Finding the Tensions using the sum of moments

[Isimanica]

Homework Statement

The tower is 70 m tall. If the tension in cable BC is
2 kN, and you want to adjust the tensions in cable
BD and BE so that the couple exerted on the tower
by the fixed support at A is zero.

A) Determine the tensions in cable BD and BE
by applying the sum of the moments about
point A .

The Coordinates in meters
B(0,70,0)
A(0,0,0)
C(-50,0,0)
D(20,0,50)
E(40,0,-40)

Homework Equations

M=Fd

The Attempt at a Solution

Well I know that I have to solve by taking the sum of the moments about A. That means that the Moment about A of Tension BC + Moment about A of Tension BD + Moment about A of Tension BE = 0
and The moment about A of Tension BC is nothing more than the Fx of Tension BC about the distance from point A to B. So That means I find the angle that Tension BC makes to the tower get my Fx and times it by the tower distance. So now what do I need to do next. Looking for the next step.

 

[PhanthomJay]

You are correct so far. Note that the moment about A from the cable tension BC is about the z axis (the axis into the plane), that is, M_z (from the force BC) is (F_x)(y). Also note that for equilibrium, the sum of all moments M_z from all the cable forces must be 0, and the sum of all moments M_x from all the cable forces about the x-axis must be zero. In general, M_z = F_x(y) + (F_y)(x), and M_x = (F_y)z + F_z(y), (and M_y = F_x(z) + F_z(x), which you don't need in this problem, since x and z are 0). So just determine those F_x and F_z components of each cable force, and solve the resulting equilibrium equations for the unknown values.

 

[Isimanica]

So what I need to do to find the Tensions of BE and BD I need to set up the Tension times the Perpendicular distance to point A?
Do I need to take the tensions and place them as if they are only in the x-z plane?
I guess what I am trying to figure out is how do I need to go about to get the distance relative to the tensions?
I can see that I have two equations from M about the z axis and M about the x axis. Requiring the use of the substitution method of one of the moments above.

 

[PhanthomJay]

So what I need to do to find the Tensions of BE and BD I need to set up the Tension times the Perpendicular distance to point A?
Do I need to take the tensions and place them as if they are only in the x-z plane?
I guess what I am trying to figure out is how do I need to go about to get the distance relative to the tensions?
I can see that I have two equations from M about the z axis and M about the x axis. Requiring the use of the substitution method of one of the moments above.

Yes, that's a good way of doing it. Look at the guyed tower in a plan view from the top. Each cable tension BE and BD will have an x and z component, which are geometrically/trigonometrically related. The equations will simplify a bit since the x and z lever arms are 0. When you solve the simultaneous equilibrium of moments equations, you'll get the x and z components of the unknown cable tensions. Then you'll have to use more geometry/trig to get the cable tensions in the guys. It's kind of tough to keep it straight. And watch your plus and minus signs.

 

[DeeNos]

I would suggest that you first draw a free body diagram of the tower to visualize the forces acting on it. This will help you determine the direction of each tension and the distances involved. Then, you can use the equation M=Fd to calculate the moment about point A for each tension. Since the net moment should be zero, you can set up an equation with the moments and solve for the tensions in BD and BE. Additionally, you can also use the equations for equilibrium (ΣF=0) to check your solution. It is important to double check your calculations and make sure they make sense in the context of the problem.