I Question about an example in Newton's Principia

[GCUEasilyDistracted]

I've started reading the Principia and have been trying to follow along with the examples. Unfortunately, I got stuck almost immediately. This example is from 'Axioms, or laws of motion', Law III, Corollary II. It is based on the following picture (everything in red is my addition):

The text states "As if the unequal radii $OM$ and $ON$ drawn from the centre $O$ of any wheel, should sustain the weights $A$ and $P$ by the cords $MA$ and $NP$; and the forces of those weights to move the wheel were required...If the weight $p$, equal to the weight $P$, is partly suspended by the cord $Np$, partly sustained by the oblique plane $pG$; draw $pH$, $NH$, the former perpendicular to the horizon, the latter to the plane $pG$; and if the force of the weight $p$ tending downwards is represented by the line $pH$, it may be resolved into the forces $pN$, $HN$." It later says "therefore if the weight $p$ is to the weight $A$ in a ratio compounded of the reciprocal ratio of the least distances of the cords $PN$, $AM$, from the centre of the wheel, and of the direct ratio of $pH$ to $pN$, the weights will have the same effect towards moving the wheel, and will therefore sustain each other." I was trying to make sense of this and derive the result.

The least distances of the cords $PN$ and $AM$ from the center $O$ should refer to the lengths of the lines $OL$ and $OK$ respectively if I'm correct. So, the "reciprocal ratio" of these ought to be $\frac{\overline{OK}}{\overline{OL}}$. The "direct ratio" of $pH$ to $pN$ should just be $\frac{\overline{pH}}{\overline{pN}}$. So the way I interpret the second statement I quoted is that the forces due to weights $A$ and $p$ will be balanced on the wheel, resulting in zero torque, if

$$\frac{m_p}{m_A} = \frac{\overline{OK}}{\overline{OL}}\frac{\overline{pH}}{\overline{pN}}$$ $$(Eq. 1)$$

Here I'm also assuming that the weight $P$ is not included under consideration.

While trying to derive this relationship, I attempted to calculate the magnitudes of the torques due to each weight.

$$\tau_A = \overline{OK}m_Ag$$

$$\tau_p = \overline{OR}T_{pN}$$

$T_{pN}$ is the magnitude of tension in the cord $pN$.

Note that $T_{pN} = m_pg\frac{\overline{pN}}{\overline{pH}}$. Therefore,

$$\tau_p = \overline{OR}m_pg\frac{\overline{pN}}{\overline{pH}}$$

Setting $\tau_A = \tau_p$ gives

$$\overline{OK}m_Ag = \overline{OR}m_pg\frac{\overline{pN}}{\overline{pH}}$$

$$\overline{OK}m_A = \overline{OR}m_p\frac{\overline{pN}}{\overline{pH}}$$

$$\frac{m_p}{m_A} = \frac{\overline{OK}}{\overline{OR}}\frac{\overline{pH}}{\overline{pN}}$$ $$(Eq. 2)$$

Equation 2 is almost the same as Equation 1, it's just that $\overline{OL}$ is replaced by $\overline{OR}$. If the statement "the least distances of the cords $PN$, $AM$, from the centre of the wheel" were replaced with "the least distances of the cords $pN$, $AM$, from the centre of the wheel" (note the lowercase $p$), then it would be identical since the least distance of $pN$ from $O$ is in fact $\overline{OR}$.

Where am I going wrong? I really appreciate any help with this.

 

[PeroK]

I'm not sure many people will be able to decipher a work of that era.