Moment of forces about the axes

  • Context: Engineering 
  • Thread starter Thread starter mingyz0403
  • Start date Start date
  • Tags Tags
    Axes Forces Moment
Click For Summary
SUMMARY

The discussion focuses on resolving moments of forces Ma and Mb into their respective y and z components. The calculations yield Ma = 1200sin(20)j + 1200cos(20)k and Mb = 900sin(20)j + 1200cos(20)k. The correct interpretation of the signs for these components is clarified using the right-hand rule, which indicates that the moment vector's direction determines its sign. Specifically, the A-torque is negative in the +y direction while the B-torque is positive in the +y direction.

PREREQUISITES
  • Understanding of vector resolution in three-dimensional space
  • Familiarity with the right-hand rule for determining vector directions
  • Knowledge of torque and moment of forces in physics
  • Basic trigonometry for calculating sine and cosine values
NEXT STEPS
  • Study the application of the right-hand rule in vector mechanics
  • Learn about torque calculations in three-dimensional systems
  • Explore vector projections onto different axes
  • Investigate the implications of sign conventions in physics
USEFUL FOR

Students and professionals in physics, mechanical engineering, and anyone involved in analyzing forces and moments in three-dimensional systems.

mingyz0403
Messages
12
Reaction score
1
Homework Statement
Shafts A and B connect the gear box to the wheel assemblies of a tractor, and shaft C connects it to the engine. Shafts A and B lie in the vertical yz plane, while shaft C is directed along the x axis. Replace the couples applied to the shafts by a single equivalent couple, specifying its magnitude and the direction of its axis.
Relevant Equations
M=RXF
I reslove Ma and Mb into y and z component.
Ma=1200sin(20)j+1200cos(20)k
Mb=900sin(20)j+1200cos(20)k
Mc=-840i

I looked at the solution and it states that the y component of Ma is negative (-1200sin(20)j+1200cos(20)k). I understand that Mc is -840i because it is Clockwise. How do you determine the sign of Ma and Mb for their y and z component.

Many times for your time answering my question.
 

Attachments

  • 1.jpg
    1.jpg
    21.5 KB · Views: 281
Physics news on Phys.org
It is best to forget about sign conventions for clockwise and counterclockwise. Rather, show the couple in its vector form using the right hand rule , and the moment vector will point in the direction of your thumb. Then you should be able to more easily find the projection and signage of the vector Components onto the y and z axes. Hint: the couple about the x-axis is negative because it points in the negative x direction.
 
  • Like
Likes   Reactions: FactChecker
The A-torque is tilted slightly down, in the -y direction. Using the right-hand rule, it is positive when looked at in the -y direction. That is the same as negative in the +y direction. The B-torque is tilted slightly up in the +y direction.
 

Similar threads

  • · Replies 5 ·
Replies
5
Views
4K
  • · Replies 1 ·
Replies
1
Views
3K
  • · Replies 22 ·
Replies
22
Views
3K
  • · Replies 4 ·
Replies
4
Views
2K
  • · Replies 9 ·
Replies
9
Views
928
  • · Replies 1 ·
Replies
1
Views
4K
  • · Replies 1 ·
Replies
1
Views
7K
  • · Replies 5 ·
Replies
5
Views
7K
  • · Replies 18 ·
Replies
18
Views
10K
  • · Replies 4 ·
Replies
4
Views
2K