I have been trying to solve for an equation that describes the torque created on a shaft as a result of friction between a mophead in rotating in contact with a hospital floor. Both coefficient of frictions are known, along with rotation speed(which as far as i gather is irrelevant) the diameter of the head is also known. Any help would be appreciated.
Eng-Tips is the largest engineering community on the Internet
Intelligent Work Forums for Engineering Professionals
-
Congratulations waross on being selected by the Tek-Tips community for having the most helpful posts in the forums last week. Way to Go!
Friction Torque on Rotating Disk
- Thread starter GalEng
- Start date
GalEng,
There should only be one coefficient of friction between the mop and floor, all affected by how wet the mop is and by the surface condition of the floor. A century old spruce plank floor would be different from a new, ceramic tile floor. Can you even assume that this is Newtonian friction?
The easiest thing to do might be to attach a wrench or vice grips to your mop head, and use a fish weighing scale to measure torque. You need to simulate the condition of the mop head as well as the contact force.
The next best thing would be to measure the linear force and calculate torque.
Torque = Force [×] Diameter / 2.
JHG
There should only be one coefficient of friction between the mop and floor, all affected by how wet the mop is and by the surface condition of the floor. A century old spruce plank floor would be different from a new, ceramic tile floor. Can you even assume that this is Newtonian friction?
The easiest thing to do might be to attach a wrench or vice grips to your mop head, and use a fish weighing scale to measure torque. You need to simulate the condition of the mop head as well as the contact force.
The next best thing would be to measure the linear force and calculate torque.
Torque = Force [×] Diameter / 2.
Wrenchbender
Mechanical
Ditto on only one COF. It sounds like you have a single disk clutch. Look in a machine design book or google it. I'm sure you will find an answer. (Reminds me of when i was in basic. The smallest guy in the platoon sat on the buffer for more normal force and a shinier floor, so maybe you want to overdesign this thing.)
btrueblood
Mechanical
Right. You know your normal force, and a value for the linear cof, you want to know the torque
Take the double intergral over r and angle theta of
P*mu*r*dA
where P is the pressure (e.g. lbs/in^2) applied to the disc, mu is friction, r is the radius coordinate to any point on the disc.
In rotational coords, dA = r*dr/dtheta
Assuming P and mu are indepdent of theta and r (no steering forces applied, the torque is:
P*mu*pi*R[sup]3[/sup]*2/3
Where R is the outer radius of the disc.
Or, where F is the normal (vertical) force applied to the disc:
Torque = 2/3*F*R
Take the double intergral over r and angle theta of
P*mu*r*dA
where P is the pressure (e.g. lbs/in^2) applied to the disc, mu is friction, r is the radius coordinate to any point on the disc.
In rotational coords, dA = r*dr/dtheta
Assuming P and mu are indepdent of theta and r (no steering forces applied, the torque is:
P*mu*pi*R[sup]3[/sup]*2/3
Where R is the outer radius of the disc.
Or, where F is the normal (vertical) force applied to the disc:
Torque = 2/3*F*R
- Thread starter
- #6
Ok thank you! I have managed to use the same method as btrueblood but have been getting the result
Torque=2/3*F*mu*R
where mu is the coefficient of friction.
And for best wrench yes i am looking into how much the normal force needs to be. The head is going to be interchangable so the floor can be mopped with a disenfectant first and then changed over to a buffer head to apply a antimicrobial polymer coating.
Torque=2/3*F*mu*R
where mu is the coefficient of friction.
And for best wrench yes i am looking into how much the normal force needs to be. The head is going to be interchangable so the floor can be mopped with a disenfectant first and then changed over to a buffer head to apply a antimicrobial polymer coating.
btrueblood
Mechanical
Sorry, you are right, dropped the mu by mistake in the last equation.
- Status
Similar threads
- Locked
- Question
- Locked
- Question