How to find Tb and Moments 3D equilibrium of rigid bodies

[mcrooster]

Homework Statement

How to find Tb and Moments 3D equilibrium of rigid bodies

Relevant Equations

Find Tb, F=4kn parallel to xy plane

Find Reaction at Ax Ay and Az

The boom is supported by a ball-and-socket joint at AA and a guy wire at BB

Hey guys, I am stuck with this question in find the Tension in B and the moments around A. I have done plenty of 2d Tension questions but not a 3D one.

 

[haruspex]

not a 3D one.

Well, yes, it's not really a 3D problem.
What is the nett of the two Fs?

 

[mcrooster]

Well, yes, it's not really a 3D problem.
What is the nett of the two Fs?

The component has a x y z, does that not make it 3d with there being 3 axis.

Anyways, the question simply said

If loads F = 4 kN lie in a plane which is parallel to the xy plane, determine the tension in the cable at B

.

 

[Lnewqban]

Welcome, rcrooster

Have you done any work so far that we could see?

Because it has a ball-and-socket joint, which is unable to counteract any moment, any force that is not contained within plane y-z would make the system non-stable.
A single guy wire can only do so much.

 

[haruspex]

The component has a x y z, does that not make it 3d with there being 3 axis.

Anyways, the question simply said .

You did not answer my question. There are two applied forces F, same magnitude but different directions. What is the resultant of those two?

 

[mcrooster]

You did not answer my question. There are two applied forces F, same magnitude but different directions. What is the resultant of those two?

Sorry for the late reply. Would Fr not be the sum of all forces which is 4cos30+4sin30+4cos30+4sin30 = 11

 

[haruspex]

Sorry for the late reply. Would Fr not be the sum of all forces which is 4cos30+4sin30+4cos30+4sin30 = 11

4cos30 is the magnitude of the X component of one of the two Fs, and 4sin30 that of a Y component. And the two X components are in opposite directions along the X axis.
You cannot add magnitudes of vectors if the vectors are in different directions. Do you not know how to add vectors?

 

[mcrooster]

4cos30 is the magnitude of the X component of one of the two Fs, and 4sin30 that of a Y component. And the two X components are in opposite directions along the X axis.
You cannot add magnitudes of vectors if the vectors are in different directions. Do you not know how to add vectors?

My bad. Fr should be 5.65.

(4cos30)^2+(4sin30)^2 and then root the answer.

 

[haruspex]

My bad. Fr should be 5.65.

(4cos30)^2+(4sin30)^2 and then root the answer.

$(4\cos(30))^2+(4\sin(30))^2=4^2\cos^2(30)+4^2\sin^2(30)=4^2(\cos^2(30)+\sin^2(30))=4^2$.
But this is still not how to add vectors.

First, resolve the vectors into a common coordinate system, paying attention to signs.
Using unit vectors based on the coordinates you can write these two F forces as
$4\cos(30)\hat x-4\sin(30)\hat y$ and $-4\cos(30)\hat x-4\sin(30)\hat y$.
Do you understand why those are the appropriate signs?

The two $\hat x$ vectors are parallel so we can add them arithmetically, but we must still honour the signs:
$4\cos(30)\hat x-4\cos(30)\hat x=0$.
And for the y coordinate
$-4\sin(30)\hat y-4\sin(30)\hat y=-8\sin(30)\hat y$.

 

[mcrooster]

I see it now. I was looking at it the wrong way. Completely overlooked the direction in which y was going so it should be a negative. As the X goes along both the positive x-axis and negative x-axis the sum will equal to zero.

So the resultant should be -8sin30y?

 

[haruspex]

I see it now. I was looking at it the wrong way. Completely overlooked the direction in which y was going so it should be a negative. As the X goes along both the positive x-axis and negative x-axis the sum will equal to zero.

So the resultant should be -8sin30y?

Yes.

 

[mcrooster]

Yes.

That gives me a y force of -4kN. This does not give me the answer to Tb. Do I now use the Z and Y axis to figure out my answer since Fx = 0

 

[haruspex]

That gives me a y force of -4kN. This does not give me the answer to Tb. Do I now use the Z and Y axis to figure out my answer since Fx = 0

Yes. As I indicated in post #2, it has reduced to being a 2D problem.