How to find the Tension in a support cable?

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To find the tension in the support cable of a wrecking ball system, it is essential to set up a free body diagram (FBD) and apply the equations for equilibrium: the sum of vertical forces (ΣFv = 0), the sum of horizontal forces (ΣFh = 0), and the sum of moments (ΣM = 0). The tension in the cable must be resolved into its horizontal and vertical components, Tcos32 and Tsin32, respectively. When calculating torques, select a pivot point and consider the moments created by all forces acting on the boom, excluding the tension itself. The angle between the boom and the cable should be correctly identified to ensure accurate calculations. Understanding these principles will help in determining both the tension in the cable and the force exerted at the hinge.
MKM
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A wrecking ball (weight = 5910 N) is supported by a boom, which may be assumed to be uniform and has a weight of 2820 N. As the drawing shows, a support cable runs from the top of the boom to the tractor. The angle between the support cable and the horizontal is 32°, and the angle between the boom and the horizontal is 48°. Find (a) the tension in the support cable and (b) the magnitude of the force exerted on the lower end of the boom by the hinge at point P.
I have been having trouble trying to finf the tension, not sure how I originally thought to just do the sum of the torques but it's not coming out. I thougit was just T = 3/4 sin 90 ? very confused?
 

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You need to construct a FBD, then do,

ΣFv = 0
ΣFh = 0
ΣM = 0

where M is a moment, or torque.
 
This is how I set up my equation Torque sum =
-2820N (L/2)sin 138
+ T(3/4)L sin 90
- 5910sin138 =0 and I get 6530 ?
What I realized later was that I think I'm supposed to use the 32 angle for the tension force in some way but I'm not sure. I hope m attachment isn;t pending for very long it's a good description
 
I think I have an idea of what your attachment will look like.
Is it like mine ?

http://img476.imageshack.us/img476/4256/mkm10sa.jpg

If so, then the FBD could be something like my 2nd attachment

http://img476.imageshack.us/img476/5760/mkm26qe.jpg
 

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wow It's exactly like your second attachment . I actually had the Tension facing outward perpendicular to the support cable but maybe I was wrong.
 
MKM said:
wow It's exactly like your second attachment . I actually had the Tension facing outward perpendicular to the support cable but maybe I was wrong.

Tension is always in the direction of the cable (or whatever else is providing the tension).
 
okay I have evrything set up but I dont; understnad how to aply the tension formula which I think is T= mv squared/ L. Also when I am looking for the sum of the torques should I just add the tension to the other forces and find Rx and Ry?
 
MKM said:
okay I have evrything set up but I dont; understnad how to aply the tension formula which I think is T= mv squared/ L. Also when I am looking for the sum of the torques should I just add the tension to the other forces and find Rx and Ry?
I don't understand your tension formula. What is the v ? There is no velocity involved.

Do what I suggested in my first post. Do a balance of forces, both horizontal and vertical. Then do a balance of moments (or torques).

ΣFv = 0
ΣFh = 0
ΣM = 0

When doing the sum of torques, Pick a point, A say, and take the moments of every force about this point. All the forces are listed in Fig2. Remember that the tension T, in Fig2, should be ignored since it has been resolved into horizontal and vertical components, Tcos32 and Tsin32 respectively. And it is these two forces that should have their moments taken, not T.
 
should the angle between the boom and the cable be 48-32=16?
 
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