Smallest possible tension and its angle

1. Sep 18, 2011

Malki92

1. The problem statement, all variables and given/known data
determine the value of Beta for which the tension in cable B is as small as possible, then find the corresponding tension. Given Alpha = 5 degrees and the weight of the object is 12kN
This is an image of it http://imageshack.us/photo/my-images/801/unledgsn.png/

2. Relevant equations
Fy=0 Fx=0 (at equilibrium)

3. The attempt at a solution
Fx= -TAsin5 + TBcos(Beta)
Fy=TAcos5 - TBsin(Beta) - 12

from Fx i get TA = (TBcos(Beta))/Sin5
and i plug it into Fy to get 12=TB(11.4cos(Beta)-sin(Beta))

but this doesnt answer my first question of what is the angle, only helps with part 2 of the question..

2. Sep 18, 2011

Spinnor

Have you learned about torque? If so measure the torque about some point along the arrow labeled tension in A, say the arrow head of tension A. In equilibrium the sum of the torques will be zero. Then it is clear what the angle beta should be so as to maximize the torque about that point due to the tension in B and thereby minimizing the tension in B.

3. Sep 18, 2011

Malki92

thanks for the reply, but well we havn't studied torque yet so we are supposed to use another way that depends on components of force vectors, we are currently studying statics in particles. so is there a way that is similar the solution Im using?

4. Sep 18, 2011

Spinnor

Let us assume what you wrote is correct,

12=TB(11.4cos(Beta)-sin(Beta))

To make TB minimum we want (11.4cos(Beta)-sin(Beta)) to be maximum. So set

d (11.4cos(Beta)-sin(Beta)) / d beta equal 0 ?

5. Sep 19, 2011

Malki92

i thought so too but then if I use -5 for beta ud get lower tension.. and for -10 you'd get about the same amount as of -5.

6. Sep 19, 2011

Spinnor

Near the angle of minimum tension (-5 degrees), tension varies slowly with change in angle.