Solve Moment Determining Homework for Pulley

[giacomh]

Homework Statement

Three forces are applied to a locked pulley. Determine the magnitude and direction of the resultant of the three forces and the perpendicular distance from the axle of the pulley to the line of action of the resultant.

**I'm unable to upload a photo of the pulley at the moment, but here's a description that should give you all you need to know:

The two cords hanging off of the pulley are 2 feet apart (so the inner circle around the pully has a diameter of 2 feet). The cords are both 20 degrees above the horizontal, and are each in the second quadrant (-x, +y direction). There is a 120 lb force on one cord, and a 40 lb force on the other. The outer circle around the pulley is 4 feet in diameter, and there is a 90 lb force hanging off of it in the 4th quadrant (+x, -y).

Overview:
Fa= 120 lb at 20 degrees above horizontal (-x,+y)
Fb=40 lb at 20 degrees above horizontal (-x, +y)
Fc=90 lb at 90 degrees below horizontal (+x,-y)2. Homework Equations

Perpendicular axis from the axle of the pulley to the line of action of the resultant:

d=M/R

The Attempt at a Solution

I already found the resultant by finding Rx and Ry and finding the magnitude. The resultant is 154.4 lb at 13.2 degrees above the horizontal. I'm having trouble with the second part, specifically finding the moment.

-120cos(20)(1)-40cos(20)(1)+(90)(2)=30

30/154.4=.192 in

1 and 2 are the radii.

The answer should be .647 in... I've tried every combination of force components to get the moment to be 100, but I'm out of luck (the moment is 100 when the distance is .647).

I'd appreciate any help!

 

[Simon Bridge]

Oh that's interesting - did you try finding the moment due to each force separately and just adding them up? Since you know the resultant, you should be able to go backwards to get the moment arm.

 

[azizlwl]

[

The Attempt at a Solution

I already found the resultant by finding Rx and Ry and finding the magnitude. The resultant is 154.4 lb at 13.2 degrees above the horizontal. I'm having trouble with the second part, specifically finding the moment.

-120cos(20)(1)-40cos(20)(1)+(90)(2)=30

30/154.4=.192 in

1 and 2 are the radii.

The answer should be .647 in... I've tried every combination of force components to get the moment to be 100, but I'm out of luck (the moment is 100 when the distance is .647).

I'd appreciate any help!

What is the unknown?

 

[giacomh]

The unknown is the moment. And yeah, I found the moment of each force. I've tried the x and y components of the two forces, and combined them with the y component of the 90 lb. No luck. I guess I really just don't understand moments...my books gives a very very vague description on knowing when to use the x or y...and the whole application of the right hand rule...

 

[Simon Bridge]

Well the moments just add up like forces do. Pick a turning direction (clockwise or anticlockwise) to be positive.

You also have the resultant force and it's direction?

Then the effective moment arm is going to be from the center of the pulley, perpendicular to the resultant force, with length $d=\mu_{tot} / F$There's probably a cunning way to combine all the individual moment arms to get a resultant moment arm but I've never done it that way.
To be specific, I'd need to see the diagram. Your description is just confusing me I'm afraid.

(I'd have thought that for a locked pulley, the net moment would be zero but I guess that's not the question?)