[alancalverd]

What will happen if the weight on the right side is increased to 20N?

The beam would rotate clockwise. So what?

[hamdani yusuf (OP)]

force x distance.

Which distance?

[hamdani yusuf (OP)]

This has nothing to do with "established science". It's just a definition, which would remain valid even if nobody had ever observed anything rotating.

When nothing is rotating, your force or tension doesn't produce torque, since the radius of rotation isn't yet defined.

[hamdani yusuf (OP)]

What will happen if the weight on the right side is increased to 20N?

The beam would rotate clockwise. So what?

What would be the normal forces at each fulcrum?

What if the direction of the force applied to the right end of the lever is upward instead?

[hamdani yusuf (OP)]

You, Hamdani, have fallen into delusions of grandeur, thinking that you are smarter than the current body of scientific discipline.

You have fallen into a logical fallacy called argument from authority.

No.
There's nothing wrong with recognising the authority of authorities.

If you're not very smart, it's better to be conservative - Jordan Peterson

But since you still have plenty of unused cognitive capacity, perhaps you can spend some time to scrutinize some current standards to see if something can be improved.

[alancalverd]

force x distance.

Which distance?

Whichever distance you choose, but if the torque is enough to initiate rotation, the obvious distance is to the center of that rotation. Have another look at my "US standard airplane moments" reply for more details.

[Bored chemist]

If you're not very smart, it's better to be conservative - Jordan Peterson

He's a Conservative and a liar.
I'm not sure he's all that bright.
But he's a very good con-man.
Maybe if you are stuck with a choice of citing him or chatGPT as your source... maybe just don't say anything.

However, his point is not absolutely unreasonable; you have just misapplied it.
Because you are not very smart, you should just listen to what the clever people say, and accept that the SI units work.

[paul cotter]

Bravo!, BC, I could not have expressed it better myself.

[hamdani yusuf (OP)]

force x distance.

Which distance?

Whichever distance you choose, but if the torque is enough to initiate rotation, the obvious distance is to the center of that rotation. Have another look at my "US standard airplane moments" reply for more details.

Before rotation actually happens, there is no radius of rotation. Radius of the rotating object doesn't always the same as radius of rotation.

[hamdani yusuf (OP)]

What will happen if the weight on the right side is increased to 20N?

The beam would rotate clockwise. So what?

What would be the normal forces at each fulcrum?

Since no one has shown their capacity to answer this question, I guess I just have to answer it myself.

If radian is removed from the formulas above, The rotational acceleration will be expressed in 1/second^2, or Hertz^2, which is erroneous. The error will affect the subsequent calculations, and produce erroneous value for the normal force at central fulcrum.

[Bored chemist]

Since no one have shown their capacity to answer this question

I have not shown you my capacity to walk and chew gum.
But it would be sensible to guess that I can.

[hamdani yusuf (OP)]

Since no one have shown their capacity to answer this question

I have not shown you my capacity to walk and chew gum.
But it would be sensible to guess that I can.

Since you haven't shown your capacity, no one will know it for sure.

[hamdani yusuf (OP)]

What if the direction of the force applied to the right end of the lever is upward instead?

[alancalverd]

The rotational acceleration will be expressed in 1/second, or Hertz,

There's nothing wrong with expressing rotation in Hz if you wish, but if your forces act via lever arms (radius) and result in rotation (arc) it's easier to calculate in radians (arc/radius). All you have done is arbitrarily added and removed radians to your units for no obvious purpose.

[hamdani yusuf (OP)]

You can add some weight to the right side of the lever without causing the lever to turn.
Do you have any idea why that's the case?

Do you have any idea why that's the case?

Of course we do; we aren't idiots.
Why did you ask?

Since you still provide no explanation, I'll explain it to you.
In ordinary cases of lever and fulcrum, the whole system stands firmly on earth, which has much larger mass and rotational inertia than the other components of the system.
In other cases where it's not the case, like on a light boat floating on water or a hot air balloon, the additional weight can cause the whole system to rotate.

[hamdani yusuf (OP)]

The rotational acceleration will be expressed in 1/second^2, or Hertz^2,

There's nothing wrong with expressing rotation in Hz if you wish, but if your forces act via lever arms (radius) and result in rotation (arc) it's easier to calculate in radians (arc/radius). All you have done is arbitrarily added and removed radians to your units for no obvious purpose.

It's not about which one is easier. The equation would produce the numerical value in the unit of 1/second^2 before any conversion, if the unit for rotational angle hasn't been specified as radian.

You will not find a single reference that claims the angle is a component of torque and I think you are being deliberately argumentative for argument's sake.

I'm proposing to change the STANDARD unit for torque in order to make it consistent with other rotational quantities. You can still use non-standard units, as long as they give you some benefits, like being easier to measure or calculate. You can use your own feet, palms, or fingers to measure length, for they are most accessible for you at some point in your lifetime. But I don't think they can be good standards.

Can you point out what's wrong with my previous post?

The table below shows the comparison between angular and linear quantities.

Here are the equations conversion, where d = arclength of the circumference corresponding to rotational angle.
θ = d * (θ/d)
ω = v * (θ/d)
α = a * (θ/d)
I = m * (d/θ)^2
τ = F * (d/θ)
L = p * (d/θ)

Regardless of the units used, the quantity measured in rotational acceleration is rotational angle divided by time squared. The quantity measured in torque, according to the list above, is force times distance of rotational motion divided by angle.

[hamdani yusuf (OP)]

Here's an example where friction is negligible, but you have reactional force which cancels out the torque up to some limit.

The limit occurred because the fulcrum cannot pull the lever down. It would be different if they are glued or bolted together.

[hamdani yusuf (OP)]

Let's learn from Newton instead. He established his laws of motion by first making statements about idealized conditions, where friction is negligible. Likewise, learning about rotational motion should start from the fundamentals. If you start with cases where frictions already predominate the equation, you are like trying to run before you can stand.

So if you want to have a strong fundamental understanding of rotational quantities, I suggest you to explore the problems with idealized coaxial pulley which I posted earlier. It introduced the core concepts without contamination of distracting quantities which make the problem unnecessarily more complicated.

[hamdani yusuf (OP)]

All of your errors have been comprehensively rebuffed by many contributors but you refuse to listen and double down on your errors.

What errors do you find in my conversion table between rotational and linear quantities?

[alancalverd]

The equation would produce the numerical value in the unit of 1/second^2 before any conversion,

Then you must be using an unconventional definiion of torque