2
$\begingroup$

I'm thinking about using a pulley system that would allow heavy loads (e.g. a rock) to be lifted with fair ease with the intended goal of harvesting the load's falling energy.

Something that could be lifted over and over.

I think most hydroelectric dams use gravity and the water flow (like a windmill but with multiple turbines instead).

Here's my question:

Would the energy generated from the falling rock be far greater than what is consumed from lifting, and would that be a viable source of energy, theoretically (provided it can be stored efficiently)?

$\endgroup$
5
  • 10
    $\begingroup$ I fail to see the point of this--it's not like you'd get any more energy than just turning a crank. The energy released from the rock falling is always going to be less than the energy used to pull it up. $\endgroup$
    – Hearth
    Jun 25, 2021 at 0:19
  • 2
    $\begingroup$ Are you confusing force and energy? $\endgroup$
    – Hearth
    Jun 25, 2021 at 0:21
  • 7
    $\begingroup$ No matter what you do, you can't get any more energy out than you put in. So there's no real point to the falling rock, just have a crank. $\endgroup$
    – Hearth
    Jun 25, 2021 at 0:42
  • 1
    $\begingroup$ Summary. In an ideal lossless system, energy in a mass m raised h metres above a surface is mgh relative to the surface. G is gravitational acceleration. The energy is the same on raising or lowering and whether using pulleys or levers or any other method. In a non ideal system energy will be lost on lifting and descending - mostly ending up as heat.energy out can NEVER exceed energy in. $\endgroup$ Jun 25, 2021 at 8:13
  • $\begingroup$ You seem to describe the classic clock mechanism of masses and an escapement. But energy in > energy out. $\endgroup$
    – Solar Mike
    Jun 27, 2021 at 4:59

5 Answers 5

2
$\begingroup$

If you are thinking about "free" energy, then that's definitely not a way to achieve it. Although with some types of pulleys you might get half, 1/3, or even less force than the weight you are pulling, you'd be needing to pull the rope, twice, 3 or ever more times.

Basically the work is constant to raise a mass at height h, and its

$$W = m \cdot g \cdot h$$

if you use a pulley that you need to apply half the force (see configuration 2).

enter image description here Figure 1: pulley configurations

then you need to pull a distance of 2h.


However, if your question is

can I use a low power input and slowly raise the weight (using a fraction of the weight of the mass that is being raised), so that I can later on obtain the energy at a much greater rate back?

Then yes. You can use for example a small photovoltaic cell and power a small motor to slowly raise a weight. Then after the weight reached a height you can release the weight and obtain the potential energy that has been accumulated.

$\endgroup$
5
  • $\begingroup$ Amazing answer, by the way. To which I'd quickly follow with: how would the potential energy accumulated compare to the initial energy used by the photovoltaic cell (i have not googled that yet) $\endgroup$
    – OakDEV
    Jun 28, 2021 at 1:34
  • $\begingroup$ I'm not looking to make a breakthrought, just maybe learn two or three concepts that could be usefull for bushcraftiness $\endgroup$
    – OakDEV
    Jun 28, 2021 at 2:20
  • $\begingroup$ Also, I didn't understood that pullies makes it easier but you need to pull for a greater distance. This makes way more sense now and I see the inutility of my question. $\endgroup$
    – OakDEV
    Jun 28, 2021 at 2:24
  • $\begingroup$ @OakDEV the potential energy accumulated will always be less than the initial energy used by the photovoltaic cell. How much less is always a matter of implementation. It can be as high as 90% or more and as low as 1% or less. $\endgroup$
    – NMech
    Jun 28, 2021 at 6:27
  • $\begingroup$ " two or three concepts that could be usefull for bushcraftiness" The problem is that gravity battery can't hold much energy, so unless you have a long rope/string and a very high tree or cliff and/or a gear system with a high ratio you'd probably be better off to "power it up" manually (see GravityLight ) . $\endgroup$
    – NMech
    Jun 28, 2021 at 6:31
4
$\begingroup$

You are describing what is called a gravity battery. See link here:

https://en.wikipedia.org/wiki/Gravity_battery

Short answer no, and the simplest way to understand it is that you can never create energy, you can harvest it, store it, and use it, change its form, but never create it. This is called the law of conservation of energy:

https://en.wikipedia.org/wiki/Conservation_of_energy

Therefore you can never put energy into lifting a rock and expect to somehow get energy out that is greater than you put in. If you have a perfectly frictionless system then at best you can hope to get the same energy out you put in, but due to friction losses you will have some loss.

This is why a gravity battery is called a battery, it effectively will allow you to store energy in it at best.

Now here is the catch, you mention a pulley system, which does not change anything I said above and there is some real simple intuition you can use to understand why. A pulley does indeed make it easier to lift the rock, but how easy it is isn't really telling you anything about energies, that is more about force. If you have a single pulley then you might reduce the force needed by half, but you also now need to pull twice as much rope to achieve the same height. So while it feels easier you have to do it for longer, and ultimately these two balance out to mean you're going to have to put in the same energy regardless of how many pulleys you use to get that rock to the same height.

So no, a pulley wont in any way change the amount of energy your putting into that rock per meter you get it off the ground.

$\endgroup$
2
  • 1
    $\begingroup$ Thats why i used the word "gather". I just assumed that since lifting a 100 pound rock with pulleys takes little physical effort, it'd be a great way to use gravity and redirection of force during pulling to harvest the velocity/force generated from the fall of that object. I remember seing the classic "guy on a stationary cycle generating energy" in movies. Energy is the biggest market worldwide, so I'd be a fool to think I found the golden formula. I'm just trying to understand how is it impossible to use pulleys to my advantage when harvesting gravitational energy $\endgroup$
    – OakDEV
    Jun 25, 2021 at 2:19
  • 1
    $\begingroup$ You can use pulleys to your "advantage" when harvesting energy.. what you cant do is get more energy out of the rock than you put into the pulley. If it is a human pulling the rock up into the air then we would probably use less food energy with a reasonable number of pulleys (there is likely some magic number). But that isnt because the pullies are adding more energy into it, but rather the human isnt wasting as much energy. Consider that holding a rock at a constant height takes energy from a human holding a rope, but he isnt putting energy into the rock just holding it there, its wasted. $\endgroup$
    – Jeffrey Phillips Freeman
    Jun 25, 2021 at 2:26
2
$\begingroup$

Would the energy generated from the falling rock be far greater than what is feeled when lifting AND would that be a viable source of energy, theoretically (Provided it can be store efficiently)?

Lifting a heavy rock with a pulley system would make it 'feel' much lighter, and could also be more efficient than trying to lift it directly. However the energy you extract from dropping it will always be less than the energy you expend to lift it - a lot less.

The efficiency of human muscle is around 18% to 26% depending on how it is used. That may seem low, but its actually about the same as a typical car engine. A pulley system will lose some energy to friction etc., but may be a net gain if it allows your muscles to work more efficiently and/or without getting tired. On the extraction side an electrical generator could be 80% to 90% efficient, so it would be a minor contributor to the total loss.

As to whether this system could be a viable source of energy, that depends on the intended application. If you have a situation where a human is able to supply the required energy but it needs to be stored and used at some later time (perhaps more or less rapidly), and you have sufficient space etc. to house the contraption, and it is safe, then it could be useful.

No stored energy system can ever be 100% efficient because there are always losses, and of course you can never get more out than you put in. But efficiency is often less important than other factors such as cost and practicality. If the device costs more to manufacture and operate than the cost of the energy it produces, people may be more inclined to use cheaper and more convenient alternatives. However they may be attracted to it for other reasons, such as being carbon-neutral, or using energy that would otherwise be wasted (eg. an exercise machine), or just for the 'cool' factor.

$\endgroup$
2
$\begingroup$

I'm thinking about using a pulley system that would allow heavy loads (e.g. a rock) to be lifted with fair ease with the intended goal of harvesting the load's falling energy.

enter image description here

You seem to have forgotten that while the force required to lift the mass is halved by the pulley that the distance the rope has to be pulled has been doubled because of the same pulley.

If there were no friction losses then the recoverable energy = the energy required to lift the weight.

  • Energy gained by raising the weight $ = mgd $.
  • Energy used on pulling rope $ = \frac m 2 \times g \times 2d = mgd $.

"You can't win. You can't even break even!"

$\endgroup$
5
  • 1
    $\begingroup$ It seems the OP was asking the energy spent to lift the rock (no matter using the pulley or not), and the energy gained from dropping it from the lifted height. Will the energy produced by the free fall and impact be greater than the energy spent? (similar to the falling water that pushes the turbine blades to generate power) $\endgroup$
    – r13
    Jun 26, 2021 at 13:55
  • 1
    $\begingroup$ @r13, let's see if the OP comes back. $\endgroup$
    – Transistor
    Jun 26, 2021 at 14:34
  • 1
    $\begingroup$ Ok. I think his case is identical to the hydroelectrical pumped-storage powerhouse, which generating power, during the high power usage period in a day, using water stored in the upper reservoir, then lifts the used water from the lower reservoir back to the upper reservoir during the low demand period. I know the advantage is based on economic consideration, but don't know the difference between the power generated and consumed. Maybe one of your guys can shed the light on it. $\endgroup$
    – r13
    Jun 26, 2021 at 14:52
  • $\begingroup$ To answer to both of you, you're both right and wrong. I meant both, so it was originally just about pullies but then I also wanted to know about the gravitational aspect of it. My case is exactly identical to the hydro-electricity powerhouses. Just on a more practical scale with a rock and maybe a small river. The real goal is to potentially generate the maximum amount of power from the surrounding area where I have a couple acres in the woods but I had completely forgotten that the pulling distance is not magically the same, it's just ditributed for a smoother pull. $\endgroup$
    – OakDEV
    Jun 28, 2021 at 2:30
  • $\begingroup$ Then again, mixing all that, I still don't know if, say a cow gradually moving a big rock up a thing for 10 minutes (or maybe with water flow) and then free fall + impact energy collected by X method would be worth the setup. If I can generate more power with a couple solar panels I guess that'd be smarter thing. $\endgroup$
    – OakDEV
    Jun 28, 2021 at 2:34
1
$\begingroup$

Joule performed an experiment which demonstrated the equivalence of work and heat. The First Law of Thermodynamics is based on the equivalence of work, energy, and heat as follows:

$$W = \delta E + Q$$

where in this case work W is done to raise the weight against the force of gravity, the change in energy of the system is the change in gravitational potential energy, and heat is dissipated in the pulley system as a quantity of heat Q.

Gravitational potential energy near the surface of Earth:

$$GPE = mgh$$

where this is the mass of the object times the standard acceleration of gravity times the height above a zero reference.

The minimum force necessary to lift the object against gravity is the given by the weight:

$$F = -w = -mg$$

where the weight is a downward force so the pulley system must exert this upward force to lift the weight. Mechanical work is force times displacement so this means the minimum amount of work performed to lift the weight would be the same as the GPE stored in the weight at a specified height h. Friction in the pulley would mean more work is performed to lift the weight than can be recovered by dropping the weight and attempting to store the energy in some other form.

Suppose the pulley system enables a user to exert 1/10 the weight force. The pulley system must still provide an output lift force equal to the weight. If the weight goes up 1 unit of distance then the user must exert 1/10 the force for 10 units of distance. No gain in work or energy is made via use of the pulley system. Friction causes a loss of energy so extra work must be done to generate heat dissipated via friction in the pulley system and this non-conservative work (against friction) cannot be recovered by lowering the weight.

If heat loss Q = 0 for an ideal system then one would perform the same amount of work to increase the GPE of the system as the system would perform when lowering the weight to store some other form of energy. No work or energy could be gained by using the pulley system except to make it possible to lift a heavy object using reduced force acting through a greater distance. Mechanical work is force acting through a distance. The pulley or other machine does not provide any gain of energy or power from input to output.

$\endgroup$
1
  • 1
    $\begingroup$ I can't grasp certain notions you mentionned on the fly. I'll read about it. If my logic isn't too buggy, this means that if someone was to manually lift a rock without the help of tools... the energy used to lift it would be far greater than was is generated when falling/dropping. Otherwise there wouldnt be a resounding "no" surrounding the actual energy input/output and the loss of energy via frition. (?) $\endgroup$
    – OakDEV
    Jun 25, 2021 at 1:50

Not the answer you're looking for? Browse other questions tagged or ask your own question.