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 degree angle with respect to bar and angle 50 degree with respect to wall. The acceleration of gravity is 9:8 m/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=m*g*cos(50)

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

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Answer 2

I accept as true with John's reasoning yet not his calculation: (4.2 + a million.4) * .3 = a million.sixty 8 N, i think of. How could desire to it take 18 Newtons to tug 2 tiny weights including 4.2 and a million.4 Newtons? How could desire to the strain mandatory be greater suitable than the burden of the gadgets?