still looking at it whilst doing other things.
Again if I have made a mistake then zekeman as a fair point that I shouldn't rush to make hasty posting, that said if he had followed his own advice the last line of his post wouldn't be nessecary.

desertfox

 

[desertfox]

Hi tfry200

Just noticed in your calculations that on the left wedge there is no allowance for friction between the wedge and the cross hatched vertical wall, on the right hand side you have labelled it F4 but there is no equivalent on the left side.

desertfox

 

[desertfox]

Hi tfry200

Looking at the case for friction on the lefthand wedge it appears that zekeman and yourself have neglected friction between the angled surface and the ball.
Looking at zekemans triangle of forces he as drawn the 600N
force vertical and the corresponding top vector at 11 degrees from horizontal which I concurr with however the vector reaction from the ball is still drawn at 15 degrees from the horizontal which is incorrect when friction is being considered it should be drawn with an angle of 26 degrees from the horizontal which would yield a force of around 990N and not the 1326N using your figures presented.

desertfox

Let He who Is Without Sin Cast The First Stone

 

[tfry200]

Desertfox,

Are you talking about friction between the left wedge and the ball or the left wedge and the grounded vertical plane? for the wedge/ball I think I have them all in there? For the wedge/wall combination I see that I do not. (Also, it's not labeled for the wedge/wall on the right, but it is in the calculations.)

This has been an interesting problem, and I'm thankful for all the help and the opportunity I've gotten on it.

 

[desertfox]

hi tfry200

In your post dated 23rd you have not accounted for friction between the lefthand wedge and the vertical wall that it slides on, I'll post a solution shortly to show you what I mean exactly.

desertfox

 

[desertfox]

hi tfry200

I have uploaded a file now showing my details for your wedges using a coefficient of friction of 0.2 throughout.
I believe I have taken into account friction on both wedges and the ball.
I get a vertical force at Fy to be about 640N.

desertfox

 

[zekeman]

But the problem with using the vector triangles on the ball(unlike the wedges) is that the vector "solution" erroneously allows for a residual moment on the ball. Think about it--
Consider the result of the vector solution.
If you break the two sets of vector forces on the ball into the normal force set and the friction set, you see that the normal set would have zero moment on the ball since all forces pass thru the center. Now the friction set under the assumption of 0.2 coefficient would yield a triangle geometrically similar to the normal set and thus satisfy the linear force equilibrium condition, BUT, unless the linear sum of the bottom two sides of the triangle is equal the top side, it would not be in moment equilibriume.
There are further problems with this friction model, notably the way they manifest themselves.
So,we need a more sophisticated friction mechanism to get the proper answer.
In that absence I used zero friction,probably not the best assumption but at least it satisfies the moment equation,LOL.
I suppose somebody ought to open another thread to address this ball equilibrium problem.

 

[desertfox]

Well I have just done a moment balance on the ball from the figures in my last post and because each force acts at a different angle on the ball, I calculated the distance at which each horizontal and vertical force acts from the ball centre and for me the moments cancelled within fractions of a Nmm, in addition there is a slight error probably due to scaling error of my vectors on force equilibruim but well within the accepted range of 10%.
So I can see no reason in the calculations other than an error why on the lefthand wedge at the interface with the ball and the 15 degree face that friction was not included and looking back at zekemans triangle of forces which was posted on the 22nd the only place that included friction was the lefthand wedge between the vertical wall and flat bottom of wedge under the heading of friction coeff 0.2.
Zekeman I can post the moment equilibrium of the ball if you require and unless I am reading your post incorrectly your saying that a ball under three resultant loads acting concurrently cannot be moment balanced when those forces are broken into there respective horizontal and vertical components, if I have misinterpreted your last post then perhaps you can expand on it.

desertfox

 

[zekeman]

No, I don't want to break up the forces into rectangular coordinates, since you can easily lose sight of what I am saying. All I'm saying is that if you, instead break it up into radial and tangential forces, the set of normal forces produces NO moment since they all pass through the ball center. Now you are left with the 3 friction forces , all tangent, each proportional the corresponding normal (assuming same friction coefficient) force to the ball and each having a moment arm equal to the ball radius, so the total moment contribution is mu*(F2+F3-F1)times the ball radius. Since that sum is not zero (comes from the fact that the third side of any triangle cannot be equal to the sum of the other two sides), the moment it produces cannot be zero. Also note that what you do in developing the force diagram speaks only to 2 of the 3 conditions necessary for static equilibrium, and NOT the third, the moment equation.

 

[desertfox]

Well zekeman your wrong the requirement of equilibrium is that all external forces and moments sum to zero,the friction forces you speak of are known as reaction forces and do not generate movement by themselves, the external resultant forces as calculated in my earlier post today do indeed meet the above requirement for equilibrium.
So therefore the friction coefficient of 0.2 should have been carried through all the components including the ball as stated earlier.

desertfox

 

[zekeman]

I am shocked that you call the friction forces irrelevant on the moment equation. I guess you have repealed the law of static equilibrium.
Rather than choosing to call me wrong, show me where the the closure of the force vectors account for the moment equation.
Remember, as I have repeatedly mentioned it DOES NOT include the moment condition.
For examples if you take the ball triangle after you have determined one of the three vectors from the left wedge, you are left with 2 unknowns, namely the absolute values of F2 and F3 and you now have three equations to get these, namely the summation of forces in the x and y directions and the moment equation.3 equations 2 unknowns. How do you do it?
I think we have gone 15 rounds already and should close our private discussion.
Maybe someone else can arbitrate this one. If not...
Its been fun.

Best regards
Zeke

 

[desertfox]

Here is an example similiar to our situation but not quite the same:-

http://www.scribd.com/doc/2289233/Mechanics-Friction

It involves a cylinder sat in a vee and given its mass you have to calculate the amount of torque to be applied to rotate it. Due to its mass it as reactions and coefficients of friction are given and from this information you calculate the reactions both on the cylinder and wall.
This is no different to what we have done with the wedge in the original post.
Now if I remove the torque in the example of the cylinder (link above)then the cylinder stops rotating however it still as the reactions acting on it due to its mass and wall contact. As stated by zekeman if you look at the moments generated by the friction then they don't cancel however the cylinder doesn't rotate either and thats because the forces calculated are those that oppose motion and only an active force which which exceeds these values will allow rotation to begin. Not anywhere does the book state that we need a better model for friction as suggested by zekeman. Perhaps zekeman can now show us a site that actually supports his claim we need a better model for friction and that ignoring friction on the ball and on both wedges was not a mistake.

desertfox

 

[desertfox]

Scroll down the link till you get to page 12 of 39.

desertfox

 

[zekeman]

OK, one more time, but it is doubtful whether we will settle this anyway.
Since you brought it up,let's take your cylinder and both of us determine the forces on it under its own weight only. Let's say it weighs 100N.Assume static friction coefficient of 0.3.
My solution, since I claim there are virtually no friction forces on it, is:
left oblique wall normal force : 100/cos(30)
right wall normal force : 100* tan(30)
No friction forces.

What's your solution?

 

[desertfox]

Well firstly if you read the question properly the cylinder weighs 300N and furthermore the solution is irrelevant but more importantly it shows what you were saying yesterday all I am now waiting for is for you to show some evidence to support what you stated earlier:-

"so we need a more sophisticated friction mechanism to get a better answer"
"In that absence I used zero friction,probably not the best assumption but at least it satisfies the moment equation,LOL"

My solution I posted yesterday follows the text book but I'll wait see what you produce in light of your statements.

As far as I can see you also left friction off the tapered face of the left wedge and completely off the right wedge which I believe was an oversight which isn't a problem because we all make mistakes.

So have you any evidence to back your claim or not?

 

[zekeman]

Just do the new problem I proposed without the blather.
I gave you my answer. What is yours?

 

[desertfox]

The answer I asked for three times now is to back up your statements as posted in my previous post hopefully you can do this time without the blaher.



resultant force Na=237N

resultant force Nb= 312N

friction force from Na= 71N

friction force from Nb= 93.6N

 

[zekeman]

First off, I never said there was zero friction for the wedge problem ( which I assumed just on the ball , only that your use of friction does not square with satisfying the zero moment condition necessary and not having a good answer I chose zero for friction against the ball. Everywhere else I used friction. I don't have an answer for the actual friction distribution.
Now, the reason I asked that we both solve your cylinder problem is to show that your answer violates the zero moment condition. That paper correctly shows the amount of torque the cylinder can sustain without slipping . In the absence of that torque you cannot use a friction force because of the moment violation take moments about the center). Moreover, even if you choose friction, how can you possibly use the static coefficient since that yields the maximum value before slippage . As an example suppose you had a weight resting on an inlined plane with an angle of 10 degrees, the friction force would be just sufficient to prevent sliding, an effective coefficient of tan(10)=.18; by your method the friction coefficient would incorrectly be 0.3 and the weight would slide up the ramp.Stop.
Why don't we bring this discussion to a humane conclusion and stop now.

 

[desertfox]

If you read my post properly, the one with the problem link you will see I stated that it confirms what you said about moment equilibrium and the friction forces not balancing, but those forces on there own are reactive forces and do not generate any movement as stated in an earlier post by me.
Secondly your post of the 21st actually suggests using sliding friction and a coefficient of 0.2,no mention of using zero friction on the ball, that only came when you realised that you hadn't accounted for it in your "with friction calculations".
If you look at your force triangles there is no allowance made on the 15 degree face of the left hand wedge and no allowance for friction made on any part of the righthand wedge irrespective of the ball and I have not stated that you said there was zero friction on the wedges.
Yes your right if you assume a value of friction coefficient, then what you get is a maximum force before slippage but the actual force on the parts could be anywhere from zero to that maximum but again this is a part of engineering where one as to make assumptions.
The reason I posted that problem with the link is that it demonstrates that it is acceptable to use a static coefficient of friction of 0.2 or whatever value you choose on a sphere or ball and follow it through to a conclusion as I did previously with the original problem.

your statement from your post dated 24th:-

There are further problems with this friction model, notably the way they manifest themselves.So,we need a more sophisticated friction mechanism to get the proper answer.In that absence I used zero friction,probably not the best assumption but at least it satisfies the moment equation,LOL.I suppose somebody ought to open another thread to address this ball equilibrium problem.

Your statements imply that the method I used with friction isn't valid and yet your method of using zero friction on the ball is a better method, if thats the case then thats what they would show in text books and teach in college but as the example cylinder shows they don't.

I think we can call it a day now.