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.

 

[IRstuff]

What is the surface finish of the steel?

TTFN

FAQ731-376

 

[aafuni]

Maybe I am not understanding something, but I am still confused as to why you would use the rolling resistance equation, or need to look at torque as cmcbain suggested. You mentioned that the wheel can move the parts, but it often spins out and does not work. To me it seems your analysis should revolve around whether the wheel spins out or not, not the rollinh resistance imparted on the wheel. Like I said before I would look at the force of friction between the plate and the rubber. I would use:

mu * N = W / g * a

This however neglects the friction of the plate on the surface you have it on so you might want to use:

mu * N = W / g * a + mu * (W - N)

In this the first mu is rubber on steel and the second is steel on whatever your slide is. The (W-N) term is the normal force between your plate and the slide it is on, assuming that the effective wieght will be the weight of the plate minus the upward force from the wheel.

If the mu * N term is not as great as the W / g * a + mu * (W - N) term, the wheel will spin out. That I thought was what you wanted. If the wheel spins out, increase the upward force, there by increasing the friction from the wheel and reducing the friction on the slide, or slow things down by using less torque from your motor to start up, decreasing the inertial term (W/g*a).

 

[zekeman]

Afun,
Look at your equation which is a fair representation of the system.
"mu1 * N = W / g * a + mu2 * (W - N)
I took the liberty of making the two coefficients (static) different (mu1>mu2 since rubber to steel>steel to floor)
Isn't it clear that if "a" (acceleration) gets too large it will slip, and if the overhang weight W-N is too great it will also slip, even if the a is near zero.

So, instead of focusing on increasing weights on the roller, i believe that control of the plate with a measured slow acceleration of the roller could do the job. When it comes to speed, the second term mu2 which is <mu1 to start with, gets even smaller when sliding.

 

[aafuni]

zekeman,
Yes, I agree that that would be a good solution to the problem, and probably the easiest to achieve as it likely requires no change to the design. The only issue is if this is part of a process in which the time of the operation is critical.

 

[zekeman]

BTW, there should be an inequality replacing the equal sign to your equation, namely
mu1 * N >W / g * a + mu2 * (W - N)
which states that the static coefficient, mu1 exceeds the requirement of moving the plate and the actual friction force would be less than mu1*N

 

[zekeman]

" The only issue is if this is part of a process in which the time of the operation is critical."

Of course and perhaps " slow" may also be hard to control.

 

[BobM3]

Are the plates oily? Cleaning might help. Or spraying them with something that would increase the coefficient of friction (I don't know what).

Or, if you really want to get fancy and spend money, make a special wheel with radial ports in it and pull a vacuum through the ports to try and pull the plate down on to the wheel.

How about a studded wheel? Hard, short, sharp studs that would dig into the plates. Of course you'd be left with small indentations in the steel.

 

[desertfox]

Hi mechengjohn

On your diagram of the set up the wheel comes up under the plate, but does not appear to be central relative to the plate, that means the plate is lifted off centre and I presume that the plate is supported further to the right of the wheel is that correct?
If that is the case, then the reaction on the wheel from the plate is less then the full mass of the plate, which means the force to overcome friction is reduced,thereby making the situation worse. Is there any control on the position of the wheel relative to say one edge of the plate or its centre, if not the reaction of the plate on the wheel will vary each time,even if the plates have equal mass and size.
Reducing the speed of the wheel would help as others have stated, but I guess your motor is a fixed entity which cannot be changed.

desertfox

 

[EspElement]

Couldn’t he test his coefficient of friction by actuating the cylinder and fixing the tire (not allowing it to rotate), then by pulling the plate manually, recording the pulling force required to slide the plate across the tire, then he can take the ratio of the pushing force (of cylinder) and how much it took to pull it to determine his coefficient of friction. Now he would have to determine if his surface force of the tire (if the tire was a 1" radius and the torque was 10 in/lbs then that would be 10 lbs of force) is indeed less than its friction, which would be required for it to function properly

Also to prove the other theories mentioned, you can let air out of the tire and judge the difference in the force it takes to slide the plate.

 

[unclesyd]

How about just changing the compound, softer, in the tire itself.
If there is room change the tire diameter to change the angle.

 

[mechengjohn]

aafuni / zekeman,

i think you may be right. I will try experimenting with slower acceleration speed to start. Although this may prove tricky with our setup at present.

desert fox - yes the plate is supported by rollers along it's travel path. So you are correct about the reaction being less than the full mass of the plate, however the plate is very heavy and at the moment the wheel is not close to being strong enough to lift the plate. I can increase the force by some margin before it starts to lift.

Thanks to everyone for the responses, all have been helpful.

 

[desertfox]

Hi mechengjohn

You need to ensure the wheel takes as much of the weight reaction as possible, but without the wheel lifting the plate above horizontal, if it lifts the plate above horizontal you will lose reaction.

desertfox

 

[desertfox]

Hi mechengjohn

One way to slow the wheel down would be to reduce the wheel radius given that

V=W*r

V=linear velocity

r= wheel radius

W= angular velocity in radians/sec

desertfox

 

[flash3780]

Why not support the rest of the plate on steel rollers?

I don't think you'll find a much better material than rubber for your drive wheel, so it seems to me that the knobs you can turn are as follows:
-Increase the force from the hydraulic cylinder - you'll have to keep it under 50% of the weight of the plate so you don't end up lifting it up.
-Decrease the friction between the plate and the supporting structure.
-Decrease the acceleration of the drive wheel (might be able to solve it with the control system).
-Add another drive wheel

Put together a free-body diagram of your steel plate, to include friction forces and inertia forces and solve for the drive force. Compare your required driving force to the slipping force and go from there.