Varignon, Pierre
,
Projet d' une nouvelle mechanique : avec Un examen de l' opinion de M. Borelli sur les propriétez des poids suspendus par des cordes
Text
Text Image
Image
XML
Thumbnail overview
Document information
None
Concordance
Notes
Handwritten
Figures
Content
Thumbnails
List of thumbnails
<
1 - 10
11 - 20
21 - 30
31 - 40
41 - 50
51 - 60
61 - 70
71 - 80
81 - 90
91 - 100
101 - 110
111 - 120
121 - 130
131 - 140
141 - 150
151 - 160
161 - 170
171 - 180
181 - 190
191 - 200
201 - 210
>
81
(55)
82
(56)
83
(57)
84
(58)
85
(59)
86
(60)
87
(61)
88
(62)
89
(63)
90
(64)
<
1 - 10
11 - 20
21 - 30
31 - 40
41 - 50
51 - 60
61 - 70
71 - 80
81 - 90
91 - 100
101 - 110
111 - 120
121 - 130
131 - 140
141 - 150
151 - 160
161 - 170
171 - 180
181 - 190
191 - 200
201 - 210
>
page
|<
<
(56)
of 210
>
>|
<
echo
version
="
1.0RC
">
<
text
xml:lang
="
fr
"
type
="
free
">
<
div
xml:id
="
echoid-div162
"
type
="
section
"
level
="
1
"
n
="
108
">
<
p
>
<
s
xml:id
="
echoid-s1466
"
xml:space
="
preserve
">
<
pb
o
="
56
"
file
="
0082
"
n
="
82
"
rhead
="
NOUVELLE
"/>
<
note
position
="
left
"
xlink:label
="
note-0082-01
"
xlink:href
="
note-0082-01a
"
xml:space
="
preserve
">DES POIDS
<
lb
/>
ſoutenus ſur
<
lb
/>
des ſurfaces.</
note
>
point A dans le parallelogramme fait ſous ces deux
<
lb
/>
lignes; </
s
>
<
s
xml:id
="
echoid-s1467
"
xml:space
="
preserve
">c’eft-à-dire, ſous AC & </
s
>
<
s
xml:id
="
echoid-s1468
"
xml:space
="
preserve
">CD, étant alors dif-
<
lb
/>
férente de AD, elle ne ſeroit pas perpendiculaire à
<
lb
/>
ce plan: </
s
>
<
s
xml:id
="
echoid-s1469
"
xml:space
="
preserve
">ainſi ce poids rouleroit alors (n. </
s
>
<
s
xml:id
="
echoid-s1470
"
xml:space
="
preserve
">3. </
s
>
<
s
xml:id
="
echoid-s1471
"
xml:space
="
preserve
">Demonſt.)
<
lb
/>
</
s
>
<
s
xml:id
="
echoid-s1472
"
xml:space
="
preserve
">du côté de L. </
s
>
<
s
xml:id
="
echoid-s1473
"
xml:space
="
preserve
">2°. </
s
>
<
s
xml:id
="
echoid-s1474
"
xml:space
="
preserve
">Mais ſi CD peut atteindre juſqu’en
<
lb
/>
quelque point D de la ligne AD, achevez le paral-
<
lb
/>
lelogramme BC, & </
s
>
<
s
xml:id
="
echoid-s1475
"
xml:space
="
preserve
">placez la puiſſance R ſuivant
<
lb
/>
AB: </
s
>
<
s
xml:id
="
echoid-s1476
"
xml:space
="
preserve
">alors elle ſoutiendra ce poids ſur ce plan.</
s
>
<
s
xml:id
="
echoid-s1477
"
xml:space
="
preserve
"/>
</
p
>
</
div
>
<
div
xml:id
="
echoid-div164
"
type
="
section
"
level
="
1
"
n
="
109
">
<
head
xml:id
="
echoid-head109
"
xml:space
="
preserve
">
<
emph
style
="
sc
">Demonstration.</
emph
>
</
head
>
<
p
>
<
s
xml:id
="
echoid-s1478
"
xml:space
="
preserve
">Puiſque (Hyp.) </
s
>
<
s
xml:id
="
echoid-s1479
"
xml:space
="
preserve
">cette puiſſance eſt à ce poids com-
<
lb
/>
me CD, ou AB qui lui eſt égale, eſt à AC; </
s
>
<
s
xml:id
="
echoid-s1480
"
xml:space
="
preserve
">leur con-
<
lb
/>
cours d’action doit le pouſſer (Lemm. </
s
>
<
s
xml:id
="
echoid-s1481
"
xml:space
="
preserve
">4. </
s
>
<
s
xml:id
="
echoid-s1482
"
xml:space
="
preserve
">Cor. </
s
>
<
s
xml:id
="
echoid-s1483
"
xml:space
="
preserve
">2.)
<
lb
/>
</
s
>
<
s
xml:id
="
echoid-s1484
"
xml:space
="
preserve
">ſuivant AD perpendiculaire (Hyp.) </
s
>
<
s
xml:id
="
echoid-s1485
"
xml:space
="
preserve
">au plan GM,
<
lb
/>
& </
s
>
<
s
xml:id
="
echoid-s1486
"
xml:space
="
preserve
">qui paſſe auſſi (Hyp.) </
s
>
<
s
xml:id
="
echoid-s1487
"
xml:space
="
preserve
">par la baſe de ce poids: </
s
>
<
s
xml:id
="
echoid-s1488
"
xml:space
="
preserve
">
<
lb
/>
Donc (Cor. </
s
>
<
s
xml:id
="
echoid-s1489
"
xml:space
="
preserve
">2.) </
s
>
<
s
xml:id
="
echoid-s1490
"
xml:space
="
preserve
">il doit demeurer deſſus en équilibre
<
lb
/>
avec cette puiſſance. </
s
>
<
s
xml:id
="
echoid-s1491
"
xml:space
="
preserve
">Ce qu’il F. </
s
>
<
s
xml:id
="
echoid-s1492
"
xml:space
="
preserve
">D.</
s
>
<
s
xml:id
="
echoid-s1493
"
xml:space
="
preserve
"/>
</
p
>
</
div
>
<
div
xml:id
="
echoid-div165
"
type
="
section
"
level
="
1
"
n
="
110
">
<
head
xml:id
="
echoid-head110
"
xml:space
="
preserve
">
<
emph
style
="
sc
">Corollaire</
emph
>
I.</
head
>
<
p
>
<
s
xml:id
="
echoid-s1494
"
xml:space
="
preserve
">Il eſt clair que ſi la puiſſance R ceſſoit de retenir le
<
lb
/>
poids EO, il couleroit le long de OL.</
s
>
<
s
xml:id
="
echoid-s1495
"
xml:space
="
preserve
"/>
</
p
>
</
div
>
<
div
xml:id
="
echoid-div166
"
type
="
section
"
level
="
1
"
n
="
111
">
<
head
xml:id
="
echoid-head111
"
xml:space
="
preserve
">
<
emph
style
="
sc
">Corollaire</
emph
>
II.</
head
>
<
p
>
<
s
xml:id
="
echoid-s1496
"
xml:space
="
preserve
">Si CD eſt la plus petite ligne qui puiſſe atteindre
<
lb
/>
de C juſqu’en AD; </
s
>
<
s
xml:id
="
echoid-s1497
"
xml:space
="
preserve
">c’eſt-à-dire, ſi l’angle ADC eſt
<
lb
/>
droit, l’angle BAD le ſera auſſi; </
s
>
<
s
xml:id
="
echoid-s1498
"
xml:space
="
preserve
">& </
s
>
<
s
xml:id
="
echoid-s1499
"
xml:space
="
preserve
">par conſéquent
<
lb
/>
cette puiſſance eſt la plus petite (Cor. </
s
>
<
s
xml:id
="
echoid-s1500
"
xml:space
="
preserve
">14.) </
s
>
<
s
xml:id
="
echoid-s1501
"
xml:space
="
preserve
">qui puiſ-
<
lb
/>
ſe ſoutenir ce poids ſur ce plan, & </
s
>
<
s
xml:id
="
echoid-s1502
"
xml:space
="
preserve
">elle ne l’y poura
<
lb
/>
ſoutenir non plus que (Cor. </
s
>
<
s
xml:id
="
echoid-s1503
"
xml:space
="
preserve
">16.) </
s
>
<
s
xml:id
="
echoid-s1504
"
xml:space
="
preserve
">ſuivant cette ſeule
<
lb
/>
direction.</
s
>
<
s
xml:id
="
echoid-s1505
"
xml:space
="
preserve
"/>
</
p
>
</
div
>
<
div
xml:id
="
echoid-div167
"
type
="
section
"
level
="
1
"
n
="
112
">
<
head
xml:id
="
echoid-head112
"
xml:space
="
preserve
">
<
emph
style
="
sc
">Corollaire</
emph
>
III.</
head
>
<
p
>
<
s
xml:id
="
echoid-s1506
"
xml:space
="
preserve
">Si CD n’eſt pas la plus petite qui puiſſe atteindre
<
lb
/>
depuis C juſqu’en AD, mais qu’elle ſoit cependant
<
lb
/>
cncore moindre que AC; </
s
>
<
s
xml:id
="
echoid-s1507
"
xml:space
="
preserve
">cette même puiſſance </
s
>
</
p
>
</
div
>
</
text
>
</
echo
>