Final wire force in a pulley system

[gtavictor]

Hello people, this is my first post here in PF and I have a problem with the modeling of a pulley system considering the friction between wire and pulley. I have a similar system as show in the image below and I have a friction factor of 0.25 between wire and pulley, my question is, how can I model an equation to find the force at the open end of the wire to move the weight?

http://img52.imageshack.us/img52/9240/534pxpulley3asvg.jpg

thanks in advance!

EDIT: Please at least do not consider the value of the friction, I just want to make the modeling of the force necessary to move the weight due to friction losses!

 

[stewartcs]

Hello people, this is my first post here in PF and I have a problem with the modeling of a pulley system considering the friction between wire and pulley. I have a similar system as show in the image below and I have a friction factor of 0.25 between wire and pulley, my question is, how can I model an equation to find the force at the open end of the wire to move the weight?

http://img52.imageshack.us/img52/9240/534pxpulley3asvg.jpg

thanks in advance!

EDIT: Please at least do not consider the value of the friction, I just want to make the modeling of the force necessary to move the weight due to friction losses!

I can't see your picture...can you attach it instead of linking it?

CS

 

[gtavictor]

Ok here's the attachment. Thanks for your help in advance!

 

[stewartcs]

Ok here's the attachment. Thanks for your help in advance!

I use the method given in API RP 9B. This method gives the efficiency of the pulley system.

The governing equation is:

\eta = \frac{K^N - 1}{K^SN(K-1)}

N = number of line parts supporting the load
S = number of sheaves (or pulleys in your case)
K = sheave friction factor (pick from table in API or get from manufacturer)

For roller bearing sheaves, K = 1.04
For plain bearing sheaves, K = 1.09

So for your case N = S = 4, gives an efficiency of 0.907. Hence, your applied force to lift the load will be F_a = \frac{L}{0.907}

I suggest referencing the given document for more information.

Hope this helps.

CS

 

[gtavictor]

I use the method given in API RP 9B. This method gives the efficiency of the pulley system.

The governing equation is:

\eta = \frac{K^N - 1}{K^SN(K-1)}

N = number of line parts supporting the load
S = number of sheaves (or pulleys in your case)
K = sheave friction factor (pick from table in API or get from manufacturer)

For roller bearing sheaves, K = 1.04
For plain bearing sheaves, K = 1.09

So for your case N = S = 4, gives an efficiency of 0.907. Hence, your applied force to lift the load will be F_a = \frac{L}{0.907}

I suggest referencing the given document for more information.

Hope this helps.

CS

Hey thanks for your help. I was doing some research on internet and I was not able to find a copy to download of this API RP 9B, I only found its title "RECOMMENDED PRACTICE ON APPLICATION CARE, AND USE APPLICATION CARE ROPE FOR OIL FIELD SERVICE" at api.org but I have to pay about $111.00 to read it, well in the end it's a standard.

The equation seems reasonable but my concern is about the modeling of the pulley system, for example I come up with this equation:
F_a = \frac{W}{4}+4F_f
where:
F_a= Applied force.
\frac{W}{4}= Weight of the load seen at the open end of the cable due to ideal mechanical advantage of the pulley system.
4F_f=4(\mu N)= Friction force multiplied by four because I have 4 sheaves. N is the normal force, where N=\frac{W}{2}= Weight seen on each pin of each sheave.

In my case I have two frictions on each pulley:
\mu_{sh}=0.1= Friction coefficient between pin and bearing.
\mu_{wsh}=0.25= Friction coefficient between wire and sheave.

You provided me with a very high fiction coefficient factor, really high indeed and the mechanical advantage is reduced a lot, and makes no sense to implement this pulley system on a block and tackle system.

What I don't know about my equation is if I have sum both friction factors so my final equation for the applied force at the end of the wire is:
F_a=\frac{W}{4}+4(\mu_{sh}+\mu_{wsh})\frac{W}{2}

At the end I don't know if doing right anyway, but if you can get me on the right track or someone else I would be very thankful!

 

[stewartcs]

Hey thanks for your help. I was doing some research on internet and I was not able to find a copy to download of this API RP 9B, I only found its title "RECOMMENDED PRACTICE ON APPLICATION CARE, AND USE APPLICATION CARE ROPE FOR OIL FIELD SERVICE" at api.org but I have to pay about $111.00 to read it, well in the end it's a standard.

The equation seems reasonable but my concern is about the modeling of the pulley system, for example I come up with this equation:
F_a = \frac{W}{4}+4F_f
where:
F_a= Applied force.
\frac{W}{4}= Weight of the load seen at the open end of the cable due to ideal mechanical advantage of the pulley system.
4F_f=4(\mu N)= Friction force multiplied by four because I have 4 sheaves. N is the normal force, where N=\frac{W}{2}= Weight seen on each pin of each sheave.

In my case I have two frictions on each pulley:
\mu_{sh}=0.1= Friction coefficient between pin and bearing.
\mu_{wsh}=0.25= Friction coefficient between wire and sheave.

You provided me with a very high fiction coefficient factor, really high indeed and the mechanical advantage is reduced a lot, and makes no sense to implement this pulley system on a block and tackle system.

What I don't know about my equation is if I have sum both friction factors so my final equation for the applied force at the end of the wire is:
F_a=\frac{W}{4}+4(\mu_{sh}+\mu_{wsh})\frac{W}{2}

At the end I don't know if doing right anyway, but if you can get me on the right track or someone else I would be very thankful!

Sorry I wasn't clear before, the mechanical advantage should be included as well. So if your MA is 4 then you'll only need L/4 as your applied force plus the increase due to the efficiency factor.

F_a = \frac{L/4}{0.907}

So if your load is 100-lbf and your MA is 4 you'll need about 27.6-lbf applied to lift the load.

CS

 

[gtavictor]

Ok, now that makes sense, but I don't have this standard you have, and this is a project for some courses of my master, project where I'm being provided with some friction factors as I posted before and I don't think I can offer the formula you offered without modeling the system. Perhaps you can guide me regarding if the formula I applied is OK, also, I will try to search this standard because some statements there may be used as good references, or maybe you can provide me with the standard itself, but if not your help and suggestions about my equations would be great!

 

[stewartcs]

Ok, now that makes sense, but I don't have this standard you have, and this is a project for some courses of my master, project where I'm being provided with some friction factors as I posted before and I don't think I can offer the formula you offered without modeling the system. Perhaps you can guide me regarding if the formula I applied is OK, also, I will try to search this standard because some statements there may be used as good references, or maybe you can provide me with the standard itself, but if not your help and suggestions about my equations would be great!

If you want to use a friction coefficient instead then you can model it as a plain bearing. For example if you had a single pulley with a friction coefficient of 0.20 and a pulley diameter of say 4-in and a shaft diameter 2-in and load of 500-lbf the total force required to lift the 500-lbf load would be about 611-lbf. I found this by simply summing the moments about the point of application of the reaction force on the shaft: (2.20)(500) - (1.80)(P) = 0.

The radius of the circle of friction in this case is r \mu which gives (1-in)(0.20) = 0.20-in. Since the radius of the pulley is 2-in, 2+0.2 = 2.20-in on the left side for moment arm and 1.80-in on the right. (Note that the 1-in is the radius of the shaft.)


Thus the total additional force you have to apply to overcome friction in this example is 111-lbf, or a total force to lift the load of 500 + 111 = 611-lbf.

For multiple pulleys you will end up with same additional total friction force since the reaction is smaller at each pulley (which means the friction force will be smaller per pulley) but you have more pulleys. So it's a wash.

However, with an MA of 4 the total required force to lift the load won't be 611-lbf. It will be 500/4 + 111 = 236-lbf. Note that the total friction is the same.

Caveat: This is for dry or slightly lubricated plain (journal) bearings.

All of that being said, I still recommend using the other formula from API.

CS

 

[gtavictor]

If you want to use a friction coefficient instead then you can model it as a plain bearing. For example if you had a single pulley with a friction coefficient of 0.20 and a pulley diameter of say 4-in and a shaft diameter 2-in and load of 500-lbf the total force required to lift the 500-lbf load would be about 611-lbf. I found this by simply summing the moments about the point of application of the reaction force on the shaft: (2.20)(500) - (1.80)(P) = 0.

The radius of the circle of friction in this case is r \mu which gives (1-in)(0.20) = 0.20-in. Since the radius of the pulley is 2-in, 2+0.2 = 2.20-in on the left side for moment arm and 1.80-in on the right. (Note that the 1-in is the radius of the shaft.)


Thus the total additional force you have to apply to overcome friction in this example is 111-lbf, or a total force to lift the load of 500 + 111 = 611-lbf.

For multiple pulleys you will end up with same additional total friction force since the reaction is smaller at each pulley (which means the friction force will be smaller per pulley) but you have more pulleys. So it's a wash.

However, with an MA of 4 the total required force to lift the load won't be 611-lbf. It will be 500/4 + 111 = 236-lbf. Note that the total friction is the same.

Caveat: This is for dry or slightly lubricated plain (journal) bearings.

All of that being said, I still recommend using the other formula from API.

CS

Thanks for your help, sorry to answer many days later, but I stopped this project a little bit, anyway you may check the new thread I started where I talk about the detailed modeling of this system!