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

 

[bvanhiel]

Gentlemen,

I believe you are both incorrect. Multiplying the friction coefficient times the normal force gives you the friction during sliding. When the parts are not sliding the friction force may be less.

In tfry200's problem, the ball will clearly slide about two of the interfaces and roll about the third. The friction force at the rolling contact will balance the moments. By looking at the frictionless solution you can determine the point with the highest normal force. This is the interface that will have rolling contact.

zekeman: you're solving by assuming 0 friction and that the ball will move freely. I suppose ignoring the problem counts as a solution of sorts...

desertfox: you are assuming that the friction force is always proportional to the normal force. This is also incorrect. The moment equation will not balance, as zekeman has pointed out. Your ball is magically accellerating.

I believe the groove problem that you've each "solved" is statically indeterminate when you consider friction.

-b

 

[zekeman]

OK, under duress, I will abandon my zero friction assumption on the ball and propose the following:
Looking at the motion of the ball as the 600 N force pushes up , it appears that the ball is in ROLLING contact with the bottom surface and sliding contact with the other 2 surfaces The F1 vector is previously determined from the force triangle on the left wedge, so the additional forces on the wedge are the normals at F2 and F3, the friction at F3 equal to 0.2*F3 and the unknown friction at F2. We now have 3 unknowns, namely F2, F3, and Xf, the friction at the rolling contact at F2 and can write the three equations for static equilibrium and look at the friction Xf and as long as it is less than the coefficient of STATIC friction I would suggest that it is valid.
I invite your comments .

 

[bvanhiel]

zekeman,

Correct... with the exception that I would pick the interface with the largest normal force as the rolling contact. This may indeed be the bottom interface, but it's unclear to me how you came to that conclusion.

-b

 

[desertfox]

Well I don't quite agree with the statement "multiplying the friction coefficient with the normal force gives you friction during sliding"
My understanding is that the normal force multiplied by coefficient of friction gives the limiting force after which that force is exceeded sliding begins. Further once sliding as begun the friction force actually drops down in magnitude as we move from static to dynamic friction coefficients.

Regarding the groove problem that was solved that was from a book link :-

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

(page 12)
So your saying that the book is wrong? Because that also shows that if you try a moment balance as previously mentioned it also contravenes equilibrium.

If our models are wrong then what is the correct method to solve the problem?

 

[zekeman]

"Well I don't quite agree with the statement "multiplying the friction coefficient with the normal force gives you friction during sliding"
My understanding is that the normal force multiplied by coefficient of friction gives the limiting force after which that force is exceeded sliding begins. Further once sliding as begun the friction force actually drops down in magnitude as we move from static to dynamic friction coefficients."

I finally agree with your statement since using the sliding friction presumes that we are pushing upward and have broken away from the static friction case and entered the sliding friction domain;the result we have, further assumes that the motion is upward. If we then decide to push downward on the upper surface I believe we will get a different result owing to the reversal of the friction force vectors.
You make a very good point about static friction and we should first verify that the system can indeed "slide" before we invoke sliding. To do this we should use the static coefficient first and if the resultant force, F5 is positive(downward)then we can choose the sliding condition; otherwise, the system will be static and the resulting F5 will be zero.

 

[bvanhiel]

Gentlemen,

In Tfry200's problem I was assuming sliding friction and that the parts were moving as a result (and in the direction) of the input force. Obviously if the direction of motion is reversed then all of the friction vectors need to be reversed. Also, as zekeman points out, if the system freezes up then F5=0. Checking for freeze up is basically the same problem except we're using the static coefficient.

If our models are wrong then what is the correct method to solve the problem?

I outlined the correct method in my post. Zekeman seems to understand it. Please review.

Regarding the groove problem that was solved that was from a book link :-

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

12)So your saying that the book is wrong? Because that also shows that if you try a moment balance as previously mentioned it also contravenes equilibrium.

The problem as posed in the book, calculating the torque required to rotate the cylinder, is solvable and the calculated forces, by definition, satisfy the equilibrium equations. The problem as posed to you by zekeman, calculating the forces of the cylinder resting in the groove, is statically indeterminate.

-b

 

[zekeman]

"The problem as posed in the book, calculating the torque required to rotate the cylinder, is solvable and the calculated forces, by definition, satisfy the equilibrium equations. The problem as posed to you by zekeman, calculating the forces of the cylinder resting in the groove, is statically indeterminate."
I have to say you are correct and the only reason I chose zero is that it satisfies the momentum eondition and is a solution; of course depending on how the cylinder is placed on the ramp there are an infinity of friction forces possible that will close the force loop but since you must satisfy the moment condition, the two friction forces must have the same amplitude.; the ideterminancy occurs because there are 3 equilibrium equations and 4 variable.( 2 normal forces and 2 friction forces) so you can specify one (within some limits)) arbitrarily to get the remaining 3, making sure that the 2 friction force vectors are of equal length.

 

[zekeman]

"The problem as posed in the book, calculating the torque required to rotate the cylinder, is solvable and the calculated forces, by definition, satisfy the equilibrium equations. The problem as posed to you by zekeman, calculating the forces of the cylinder resting in the groove, is statically indeterminate."
I have to say you are correct and the only reason I chose zero is that it satisfies the momentum eondition and is a solution; of course depending on how the cylinder is placed in the groove there are an infinity of friction forces possible that will close the force loop but since you must satisfy the moment condition, the two friction force vectors must have the same amplitude.; the ideterminancy occurs because there are 3 equilibrium equations and 4 variables.( 2 normal forces and 2 friction forces) so you can specify one (within some limits)) arbitrarily to get the remaining 3, making sure that the 2 friction force vectors are of equal length.

 

[desertfox]

bvanhiel
Oh right I see the rolling friction on the face with the highest normal force, okay I thought you might be posting some working out for us all to see.
The reason I asked for some working out is that I cannot see how it will solve the out of balance moment on the ball as you claim, so it would be of interest and help the OP too. Perhaps you would be kind enough to post your full solution and then the thread can be closed.

zekeman
Glad we can agree on something, anyway from our initial figure of 0.2 for sliding friction and the 600N vertical force on the wedge it would seem that motion was just impending however if we have rolling friction at that point I suppose that could change things dramatically.

 

[bvanhiel]

I'll work on a solution when I get a chance. For a complete solution there are actually four cases to be solved: static and sliding friction for each direction. The difference between the static and sliding cases is just the coefficent of friction used.

-b

 

[gwolf2]

OK, if the number of posts exceeds 75 without the right answer appearing I promise I'll get my pencil out. MGIH!

 

[desertfox]

gwolf2

Why wait, just go ahead and put your solution in.

desertfox

 

[zekeman]

Wolf,
"There is only one way to do this properly and no-one has yet.
I can't be bothered with this anymore - eject."

I believe you quit 34 posts ago.
No wonder we haven't nailed it.

Cheers

 

[gwolf2]

Sorry to appear evasive but I'm rather busy atm with a lot of work related travel and projects. I thought someone would have nailed it by now. Also it is a tips forum not a free maths service. Maybe next weekend.

Regards,

gwolf

 

[desertfox]

Hi gwolf2

Not so much a question of nailing it as agreeing which solution/method is the correct one.
We seem to be at odds as to whether we have moment equilibruim on the ball?, rolling or sliding friction or both.
So what were looking for if the solutions are wrong why are they wrong and how to put them right and if someone can produce the perfect solution then lets have it.