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Load force vs. Effort force

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edy:
I wouldn’t necessarily argue with the teacher, but I would teach my kids that there are other ways to think about the problem. The main thing is to get the answer correct and make sure I can follow that their reasoning is correct. I can’t stand when they get some math homework and they are told to do things a certain way that doesn’t really teach them anything they can apply to anything else. The idea of learning fundamentals is that understanding that lets you construct all the more complicated systems that depend on it. If they learn some strange way to do something that they can’t reason from more fundamental building blocks then they are just robots doing things they can’t understand and will eventually get in trouble obtaining the correct answers when they hit some unexpected situation.

edy:

--- Quote from: IanB on April 01, 2022, 07:33:40 pm ---To illustrate how complex this subject can be, let me offer an example.

Suppose we have a 1 kg weight resting on a table. We can say it is exerting a downward force on the table of 9.8 N. But at the same time, the table is pushing up on the weight with a force of 9.8 N. Since the two forces are equal and opposite, we can deduce that the net force on the weight is zero. We can verify that the net force is zero since we know that force equals mass times acceleration. Since the weight is clearly stationary and not accelerating, the net force on the weight must be zero.

Now let's suppose we lift the table very, very slowly through a height of 1 m. We could say that work was done on the weight, and it has gained a potential energy of 9.8 Nm (force times distance). But we also know the net force on the weight was zero, and we know that since we moved the table ever so slowly there was no appreciable acceleration. So if the net force on the weight was zero, and zero was moved through 1 m, how was any work done? Where did the potential energy come from?

Let's suppose we want to explain this to a child. How could we do that?

--- End quote ---

The net force on the weight is not zero because it is moving up. The force that the table exerts on the weight may be 9.80001 N, even though the weight on the table is 9.8 N. That differential means the weight translates vertically. By the time you stopped moving the table you are back to 9.8 N and it equals out the weight again so no movement, but you are 1 m higher than where you started. It was close to zero but not quite zero, and over time that small amount integrated itself to the full work done. Same if you drop the weight slowly, by moving table down you are supporting the weight by 9.7999999 N while it still pushes down with 9.8 N, so that 0.0000001 N differential is enough to make the weight drop.

Now I know at the beginning and ends of movement there is more and less force felt by the weight due to acceleration/deceleration but during steady state movement (most of the time) there is a stable force differential leading to steady translation at a fixed speed of the weight up and down. The acceleration/deceleration effects would cancel out and appear as the “limit” case in which almost no acceleration and deceleration occur and it is just moving…. Force x distance. At least that’s how I understand it. Is it too hard for a kid to understand this? Depends.

IanB:

--- Quote from: edy on April 01, 2022, 10:09:26 pm ---The net force on the weight is not zero because it is moving up. The force that the table exerts on the weight may be 9.80001 N, even though the weight on the table is 9.8 N. That differential means the weight translates vertically. By the time you stopped moving the table you are back to 9.8 N and it equals out the weight again so no movement, but you are 1 m higher than where you started. It was close to zero but not quite zero, and over time that small amount integrated itself to the full work done.
--- End quote ---

However, the formula for work done is the integral of force over distance, not the integral of force over time. If we have an infinitesimal force of 0.0001 N integrated over a distance of 1 m, that is 0.00001 joules not 9.8 joules.


--- Quote ---Same if you drop the weight slowly, by moving table down you are supporting the weight by 9.7999999 N while it still pushes down with 9.8 N, so that 0.0000001 N differential is enough to make the weight drop.

Now I know at the beginning and ends of movement there is more and less force felt by the weight due to acceleration/deceleration but during steady state movement (most of the time) there is a stable force differential leading to steady translation at a fixed speed of the weight up and down.
--- End quote ---

But Newton's laws of motion say that a body moving at a constant velocity experiences no inertial forces, so in steady state movement the net force on the weight is zero.


--- Quote ---The acceleration/deceleration effects would cancel out and appear as the “limit” case in which almost no acceleration and deceleration occur and it is just moving…. Force x distance. At least that’s how I understand it. Is it too hard for a kid to understand this? Depends.

--- End quote ---

Clearly gravity comes into play here, but now we have to deal with the physics of conservative fields and forces on bodies moving within those fields.

One can say the weight is experiencing a force due to gravity, but if that is the case, why isn't it accelerating? Why is it sitting in stationary fashion on the table?

Is there a force, or is there not a force?

To go back to the top of the thread, if I am pulling on the rope of a pulley, am I pulling down, or is the rope pulling up?

RJSV:
EDY:
   Immediately, THIS image popped into my head:
Suggest:. Place yet another 'U' shaped rope, attached the same way, so now you have Weight / 3 on each rope. But that in itself supports your argument, as that 'shared' weight is meaningless, by itself, and simply does not work, as that third rope with pulley becomes immediately separated, as you lift, and would just sit there. I suppose, maybe, that 'two handed pull' done by grabbing BOTH free ends and lifting, perhaps would be two-thirds perceived lifting, while the...yeah uh no...
I'm thinking, a second rope/pulley would not change any of the ratios ! They are, essentially in parallel.
(This is gonna bug me).
You're on the right track though, it's needing the distance covered, while pulling, to fill out the 'Work' calculation properly.

Someone:

--- Quote from: edy on April 01, 2022, 03:49:41 pm ---Questions are 1. What is a load force? 2. What is an effort force? 3. How much effort force is needed to lift 300 kg weight in fixed pulley system, 4. …movable pulley system? 5. In a movable pulley system, where does the other effort come from?
--- End quote ---
Is that verbatim? Perhaps a photo/scan of the exact question and any illustrations/images it has with it would help.

Based on precisely what you wrote: Answer 5) Effort /= Force

The terminology is consistent and commonplace:
https://www.engineeringtoolbox.com/pulleys-d_1297.html


--- Quote from: IanB on April 01, 2022, 06:33:11 pm ---We have to remember that some things are not really easy to understand at 9 or 10 years old, and there is no point getting too technical before students have enough knowledge to appreciate the details.
--- End quote ---
Some 10 year olds will happily grasp statics and vectors, but they're not concepts most would be able to understand. Hence the baby steps. I checked some physics texts, and they completely ignore statics/levers (leave that for the engineering texts) and start with dynamics instead.


--- Quote from: edy on April 01, 2022, 03:49:41 pm ---Thank you all for your thoughts on this primary school handout and please let me know if I am just over-thinking this or is this just going to confuse my kids and they will have to unlearn this so they can understand it properly later on? I want to teach it properly in a way that can be generalized to all types of machines and my kids understand the fundamentals properly, not just memorize a bunch of stupid examples that give half-arsed explanations that confuse them.
--- End quote ---
You are over thinking it! A question is from it's defined assumptions/definitions (which we have not seen in full). Yes, using the word work or effort is muddling the water but only if you already associate specific technical definitions to those words, introduced much later in the education system. Its not practical to avoid all possibly technical words when explaining basic ideas to beginners.

I checked some texts here that label similar problems with simple letter designations, and talk about differentiating forces (for statics) based on applied/inherent external/internal acting/reacting.

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