Solving Horizontal Tension: Step-by-Step Guide

[fldk31]

Hi guys, I came across this question and I was wondering if someone could explain how we get:
TLsinO - mg(1/2LcosO) = 0
Your help would be much appreciated.

 

[axmls]

Do you know how to calculate a torque?

 

[fldk31]

t=fd and t-wa ?

 

[axmls]

What does d represent there?

 

[fldk31]

distance...

 

[axmls]

Not just any distance. It's specifixally the perpendicular distance from the wall. Since the beam is at an angle, you need the trig equations to find the perpendicular distance.

 

[fldk31]

If you understand this, what's the problem to explain it in a couple sentences instead of playing games? If you don't want to explain, it's totally fine.
I am so confused what these two parts mean:
1) TLsinO
2) mg(1/2LcosO)

 

[haruspex]

If you understand this, what's the problem to explain it in a couple sentences instead of playing games? If you don't want to explain, it's totally fine.
I am so confused what these two parts mean:
1) TLsinO
2) mg(1/2LcosO)

If you are familiar with torque then you should know this: if a force F acts through a point P, the torque the force has about a point Q is F x distance PQ x sin(angle between PQ and the direction of the force).
Can you relate that to the diagram and see how the expressions you quote correspond to the torque exerted by T about the hinge, and the torque the weight of the arm exerts about the hinge?