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A homogenous beam $OA$ with the length $4a$ and the mass $m$ can rotate in the vertical plane along a horizontal axis through $O$ and is kept in equilibrium by a wire running from the points A and C through a castor in B. Determine the straining force $S$ in the wire and the reaction forces from the axis on the beam at the point $O$. Figure $$\begin{cases} (1) \, |\vec{A}|\sin 60+|\vec{C}|=|\vec{R_y}|+|m\vec{g}| \\ (2) \,|\vec{A}|\cos60 = |\vec{R_x}| \\ (3) \, a|\vec{C}| \cos 30 + \frac{4a}{\sqrt3}|\vec{A}|=|m\vec{g}|a(\sqrt{3}-\frac{\sqrt{3}}{2}) \end{cases}$$ Where $(1)$ is the equation for the forces in the vertical axis, $(2)$ the horizontal and $(3)$ the moment equation to the point $O$. But as you can see, I have 4 variables ($\vec{A}, \vec{C}, \vec{R_x}, \vec{R_y}$) and 3 equations (the $a$ disappears in $(3)$). Is there something I have missed or how can I solve this problem?

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    $\begingroup$ Nice question - thanks for taking the effort to show your reasonning clearly - I will add the homework and exercises tag as it fits into that category, but well done for making it so clear $\endgroup$ – tom Apr 28 '15 at 11:00
  • $\begingroup$ Although the effort shown is laudable, this question does not ask any question except for "How to solve this problem?" and is thus off-topic for not asking a conceptual question. $\endgroup$ – ACuriousMind Apr 28 '15 at 15:19
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    $\begingroup$ @AcuriousMind Well, I'm really asking how I can solve the problem of having 4 variables and 3 equations. I thought I had missed an equation or something. Did not know this qualified as off-topic. $\endgroup$ – Lozansky Apr 28 '15 at 18:31
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If the wire used around the pulley is considered ideal and massless, tension in the wire is same at each point and hence vector A and vector C are equal in magnitude. This reduces one variable and you will be left with 3 equations and 3 variables.

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    $\begingroup$ nice answer - conveys well the key point needed to solve the problem without doing the work $\endgroup$ – tom Apr 28 '15 at 10:59
  • $\begingroup$ @Edark If you look at the figure, you can easily calculate that the perpendicular distance between $\vec{C}, m\vec{g}, \vec{A}$ and the point $O$ are $$a \frac{\sqrt{3}}{2}, 2a \frac{\sqrt{3}}{2}, a \frac{4}{\sqrt{3}}$$ and if $S$ denotes the tension in the wire then $$S(a \frac{\sqrt{3}}{2}+a \frac{4}{\sqrt{3}}) = mg (2a \frac{\sqrt{3}}{2}) \leftrightarrow S(\frac{\sqrt{3}}{2}+\frac{4}{\sqrt{3}}) = mg \sqrt{3} \leftrightarrow S = mg \frac{\sqrt{3}}{\frac{\sqrt{3}}{2}+\frac{4}{\sqrt{3}}} = 6/11$$ The answer sheet says that $S = \frac{2\sqrt{3}}{4+\sqrt{3}}$. What is wrong? $\endgroup$ – Lozansky Apr 28 '15 at 11:11
  • $\begingroup$ Thanks @tom. My answer to another question got deleted for giving out the solution, so I tried not to over here. $\endgroup$ – Edark May 1 '15 at 6:14
  • $\begingroup$ @Lozansky How did you calculate perpendicular distance for A? The perpendicular distance is not $ \frac{4a}{\sqrt3} $. The other distances are correct. Note: You can use the definition of torque too. P.S: The answer for S given by the sheet is correct. $\endgroup$ – Edark May 1 '15 at 6:35
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    $\begingroup$ @Edark Yes, you are correct, I miscalculated that distance. I now got the correct answer and I have upvoted and accepted your answer. $\endgroup$ – Lozansky May 7 '15 at 15:58

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