Rod and cord problem

An uniform rod is hinged in A in equilibrium by a light cord, as shown in the figure. Determine the vertical and horitzontal forces on the rod exerted by the hinge and also the d from the aproppiate torque equation.

Here's a link to the code Mathjax of the quadratic equation

The solution of the problem step by step is as follows

a)

b)

\begin{align} ΣF_x=0 =&\cssId{Step1}{-Fhx+Tx=0}\\[3px] &\cssId{Step2}{Tx=Fhx}\\[3px] &\cssId{Step3}{Fhx=sin(β)·T}\\[3px] \end{align}
  1. In the first step we make the equation summation of forces on the horitzontal axis
  2. In the second step we isolate the hinge force in the horitzontal axis
  3. In the third step we substitute the horitzontal component of the tension by the data we know by trigonometry

Please enter the following data to calculate the previous equation

Enter β

Enter tension

Equation 1 result is


Comment:


\begin{align} ΣF_y=0 =&\cssId{Step4}{Ty-Fhy-W-mg=0}\\[3px] &\cssId{Step5}{Ty-W-mg=Fhy}\\[3px] &\cssId{Step6}{Fhy=T*cos(β)-W-mg}\\[3px] \end{align}
  1. In the first step we make the equation summation of forces on the vertical axis
  2. In the second step we isolate the hinge force in the vertical axis
  3. In the third step we substitute the vertical component of the tension by the data we know by trigonometry

Please enter the following data to calculate the previous equation

Enter β

Enter tension

Enter weight

Enter mass

Enter gravity

Equation 2 result is


Comment:


\begin{align} Fh= &\cssId{Step7}{√(Fhx + Fhy)}\\[3px] \end{align}
  1. In this step, we calculate the module of Fh, using Fhx and Fhy from the previous equations.

Please enter the following data to calculate the previous equation

Enter Fhx

Enter Fhy

Equation 3 result is


Comment: