Coefficient of friction of the two surfaces...

[ALEng2]

Hi, I have what should be a simple problem but have already had multiple different answers from qualified engineers.

Looking at the attached image the problem is as follows:

The green wedge is pushed downwards by force F.

Due to the shape of the wedge and the retaining jaws there will be horizontal forces pushing the jaws outwards – R1 and R2. As the geometry is the same each side, R1 will equal R2.

The jaws are retained and cannot move.

In terms of F, what does R1 (and R2) equal?

The basic equation is answered in my post below, but how does the coefficient of friction affect this?

Thanks.

 

[3DDave]

Did those multiple engineers show you how they got their answers?

 

[ALEng2]

Sorry missed an important bit out of my OP. The consensus seems to be that the basic equation can be calculated by the angle of the normal force, then using basic trigonometry, the question is how the coefficient of friction of the two surfaces (steel on steel) affects that calculation.

The basic answer is R1 = R2 = (F/2)tan60
(this is due to the angle of the normal force (hyp) which we know the angle of, it then creates a force triangle where the adjacent is F/2 so the opp is R1.)

But how does the coefficient of friction affect this?

ôAn optimist will tell you the glass is half-full; the pessimist, half-empty; and the engineer will tell you the glass is twice the size it needs to beö

 

[hydtools]

Look into the component forces of R1 and R2. One component perpendicular to the surface, the normal force. One component parallel to the surface, the friction force. The friction force will be in the direction resisting motion.

Ted

 

[TimSchrader2]

In order for the block to move (a bit) it would have to bend the supports and overcome any friction from the normal force between the two surfaces. So the a friction force component is also opposing the F force.

 

[BrianE22]

The basic answer is R1 = R2 = (F/2)tan60
(this is due to the angle of the normal force (hyp) which we know the angle of, it then creates a force triangle where the adjacent is F/2 so the opp is R1.)

I'd like to see your FBD. I don't see how you cancel out the coefficient of friction.

 

[IRstuff]

I don't see why would friction enter into it? Nothing is moving, and nothing is being kept from moving by friction.

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[SwinnyGG]

But how does the coefficient of friction affect this?

The coefficient of friction doesn't alter the angle of the applied force, the normal force(s) between the parts, or the direction of the friction force. So for the purpose of calculating what force components are in the system, you don't need coefficient of friction at all.

When you get to the next step and you try to determine if your resolved forces will cause slip between the two faces, THEN you need coefficient of friction to know if the faces will slide or not.

 

[Compositepro]

If you assume zero friction the answer is simple geometry. If friction is infinite (rough or serrated contact surface) then R1 and R2 will be zero. So friction must be a factor. In addition, if there is any deflection, the contact point changes so, although you may calculate the magnitude of R you don't really know where it is applied. A good guess would be at the edge of the mating wedge surfaces (not the centroid).

 

[Denial]

Let's start with a really trite statement.[ ] The wedge will be slipping downwards, or it will be slipping upwards, or it won't be slipping at all.

Let t be the slope angle (60° in the sketch), and m be the coefficient of friction.[ ] Then...

If it is slipping downwards R[sub]d[/sub] = (F/2)*[sin(t)-m*cos(t)]/[cos(t)+m*sin(t)]

If it is slipping upwards R[sub]u[/sub] = (F/2)*[sin(t)+m*cos(t)]/[cos(t)-m*sin(t)]
but with an obvious lower bound of zero.

Any value of R between these two limits is possible, depending upon the loading history among other things.[ ] So, if you are getting different answers from different engineers it might be because different answers ARE possible.

 

[chicopee]

R1 and R2 should be normal to the conical surfaces; Friction forces Ff1 and Ff2 will have vector lines parallel to the conical surfaces. Interestingly enough Ff1 and Ff2 can also be viewed as shear forces on the conical surfaces. See attached FBD

 

[SwinnyGG]

Assuming a static approach, which is all the OP has appeared to ask for as far as I can tell, the magnitude of the normal force has absolutely nothing to do with the ultimate force holding capability between the two faces. This is a geometry problem.

Once you step into the real world (or at least start to move in that direction), which again, appears to me to be out of the scope of what the OP has asked about, friction matters a great deal. But we're not talking about the real world. We're talking about a first or maybe second year homework problem.

 

[chicopee]

Oops to much distraction around my work environment. See attached sketch

 

[desertfox]

Hi ALEng2

Friction will not have any part to play in your question because the block isn't moving, so the problem is solved as you have already stated by the triangle of forces. Now if a load is applied into the page on the cross section shown then the block will slide when that force exceeds the friction force generated by R1 an R2, the friction force being equal to (R1+R2)*coefficient of friction.

“Do not worry about your problems with mathematics, I assure you mine are far greater.” Albert Einstein

 

[TimSchrader2]

Hello

There is alawys a friction force with a normal force, static before moving fd after. So it is a component in the free body triangles. a brain twister that may have been solved Denial, in the complicated generic form. I think m* is the friction factor?

The question was to solve for R1 and R2 in terms of F

If down is Y and F is all Y. A simpler formula should be.

F= (R1y x 2) + (friction in y x 2)

and Friction = R1 x frict factor.

 

[chicopee]

Those who deny the lack of friction, have ever heard about static coefficient of friction?

 

[SwinnyGG]

F is a fixed value.

The angles between F and each component vector are all fixed.

This is a triangle with one known length, and three known angles- meaning that the lengths of the other two sides are, by definition, fixed. The magnitude of the vector representing the force component parallel to the bearing faces cannot change based on the coefficient of friction.

 

[BrianE22]

If you considered this as an axially loaded beam, wouldn't there be shear stress acting on the element rotated at 60 degrees? If there is (I'm rusty on stresses in beams), wouldn't that shear stress be carried via friction along the interface surface?

 

[Denial]

Yes, TimSchrader2, m is what I call the "coefficient of friction" and what I believe you are calling the "friction factor".

 

[desertfox]

Hi

Take a look at this link

https://study.com/academy/lesson/what-is-friction-definition-formula-forces.html

scroll down till you get to the sub heading Static Friction. It states that if there is no force trying to cause motion then the friction force is zero. The OP clearly states the jaws are retained and do not move, if there are no forces acting to slide the block along the jaws horizontally then friction isn't involved.

“Do not worry about your problems with mathematics, I assure you mine are far greater.” Albert Einstein