What Are the Torques on a Rectangular Plate in Equilibrium?

Click For Summary
SUMMARY

The discussion focuses on calculating torques on a rectangular plate in equilibrium, specifically around points A, B, and C. It is established that to achieve equilibrium, one screw must exert a pulling force while the other exerts a pushing force with equal horizontal magnitude. The torque around point C, located halfway up the pole, is calculated as ½bmg, demonstrating that any point can serve as an axis for torque calculations. This reinforces the principle that torques can be evaluated from multiple axes without altering the fundamental values of the forces at points A and B.

PREREQUISITES
  • Understanding of torque and its calculation in physics
  • Familiarity with the concepts of equilibrium and forces
  • Basic knowledge of rotational dynamics
  • Ability to interpret diagrams and apply them to physical scenarios
NEXT STEPS
  • Study the principles of static equilibrium in mechanics
  • Learn about the calculation of torques in complex systems
  • Explore the concept of center of mass and its effect on torque
  • Investigate the application of torque in real-world engineering problems
USEFUL FOR

Students of physics, mechanical engineers, and anyone interested in understanding the principles of torque and equilibrium in physical systems.

zezima1
Messages
119
Reaction score
0
Consider the system on the attached picture showing a rectangular plate attached to a pole by two screws - one in A and one in B. Now to figure out the condition for equilibrium one would calculate the torques around point A and B to find that one screw must be pulling and the other one pushing the plate with a horizontale force of equal magnitude.

Now my problem is that I have never really understood this whole idea, that you should be able to get zero torque around every possible rotation axis. Does that then mean that you can also calculate the torque around for instance a point C halfway up the rod?
If so let's try do that. I marked C on the picture. Then you get a torque equal to ½bmg. But wouldn't that then mean that the force in point A is creating an extra torque as seen from a point of view where C is that axis of rotation?
 

Attachments

  • Unavngivet.png
    Unavngivet.png
    4.1 KB · Views: 667
Physics news on Phys.org
zezima1 said:
Now my problem is that I have never really understood this whole idea, that you should be able to get zero torque around every possible rotation axis. Does that then mean that you can also calculate the torque around for instance a point C halfway up the rod?
Sure. You can use any point for calculating torques. (Some may prove easier than others.)
If so let's try do that. I marked C on the picture. Then you get a torque equal to ½bmg.
That's the torque about C due to the weight of the plate. Now include the torques due to the forces at A and B.
But wouldn't that then mean that the force in point A is creating an extra torque as seen from a point of view where C is that axis of rotation?
Not sure what you mean. You'll get the same values for the forces at A and B as you would have using any other point as your axis.
 
Hmm I got it now, I wasn't thinking properly. Thanks
 

Similar threads

  • · Replies 8 ·
Replies
8
Views
2K
Replies
6
Views
3K
  • · Replies 7 ·
Replies
7
Views
2K
  • · Replies 3 ·
Replies
3
Views
2K
  • · Replies 5 ·
Replies
5
Views
2K
  • · Replies 11 ·
Replies
11
Views
2K
  • · Replies 42 ·
2
Replies
42
Views
4K
  • · Replies 25 ·
Replies
25
Views
2K
Replies
6
Views
2K
  • · Replies 14 ·
Replies
14
Views
3K