# Why is this length included when finding moment about an axis

1. Nov 30, 2012

### wahaj

1. The problem statement, all variables and given/known data
I have to find reaction at the joints but that's not important. I'm having trouble understanding what the book is doing but first allow me to explain the diagram so there is no confusion because 3D pictures are kinda hard to draw. In the first diagram there is a crank shaft running along the y axis. 0.1 m after point B it goes down 0.2 m along the z axis and then 0.1 m along the y axis.A is a thrust bearing and B is a smooth journal bearing

2. Relevant equations

Statical equilibrium equations apply here

3. The attempt at a solution
When finding the moment about the z axis the equation should look something like this according to me.
Bx(0.5+0.6)- ux x rCA x F = 0
rCA=-0.1i - 1.3j + 0.2k
and ux = i
but here is how the book does it.
Bx(0.5+0.6) - F(0.2+0.1+0.6+0.5) = 0
Now I understand to find the moment about an axis I have to take the force and multiply it by the perpendicular distance between the force and a point on the axis. This works if both the axis and the force lie on one plane. When dealing with multiple planes I have to take the cross product. But I really have no idea why the book is doing what its doing. Can someone tell me if my equation is right or not and also explain why the book simply takes all the distances involved in getting from C to A and adds them without any regard of which axis they follow. Ask for clarification if I did a bad job of explaining.

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2. Nov 30, 2012

### PhanthomJay

Your dimensioning seems off, and I don't know what happened to the 75 kg load, but in any case you can solve by noting that in general , summing all moments about an axis
Mx = Fy(z) + Fz(y), My = Fz(x) + Fx(z), and Mz = Fx(y) + Fy(x). x, y , and z are perpendicular distances. Looks like for solving Bx, solution used last equation with x = 0. No need to fool with cross product.

3. Nov 30, 2012

### wahaj

yes like I said 3d images are hard to draw. don't worry about the load its not important to my question because it doesn't have an x or y component to cause rotation about the z axis. I know how to find the moment about an axis I'm just having trouble with the distances. There are two forces that produce a moment about z axis. The force F and the reaction force at B. I understand how the distance for the reaction force was calculated. but not for F. using the equation you gave me
Mz = Fx(y) + Fy(x) (I'm assuming this is how you meant it)
Since the force F acts only in the x direction Fy is 0 as a result Fy(x) is 0. this leaves me with Fx = F and y = 0.2
so Mz = 0.2F
this is still different than
F(0.2+0.1+0.6+0.5)

4. Nov 30, 2012

### PhanthomJay

It is not a good idea to denote an applied force as F, since F is more of a general term to designate a Force; you should use a different letter like P for example to designate the applied force. But anyway, apparently moments are being summed about joint A to solve for Bx, so the y distance for the applied force F (or P if you will) is the distance along the y axis from P to A, which is difficult to determine from your sketch (looks to me like 0.6 + 0.5 + 0.1 + 0.1, the way you have described it, not drawn it). y is certainly not 0.2, that is a z dimension yielding a moment about the y axis.

Last edited: Nov 30, 2012
5. Nov 30, 2012

### wahaj

Yes the 0.2 is on the z axis my bad I'm kinda frustrated with this. and you got the numbers just right. The moments are being summed about the z axis. The reason A isn't included is because the Z axis passes through it so A can't produce any moment about z. It seems my drawing is more confusing than I thought. I'll wait till monday and ask a professor instead. thanks for trying to help.