Answer:
The Tension [tex]T = \frac{mgL}{2xsin \theta}[/tex]
The horizontal component [tex]\sum F_x[/tex] = [tex]\frac{mgL}{2x} cot \theta[/tex]
The vertical component [tex]\sum F_y[/tex] = [tex]mg(1 - \frac {L}{2x})[/tex]
The angular acceleration ∝ = [tex]\frac{3g}{2L}[/tex]
Explanation:
The illustration of what the question depicts is shown in the diagram below;
SO; Equilibrium
[tex]\sum Fx = 0 = \sum Fy[/tex]
although the net torque is zero about any point ;
Now;
[tex]mgL/2 = Tsin \theta .x\\\\Tension \ T = \frac{mgL}{2x sin \theta}[/tex]
The horizontal component is :
[tex]\sum F_x = 0 \\\\F_{11} = T cos \theta \\\\\\= \frac{mgL}{2x} cot \theta[/tex]
The vertical component is :
[tex]\sum F_y = 0[/tex]
F⊥ = mg - T sin θ
= [tex]mg(1 - \frac {L}{2x})[/tex]
b) Torque about point 0 =
[tex]mg\frac{L}{2}= I \alpha \\\\ \alpha = \frac{mgL}{2I}\\\\\alpha = \frac{mgL}{2 \frac{mL^2}{3}}\\\\\alpha = \frac{3g}{2L}[/tex]
