Trouble with elasticity and torque on a body in equilibrium

AI Thread Summary
The discussion revolves around the analysis of forces and torques on a U-shaped metal bar fixed in the ground when a force is applied to widen it. The poster struggles to reconcile the presence of resultant torque with the bar being in equilibrium. Participants suggest that a diagram would clarify the situation and emphasize that not all forces may have been considered. It is noted that the portion of the bar in the ground cannot be treated as a single point, as it contributes to torque regardless of its shape. The conversation highlights the complexity of analyzing equilibrium in systems with elastic deformation.
FG_313
Messages
17
Reaction score
1
Supose we have a ¨U-shaped" metal bar, stuck in the ground, and I aply a force on the top of the part on the right, trying to make it wider. If I were to avaliate the forces acting on a specific part of the bar that includes the point that I´m applying the force, let`s say the portion from that point to half the height of the U-shaped bar, there would be 2 forces on the horizontal direction (ignoring the weight): my force and an elastic force from the metal bar (It will deform a little bit until it matches my force). But, if I were to avaliate Torques acting on that portion of the U-bar, and If I took the reference point on any of the points which the one of the two forces are acting, there would be a resultant torque, which is impossible, since it`s in equilibrium. Where have I got this wrong? And why?
 
Physics news on Phys.org
I think you need a diagram. I'm not sure I understand your problem.

If you have no special software, you could use a simple Paint program and attach a .png file to your post.

avaliate - I think you mean evaluate

I think you are not considering all possible forces. I've quickly added more on a diagram,( though I can't say they are accurate or complete.)
Ubolt.png
 
+1 about a diagram.

I believe the answer is that the part in the ground is not a single "point" that you can take as the origin. Rather there is a "length" of pipe in the ground. What ever point you choose as origin there will be a force acting on the bit in the ground to provide a torque.

This is true even if the bit in the ground is curved like the bottom of the letter u.
 
I have recently been really interested in the derivation of Hamiltons Principle. On my research I found that with the term ##m \cdot \frac{d}{dt} (\frac{dr}{dt} \cdot \delta r) = 0## (1) one may derivate ##\delta \int (T - V) dt = 0## (2). The derivation itself I understood quiet good, but what I don't understand is where the equation (1) came from, because in my research it was just given and not derived from anywhere. Does anybody know where (1) comes from or why from it the...
Back
Top