Why do I obtain a zero net force in tension for a torque happening on a bar supported on its center?

Question

The problem is as follows:

A cable goes across points A and B such that the bar will not slide as indicated in the figure from below. The mass of the bar is $2.5\,kg$. Assuming that the surface has no friction, find the tension on the wire so that the system remains in mechanical equilibrium.

The alternatives given in my book are as follows:

$\begin{array}{ll} 1.&12\,N\\ 2.&10\,N\\ 3.&14\,N\\ 4.&16\,N\\ 5.&18\,N\\ \end{array}$

What I attempted to do to solve this problem was to put the reaction force due the weight on the midpoint and using the point on $C$ as rotation axis. Therefore taking the distance for each force to that point.

Because the Normal force is not given I assumed that it was the resultant from the weight and the wire pulling to the wall as indicated in the diagram. Considering the direction of such force I did the following:

$N=w\sin 53^{\circ}+T\sin 37^{\circ}$

Then:

$-5N+10\left(T\sin37^{\circ}\right)+5\left(w\sin 53^{\circ}\right)=0$

$-5\left(w\sin 53^{\circ}+10T\sin 37^{\circ}\right)+5\left(w\sin 53^{\circ}\right)=0$

$-5\frac{4w}{5}-5\frac{3T}{5}+10T\frac{3}{5}+5\frac{4w}{5}=0$

Because $w= 2.5 \times 10 = 25$

The preceding equation would become into:

$-4 \times 25 - 3T + 6T + 4 \times 25 =0$

$3T=0$

$T= 0\,N$

Therefore I end up obtaining a zero force for the tension. What could I be doing wrong?. Can somebody help me with the appropiate calculation for the tension using my approach?.

Answer 1

With A as the pivot point, establish the equilibrium equations below,

$$T = N\cos 53,\>\>\>\>\>N = mg \sin 53$$

to obtain the tension

$$T = mg \cdot\cos 53\cdot\sin 53 = (2.5\times10)\cdot \frac 35\cdot\frac45 = 12N$$

Answer 2

Characterizing

$$ \cases{ W=(0,-m g)\\ T=t_0(1,0)\\ V = v_0(0,1)\\ F_C=f_0(-\sin\alpha,\cos\alpha)\\ x_C=\frac{h}{\tan\alpha}\\ y_C = h\\ x_G = \frac L2 \cos\alpha\\ y_G = \frac L2 \sin\alpha\\ G=(x_G,y_G)\\ C=(x_C,y_C)\\ O=(0,0) } $$

we have now the equilibrium equations

$$ \cases{ V+T+W+F_C=0\\ (C-O)\times F_C + (G-O)\times W = 0 } $$

Three equations and three unknowns $t_0,v_0, f_0$