Pulley system (ideal pulley) Find the angle with vertical line

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Homework Help Overview

The discussion revolves around an ideal pulley system involving two blocks, where the original poster seeks to determine the angle theta with respect to the vertical line, as well as the accelerations and tensions in the system. The problem is set within the context of classical mechanics, specifically focusing on forces and tensions in a pulley system.

Discussion Character

  • Exploratory, Assumption checking, Mathematical reasoning

Approaches and Questions Raised

  • The original poster presents equations relating tensions and forces, questioning their correctness and what might be missing. Some participants discuss the vector nature of tensions and suggest eliminating variables to find accelerations.

Discussion Status

Participants are actively engaging with the problem, with some providing guidance on the vector addition of tensions. There is an exploration of relationships between the angles and tensions, but no explicit consensus has been reached regarding the final values or methods.

Contextual Notes

The original poster mentions needing to express accelerations and tensions in terms of mass and gravitational acceleration, indicating a focus on deriving relationships rather than numerical solutions. There is also a potential assumption regarding the angle being 45°, which is under discussion.

thirteenheath
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Ideal pulley and strings, no friction.The pulley and the second string don't move.
I need to find the numerical value of theta, the acceleration of the two blocks and the tensions.Accelerations and tension must be given in terms of m and g.I guess
2T1=T2
T1=m2*a
m1*g-T1=m1*aTo find theta I need to know the angle between T2 and T2*cosθ.
Are these equations correct?What is missing?
 

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Hi thirteenheath, welcome to PF.

T2, the tension in chord 2 is not parallel with the forces of tension (T1, one horizontal and one vertical), chord 1 exerts on the pulley. Add them as vectors to get the force on the pulley. As the pulley does not move, the resultant of the tensions must be zero.
You can eliminate T1 from the last two equation and get the acceleration of the blocks.

ehild
 
Thanks ehild!

Oh I see...So if I understood we'll have

T1=T2*cosθ

T1=T2*sinθ

Then cosθ=sinθ and θ=45°.

Plus, through the last two equations in the previous post we find that

a=m1*g/(m1+m2).

Now I just need to replace the acceleration in any of the previous equations to find the values of T1 and T2.

Is that it?

Thanks again. =)
 
thirteenheath said:
Thanks ehild!

Oh I see...So if I understood we'll have

T1=T2*cosθ

T1=T2*sinθ

Then cosθ=sinθ and θ=45°.

Plus, through the last two equations in the previous post we find that

a=m1*g/(m1+m2).


Now I just need to replace the acceleration in any of the previous equations to find the values of T1 and T2.

Is that it?

Thanks again. =)

Yes, it will be all right.

ehild
 

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