[vkrmvkrm 11 (OP)]

As per sketch there is a box having a 10 kg.ball and counterweight mass is 10.2. This seesaw is slightly tilted towards counterweight. Now I want to know how much energy will be required as a input if I liftup the arm of 10 kg ball at 90 degree angle but remember the ball will be released only after when the arm get a certain height at 90 degree angle. The box is mounted at 45 degree angle Counter weight is permanently mounted The arm is 1 meter long The box length is 1 meter.
I would like to tell that the ball will get back it's initial position due to the seesaw so there will be no need of extra energy in it.

[guest39538]

As per sketch there is a box having a 10 kg.ball and counterweight mass is 10.2. This seesaw is slightly tilted towards counterweight. Now I want to know how much energy will be required as a input if I liftup the arm of 10 kg ball at 90 degree angle but remember the ball will be released only after when the arm get a certain height at 90 degree angle. The box is mounted at 45 degree angle Counter weight is permanently mounted The arm is 1 meter long The box length is 1 meter.
I would like to tell that the ball will get back it's initial position due to the seesaw so there will be no need of extra energy in it.

I have no idea by your explanation or drawings what you are trying to do. Please explain.

[vkrmvkrm 11 (OP)]

] ]https://youtu.be/NTKw2Hp6tPw  [https://youtu.be/9yQ9Ln7jBQ8youtube]

See these two links.
In the video you can see a seesaw having two ball each side .the red color ball side arm of seesaw is slightly heavier than green color side arm.the weight of each ball is 30 gram and weight of each box is 20 gram.
The red ball is a counterweight and green ball will work to create energy.
The green ball will be free to move but counterweight will be permanently mounted.
Now I explain:
When I lift up the arm of green color ball up to a certain height and then you can see the movement of ball in the video that ball is moving due to a slope .
But when I left the arm then the ball is getting back it's previous position or in other words the seesaw is getting back it's horizontal position due to overunity.
It is totally against physics laws as ,as per physics laws the seesaw shouldn't get back it's horizontal position but in the video you can see it clearly.
Now replace this ball with 10 kg.ball and counterweight 10.200 kg.and calculate input and output using potential energy formula.
In this mechanism don't think about input as input will be very minimal but think about output.
The movement of ball be in the box will work as a output so the length of arm and length of box will be also important.
The ball mass=10 kg.
The length of arm is=1 meter
The length of box is=80 cm.
The width will be depend on the diameter of ball.
Now calculations:
Input is very minimal
Output if a 10 kg ball will be rolled down from 10 cm height(though the height will be more) than using potential energy formula
10*10*.10=10 joule
So the output will be 10 joule but input will be very minimal.
I would like to insist on some following points.
(1)the angle of box will be 120 degree downside(see the Sketch)
(2) two piston generator will be mounted on the sidewall of the box.
If still you have any doubts then pls tell me.
(3)there is overunity in this mechanism otherwise seesaw wouldn't get back it's horizontal position.
I request you that if you any doubt regarding to understanding then pls tell me as I am ready to clear your all doubts
In this video you can see easily how ball will fall down from a certain height.i have used a pin to hold the ball but changes in the design of box will work to hold and drop the ball after getting a certain height.
There are three mechanical advantage in this mechanism.
(1)The seesaw will itself get it's initial position.
(2) when ball will hit with piston generator then the generator will also hit it back in response (3) when ball will fall down on generator the the seesaw will be tilted further and so counterweight but this counterweight will hit with a spring which will work to bounce back the seesaw.
In this way this mechanism will work.

[guest39538]

https://youtu.be/NTKw2Hp6tPw  [https://youtu.be/9yQ9Ln7jBQ8youtube]

your video link does not work . It works if you click my quoted.

 i will let you know what i think after watching it

[guest39538]

Ok after watching the video you have achieved nothing but I give you some respect for efforts.

The container on the right is offset so of course the ruler will find the center of gravity.  Congratulations though in discovering balance for yourself although it is nothing knew to counter balance a system.

[vkrmvkrm 11 (OP)]

I think you are in a hurry to prove it wrong.
Now I tell you where is overunity in it.
At first there will be need of only 3 joule energy as an input in this mechanism for a 10 kg.mass ball to and to turn the arm of this seesaw at 90 degree angle.the box with ball is mounted at 45 degree angle when it is in rest position butwhen seesaw tilts then the box with ball will be positioned at almost 90 degree.(you can see in next video).
Now you will raise a question that input will be same as output but no there will be much more difference between input and output.
When the arm of seesaw will be lifted up the the ball in container will be not rolled down (seesketch of box).the ball will be released after getting a certain height.this is the main advantage in this mechanism.
Now calculate.
At first input
There will be need of only 3 joule energy maximum as an input.
Output
If a ,10 kg ball fall down from 80 cm height then using mgh formula
10*10*.80=80 joule
But interestingly the ball will get back it's initial position due to the mechanism of seesaw.
You can see it in second video.
If you have still any doubts then tell me.
I'm using my smartphone to reply you .my request to you please also open the second as you did first link

[Colin2B]

If a ,10 kg ball fall down from 80 cm height then using mgh formula
10*10*.80=80 joule

But the ball is not free falling, its rate of decent is constrained by the weight on the other side of the seesaw, so it is doing work to raise this weight up 80 cm. You need to take this into account in your calculations.
You can check this by attaching a pulley and weight to the green end and seeing how much weight the descending arm will lift, it won't do 80J of work.

[vkrmvkrm 11 (OP)]

Dear Sir,
The ball will fall freely in this mechanism.
I explain it.
The seesaw is slightly imbalanced towards counterweight  and ball Wis being lifted up (remember ball is not rolling along with lifting arm.it is still in rest position) so the input will be very minimal (3joule).
Now after getting a certain angle by the arm the box is positioned at90 degree angle .
Now ball will be released from the rest due to the design of box and fall freely on the wall of piston generator generating 80 joule.
When ball is in air in the box between releasing point and piston generator wall the seesaw will be in tilting position towards counterweight as there will be no mass up to that time on the green color box arm.
So in this way the ball will fall down freely.
There is nothing to do doubt about the feasibility of this device.
It is clear cut an overunity device.

[Colin2B]

Dear Sir,
The ball will fall freely in this mechanism.

sorry, I should have made it clear I was commenting on your video, which does not show an overunity device..
However, looking at your 2 diagrams, you have the same problem of lifting the counterweight, you need to factor that into your calculations.

[vkrmvkrm 11 (OP)]

I explain and clear your doubts regarding input.
The mass of ball is 10 kg.and counterweight is 10.200 so the seesaw will tilted slightly towards counterweight.
Now I lifting up the arm of ball10kg so I will have to apply very minimal input.but remember and I insist on this point that the will be released only after the arm gets 110 degree angle or box gets 90 angle after lifting up.
I have shown it in my second video.
The ball will not roll down at the the time of lifting up the box .it will be in stagnant position .
If possible you can try it with a very simple Experiment.
If there is big seesaw in nearby park then at first balance the seesaw with two children .just tell them sit on the seesaw and now lift up any arm of seesaw.
You will easily understand the input as it will be very minimal.
I have done it by sitting two children.
Please do it.to clear your doubts regarding input energy.

[Colin2B]

You will easily understand the input as it will be very minimal.
I have done it by sitting two children.
Please do it.to clear your doubts regarding input energy.

Yes, I am very familiar with the seesaw principle.
The input energy is minimal because lifting is assisted by the counterweight and this should be allowed for in the input energy calculation.
However, the item is question is not the input energy, but the output energy for which your calculation is incorrect.
Your potential energy calculation is correct for a non-counterbalanced weight, but the falling weight also has to lift the counterbalance weight and the energy required to do this should be subtracted from your calculation in order to show the minimal amount of output energy available from the seesaw.
Do try the pulley weight experiment I suggested.
This is my last post on this subject.

[vkrmvkrm 11 (OP)]

You are forgetting one thing that the ball is positioned in a BOX.so when ball  will fall down from a certain height then it will be again in the box and you know very well  that if this ball gets back it's initial position back in the back then the system will get balanced or will work to lift up counterweight.
See again my videos carefully.you will get convinced.