Why this seesaw is working against physics laws

[vikram_gupta11]

Dear Sir,
As per sketch if the long tube is mounted very near to fulcrum with 10 kg weight and counterweight(13kg) is mounted 1 meter from fulcrum then I think input energy will be less.
There is a spring based brake system to store the energy of counterweight as well as to provide it momentum.
So input energy will be only 30 Joule if counterweight is 13 kg.

[sm0ky2]

I am having some issues with my test rigs


First: when the ball falls to the right, most of the weight is balanced
         by the counterweight.
So the force it creates is very small.


Next:
When the ball falls to the left, it has more force but
We spend that by either compressing a spring or lifting the weight.
To get it back the other way.


Because the weight has moved, it becomes no longer balanced.


If we make the ball heavy on the right side,
and balanced on the left, so it resets itself
We spend energy to lift it from right to left.

[sm0ky2]

Perhaps the standard "wheel analysis" is not valid here,
because it does not make a complete circle.


we might need to look at this from a perspective of 'absolute height'


i.e.: ball initial height and ball final height
This would give a more accurate energy diagram

[vikram_gupta11]

I am having some issues with my test rigs


First: when the ball falls to the right, most of the weight is balanced
         by the counterweight.
So the force it creates is very small.


Next:
When the ball falls to the left, it has more force but
We spend that by either compressing a spring or lifting the weight.
To get it back the other way.


Because the weight has moved, it becomes no longer balanced.


If we make the ball heavy on the right side,
and balanced on the left, so it resets itself
We spend energy to lift it from right to left.

Dear Sm0ky2 Sir,
See the sketch.if we use a lock mechanism under counterweight to prevent the falling of counterweight then the force will be increased very much but there will be no need of extra energy in the lock mechanism.
Second if we setup the device as per the sketch then there will be need of very less input .for example,if the entire mass of device is 25 kg then we will have to tilt it just 10 degree more so that the ball could fall down.
Please consider these two points.
First lock mechanism and second rest position of device in this particular angle as per sketch.
If counterweight is also a ball in a tube then it will work to provide momentum due to impact of weight.
A spring will work to bounce back the device after hitting with counterweight side so that the device could get back it's initial position.
Thank you Sir.

[vikram_gupta11]

Dear Sir,
If the device is set at this angle in initial position then the input will be very minimal even if we calculate it's entire mass as there will be need of just 10 degree more to tilt it.
If the entire mass is 25 kg then input is  only 25 joule to tilt it 10 degree more.