How do I calculate force required to generate grip? 2

[mechengjohn]

Hi guys, hopefully someone here knows how to tackle this. I've been trying to work it out for a while.

We have a machine which moves heavy plates with a spinning wheel. The wheel presses the plate from under then spins to drive the plate onto a rack. Sometimes the wheel doesn't have enough grip on the plates and just spins without moving them. So I'm trying to calculate the amount of force required to generate enough grip (without lifting the plate!).

I thought I could just use a rolling resistance equation, F= (f.W)/r .

F = force
f = resistance
W = weight
r = radius of wheel

Where usually this is done to work out the force required to overcome the rolling resistance, I have a given force (the torque of the motor driving the wheel) and could work out W which would be the minimum force. ie less than this and it would overcome rolling resistance and slip. But this method isn't producing a sensible answer.

I can upload a picture or FBD to make it more clear.

Any ideas? Sorry if something like this has come up before, I did try the search fucntion but couldn't find anything.

 

[swertel]

My first intuition, not seeing the details of the geometry, would be to increase the coefficient of static friction.

What materials are slipping?
Can you change materials?
Can you change the surface finish of one or both of the materials (roughen them up)?
Can you use a coating, or skateboard tape, to increase friction without damaging the product?


--Scott

http://wertel.eng.pro

 

[desertfox]

Hi mechengjohn

Yes lets have a picture or sketch with some dimensions on and we should be able to help you further.

regards

desertfox

 

[mechengjohn]

Thanks for the reply.

The wheel is made of a patterned rubber, like a car tyre. The plate is steel. I can't do anything to the plate but could change the wheel tyre, however I don't know if that would be much help.

 

[MintJulep]

Assuming that you are driving the wheel with an electric motor you DO NOT HAVE A KNOWN DRIVING FORCE.

The torque that may be written on the name plate of the motor is the rated torque of the motor. Connect the motor to a load that requires more torque than than and you'll kill the motor. Otherwise, the torque that the motor produces will be proportional to the load it is driving.

The maximum amount of force that the wheel can impart on the plate is simply the coefficient of adhesion x the normal force on on the wheel. Unless the entire weight of the plate is balanced on the wheel the normal force will be less than the weight of the plate.

The forces resisting movement are the remaining weight of the plate x the coefficient of friction against whatever it is resting on and its inertia.

 

[aafuni]

The equation you have there, I believe, is for rolling resistance on a moving object. The F term would not be torque but would be the force resisting forward motion. The f term would be a coefficient of rolling resistance in length units (normally given as b). Then you are right about the W being weight (technically the normal force N).

You would likely be better off using an analysis of the point of contact using the standard static friction analysis. F = mu * N, Look up tyre on metal and that will give you the amount of friction you are generating.

 

[mechengjohn]

See if this helps at all.

Sorry for the quality but there's only so much you can do with ms paint...

 

[zekeman]

The problem may be 2 fold
1) there is some mechanical resistance other than inertia
2) more likely:the acceleration of the wheel is too great to sustain a static friction force and thus it goes into sliding or slipping and gets worse as the interface heats up.
This is readily explained with the startup equation which states that to sustain static friction you must have

u*W>W/g*a
where
u= static friction
W plate weight
g gravity
a acceleration of the wheel periphery
So solving for a
a<u/g

 

[zekeman]

Error
As usual.The necessary wheel periphery acceleration should be
a<u*g

 

[aafuni]

Your easiest fix may be to add a roller on the top surface of the plate. With a roller above the plate, the normal force can be increased until your problem is solved. As it is the normal force can only ever be a fraction of the plates weight, without moving the plate.

Does the wheel only start spinning once it is in contact with the surface? If it is constantly spinning the force is going to be reduced (it will be kenetic friction rather than static).

 

[swertel]

In simplest terms, slipping is due to friction.
Friction Force = (Normal Force) * (Coefficient of Friction)

If we can't lift the plate, then we can add any more Normal Force (F=ma + mg) which means we have to change the coefficient of friction.

I have to double check if distributed loading would make a difference on the normal force. For example, make your wheel smaller to get a higher pressure. The resultant force is the same, but the local force per area is higher (pressure) and may increase the systems "sticking power."

--Scott

http://wertel.eng.pro

 

[swertel]

Also my bad. Got distracted while typing and didn't see the attachment or other posts.

For some reason, I've been picturing a spinning disc although you've clearly stated a wheel.

--Scott

http://wertel.eng.pro

 

[mechengjohn]

Thanks for all the replies folks.

The wheel is only activated after contact is made. A roller on the top is tricky for various other reasons.

Am I right in thinking that a solution would have to account for contact area of the wheel? As surely a wider tyre would have more grip.

 

[aafuni]

Wider tyre would increase the coefficient of friction. Additionally lowering the tyre pressure would help increase this coefficient, and wouldnt need a redesign.

Like swertel said, you will either need to increase the coefficient of fiction if you cant increase the normal force, or you will need to slow down the wheel. As was said you have a system where m * a = mu * N, so if you keep the friction force part (mu * N) the same but decrease the acceleration on the wheel/plate then you may be able to avoid slipping out and move your plate. Is your process dependant on the current speed or can you try lowering the wheels acceleration?

 

[mechengjohn]

What I was actually aiming for was to increase the normal force. ie to use a larger cylinder.

I was just looking for a way to determine how much force is required.

I'll try playing with the equations mentioned.

Thanks again.

 

[MintJulep]

As surely a wider tyre would have more grip.

Wider tyre would increase the coefficient of friction.

Let's recall the very simplified model of friction force that they taught us in school.

F = mu * N

Force = coefficient of friction * normal force.

Notice that there is no term for area in this equation.

Now in the real world friction isn't this simple. Friction is often a function of contact pressure, temperature, slip or creep and an bunch of other things.

Contact pressure is obviously a function of area. However friction does not necessarily always increase with increasing contact pressure. Also, to further complicate things non-uniform contact pressure will probably result in non-uniform temperatures.

So it's not at all certain that increasing the tire size will increase the available grip.

I was actually aiming for was to increase the normal force. ie to use a larger cylinder.

If there is nothing other than gravity holding the plate down the maximum possible normal force is its weight. Since it seems unlikely that the place will ever be supported solely by the wheel, the normal force will always be less than the weight.

 

[mechengjohn]

Thanks for the input MintJulep.

When I first tried this problem, I converted my motor torque to force. (T = F*r) where r is radius of the wheel.

Then using the equation in the OP with an f value of 0.0077m (which is for hard rubber on steel I believe) worked out the minimum Normal force needed.

As has been stated the maximum Normal force cannot exceed the weight of the plate. So I thought that the ideal Normal force would be anywhere between these two values.

This method still seems correct in my mind, however the values for normal force I was getting were a factor of 30 over what we have been using so I assumed I had made an error in my method somewhere.

 

[cmcbain]

Mechengjohn,
I believe that if you use the method above, all your answer will tell you is the normal force required to prevent the wheel from slipping under any possible condition.

If it were me, I would attempt to find out the resistance to movement that the plate has. That could be difficult to do by analytical means, so an experiment might be the quickest way.

Then, once I had that force (for the force required to start the plate moving), I would check to make sure my motor could provide that much torque. Then, I would plug it into the friction equation, and see how much normal force is required on the wheel.

Finally, I might use the rolling resistance equation with a couple iterations to try to get a more exact value.

You might look in a fluid power handbook. Some of them have much more detailed rolling resistance equations that take area of contact, normal force, etc. into account

 

[Occupant]

How is the wheel accelerated assuming it is at standstill before it hits the plate? If you accelerate it too much, it will spin without moving the plate since the inertia of the plate will prevent it from moving. Once the wheel slips you're into dynamic friction and you have lost the ballgame.

 

[desertfox]

Hi mechengjohn

I think your on the right lines with rolling friction, your difficulty will be obtaining a correct coefficient for your situation.
Increasing the normal force will help, but at present the only normal force you have is the weight of the plate sitting on the wheel as stated by others.
You will get some benefit by increasing the width of the tyre or decreasing tyre pressure, because rubber is a compliant material, so its true area of contact is very close to its nominal area of contact, whereas normally the true area of contact is usually much less than the norminal area of contact when two harder materials are placed on one another, which why it is stated that the true area in the latter case is proportional to the normal force pushing them together. See this site scroll down till you see this heading:-
The microscopic origin of friction forces

http://www.engin.brown.edu/courses/en3/Notes/Statics/friction/friction.htm

Also have a look at this site it as loads of information on friction and links to other information on other sites

http://www.roymech.co.uk/Useful_Tables/Tribology/co_of_frict.htm#sliding

goodluck

desertfox