Componets of Net Torque

[guyvsdcsniper]

Homework Statement:

Problem 12.3 A uniform beam, 2.20 m long with mass m = 25.0 kg, is mounted by a small hinge on a wall as shown in the figure The beam is held in a horizontal position by a cable that makes an angle 0 = 30.0° . The beam supports a sign of mass M = 28.0 kg suspended from Store its end. Determine the components of the force h that the (smooth) hinge exerts on the beam, and the tension FT in the supporting cable.

Relevant Equations:

ΣT=0

I am reviewing this problem from my textbook and am having some trouble with it. When solving for ΣTorque I chose the hinge to be the axis of rotation. I understand that from this point if the board is being held up and it were to fall after removing support, mg (the center of mass of the board) would fall in the clockwise direction making it negative, Mg (the sign/end of the board) would fall in the clockwise direction making it negative as well. What is confusing me is how Fty is positive/being added to the Σtorque equation. For a visual, please refer to the "alternate solution" section at the bottom of this image. This is how my professor approached this problem and was hoping someone could explain why Fty is positive? Is it because it is being supported by FT?

 

[kuruman]

The way I prefer to think about the sign of a particular torque is to say to myself, "If the force that is generating it were the only force acting on the system, will the system rotate clockwise or counterclockwise?" Here F_Ty is up, therefore the torque it generates about the hinge is counterclockwise or positive.

 

[guyvsdcsniper]

The way I prefer to think about the sign of a particular torque is to say to myself, "If the force that is generating it were the only force acting on the system, will the system rotate clockwise or counterclockwise?" Here F_Ty is up, therefore the torque it generates about the hinge is counterclockwise or positive.

That makes sense! Didnt think about Fty being up. I missed that. Thank you!