Which of these rods are in equilibrium?

[haruspex]

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

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?

 

[ymnoklan]

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?

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).
If the normal force was less than the weight, the block would sink through the floor.
Is this what you mean? Or have I misunderstood something again?

 

[haruspex]

Newton’s first law says that the sum of the forces must be zero (as it is at rest).

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.

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).

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.

If the normal force was less than the weight, the block would sink through the floor.

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?
It might help to think in a bit more detail about the process. All solid bodies have some springiness. Suppose you were to hold the block so that it is just touching the floor, no forces yet, then let go. It will descend a tiny bit, compressing the surfaces of both floor and block, but soon stops. How were the forces changing as it was making that descent?

 

[ymnoklan]

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.

I'm confused. Doesn't the sum of forces HAVE to be zero if the box 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.

Good point. Thank you for pointing this out.

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?

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]

I'm confused. Doesn't the sum of forces HAVE to be zero if the box is at rest?

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.

The solid is rigid compared to the liquid?

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.

 

[ymnoklan]

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.

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?

 

[jbriggs444]

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?

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.

 

[haruspex]

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?

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

 

[ymnoklan]

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.

That was very helpful - thank you!

 

[ymnoklan]

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

Thank you for your help!