Galilei, Galileo, Les méchaniques, 1634

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<p type="main">
<s id="s.000273">
quelle RM repreſente le plan horizon­
<lb/>
<figure id="id.047.01.081.1.jpg" xlink:href="047/01/081/1.jpg" number="26"/>
<lb/>
tal, ſur lequel ie
<lb/>
ſuppoſe que le plan
<lb/>
PM eſt eleué de 30.
<lb/>
degrez, & conſe­
<lb/>
quemment qu'il
<lb/>
fait 60. degrez auec
<lb/>
le plan perpendi­
<lb/>
culaire BC. </s>
<s id="s.000274">Or il eſt certain que la
<lb/>
force qui retient le poids, ou le globe
<lb/>
BSA ſur le plan incliné eſt audit poids,
<lb/>
comme la perpendiculaire PR eſt à
<lb/>
l'hypotenuſe PM: & parce que cette
<lb/>
hypothenuſe eſt double de la
<expan abbr="perpẽdi-culaire">perpendi­
<lb/>
culaire</expan>
, vne force vn peu plus
<expan abbr="grãde">grande</expan>
que
<lb/>
ſouz double le leuera, de ſorte que ſi le
<lb/>
globe peſe 2. liures le poids P, ou O
<expan abbr="peſãt">peſant</expan>
<lb/>
vne liure, & vn grain le pourra tirer. </s>
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<p type="main">
<s id="s.000275">Il faut encore remarquer que la force
<lb/>
qui doit empeſcher que le poids ne
<lb/>
coule & ne peſe point ſur le plan PM
<lb/>
doit eſtre au poids, comme la baſe RM
<lb/>
à l'hypotenuſe PM. </s>
<s id="s.000276">Or quand on veut
<lb/>
tirer le poids ſur le plan incliné, il faut
<lb/>
mettre vne poulie au haut du plan,
<lb/>
comme l'on void en D. </s>
</p>
<p type="main">
<s id="s.000277">Où l'on doit conſiderer la force qui­
<lb/>
ſouſtient le poids dans la ligne </s>
</p>
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