Tension

can you give the angles

angle made by the rope on either side with the vertical is theta

t=mg/2cosθ

i'm getting the ans mg/2tan(theta)

no the ans is in tanθ

@philip i gave u the wrong angle sry

@prateek ur ans is correct can u post the method

@ subhash
i don't think the θ given by u earlier was wrong....

if we consider the right portion of the wire....
the vertical component of tension at the point where the string is hanged by the wire is Tsinθ and the horizontal component is Tcosθ(towards right).....
where T=tension on the wire due to wall.....

NOW,
Let, the tension at the middle point of wire will be T'....
and the direction of tension will be horizontal towards left....
since it is equilibrium....

Tsinθ=mg/2

Tcosθ=T'
Solving the two eqⁿ we can calculate T'.....

check this out.....
is this not to be pinked....
guys many of my posts are not been pinked though they r useful....