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!.

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!.

Last edited: