# How to find the modulus of the tension of a wire when a sphere is over a bar?

The problem is as follows:

The figure from below shows a system which is at equilibrium. The sphere is frictionless, homogeneous and its mass is $$5\,kg$$. The homogeneous bar has a mass of $$10\,kg$$. Find the modulus of the tension of the wire. Assume that the acceleration due gravity is $$10$$ meters per second square. M is the midpoint of the bar.

The alternatives are as follows:

$$\begin{array}{ll} 1.&62N\\ 2.&71.25N\\ 3.&80.05N\\ 4.&92.25N\\ 5.&100N\\ \end{array}$$

Can someone help me here?. I'm not exactly how to apply the condition of equlibrium for this problem. I'm confused exactly how should I use the information of the angle given at one end of the wire.

What I've attempted to do was the following:

$$-Tl+100\sin 53^{\circ}\left(\frac{l}{2}\right)+50\sin 53^{\circ}\left(\frac{l}{2}\right)=0$$

But this did not solved the problem. The part where I'm strugglin the most is where are the forces acting. Therefore I need help with the free body diagram of the forces here. Can someone help me here please?.

Let the sphere exert a force $$F$$ on the rod perpendicular to it. We know that the vertical component of this force will be we equal to the weight of the sphere. $$F\times\cos37^o=10\times 9.8$$ Now, let the component of the weight of the rod in perpendicular direction be equal to $$F'$$. $$F'=5\times 9.8\times\cos37^o$$ We know that the system is in equilibrium. This implies that $$\tau_{net}=0$$ on the rod. This gives us our final equation which gives the value of tension $$T$$. $$\frac {F+F'}2=T$$