Magnitude of moment about axis (Space Truss)

Click For Summary
SUMMARY

The discussion focuses on calculating the moment of force F = {40i -30j -100k} N about the axis CA of a tripod ABCD. Tim attempts to find the unit vector UCA and the moment MCA using the position vector rCA, but questions the accuracy of his position vector rDC. A participant points out that Tim's rCA calculation is incorrect, emphasizing the correct coordinates for points A and C in relation to the origin.

PREREQUISITES
  • Understanding of vector mechanics and moments
  • Familiarity with unit vectors and their calculations
  • Knowledge of position vectors in three-dimensional space
  • Proficiency in using vector notation and operations
NEXT STEPS
  • Review the calculation of unit vectors in three-dimensional mechanics
  • Study the principles of moments and their applications in static equilibrium
  • Learn how to derive position vectors from given coordinates
  • Explore examples of calculating moments about different axes in truss structures
USEFUL FOR

Students and professionals in engineering, particularly those studying statics and dynamics, as well as anyone involved in structural analysis of trusses.

U1021981
Messages
3
Reaction score
0

Homework Statement



Determine the magnitude of the moment of the force F about the axis CA of the tripod
ABCD, if F = {40i -30j -100k}N.

Image attached.

2. The attempt at a solution

Based on a similar question I found in the textbook my method to solve this question:

Find UCA = rCA/rCA = (-2/√8)i + (2/√8)j

Find MCA = UCA.(rDC?? * F)

However, I am unsure if the finding the position vector rDC is correct?

Am I going about this the right way?

Please help.

Thanks,

Tim
 

Attachments

  • Q6.35.jpg
    Q6.35.jpg
    12.7 KB · Views: 928
Physics news on Phys.org
U1021981 said:

Homework Statement



Determine the magnitude of the moment of the force F about the axis CA of the tripod
ABCD, if F = {40i -30j -100k}N.

Image attached.

2. The attempt at a solution

Based on a similar question I found in the textbook my method to solve this question:

Find UCA = rCA/rCA = (-2/√8)i + (2/√8)j

Find MCA = UCA.(rDC?? * F)

However, I am unsure if the finding the position vector rDC is correct?

Am I going about this the right way?

Please help.

Thanks,

Tim
Here's your image:

attachment.php?attachmentid=50244&d=1346128205.jpg


First of all your rCA must be incorrect. Look closely. Point A is 1.5 m from the origin on the y-axis. Point C is 2 M from the origin on the x-axis.
 

Similar threads

  • · Replies 9 ·
Replies
9
Views
2K
  • · Replies 4 ·
Replies
4
Views
3K
  • · Replies 5 ·
Replies
5
Views
7K
  • · Replies 4 ·
Replies
4
Views
2K
  • · Replies 2 ·
Replies
2
Views
3K
Replies
2
Views
2K
  • · Replies 1 ·
Replies
1
Views
2K
  • · Replies 48 ·
2
Replies
48
Views
8K
  • · Replies 2 ·
Replies
2
Views
3K
  • · Replies 1 ·
Replies
1
Views
4K