Archimedes Derivation: Law of the lever

[christian0710]

Hi, I'm watching a talk about Archimedes Law of the lever, and I was wondering: Does anyone know how the formula d=w/2w/b-a) was derived from the lecture drawing at 9.00 min in? The speaker just skips the algebraic derivation.

 

[BvU]

With "d=w/2w/b-a)" you mean d=W/(2w) (b-a) ?

It's a simple torque balance (see at 8:51).

On the left you have $d \times w + {a\over 2}\times pW$, on the right ${b\over 2}\times qW$,

and $\ \ a = {p\over l}\ \$ and $b = {q\over l}\ \$ so that $$
d \times w = \left ( {b^2 \over 2l} - { a^2 \over 2l} \right ) \times W.
$$Now use $\ \ b^2-a^2 = (a-b) (a+b) = (a-b)\; l\ \$ and there you are !

 

[christian0710]

With "d=w/2w/b-a)" you mean d=W/(2w) (b-a) ?

It's a simple torque balance (see at 8:51).

On the left you have $d \times w + {a\over 2}\times pW$, on the right ${b\over 2}\times qW$,

and $\ \ a = {p\over l}\ \$ and $b = {q\over l}\ \$ so that $$
d \times w = \left ( {b^2 \over 2l} - { a^2 \over 2l} \right ) \times W.
$$Now use $\ \ b^2-a^2 = (a-b) (a+b) = (a-b)\; l\ \$ and there you are !

Thank you very much! There is one more thing i can't seem to understand: I understand that little w is the weight of the object on the lever, but right next to little w there is a big W pointing down (on the left side). What does the big W mean? And what is it doing on the left side?

 

[BvU]

W is the weight of the beam on which the weight small w is resting.
A fraction ${a\over l}$ of W is on the left side, and a fraction ${b\over l}$ of W is to the right of the balance point