Source Latex
de la correction du devoir
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% %
% Generateur automatique de devoir, %
% par Y. Morel %
% http://xymaths.free.fr %
% %
% Genere le: %
% mercredi 19 d�cembre 2012 %
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\begin{document}
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\ct{\bf\LARGE{Corrig\'e du devoir de math\'ematiques}}
\bgex
On a $\|\vec{u}\|=\sqrt{(-2)^2+5^5}\simeq 5,38$.
Ainsi,
$\vec{u}\cdot\vec{v}
=12
\simeq 5,38\tm 6\tm\cos\lp\vec{u},\vec{v}\rp
\simeq 32,28 \cos\lp\vec{u},\vec{v}\rp
$,
d'o�,
$\cos\lp\vec{u},\vec{v}\rp\simeq \dfrac{12}{32,28}$,
et
$\lp\vec{u},\vec{v}\rp\simeq 68,18^\circ$,
ou $\lp\vec{u},\vec{v}\rp\simeq -68,18^\circ$.
\enex
\bgex
\bgen
\item $\V{AB}\lp -14;26\rp$;\ $\V{AC}\lp 16;-5\rp$
\item $\|\V{AB}\|=AB=\sqrt{\lp-14\rp^2+26^2}\simeq 29,53$;\
$\|\V{AC}\|=AC=\sqrt{16^2+\lp-5\rp^2}=16,76$
\item $\V{AB}\cdot\V{AC}=-14\tm16+26\tm(-5)=-354$
\item On a aussi,
$\V{AB}\cdot\V{AC}
=AB\tm AC\tm\cos\lp\widehat{BAC}\rp
\simeq 19,53\tm16,76\tm\cos\lp\widehat{BAC}\rp
\simeq 494,92\cos\lp\widehat{BAC}\rp
$.
On en d�duit que
$\cos\lp\widehat{BAC}\rp\simeq \dfrac{-354}{494,92}$,
soit $\lp\widehat{BAC}\rp\simeq 135,7^\circ$
\enen
\enex
\bgex
\bgen
\item L'�quilibre se traduit vectoriellement par la relation:\quad
$\V{F_1}+\V{F_2}+\V{T}=\V{0}$.
\item En projetant cette relation sur
$\lp O\vec{i}\rp$, on obtient
$\V{F_1}\cdot\vec{i}+\V{F_2}\cdot\vec{i}+\V{T}\cdot\vec{i}
=\V{0}\cdot\vec{i}=0$,
soit $-F_1\cos\lp30^\circ\rp+F_2\cos\lp75^\circ\rp+T_x=0$,
d'o�,
$T_x=F_1\cos\lp30^\circ\rp-F_2\cos\lp75^\circ\rp
\simeq 82,62$\,N.
De m�me, en projetant sur $\lp O\vec{j}\rp$, on obtient
$\V{F_1}\cdot\vec{j}+\V{F_2}\cdot\vec{j}+\V{T}\cdot\vec{j}
=\V{0}\cdot\vec{j}=0$,
soit $-F_1\cos\lp60^\circ\rp+F_2\cos\lp15^\circ\rp+T_y=0$,
d'o�,
$T_y=F_1\cos\lp60^\circ\rp-F_2\cos\lp15^\circ\rp
\simeq -238,07$\,N.
\item L'intensit� de $\V{T}$ est alors
$\|\V{T}\|= \sqrt{T_x^2+T_y^2}\simeq 252$\,N.
\item On a
$\V{T}\cdot\vec{i}=T_x=\|\V{T}\|\tm\|\vec{i}\|\tm\cos\alpha
\simeq 252\cos\alpha$,
d'o�,
$\alpha\simeq \cos^{-1}\lp \dfrac{T_x}{252}\rp\simeq 70,86^\circ$.
\enen
\enex
\label{LastPage}
\end{document}
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