What is the tension in the rope DE?

[Suraj M]

Homework Statement

AB = AC =1.6m, hinged at A. a rope DE of length 0.5m is tied half way up.
A weight of 40Kg is suspemded from F 1.2m from A. Assuming frictionless floor and negligible mass of the ladder Find the tension in the rope DE.

Homework Equations

torque[/B] = Force × perpendicular distance

The Attempt at a Solution

$N_band N_c$ are the normal forces at A and B.
if i take the net moment about A:
$N_b d_1 - 400d_2 = N_cd_1$
also
$N_b +N_c = 400N$
Solving these equations an using the ratio $\frac{d_1}{d_2}$
I get $N_b = 250N and N_c=150N$
this corresponds to the answer given for $N_b and N_c$
But how do i find out the tension, their moments get canceled out!

 

[SteamKing]

Once you have solved for N_b and N_c, you should be able to construct free body diagrams for the legs AB and AC.

Since the whole ladder is in equilibrium, each leg must be in equilibrium as well.

 

[Suraj M]

Rotational equilibrium?? i tried that and got 2 equations with 3 variables.
EDIT: Got it thank you!