- #36
Jeroen Staps
- 36
- 0
OK, I seeharuspex said:I meant to write
OK, I seeharuspex said:I meant to write
But then there are still two unknowns, namely θ and φ.haruspex said:I meant to write
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.Jeroen Staps said:But then there are still two unknowns, namely θ and φ.
Alright I'll try to find the relationship then.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.
Was on holiday ;) , I just don't understand how I can solve this problem when there are more than one unknownsBvU said:Been a while. Remind us what force balance equations you have gathered so far and which one you don't understand
First, find all the angles of triangle ACD in terms of α, θ, φ.Jeroen Staps said:Still stuck on finding the relationships :(
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 φJeroen 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
That’s good. You know the lengths of two sides of ACD. What equation can write connecting those with two of its angles?Jeroen Staps said:
I now have the following equation that seems to be correct: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?
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.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
As I posted, you can use angles θ and φ to write a force balance equation.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?
I used the cosine rule twice to come up with thisharuspex said:I was thinking of something simpler, though it may be equivalent: the sine rule.
So does my post #45 make sense?haruspex said:As I posted, you can use angles θ and φ to write a force balance equation.
I believe it is a statics question, how the equilibrium position depends on the applied tension.BvU said:Adding constant acceleration should be easy.
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.haruspex said:I believe it is a statics question, how the equilibrium position depends on the applied tension.
Which is the same as I wrote.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.