The discussion centers on the equilibrium of rods in various configurations (A, B, C, D) supported by cylinders. Configuration A is definitively not balanced due to the net clockwise torque about the pivot. In configuration B, the rod is supported as indicated by the condition x>d. Configurations C and D require mathematical analysis to determine stability, with D being more likely to achieve equilibrium due to the shifting center of mass. The participants emphasize the importance of using torque calculations and LaTeX for clarity in mathematical expressions.
PREREQUISITESStudents and professionals in physics, engineering, and mechanics who are analyzing the stability of structures and systems involving forces and torques.
If the normal forces from the floor were to exceed the weight the block would rise, yes? What physical law says a block placed gently on the floor won’t, in general, rise off it?ymnoklan said:I don’t think I get what you mean. Could you give me a hint? I mean if the sum of the forces weren’t zero, the system would accelerate, which is NOT what we want here
Newton’s first law says that the sum of the forces must be zero (as it is at rest). Newton’s second law says that the acceleration must be zero (as it is not accelerating) and Newton’s third law says that the opposite reaction to the block’s weight (the normal force from the floor to the block), must be equal in size (but upwards).haruspex said:If the normal forces from the floor were to exceed the weight the block would rise, yes? What physical law says a block placed gently on the floor won’t, in general, rise off it?
What if it were less than the weight?
But why is at rest? You cannot use the fact that the sum of forces is zero because you are using its being at rest to prove that, so your argument becomes circular.ymnoklan said:Newton’s first law says that the sum of the forces must be zero (as it is at rest).
Newton III says the force the block exerts on the floor and the force the floor exerts on the block are equal and opposite. It does not say the weight of the block (the force gravity exerts on the block) equals either of these.ymnoklan said:Newton’s third law says that the opposite reaction to the block’s weight (the normal force from the floor to the block), must be equal in size (but upwards).
Right. Indeed, if the floor were liquid water that would happen. So why doesn’t it happen with a stone or wooden floor? What is different?ymnoklan said:If the normal force was less than the weight, the block would sink through the floor.
I'm confused. Doesn't the sum of forces HAVE to be zero if the box is at rest?haruspex said:But why is at rest? You cannot use the fact that the sum of forces is zero because you are using its being at rest to prove that, so your argument becomes circular.
Good point. Thank you for pointing this out.haruspex said:Newton III says the force the block exerts on the floor and the force the floor exerts on the block are equal and opposite. It does not say the weight of the block (the force gravity exerts on the block) equals either of these.
The solid is rigid compared to the liquid? While water flows and is easily deformed (because the molecules are not held in fixed positions), the solid is made up of molecules that are held tightly together in a fixed structure. I don't think this is what you are hinting to though.haruspex said:Right. Indeed, if the floor were liquid water that would happen. So why doesn’t it happen with a stone or wooden floor? What is different?
Yes, it is at rest if and only if the forces are in balance, but you cannot use that to prove one is true unless you have first proved the other is true by some other means. Otherwise you could apply that to the block on water case and say that should be at rest too.ymnoklan said:I'm confused. Doesn't the sum of forces HAVE to be zero if the box is at rest?
Right.ymnoklan said:The solid is rigid compared to the liquid?
Okay, I think I get it now. Thank you so much for all of your help. I have a similar problem:haruspex said:Yes, it is at rest if and only if the forces are in balance, but you cannot use that to prove one is true unless you have first proved the other is true by some other means. Otherwise you could apply that to the block on water case and say that should be at rest too.
Right.
The block pressing on the floor deforms it. The floor, being rigid, resists deformation and develops an increasing resistance until it is sufficient to prevent further ingress of the block.
In short, we can define the normal force that rigid bodies exert on each other as the minimum force required to prevent interpenetration.
Applying this to case D, it should be clear that if the mass centre of the rod descends significantly then something has to give: bend, break or be crushed. You could make that rigorous through geometry; no need to analyse forces and torques.
This is is the argument I was hinting at in post #10.
https://en.wikipedia.org/wiki/Statically_indeterminateymnoklan said:Now, we introduce a third support somewhere between the two others, for example, at a distance a from the left end and b from the right end (see figure). Is it possible to determine the forces exerted on the rod by each of the three supports?
In the idealisation of rigid bodies and straight lines, it's statically indeterminate: https://en.wikipedia.org/wiki/Statically_indeterminate.ymnoklan said:Okay, I think I get it now. Thank you so much for all of your help. I have a similar problem:
A rod with mass MMM is supported by two supports, one at each end of the rod. Each support exerts a force of Mg/2 on the rod, keeping it at rest.
Now, we introduce a third support somewhere between the two others, for example, at a distance a from the left end and b from the right end (see figure). Is it possible to determine the forces exerted on the rod by each of the three supports?
When I try to solve it, I only get these equations:
F_2*a + F_3*(a+b) = M*g*L/2 (torque balance around left support)
F_1 + F_2 + F_3 = M*g (force balance)
But it stops there?
That was very helpful - thank you!jbriggs444 said:https://en.wikipedia.org/wiki/Statically_indeterminate
If you know the shape of the rod when it is relaxed, its dimensions and the characteristics of the material from which it is constructed then you may be able to compute how it will sag over the three supports and, accordingly, how its weight is distributed over them.
There are well known formulas for the stiffness of a beam that may make this fairly straightforward. However, I am untrained in that area of mechanical engineering.
If one idealizes the rod as being straight and perfectly rigid then the situation is statically indeterminate.
Thank you for your help!haruspex said:In the idealisation of rigid bodies and straight lines, it's statically indeterminate: https://en.wikipedia.org/wiki/Statically_indeterminate.
To solve it you would need much more detail:
- any extent to which the rod is not exactly straight or the supports are not exactly in line
- Young's modulus of the rod
- hardnesses of the surfaces in contact