Solve Mechanics Question: Hanging Thread, Gravity, Angle @

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Homework Help Overview

The problem involves a thread hanging between two hinges under the influence of gravity, with a focus on determining the tension at the midpoint of the rope. The angle formed by the tangent to the rope at one hinge and the line connecting the hinges is also a key factor in the discussion.

Discussion Character

  • Mixed

Approaches and Questions Raised

  • Participants explore the relationship between tension and gravitational forces, with initial attempts to express tension in terms of angles and mass. There are discussions about the distribution of weight along the rope and the application of trigonometric relationships.

Discussion Status

Some participants have provided insights into the relationships between the forces acting on the rope, while others have confirmed the validity of certain expressions derived from their reasoning. The conversation reflects a collaborative effort to clarify the mechanics involved without reaching a definitive conclusion.

Contextual Notes

There is an acknowledgment of the need for specific values, such as the mass of the rope, to fully resolve the problem. Participants also note the importance of free body diagrams in analyzing the forces at play.

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mechanics question..please help!

Homework Statement


a thread hangs between 2 hinges A and B. gravity is acting. The angle(acute) made by tangent to rope at point A and the line AB is @. find the tension at the lower most point..i.e the mid point of the rope?


Homework Equations



don know!

The Attempt at a Solution

 
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You need to know the mass of the rope in order to calculate the tension. Assume the mass is M. Then you must realize that the vertical force and tension at A are related by basic trig. You must show some attempt at a solution for further assistance.
 


ok ...i take the tension to be t1 at point A...and t2 at bottom most point then
t1sin@=mg...then how to proceed??
 


EDIT: [strike]Yes, that's OK. [/strike]Sorry, I was half asleep this morning, that is not OK. The weight of the rope is divided equally amongst points A and B. Sorry about that . Now the direction of the tension t2 at the bottom most point is in the horizontal direction, right? So try taking a free body diagram of the left half of the rope and then what does Newton 1 tell you when looking in the x direction at point A?
 
Last edited:


oh ...okkkk...so i do this
for left half of the rope ...
in x direction ...
its ... t1cos@=t2..(1
then t1sin@=mg/2...(2

so i divide 1/2 and
cot@=2*t2/mg..
so t2=(mg cot@)/2...!
right??!
 


twinklealices said:
oh ...okkkk...so i do this
for left half of the rope ...
in x direction ...
its ... t1cos@=t2..(1
then t1sin@=mg/2...(2

so i divide 1/2 and
cot@=2*t2/mg..
so t2=(mg cot@)/2...!
right??!
Oh yes, right you are! :smile:
And welcome to PF!:smile::smile:
 


oh ..thank you!...this is an awesome site for doubt clarification!
 

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