I did a thing

[grayson]

So, I was playing around with SI units and came up with the equation in the picture.

where v is velocity, F₁ is the force exerted on an object, m₂ is the mass of the object the force was asserted on and d₂ is displacement. Let me go through how I came up with this. First, I broke down force into its base components. A newton is equal to one kg*m*s^-2. So, I divided it by kg to remove the kg on the top. We are now left with m*s^-2. This is almost velocity but not quite. So, I squared it, so it is now sqrt(m)*s^-1. I just multiplied it by m to get m/s. Just to avoid confusion, kg is weight, m is length and s is time. So, I put this into an equation and suddenly I got this. Basically, it is just velocity determined by force and weight and displacement. And because of newtons laws, the displacement should apply to both of the objects, so d₂ is kind of unnecessary.

[Genady]

14 minutes ago, grayson said:

So, I was playing around with SI units and came up with the equation in the picture.

where v is velocity, F₁ is the force exerted on an object, m₂ is the mass of the object the force was asserted on and d₂ is displacement. Let me go through how I came up with this. First, I broke down force into its base components. A newton is equal to one kg*m*s^-2. So, I divided it by kg to remove the kg on the top. We are now left with m*s^-2. This is almost velocity but not quite. So, I squared it, so it is now sqrt(m)*s^-1. I just multiplied it by m to get m/s. Just to avoid confusion, kg is weight, m is length and s is time. So, I put this into an equation and suddenly I got this. Basically, it is just velocity determined by force and weight and displacement. And because of newtons laws, the displacement should apply to both of the objects, so d₂ is kind of unnecessary.

In other words, you have found that \(v=\sqrt {ax}\), where \(v\) is speed, \(a\) is acceleration, and \(x\) is displacement. There is a known kinematic equation for this: if a body starts moving from a rest with a constant acceleration \(a\), when it moves the distance \(x\) its speed is \(v=\sqrt {2ax}\).

[KJW]

27 minutes ago, Genady said:

In other words, you have found that v=ax−−√, where v is speed, a is acceleration, and x is displacement. There is a known kinematic equation for this: if a body starts moving from a rest with a constant acceleration a, when it moves the distance x its speed is v=2ax−−−√.

Maybe one can get rid of the "2" under the square root by some form of pulley mechanism. The problem with using dimensional analysis to derive formulae is that one can miss out on dimensionless constants, as has happened here.

 

Edited December 10, 2023 by KJW

[Genady]

2 minutes ago, KJW said:

The problem with using dimensional analysis to derive formulae is that one can miss out on dimensionless constants, as has happened here.

Right. I've expected a question from OP regarding this \(2\).

Edited December 10, 2023 by Genady

[grayson]

2 hours ago, Genady said:

Right. I've expected a question from OP regarding this 2 .

Yah, I was wondering about that.

[Genady]

2 minutes ago, grayson said:

Yah, I was wondering about that.

Dimensional analysis does not help you to find out dimensionless factors.