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

 

[gwolf2]

You need to put a normal contact force N at each contact and an associated friction force mu.N at the same point. For the side wedge/track contact assume two contact points top and bottom and again add N and mu.N.

Then treat each wedge and the ball as a free body in which forces are in equilibrium. This will give you some simultaneous equations which you solve by elimination.

Don't forget the moment equilibrium in the free body force balance.

Try to get rid of the contact at the circular support at the bottom - I'm not 100% sure but I think the three point contact on the ball makes it unsolvable (Newton's three ball problem)

Don't forget to add gravity M.g to each component.

Hope this helps, maybe someone will do it in detail for you.

gwolf.

 

[tfry200]

Hmmm...this is a pretty small system (the ball is 5mm dia) compared to the forces involved, so I've been ignoring the weight of the elements. Also, would it be possible to treat this as a frictionless problem?

 

[desertfox]

Hi tfry200

The 3rd support makes it difficult to analyse however if you assumed angle for the bottom support reaction one could solve it graphically with a polygon of forces diagram I think.
Also according to my calculations the force to the left of the ball is 600N * Cos 75 deg = 155.29N.

I have uploaded a sketch of what I think the free body diagram would look like see if it helps.

desertfox

 

[israelkk]

desertfox

The ball free body diagram is not enough for analysis, the left wedge free body diagram should be considered too because it rubs on the left side wall and the friction between the left wedge with the wall will affect the results.

 

[tfry200]

Okay I follow you with the forces on the ball. The way the ball sits into that curved support on the bottom is the only part of the system i'm not completely sure about, but I think ideally the actual position is more like this (see new attachment).

The reason I think the force is 600/cos(75) is because it's acting like an inclined plane system, so the force should increase due to the mechanical advantage. (As I write this I double checked myself and I think I had it wrong. Mechanical advantage is length/rise, i.e. hypotenuse/opposite or 600*1/sin(75)=621.2)

 

[desertfox]

Hi israelkk

Yes I was using the assumption that the thing was frictionless as was mentioned in the OP's second post which of course in reality it isn't, I am still scanning my books to see what else I can find.

desertfox

 

[desertfox]

Hi tfry200

I presume what were looking at is a static analysis? the vertical force of 600N acting that acts vertical,if you manage that the wedge didn't have the 15 Degree face and was just vertical then there would be zero force acting on the ball assuming the face touch tangential and forgetting friction. Now as you taper the face a small proportion of that vertical force exerts some load on the ball so I think in this case the 600N * Cos 75 deg is correct.
We also need dimensions from the ball centre to the support points I have assumed in my sketch that the angles of the faces are the same off the centre of the ball they may not be you need to confirm.
I don't know what sort of accuracy your looking for and whether ignoring friction matters assuming it doesn't then I have drawn a rough sketch of what I think the force diagram would look like.

 

[zekeman]

If you are only interested in Fy, then a quick and dirty graphical method is available using the virtual work principle, useful here assuming no friction.
First layout the system with reasonable accuracy, like you have done. Now graphically move the left wedge vertically, say delta x=1/4 inch on the drawing and allow the system to expand . Now note the vertical displacement, delta y at the Fy location.
From virtual work
600*delta x =Fy* delta y
so
Fy=600*delta x/delta y
You can also get the delta y analytically, involving some trig.

 

[tbuelna]

tfry200,

You should not ignore friction effects, for many reasons:

- The wedge surface contacts with the spherical intermediate member don't result in a pure rolling motion, and must allow some relative sliding. Which would imply a high friction loss at those contacts.

- The small included angle (15deg)of the driving wedge member surface would produce a high normal force at the wedge's reaction point, for a given axial force (ie. 600N). The wedge must move vertically to produce a continuous force on the sphere, which means it must be able to slide relative to the outer surface it bears against. If it can't slide, the the driving load will balance out once the elasticity of the mechanism's structure achieves equilibrium with the resisting friction forces.

- The negative 7.5deg wedge angle on the reacted side of the output wedge element would also produce lots of friction losses.

I'm too lazy to do a vector analysis of your mechanism. But at a glance, that device appears that it would be self-locking with your 600N input force, even if you apply a liberal Mu for greasy metal surface contacts (ie. <0.2). So your Fy value would thus be zero.

.....Unless of course, you wish to ignore friction.

Regards,
Terry

 

[gwolf2]

You must include friction to get your free body diagrams to work properly. Then you do force and moment equilibrium, then solve the (very simple) simultaneous equations and voila. There is only one way to do this right and that is to be methodical and include every load in a consistent axis system and follow through the equations.

 

[zekeman]

If you include friction, you are talking 3 free body diagrams AND an assumption of the friction coefficient ( I would initially assume the sliding friction coefficient). Now starting with the left wedge, you draw the closed triangular vector loop where the vector from the left side of the wedge is at angle whose tangent to wall normal is the assumed coefficient. Loop closure yields F1; you next close the loop containing the ball, which yields F3; finally the right wedge is done where the vector force from the right wall is treated similar to the first wedge.
This then solves the problem assuming sliding friction which could be a problematical assumption.

 

[zekeman]

Pursuant to my previous post on 11/20, I worked out the trig and found
delta y/delta x=tan(15)*tan(37.5)=.206.
so
Fy=600/.206=2918 lb
without friction

I also did this with the 3 triangle loops and got
Fy= 3020 lb, more or less confirming the above.

With a friction coefficient of ,2 using the three triangular approach of my previous post, I got
Fy= 1547 lb

 

[desertfox]

Hi tfry200

Well your post is stirring quite a bit of interest.
I post my working out for the vertical force on the righthand side of your wedge diagram.
I get a vertical force of about 219N having resolved the 600N force on the lefthand side wedge to parallel and at right angles to it see what you think.
I have ignored friction by the way.

zekeman your answer's are in lbs I think they should be Newtons.

regards

desertfox

 

[zekeman]

Fox,
Why don't you look at other solutions, before hastily posting and avoid the apologies later.
Your value for F1 is obviously wrong. The frictionless case already has shown F1 be
600/cos(75) , a lot higher than yours.
Sorry about my Newton /lb mixup, but the answers I gave are numerically correct since I started out using lbs.

 

[desertfox]

hi zekeman

Well all I have is your numbers where is the working out where have I gone wrong?

desertfox

 

[gwolf2]

tfry200,

See what happens if you are not careful and methodical and include all forces including friction!

There is only one way to do this properly and no-one has yet.

I can't be bothered with this anymore - eject.

 

[zekeman]

Fox,
I have no idea if I can send this but I did the triangles with and without friction

 

[desertfox]

zekeman

I cannot open that can you send it as a pdf.

desertfox

 

[zekeman]

Well, I did it in Excel. Let's see if you can get it.
With and without friction, Results are close (but not exactly as my previous answers)