Torque, beam problem.

Question

The length of the beam is L, mass of the beam m, there is a hinge at the origin O

Equilibrium conditions gives $\sum\limits_{i=1}^n \textbf{F}_i=0$ and $\sum\limits_{i=1}^n \Gamma_i=0$
Position vectors are $\textbf{r}_R=0$, $\textbf{r}_W=\frac{1}{2}L\textbf{i}$, $\textbf{r}_T=L\textbf{i}$

$\textbf{W}=-mg\textbf{j}$
$\textbf{R}=R_i\textbf{i}+R_j\textbf{j}$
$\textbf{T}=|\textbf{T}|\cos(\alpha)\textbf{i}+|\textbf{T}|\sin(\alpha)\textbf{j}$

Torques are
$\Gamma_R=\textbf{0}$
$\Gamma_W=\textbf{r}_W\times\textbf{W}= \frac{1}{2}L\textbf{i}\times(-mg\textbf{j})=-\frac{1}{2}Lmg\textbf{k}$
$\Gamma_T=\textbf{r}_T\times \textbf{T} = L\textbf{i}\times (|\textbf{T}|\cos(\alpha)\textbf{i}+|\textbf{T}|\sin(\alpha)\textbf{j})=L|\textbf{T}|\sin(\alpha)\textbf{k}$

Resolving the second equilbrium condition in the $\textbf{k}$-direction gives $L|\textbf{T}|\sin(\alpha)-\frac{1}{2}Lmg=0$, hence
$|\textbf{T}|=\frac{1}{2}mg \csc(\alpha)$

Resolving the first equilibrium condition in the $\textbf{i}$ and $\textbf{j}$-direction gives
$R_i+|\textbf{T}|\cos(\alpha)=0$
$R_j+|\textbf{T}|\sin(\alpha)-mg=0$
Hence
$R_i=-|\textbf{T}|\cos(\alpha)=-\frac{1}{2}mg\cot(\alpha)$
$R_j=mg-|\textbf{T}|\sin(\alpha)=mg-\frac{1}{2}mg=\frac{1}{2}mg$
Therefore $\textbf{R}=-\frac{1}{2}mg\cot(\alpha)\textbf{i}+\frac{1}{2}mg\textbf{j}$

From the first equilibrium condition, the tension in the string is
$\textbf{T}=-\textbf{R}-\textbf{W}=\frac{1}{2}mg\cot(\alpha)\textbf{i}-\frac{1}{2}mg\textbf{j}+mg\textbf{j}=\frac{1}{2}mg\cot(\alpha)\textbf{i}+\frac{1}{2}mg\textbf{j}$

Ive convinced myself that I have the force $\textbf{T}$ wrong, can anyone confirm that?

Answer 1

Given the forces an the respective application points

$$ \cases{O\to \vec R\\ A\to \vec W\\ B\to \vec T} $$

the static equilibrium needs

$$ \cases{ \vec R + \vec W + \vec T = 0\\ (A-O)\wedge\vec W+(B-O)\wedge \vec T = 0 } $$

Here

$$ \cases{ O = (0,0)\\ A = (\frac L2,0)\\ B = (L,0)\\ \vec R = (R_x,R_y)\\ \vec W = mg(0,-1)\\ \vec T = T_0(\cos\alpha,\sin\alpha) } $$ resulting into the linear system

$$ \cases{ R_x+T_0\cos\alpha = 0\\ R_y-m g+T_0\cos\alpha = 0\\ T_0\sin\alpha-\frac 12 mg = 0 } $$

three equations and three unknowns $R_x,R_y,T_0$

and consequently

$$ \cases{R_x = -\frac 12 m g \cot\alpha\\ R_y = \frac 12 m g\\ T_0 = \frac 12 \frac{m g}{\sin\alpha} } $$