# Tension in a connecting rope

1. Feb 24, 2015

### Suraj M

1. The problem statement, all variables and given/known data

AB = AC =1.6m, hinged at A. a rope DE of lenght 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.

2. Relevant equations
torque
= Force × perpendicular distance

3. 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 cancelled out!

2. Feb 24, 2015

### SteamKing

Staff Emeritus
Once you have solved for Nb and Nc, 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.

3. Feb 24, 2015

### Suraj M

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