How to find the tension in a wire connecting three spheres?. The problem is as follows:
The diagram from below shows three spheres identical in shape and weigh $6\,N$. The system is at static equilibrium. Find the tension in Newtons ($\,N$) of the wire connecting $B$ and $C$.

The alternatives given are:
$\begin{array}{ll}
1.&\frac{\sqrt{3}}{2}\,N\\
2.&\sqrt{3}\,N\\
3.&2\sqrt{3}\,N\\
4.&3\sqrt{3}\,N\\
5.&4\sqrt{3}\,N\\
\end{array}$
I'm not sure exactly how to draw the FBD for this object. Can someone help me here?. I'm assuming that the weight of the top sphere which is $A$ will generate a reaction and a tension making a triangle.
Since the weight is $6\,N$ then using vector decomposition it can be established that: (Using sines law)
$\frac{6}{\sin 30^{\circ}}=\frac{T}{\sin 60^{\circ}}$
Therefore:
$T=6\sqrt{3}$
But this doesn't check with any of the alternatives. I'm confused exactly where the Reaction is happening and why?. Help here please!.
 A: Equivalent static problem:

by imposing the global equilibrium:

therefore by extracting the hinge A:

and placing it in equilibrium:
$$
\begin{cases}
N_{AB} + N_{AC}\,\cos\alpha = 0 \\
\frac{P}{2} + N_{AC}\,\sin\alpha = 0
\end{cases}
\; \; \; \; \; \; \Leftrightarrow \; \; \; \; \; \;
\begin{cases}
N_{AB} = \frac{P}{2 \tan\alpha} \\
N_{AC} = -\frac{P}{2 \sin\alpha}
\end{cases}
$$
from which, for symmetric issues, it can be deduced that:


*

*$AC$ and $BC$ are subject to compression of intensity $\frac{6\,N}{2 \sin(30°)} = 6\,N$;

*$AB$ is subject to a traction of intensity $\frac{6\,N}{2 \tan(30°)} = 3\sqrt{3}\,N$ (answer to the question).


It's now clear that, thanks to symmetry, it's sufficient to refer to the following diagram:

from which:
$$
\tan\alpha = \frac{P/2}{T} \; \; \; \Leftrightarrow \; \; \; T = \frac{P}{2 \tan\alpha} = \frac{6\,N}{2 \tan(30°)} = 3\sqrt{3}\,N\,.
$$
A: This is a problem where the method of Virtual work .. works at best.
Keep the sphere B fixed, cut the wire, and take that the segment BA moves by an angle $d\alpha$ CW.
Then the work , positive, done by $P_A$ would be
$$
\left. {P_{\,A} \,2r\;d\left( {\sin \alpha } \right)\;} \right|_{\,\alpha  = 30^\circ }
  = \left. {12r\;\cos \alpha \;d\alpha } \right|_{\,\alpha  = 30^\circ }  = 12r\;{{\sqrt 3 } \over 2}\;d\alpha
$$
while the work, negative, done by the tension in C will be
$$
\left. {T_{\,C}  \cdot 2 \cdot 2r \cdot \left( { - d\left( {\cos \alpha } \right)} \right)\;} \right|_{\,\alpha  = 30^\circ }
  = \left. {T_{\,C} 4r\;\sin \alpha \;d\alpha } \right|_{\,\alpha  = 30^\circ }  = T_{\,C} 4r\;{1 \over 2}\;d\alpha 
$$
Equating the two
$$T_{\,C} \; = 3\;\sqrt 3 \;$$
