Calculate Resultant Couple from 3 Couples in 3D

  • Thread starter Thread starter wannawin
  • Start date Start date
  • Tags Tags
    3d
AI Thread Summary
To calculate the resultant couple from three couples in 3D, the moments can be resolved into their force components without needing specific measurements for 'r', as couples are independent of 'r'. The resultant couple can be represented as a vector perpendicular to the plane of the couples, using the right-hand rule for direction. The calculations yield a resultant magnitude of approximately 4.5012, with angles of 0 degrees to the positive x-axis, 2.91 degrees to the positive y-axis, and 87.1 degrees to the negative z-axis. Attention to sign conventions is crucial, especially regarding the direction of the y component. The discussion emphasizes the importance of clarity in vector direction and magnitude when determining the resultant couple.
wannawin
Messages
14
Reaction score
0

Homework Statement


Replace the three couples with a single resultant couple. Specify its magnitude and direction of its axis using angles to the positive x, y, and z axes.

yphi0.jpg



Homework Equations



M = r x F (r cross F)

The Attempt at a Solution



I'm pretty sure I want to resolve the moments into their force components, but to do that I would need to set an r. But since I have no measurements other than the angle of inclines I don't see how I can do that.
 
Physics news on Phys.org
wannawin said:

Homework Statement


Replace the three couples with a single resultant couple. Specify its magnitude and direction of its axis using angles to the positive x, y, and z axes.

yphi0.jpg



Homework Equations



M = r x F (r cross F)

The Attempt at a Solution



I'm pretty sure I want to resolve the moments into their force components, but to do that I would need to set an r. But since I have no measurements other than the angle of inclines I don't see how I can do that.
The nice thing about couples is that they are independent of 'r'...the moment of a couple about any point is the couple itself. The couple can be represented by a vector pointing perpendicular to the plane of the couple, following the right hand rule method for its direction (+ or - ) perpendicular to the plane. Then solve the resultant couple as a vector located anywhere on the plane, but with a certain magnitude and direction.
 
Then I guess this question isn't that complicated at all. If you number the couples 1-3 going from left to right:
M1x=0
M1y=1.5cos20
M1z=1.5sin20

M2x=0
M2y=1.5
M2z=0

M3x=0
M3y=1.75cos25
M3y=-1.75sin25

Therefore MRx=0
MRy=4.496
MRz=-0.2265

MR has a magnitude of 4.5012

And the angles are 0 with the +x axis, 2.91 with the +y axis, and 87.1 with the -z axis (using simple trig)

Barring any issues with significant digits, I think that makes some sense since the resultant is almost vertical.
 
You may have your plus/- signs mixed...the y component points down, are you calling that the positive y axis? Otherwise, your work is very good.
 
PhanthomJay said:
You may have your plus/- signs mixed...the y component points down, are you calling that the positive y axis? Otherwise, your work is very good.

I did. Good catch and thanks a lot
 

Similar threads

Replies
2
Views
2K
Replies
6
Views
3K
Replies
1
Views
9K
Replies
1
Views
2K
Replies
3
Views
7K
Replies
1
Views
2K
Replies
2
Views
1K
Replies
3
Views
5K
Back
Top