Take Up Pulley Moment Reaction Forces

[mechman24]

Hi All,

I'm having some issues with probably a basic force and moment resolving equation. I basically have a conveyor pulley that is mounted on a trolley with with 4 wheels on a rail. The pulley has two opposing and equal forces (f1 and f2). f1 represents the tension of the belt against the pulley and f2 on the otherside is a sheave attached to a winch via rope. F1 acts eccentrically to the centre of the pulley (the belt is not tracking centrally) at a distance of d1 thereby creating a turning moment. this turning moment wants to rotate the trolley and thus created an axial reaction force acting on the four wheels. note that the rail and wheel contact is preventing the trolley from turning

i am trying to resolve the force acting on the wheels (R1 and R2) but i am not sure of my free body diagram is correct or not and if i have resolved these forces correctly.

any help is appreciated.

image of the problem below

 

[HTURKAK]

Assumptions;

- The trolley is rigid,
- the wheels are identical and free to travel along belt ( F1- F2 direction)
- So, the F1 and F2 is equal
With the above assumptions,
- The moment you have calculated is OK ( M1= F1*d1)
- The moment M1 is resisted by the wheels in transversal direction is equal for all four wheels .
- Apparently the CG of four wheels is at (d2+d3)/2 and the transversal force resisted by four wheels ;

R1=R2 ⇒ 4*R1*((d2+d3)/2) = F1*d1 ⇒ R1 = F1*d1/ (4*(d2+d3)/2)

The resisting forces will be at the opposite directions to which you have shown .


He is like a man building a house, who dug deep and laid the foundation on the rock. And when the flood arose, the stream beat vehemently against that house, and could not shake it, for it was founded on the rock..

Luke 6:48

 

[mechman24]

Is it ok to assume that moment will be distributed amongst the four wheels equally? I would think that because the lever arms are different (i.e d2 and d3) the resulting forces would also be different

 

[HTURKAK]

Yes. O.K...Consider a simply supported beam and the load is only a moment. The support reactions will not change with the moment location but internal resisting moment of the beam changes .(e.g. when the moment applied at center of the beam , very near to one of the supports..etc) . Pls look tables of structural eng., strength of materials for the reactions of simply supported beam.



He is like a man building a house, who dug deep and laid the foundation on the rock. And when the flood arose, the stream beat vehemently against that house, and could not shake it, for it was founded on the rock..

Luke 6:48

 

[mechman24]

Thanks for clearing that up HTURKAK makes a bit more sense

The only thing i still dont quite understand is the R1 = -R2 = F1*d1/((d2+d3)/2) and COG component you have used. I would think it would be F1*d1/(2*(d2+d3)). Given what you mentioned about the simply supported beams and pictue below. The reaction forces would be half.

 

[HTURKAK]

Sorry for mistake at paranthesis ..I have edited my previous post.

Just use simple equilibrium equations,

2R1=2R2

Assume the moment applied at R2 axis (or any point along F2 axis )
Use your reaction formula Ra=M/L, M= F1*d1, L= d2+d3

2R1= F1*d1/ ( d2+d3).



He is like a man building a house, who dug deep and laid the foundation on the rock. And when the flood arose, the stream beat vehemently against that house, and could not shake it, for it was founded on the rock..

Luke 6:48