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 $C$.

Sketch of the problem

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)

$\frac{6}{\sin 30^{\circ}}=\frac{T}{\sin 60^{\circ}}$



But this doesn't check with any of the alternatives. I'm confused exactly where the Reaction is happening and why?. Help here please!.

  • $\begingroup$ Tension isn't unitless so all answers are wrong. $\endgroup$ Feb 6 '20 at 20:14
  • 1
    $\begingroup$ Should be on Physics SE $\endgroup$ Feb 6 '20 at 20:15
  • $\begingroup$ @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. $\endgroup$ Feb 6 '20 at 20:19
  • $\begingroup$ You have reasoned correctly but the questioner has poorly communicated that they probably want onl one of the tension forces. $\endgroup$ Feb 6 '20 at 20:21
  • $\begingroup$ Yes. Their policy is a bit draconian. $\endgroup$ Feb 6 '20 at 20:23

Equivalent static problem:

enter image description here

by imposing the global equilibrium:

enter image description here

therefore by extracting the hinge A:

enter image description here

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:

enter image description here

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


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