# Determining straining force and reaction forces on a beam [closed]

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 $O$. $$\begin{cases} (1) \, |\vec{A}|\sin 60+|\vec{C}|=|\vec{R_y}|+|m\vec{g}| \\ (2) \,|\vec{A}|\cos60 = |\vec{R_x}| \\ (3) \, a|\vec{C}| \cos 30 + \frac{4a}{\sqrt3}|\vec{A}|=|m\vec{g}|a(\sqrt{3}-\frac{\sqrt{3}}{2}) \end{cases}$$ 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?

• 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
– tom
Commented Apr 28, 2015 at 11:00
• 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. Commented Apr 28, 2015 at 15:19
• @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. Commented Apr 28, 2015 at 18:31

• @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? Commented Apr 28, 2015 at 11:11
• @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. Commented May 1, 2015 at 6:35