NOVA Physics Torque Example 3

[Caffine]

Now, I'm not the best physics student so this question definitely had me stumped. For those of you who have the book and have made sense of that example, i'd appreciate the help. For those of you who don't have the question, here goes:

A pole of length L is connected to a hinge at point A on a vertical wall, making an angle a with the wall. A horizontal string connects to the wall at point B and the end of the pole at point C. A box of candy of mass m hangs from a string at the end of the pole.

1- what is the tension in the horizontal string?
2- what is the magnitude of the force of the wall on the pole at point
A?

*now the book used point A as the pivot and reached this answer:
T=mg(sin(a)/cos(a))
AND
F= -(mg)/(cos(a)): it goes on to explain that the F is negative because the direction chosen for force of the wall on the pole was incorrect (it was compressional and not tensile).

Can any of you that are good at physics explain this to me?? Thanks a bunch!!

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[Caffine]

If the pivot point is acting at point A, where exactly is the CCW torque? I can understand that the CW torque is provided by: Lsin(90-a)mg but am having a hard time trying to figure out where the CW torque is coming from for this to be a Torque Balancing problem.

They did something wierd:

"We choose A for the pivot, since that choice kills the force Fs but keeps T and mg. Setting the sum of the torques equal to zero gives:

LTsin(90-a) - Lmgsina = 0.
LTcosa - Lmgsina = 0,

T = mg sina/cosa

b. In order to find Fs, we set the sum of vertical forces to zero so that Fnety = 0,

Fscosa - mg = 0
Fs = mg/cosa"

 

[Caffine]

Arghhsafl20kasdfllfadf is how I feeel about this chapter.