Tension at the lowest point!

Classical Mechanics Level 2

A heavy string of mass $m$ hangs between two fixed points $A$ and $B$ at the same level. The tangents to the string at $A$ and $B$ are at an angle $\theta$ with the horizontal as shown in the figure. The tension in the string at lowest point is?

\frac{mg}{2\cot \theta }

\frac{mg}{2\cos \theta }

\frac{mg}{2\sin \theta }

\frac{mg}{2\tan \theta }

1 solution

Nishant Rai
May 19, 2015

$T \sin \theta = \frac{mg}{2}$

$T' = T \cos \theta = \frac{mg}{2} \cot \theta =\frac{mg}{2 \tan \theta }$

did it the same way !

CH Nikhil - 6 years ago

me too !!!

Vaibhav Prasad - 6 years ago

Wouldn't the tension be zero as there would be same magnitude of tension from right end

Ruchika Sharma - 2 years, 9 months ago

Isn't the tension same throughout the string Considering that I am getting T=mg/2sin

Kunal Wadhwa - 2 years, 8 months ago

Can someone please explain this more clearly?

Psi Bar Psi - 1 year, 5 months ago

Reply to Psi Bar Psi’s comment. Considering the left half of the string. For the strings to not move the forces should be balanced in all directions. Vertical force at the point of attachment should be equal to the weight of the half string. Now the tension also has some horizontal component which needs to be balanced. This horizontal force is equal to the tension at the lowest point, as the tension (at the lowest point) is completely horizontal.

Ishan Tandon - 11 months ago

Tension at the lowest point!

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