Solve Tension in Cables BC, BD for Simple Statics Problem

[papasmurf]

Homework Statement

Determine the tension in cables BC and BD, and reactions at the ball-and-
socket joint at A for the mast shown in the figure.

Answers are given by the teacher for checking purposes: TC=707 N; TD=1500 N, Ax=0 N,
AY=0 N, AZ=1500 N

Homework Equations

ƩM_x=0
ƩM_y=0
ƩM_z=0
ƩF_x=0
ƩF_y=0
ƩF_z=0

The Attempt at a Solution

Taking clockwise to be the positive moment direction about the x-axis. In other words, if you are standing at the positive side of the x-axis and look down towards the negative side, the moment will be spinning clockwise.
ƩM_x=0= -1kN(6m) + T_BDcos(tan^-1(3/6))cos(45)[kN]*(6m)
...this attempt at a solution is way wrong.

The way I thought about it was, take the cos(tan^-1(3/6)) to get your T_BD in the yz-plane. Next, I took the cos(45) to get it going strictly in the y direction, and then multiplied by the perpendicular distance from the x-axis, 6 m. The problem I'm having is knowing what to do with the given force of 1 kN. There is no specified direction or any information stating where this force is going to. I know how to use the moment and force equations if I can get these things broken up into components correctly.

 

[gneill]

From your rough figure it looks as though force F is anti-parallel to the y-axis (that is, it is parallel to the xy plane and directed towards the -y direction).

If you project all the forces onto the xy-plane then you'll have a pretty straightforward exercise in statics.

 

[rude man]

Right now, without knowing the direction of F I'm dubious as to whether the problem can be solved.

If we assumed point D is along the y axis, just for the sake of argument, then the force summations at point B would be

Fx + Tc/√2 = 0
Fy + Td/√2 = 0

(Summing moments about A gives no new information.)

But √(Fx^2 + Fy^2) = C = 1kN

So we have

Fx + Tc/√2 = 0
√(C^2 - Fx^2) + Td = 0

I can pick any Fx < C, that determines Tc, the Td is also determined. So Fx and Fy are not uniquely defined. If I reorient F in the (presumably) x-y plane, Tc and Td will be correspondingly redistributed to satisfy ƩF = 0.

I guess the question is, since point D is offset from the y-axis into the -x direction, does this impose another condition on F such that its direction is then defined? Stay tuned, got to think about it some more.

 

[rude man]

Looking at it more, I don't think the fact that D is offset into the -x direction changes the aforesaid.

Now we write
ƩFx = 0 → Fx + Tc/√2 + Td•i = 0
ƩFy = 0 → Fy + Td•j = 0
Fx^2 + Fy^2 = 1 kN

where

i,j are unit x and y vectors,
Tc = Tc*ûc
and Td = Td*ûd
Vectors in bold. ûc and ûd are unit vectors pointing from the flagpole top to C and D respectively. Tc and Td are positive scalars.

ûc and ûd are known (check your analyt geometry or calculus book).

Again, I don't see that Ʃmoment about point A will give more info. The moments are just the x and y forces about the y and x axes respectively, times L, the length of the pole.

This means again that, since there are 4 unknowns and only 3 equations, we can choose Fx arbitrarily < 1 kN.

Maybe someone else can look at this and refute my reasoning. Woukldn't be the first time ...

 

[papasmurf]

Rudeman, you had the right idea. The 1kN force is in fact in the y direction. I used the j unit vector for the -1kN force. I used the moment about A, and then I used the i, j, and k components of that to find T(bd) and T(bc). Then I summed up the force components to find the A forces. Thank you all for the helpful repiles.

 

[PhanthomJay]

This means again that, since there are 4 unknowns and only 3 equations, we can choose Fx arbitrarily < 1 kN.

There are just 3 external unknowns at A, C, and D...all members whether tension wires or compression columns are 2-force members, so the problem is statically determinate using just 3 of the 6 equations (the other 3 are superfluous, as you noted).

 

[rude man]

There are just 3 external unknowns at A, C, and D...all members whether tension wires or compression columns are 2-force members, so the problem is statically determinate using just 3 of the 6 equations (the other 3 are superfluous, as you noted).

There were 4 unknowns before it was determined that F was in the -y direction: Fx, Fy, Tc and Td. Defining F = -1kn*j resolved the indeterminacy.