What Are the Torques on a Rectangular Plate in Equilibrium?

AI Thread Summary
The discussion focuses on calculating torques on a rectangular plate in equilibrium, specifically around points A, B, and a point C located halfway up the rod. It emphasizes that to achieve equilibrium, one screw must exert a pulling force while the other exerts an equal pushing force. The participants clarify that torques can be calculated around any point, including point C, which yields a torque of ½bmg due to the plate's weight. The conversation highlights that the forces at points A and B will yield consistent torque values regardless of the chosen axis of rotation. Understanding this principle is crucial for analyzing the equilibrium of the system effectively.
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: 645
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
 
I multiplied the values first without the error limit. Got 19.38. rounded it off to 2 significant figures since the given data has 2 significant figures. So = 19. For error I used the above formula. It comes out about 1.48. Now my question is. Should I write the answer as 19±1.5 (rounding 1.48 to 2 significant figures) OR should I write it as 19±1. So in short, should the error have same number of significant figures as the mean value or should it have the same number of decimal places as...
Thread 'Collision of a bullet on a rod-string system: query'
In this question, I have a question. I am NOT trying to solve it, but it is just a conceptual question. Consider the point on the rod, which connects the string and the rod. My question: just before and after the collision, is ANGULAR momentum CONSERVED about this point? Lets call the point which connects the string and rod as P. Why am I asking this? : it is clear from the scenario that the point of concern, which connects the string and the rod, moves in a circular path due to the string...
Thread 'A cylinder connected to a hanging mass'
Let's declare that for the cylinder, mass = M = 10 kg Radius = R = 4 m For the wall and the floor, Friction coeff = ##\mu## = 0.5 For the hanging mass, mass = m = 11 kg First, we divide the force according to their respective plane (x and y thing, correct me if I'm wrong) and according to which, cylinder or the hanging mass, they're working on. Force on the hanging mass $$mg - T = ma$$ Force(Cylinder) on y $$N_f + f_w - Mg = 0$$ Force(Cylinder) on x $$T + f_f - N_w = Ma$$ There's also...
Back
Top