Smallest possible tension and its angle

[Malki92]

Homework Statement

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/

Homework Equations

Fy=0 Fx=0 (at equilibrium)

The Attempt at a Solution

Fx= -T_Asin5 + T_Bcos(Beta)
Fy=T_Acos5 - T_Bsin(Beta) - 12

from Fx i get T_A = (T_Bcos(Beta))/Sin5
and i plug it into Fy to get 12=T_B(11.4cos(Beta)-sin(Beta))

but this doesn't answer my first question of what is the angle, only helps with part 2 of the question..

 

[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.

 

[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 I am using?

 

[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 ?

 

[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.

 

[Spinnor]

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