Page:The Mathematical Principles of Natural Philosophy - 1729 - Volume 1.djvu/371

Rh the force with which the corpucle attracted by the whole phere will be as the area ANB.

And univerally if the centripetal force tending to the everal particles of the phere be uppoed to be reciprocally as the quantity V; and DN be made as $$\textstyle \frac {DE^2 \times PS}{PE \times V}$$; the force with which 5 corpucle is attracted by the whole phere will be as the area ANB.

The things remaining as above it is required to meaure the area ANB. (Pl. 23. Fig. 1.)

From the point P let there be drawn the right line PH touching the phere in H; and to the axis PAB letting fall the perpendicular HI, biect PI in L; and (by prop. 12. book 2. elem.) $$\scriptstyle PE^1$$ is equal to $$\scriptstyle PS^2 + SE^2 + 2PSD$$. But becaue the triangles SPH, SHI are alike. $$\scriptstyle SE^2$$ or $$\scriptstyle SH^2$$ is equal to the rectangle PSI. Therefore $$\scriptstyle PE^2$$ is equal to the rectangle contained under PS and PS + SI + 2SD; that is under PS and 2LD. Moreover $$\scriptstyle DE^2$$ is equal to $$\scriptstyle SE^2 - SD^2$$, or $$\scriptstyle SE^2 - LS^2 + 2LD - LD^2$$, that is, $$\scriptstyle 2SLD - LD^2 - aLB$$. For $$\scriptstyle LS^2 - SE^2$$ or $$\scriptstyle LS^2 - SA^2$$ (by prop. 6. book 2. elem.) is equal to the rectangle ALB. Therefore if intead of $$\scriptstyle DE^2$$ we write $$\scriptstyle 2SLD - LD^2 - ALB$$; the