# 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!.

• Tension isn't unitless so all answers are wrong. Feb 6 '20 at 20:14
• Should be on Physics SE Feb 6 '20 at 20:15
• @PaulChilds Sorry, I forgot to add them when I copied it down in a rush. But this shouldn't be a problem. The question as it stands would probably be not accepted in Physics SE as their current policy. And for me this more as of a vector problem. Although I'm looking for an algebraic method. Feb 6 '20 at 20:19
• You have reasoned correctly but the questioner has poorly communicated that they probably want onl one of the tension forces. Feb 6 '20 at 20:21
• Yes. Their policy is a bit draconian. Feb 6 '20 at 20:23

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:

$$\tan\alpha = \frac{P/2}{T} \; \; \; \Leftrightarrow \; \; \; T = \frac{P}{2 \tan\alpha} = \frac{6\,N}{2 \tan(30°)} = 3\sqrt{3}\,N\,.$$

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 $$T_{\,C} \; = 3\;\sqrt 3 \;$$