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# Title: Leibniz-Handschriften zur Mathematik LH 35, 13, 2c
# Folio 6r, page id 00068199:0011
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.La2 dos Xp
Siteux.)
SCax XX) = qux - 2X
semi
quaeratur quando centu gravitatis segmenti distantia
ab Axe semicicenti sit maxima. Momenti distantia
ad ab Axe AK sut sit, semisegmentum ABCA, ejus centrum
gravitatis G ejusque distantia ab AxC GF, compleatur rectangutur
BFGL, patet ex ali. se jam si L sit medium punctum ipsius BC
patet GL fore fungentem lineae G(G) seu FG maximam.
 Mromentum ex ac ipsius ABCA ex axc habetur, ad sidionas
er upplices ac aream ejus prodit FG, id ergo aequale func erit ipsi
ad sectores BC
imanguli cenoiempavitatis ex vertiae Eius elementu (B)ea
sit sectoris KACA centrm
gravitatis G sit rabu nvissimi
AC, triens CL, itaque ipsa LG
tanget curvamcentronum quodsi
Gbsciet momentum Elemerdi sectoris
seu HC(C) seu si sit normalis ad KC
Cait distantia centri G. tumina a circunferer
tra, sed video hoc nonse conturpere
Nulla datur Cenri G distantro jumima
ab AK, nam ibi foret LG incallela ipsi A.
ad eadem debet esse nonnalis ad KCC ecgo
KC fi K H er. ex ejus trente educta
axi paralleto incideret in lentum gravitatis sed
genas ACH, etiam incideret, ergo csaretur
sit AC curva, AC, curvacentrom
arcuum jungatur CG, ea curva
centroum tunget. Hinc si arcus
sit ipsi lota cicumferem
tia constal andem cenla
crar. circumferentiae
incicere incentrm
circuli K, Itaque
t
e
M
Ds
punctum medium G, cujus distantia
ab axe GF. Ex Bagatur in
E
(C)(B) hormalis Bd. Erut
elementi arta BC in Bd. id
duaatur in CG. et erit
SBC. FG. BD elementum
momenti. Porro datur
A HK: (E =F BCy AB, X BB) dx erit
jam BB):BD = HE:EK =  1: Y fiet
recta cum A sit pun
stum ullima juncta
BD = Y. B(B) et = Ydx. en et elemtntur
eut trianguli erit yYdx. Rursus FG est ad BG
it EK ad HE seu ut Y ad 1. Ergo FG - r BG- 1.
ergo Eleruentam momenti est rrgus dx. sed
y: X = p. fiet y = px. et momentum
elementr  tutmenti erit: rrppXX dx.. ipsum
s
mimentum eut = Irpp X.; sed area
E x C.
tria nguli est Tyrdx seu S pYXax = pX
et distactia sentri gravitatis trianguli ab axe
erib ZYBX. seu distantia eut ad ad y AX
ut L5 ad 1. seu sed Y = EK; HE
Ergo ut ZEK ad HE. seu distantia ent
ad y seu HE, ut Z Ek ad HE. ergo
ZEK Z. Xm. itaque si in
basi biranguli latere trianguli
erit
le sumas
trientem HN, inde ab alio latere AH, et
per N dueas parallelam ipsi AH ea
transibit per centrum gravitatis
Hiet (Eaxdx = XX dx) Sdx HCrax. X)  ((C XX - L () dx (2ax - a) = c
Ak
laaget
2C= max
Sax L(IK XX
(ex srax. nx E A ul
monentam et
c
entem qua gr. quadtantis, quod falsum
2XX = LX
1
lineam centroum
aZ = Xdoy= ydx
dx (2aX -XX)
en (2aX -XX) = axX - X
5+cax xxAx
TaX
adclores