Calculating Friction in an L-Shaped Bar System

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To calculate the coefficient of friction (CoF) in an L-shaped bar system, one must consider the forces acting on the joint, particularly the normal force, which is crucial for determining friction. The measurement involves applying a force at a distance from the joint until movement occurs, noting that the moment created by this force is key to understanding friction. It's important to maintain constant or zero velocity during the measurement to differentiate between static and dynamic friction. For accurate modeling in software like MSC Adams, understanding how the software interprets friction and the parameters it requires is essential. The discussion emphasizes the need for clarity on the joint's characteristics and the forces involved to effectively calculate the CoF.
tony1990
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Hi there,

My question is about a measurement in the following setting:

The setting:
A fixed bar, in vertical postion, in connected by a joint with another bar, in a horizontal position. So together they form an L shape.
I know the geometry and mass of both bars.

The measurement:
I try to move to horizontal bar - very slowly - , at a distance D from the joint, while I measure the amount of force that I produce. At a certain moment, the bar starts moving and I stop the force measurement.

The question:

Will I be able to calculate the coefficiënt of friction in the joint, after this measurement?
An If so, how I can do that?


Hope someone can help me.
Thanks in advance,
Tony
 
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Yes - kind of. You can make this measurement, but you will need some more information about the joint to get it in the "classical" form. The force times the distance from the joint at which it is applied will give you a moment. The magnitude of the moment when the joint slips may be all you need, depending on what you're looking for.

If you really want the coefficient of friction between the joint materials, you need to know how much force there is in the plane of the joint, or how much preload or radial force your joint is seeing. You also need to know the radius of your joint (what does your joint look like? is it a pin?), which will tell you the moment arm of the friction force. Can you figure it out from here? Is this what you're looking for?

-Kerry
 
Hi Kerry. Thanks.

It is what I am looking for, but I still cant figure it out from here.

My biggest problem is: What are the other forces, beside the applied (external) force, and how do I calculate them?

Especially since the (possible) movement is a circular motion - a rotation - (perpendicular to the gravity force), I am not able to figure what are the other forces are. I am correct to say that if I move the bar very slowly I can neglect the inertia forces?
Which forces remain then? It has to be some equivalent to the normal force during a linear motion. Note: the rotating bar move in a plane perpendicular to the gravity (force)

The joint is a planar pin joint, and if I assume (1) there is no slip in the joint (2) the bars are both rigid. The joint is passive - so no preload? - and very likely have a form of coulomb friction.

Can you point in the right direction?

Thanks!

Tony
 
Hello Tony,

Inertial forces are related to accelerations, so moving it slowly (or pressing on it without moving it, then noting the position when it finally does move) vs. moving it quickly will be the difference between coefficient of static friction and coefficient of dynamic friction. The key is constant velocity (or in the case of static friction measurement, zero velocity).

What exactly do you want to do with this data? My guess is that a "friction moment" will serve your purposes, and going through the extra work to determine a coefficient of friction will not be very useful, since you'll have to convert it back into moment form to use it anyway - the other forces will drop out. This does assume that the only friction is coulomb friction (no viscous or higher terms).

If you really do need to back out the coefficient of friction, then you are correct that you need to know something akin to a "normal" force. This force is actually normal to your pin joint, and depending on your joint it may be tricky to determine. The weight of the bar will impose a moment on your pin, which means the normal force will change with position along the pin. It may be valid to "average" the normal forces for the purposes of your friction measurement.

Is your joint a little "sloppy?" Is there freeplay in the off-axis direction? This is a sure sign that there is no preload in the joint, which will make determining the average force on the pin much easier.

-Kerry
 
Last edited:
Thanks again!

Ah ok. *In my confused state I mix up everything.*

I would like to model the setting in MSC adams software. Familiar with that software? And in order to built a valid model in MSC adams I needs the static (and dynamic) friction coefficient of the joint. A friction momen won't due the trick.

So I do need to calculate the CoF

And no the joint is not sloppy. Hmm... Then my first step will be - not to make the first calculation difficult -to calculate the CoF with the assumption that there is no preload.

Tony

PS. If i interpret your advice correctly, two vital elements of the measurement are that I:
make sure the movemevement is at constant - or zero - velocity
make sure the applied force is in the same plane as the foint DoF.
 
So then you need to know how ADAMS models the friction. Is it something like
M_f_z = \mu \left( F_x + F_y \right) r
where r is the pin radius, the pin axis is in the z-direction and Fx and Fy are the off-axis forces, or maybe it's just a friction moment threshold where it doesn't move until the limit is passed, then it applies the dynamic friction moment as an offset to whatever other moments are acing on the joint? You should check the documentation and design your test based on that. What are all of the inputs that ADAMS is asking for? Is there a pin radius parameter? I'm not familiar with a "coefficient of friction" used this way. I have seen it modeled as a percentage, though. Maybe this is what you're looking for? So if I apply a torque of 10, the torque I see is actually some percentage of 10, since the remainder goes into friction?

Hope this helps.

-Kerry

p.s. I'd be interested to hear what the documentation says, so if you figure it out, let us know! And of course, any more questions - let us know about that, too.
 
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