Describe the position of a pulley attached to a sling

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The discussion focuses on analyzing the behavior of a floating pulley attached to a sling, particularly how its position changes under a force applied at the drum. Key parameters include the drum's location, sling length, load weight, and friction percentage in the pulley. Participants suggest creating free body diagrams to visualize the forces acting on the system, emphasizing the role of tension in the cables and the impact of friction on the load's movement. The conversation also touches on the complexities of deriving equations of motion due to the unknown position of the pulley and the angles involved. Overall, the analysis aims to understand the dynamics of the pulley system under varying forces.
  • #31
haruspex said:
Consider the triangle formed by the drum, the pulley and sling's support. You know two sides and the angle one of them makes to the vertical, you can create an unknown for the angle the sling makes to the vertical. In terms of those, what angle does the rope make to the vertical?
upload_2019-2-28_13-47-8.jpeg


I don't fully understand. For example this is the situation with given parameters and CD has a length of 2.
alpha is the angle in ABC
 

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  • #32
Jeroen Staps said:
situation with given parameters
?:) ?

I repeat #22: the sketch of #7 has T1 and T3 collinear, so that ##\sum \vec F## has a component tangent to the dashed circle (from T2). ##\qquad## For a constant position of the pulley (pseudo steady state: equilibrium or constant acceleration of the load -- up or down) that component has to be zero.

If grokking isn't, try starting with a very far away drum (##\alpha## constant)
 
  • #33
Jeroen Staps said:
View attachment 239501

I don't fully understand. For example this is the situation with given parameters and CD has a length of 2.
alpha is the angle in ABC
That's a helpful diagram.
Let CD make angle θ to the vertical (angle BCD) and AD make angle φ to the vertical.
Can you get an equation relating , θ, α and the lengths AC, CD?
 
  • #34
haruspex said:
That's a helpful diagram.
Let CD make angle θ to the vertical (angle BCD) and AD make angle φ to the vertical.
Can you get an equation relating , θ, α and the lengths AC, CD?
I'll try, not sure if it is possible though.
Because α, AC and CD are fixed and θ can change.
 
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  • #35
Jeroen Staps said:
I'll try, not sure if it is possible though.
Because α, AC and CD are fixed and θ can change.
I meant to write
haruspex said:
Can you get an equation relating , θ, φ, α and the lengths AC, CD?
 
  • #36
haruspex said:
I meant to write
OK, I see
 
  • #37
haruspex said:
I meant to write
But then there are still two unknowns, namely θ and φ.
 
  • #38
Jeroen Staps said:
But then there are still two unknowns, namely θ and φ.
Doesn’t matter. At this stage we are just trying to find a relationship between them. Next step will be to involve them in force balance equations.
In fact, if you are stuck on getting that relationship, set that aside and move onto the other equations.
 
  • #39
haruspex said:
Doesn’t matter. At this stage we are just trying to find a relationship between them. Next step will be to involve them in force balance equations.
In fact, if you are stuck on getting that relationship, set that aside and move onto the other equations.
Alright I'll try to find the relationship then.
 
  • #40
Still stuck on finding the relationships :(
 
  • #41
Now I have that: AD2=AC2+CD2-2*AC*CD*cos(γ-θ), with γ the angle ∠ACB. But still don't understand the force balance equations
 
  • #42
Been a while. Remind us what force balance equations you have gathered so far and which one you don't understand :wink:
 
  • #43
BvU said:
Been a while. Remind us what force balance equations you have gathered so far and which one you don't understand :wink:
Was on holiday ;) , I just don't understand how I can solve this problem when there are more than one unknowns
 
  • #44
Jeroen Staps said:
Still stuck on finding the relationships :(
First, find all the angles of triangle ACD in terms of α, θ, φ.
 
  • #45
BvU said:
Been a while. Remind us what force balance equations you have gathered so far and which one you don't understand :wink:
upload_2019-3-7_13-26-17.jpeg

Does this make sense?
 

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  • #46
haruspex said:
First, find all the angles of triangle ACD in terms of α, θ, φ.
upload_2019-3-7_13-26-54.jpeg


Need more?
 

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  • #47
Jeroen Staps said:
Now I have that: AD2=AC2+CD2-2*AC*CD*cos(γ-θ), with γ the angle ∠ACB. But still don't understand the force balance equations
And I have that: BD2=AB2+AD2-2*AB*AD*cos(α2) when you substitute AD from the earlier post in this equation then you have an equation to solve α2 with θ as unknown and α2 is equal to α-(180-90-φ). So I can express θ in φ
 
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  • #48
Jeroen Staps said:
That’s good. You know the lengths of two sides of ACD. What equation can write connecting those with two of its angles?
 
  • #49
haruspex said:
That’s good. You know the lengths of two sides of ACD. What equation can write connecting those with two of its angles?
I now have the following equation that seems to be correct:

φ = cos-1((AC-CD*cos(γ-θ)) / √(AC2+CD2-2*AC*CD*cos(γ-θ))) + 90 - α
 
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  • #50
Jeroen Staps said:
I now have the following equation that seems to be correct:

φ = cos-1((AC-CD*cos(γ-θ)) / √(AC2+CD2-2*AC*CD*cos(γ-θ))) + 90 - α

But how do I use this to describe the location of D when there is a certain mass hanging at D and there is a certain pulling force in AD?
 
  • #51
Jeroen Staps said:
I now have the following equation that seems to be correct:

φ = cos-1((AC-CD*cos(γ-θ)) / √(AC2+CD2-2*AC*CD*cos(γ-θ))) + 90 - α
I was thinking of something simpler, though it may be equivalent: the sine rule.
 
  • #52
Jeroen Staps said:
But how do I use this to describe the location of D when there is a certain mass hanging at D and there is a certain pulling force in AD?
As I posted, you can use angles θ and φ to write a force balance equation.
 
  • #53
haruspex said:
I was thinking of something simpler, though it may be equivalent: the sine rule.
I used the cosine rule twice to come up with this
 
  • #54
haruspex said:
As I posted, you can use angles θ and φ to write a force balance equation.
So does my post #45 make sense?
 
  • #55
Yes, at least for equilibrium. Adding constant acceleration should be easy.

But I wished you wouldn't change notation every time.
And you can easily simplify ##\ \sin(\pi/2-\pi/4-\phi)\ ## and ##\ \cos(\pi/2-\pi/4-\phi)\ ## I should hope.

So how many equations with how many unknowns do you now have altogether ?
 
  • #56
BvU said:
Adding constant acceleration should be easy.
I believe it is a statics question, how the equilibrium position depends on the applied tension.
 
  • #57
haruspex said:
I believe it is a statics question, how the equilibrium position depends on the applied tension.
The question is: What is the position of the pulley when there is a given mass of the load and a given ratio between the force of gravity and the pulling force.
 
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  • #58
Jeroen Staps said:
The question is: What is the position of the pulley when there is a given mass of the load and a given ratio between the force of gravity and the pulling force.
Which is the same as I wrote.
 

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