Solve Mechanics Question: Hanging Thread, Gravity, Angle @

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The discussion focuses on solving a mechanics problem involving a hanging thread between two hinges under the influence of gravity. The key points include the need to determine the tension at the midpoint of the rope, with the angle made by the tangent at point A being crucial for calculations. The tension at point A (t1) and the tension at the midpoint (t2) are related through trigonometric relationships, specifically t1sin@ = mg/2 and t1cos@ = t2. The final formula derived for the tension at the midpoint is t2 = (mg cot@)/2. The conversation emphasizes the importance of free body diagrams and Newton's laws in solving the problem.
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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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