How to find the tension of $T_1$ and $T_2$?

Question

(Large version)

I want to find the tension of $T_1$ and $T_2$. Also researched something on internet before posting this question. I had found that

$$ T_1 = T_2 \frac {\cos(30)}{\cos(30)} = m \cdot 9.8 \, \text{m}/\text{s}^2 $$

My Kindest Regards!

Answer 1

Tension $\vec T_1$ is the easy one, because we have a static balance of forces: $$ \vec T+\vec T_1+m\vec g=0. $$ From the vertical components one then gets (I'm using the same letters without arrows to denote the magnitudes of vectors, as it is customary in kinematics): $$ T_1\cos30°=mg, \quad\hbox{that is:}\quad T_1={2mg\over\sqrt3}. $$ Tension $\vec T_2$, instead, takes place at a time when the mass is moving: even if its velocity vanishes at $B$, its acceleration $\vec a$ is not vanishing and is constrained by kinematics to be perpendicular to $\vec T_2$. We have of course $$\vec T_2+m\vec g=m\vec a,$$ and separating the components of this equation we get: $$ T_2\sin30°=ma\cos30° \quad\hbox{and}\quad mg-T_2\cos30°=ma\sin30°. $$ From these equations you can solve for $a$ and $T_2$.

Answer 2

I still do no not understand the problem. But let us try to start anyway and discuss further.

With $m$ at rest at $A$ we have a force equilibrium $$ 0 = F_G + T_1 + T_{AC} $$ In $x$-direction it is $$ 0 = 0 + \lVert T_1 \rVert \sin(30^\circ) - \lVert T_{AC} \rVert $$ and in $y$-direction $$ 0 = -mg+ \lVert T_1 \rVert \cos(30^\circ) + 0 $$ so $$ \lVert T_1 \rVert = \frac{mg}{\cos(30^\circ)} $$ and $$ T_1 = \frac{mg}{\cos(30^\circ)} \begin{pmatrix} \sin(30^\circ) \\ \cos(30^\circ) \end{pmatrix} = mg \begin{pmatrix} \tan(30^\circ) \\ \cos^2(30^\circ) \end{pmatrix} $$

Update: (After the problem was clarified)

So we have a pendulum. First fixed at $CA$ and then released.

At point $B$ we have the force equilibrium \begin{align} m \ddot{r} &= F_G + T_2 \iff \\ m \lVert \ddot{r} \rVert \begin{pmatrix} -\cos(30^\circ) \\ -\sin(30^\circ) \end{pmatrix} &= m \begin{pmatrix} 0 \\ -g \end{pmatrix} + \lVert T_2 \rVert \begin{pmatrix} -\sin(30^\circ) \\ \cos(30^\circ) \end{pmatrix} \end{align} This gives $$ \lVert \ddot{r} \rVert = \lVert T_2 \rVert \frac{\tan(30^\circ)}{m} \\ $$ and $$ -\lVert T_2 \rVert \tan(30^\circ) \sin(30^\circ) = -m g + \lVert T_2 \rVert \cos(30^\circ) \iff \\ \lVert T_2 \rVert = \frac{mg}{\cos(30^\circ) + \tan(30^\circ)\sin(30^\circ)} = mg\cos(30^\circ) $$ which finally gives $$ T_2 = mg \cos(30^\circ) \begin{pmatrix} -\sin(30^\circ) \\ \cos(30^\circ) \end{pmatrix} = mg \begin{pmatrix} -\cos(30^\circ) \sin(30^\circ) \\ \cos^2(30^\circ) \end{pmatrix} $$ So $T_1$ and $T_2$ are equal in the $y$-component, but not in the $x$-component. Aside from the direction the $T_1$ $x$-component is larger.