# How to find the tension in a wire connecting three spheres?.

#### Chemist116

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!.

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#### romsek

Math Team
doesn't this give a tension in the cord of $6\sqrt{3}~N$ ?

#### skeeter

Math Team
doesn't this give a tension in the cord of $6\sqrt{3}~N$ ?
no, see post #4 at the link

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