What is the magnitude of the horizontal force Hx on the pivot?

Question

A uniform bar of length 10 m and mass 9 kg is attached to a wall with a hinge that exerts on the bar a horizontal force Hx and a vertical force Hy. The bar is held by a cord that makes
a 90± angle with respect to bar and angle 50 degree with respect to wall. The acceleration of gravity g = 9.8m/s^2
What is the magnitude of the horizontal force Hx on the pivot? Answer in units of N.

Answer 1

Start by summing torques at the hinge to find the tension in the cord, T

T*10=m*g*cos(50)*5
T=m*g*cos(50)/2

The only horizontal forces are the horizontal component of T and the Horizontal component of the rod on the hinge
Therefore
sin(50)*T=Hx
Hx=sin(50)*cos(50)*m*g/2
=sin(50)*cos(50)*9*9.81/2

j

Answer 2

I trust John's reasoning yet not his calculation: (4.2 + a million.4) * .3 = a million.sixty 8 N, i think of. How ought to it take 18 Newtons to tug 2 tiny weights which includes 4.2 and a million.4 Newtons? How ought to the stress necessary be extra advantageous than the burden of the products?