Deux exercices supplémentaires avec figure produite avec inkscape et export du texte de la figure dans un fichier latex. Attention, modifications manuelles de ce fichier pour y mettre des commandes latex d'équations.
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@ -892,6 +892,74 @@ On lâche la première à vitesse initiale nulle. Calculez la vitesse de la seco
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\end{solos}
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\end{exos}
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\begin{exos}
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On lâche, à vitesse initiale nulle, une masse m de \unit{3}{\kilo\gram}, sur un plan incliné d'un angle \(\alpha = \unit{20}{\degree}\) sans frottements. Calculez sa vitesse après un temps t de \unit{2}{\second}. Réponse~: \unit{9,72}{\metre\per\second}.
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\begin{solos}
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Deux forces agissent ici : la réaction du plan, qui lui est normale (c'est-à-dire perpendiculaire), et le poids de la masse. Comme la réaction du plan n'a aucune composante parallèlement au plan, elle ne peut être responsable de l'accélération de la masse le long de celui-ci.
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Il faut donc trouver la composante du poids qui est parallèle au plan incliné. L'angle entre le poids et un plan horizontal est de \unit{90}{\degree}. Quand le plan est incliné, cet angle diminue de la valeur de l'inclinaison. L'angle \(\beta\) entre le plan incliné et le poids et donc \(\beta=90-\alpha\).
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Comme la projection du poids selon l'angle \(\beta\) correspond à sa composante parallèle au plan, dans le triangle rectangle composé du poids comme hypoténuse et de ses composantes parallèle et perpendiculaire au plan, la composante parallèle au plan correspond au côté adjacent. Ainsi, on peut écrire~:
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\begin{align*}
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P_{//}&=P\cdot\cos(\beta)=P\cdot\cos(90-\alpha)\\
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&=P\cdot\sin(\alpha)
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\end{align*}
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La seconde loi de Newton s'écrit donc le long du plan incliné~:
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\begin{align*}
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\sum F^{ext}=P\cdot\sin(\alpha)&=m\cdot a\;\Rightarrow\\
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m\cdot g\cdot\sin(\alpha)&=m\cdot a\;\Rightarrow\\
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a&=g\cdot\sin(\alpha)\;\Rightarrow\\
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a&=9,81\cdot\sin(20)=\unit{3,36}{\metre\per\second\squared}
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\end{align*}
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Avec une vitesse initiale nulle, pour un MRUA d'accélération calculée ci-dessus, la vitesse au bout d'un temps t=\unit{2}{\second} s'obtient par~:
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\[v=a\cdot t+v_0=3,36\cdot 2=\unit{9,72}{\metre\per\second}\]
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\medskip
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Une autre manière de résoudre le problème est de procéder avec méthode. Le système qu'on doit choisir est bien évidemment la masse m, puisque c'est de celle-ci qu'on cherche l'accélération pour en trouver la vitesse au bout de \unit{2}{\second}.
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\begin{figure}[h]
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\centering
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\caption[Plan incliné]{Le plan incliné}\label{incline}
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\medskip
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La figure \ref{incline} présente ensuite le dessin des forces extérieures et le choix du système d'axes. Remarquez que ce dernier l'a été selon l'inclinaison du plan. Il aurait pu ne pas en être ainsi, mais ce choix simplifie les calculs, car la masse étant contrainte à se déplacer le long du plan, son accélération perpendiculairement est nulle. Les équations de la seconde loi de Newton, obtenues par projection des forces extérieures et de l'accélération selon les axes, peuvent alors s'écrire :
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\begin{align*}
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\sum F^{ext}_x&=P_x=m\cdot a_x &\text{sur l'axe x}\\
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\sum F^{ext}_y&=R-P_y=m\cdot a_y=0 &\text{sur l'axe y}
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\end{align*}
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car l'accélération perpendiculairement au plan est nulle, comme déjà mentionné.
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\smallskip
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Si on considére l'angle \(\alpha\), en s'imaginant le plan incliné horizontal, on comprends qu'il se reporte entre le vecteur poids \(\overleftarrow{P}\) et sa compostante selon y \(P_y\).
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Avec le triangle rectangle formé par le poids et ses composantes et un peu de trigonométrie, on peut en déduire~:
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\begin{align*}
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P_x &= P\cdot \sin(\alpha)\\
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P_y &= P\cdot \cos(\alpha)
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\end{align*}
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Comme par ailleurs on sait que \(P=m\cdot g\), on peut réécrire les équations de Newton sur les axes comme~:
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\begin{align*}
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\sum F^{ext}_x&=m\cdot g\cdot \sin(\alpha)=m\cdot a_x\\
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\sum F^{ext}_y&=R-m\cdot g\cdot \cos(\alpha)=m\cdot a_y=0
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\end{align*}
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La première de ces équations permet de trouver l'accélération du bloc selon le plan incliné~:
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\begin{align*}
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a=a_x&=g\cdot \sin(\alpha)\\
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&=9,81\cdot\sin(20)=\unit{3,36}{\metre\per\second\squared}
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\end{align*}
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\smallskip
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Mais une information supplémentaire nous est donnée par la seconde équation, c'est la valeur de la réaction R au plan~:
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\begin{align*}
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R&=m\cdot g\cdot \cos(\alpha)\\
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&=3\cdot 9,81\cdot \cos(20)=\unit{27,67}{\newton}
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\end{align*}
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\end{solos}
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\end{exos}
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\begin{exos}
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Un exercice de test.
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\begin{solos}
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]
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Appendix M.
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8 no-tam-ment
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ommons .
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[235] [236
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\setcounter{bottomnumber}{2} % nombre maximal de flottant en bas d'une page (par défaut 1)
|
||||
%\setcounter{totalnumber}{3} % nombre maximal de flottant sur une page (par défaut 3)
|
||||
\renewcommand{\topfraction}{0.9} % proportion maximale de la page qui peut recevoir des flottants en haut de la page (par défaut 0,7)
|
||||
\renewcommand{\textfraction}{0.1} % proportion minimale d'une page qui reçoit du texte (sans les pages de flottants)
|
||||
\renewcommand{\floatpagefraction}{0.8} % proportion minimale de flottants sur une page pour qu'elle devienne une page uniquement de flottants
|
||||
\renewcommand{\bottomfraction}{0.5} % proportion maximale de la page qui peut recevoir des flottants en bas de la page (par défaut 0,3)
|
||||
|
||||
%\newcommand\relphantom[1]{\mathrel{\phantom{#1}}} % sert à afficher un = invisible
|
||||
%----------------------------------------------------------------
|
||||
|
||||
\graphicspath{{./Images/}{./Introduction/Images/}{./Cinematique/Images/}{./Dynamique/Images/}{./MecaniqueDim/Images/}{./MecaniqueDifferentielle/Images/}{./QtiteMvt/Images/}{./Energie/Images/}{./EnergieOS/Images/}{./Thermodynamique/Images/}{./ThermodynamiqueOS/Images/}{./Annexe-UnitesInternationales/Images/}{./Annexe-SystemeCoordonnees/Images/}{./Annexe-MesuresDistances/Images/}{./Annexe-TravauxPratiques/Images/}{./Annexe-Rotations/Images/}{./Annexe-MRUA/Images/}{./Annexe-ChuteLune/Images/}{./Annexe-Satellites/Images/}{./Annexe-Relativite/Images/}{./Annexe-Maree/Images/}{./Annexe-Energies/Images/}{./Annexe-Exercices/Images/}{./Annexe-Incertitudes/Images/}} % spécifie les dossiers dans lesquels sont les images
|
||||
|
||||
%--------------------------------------------------------------
|
||||
% définition de l'image de la page de titre
|
||||
\usepackage{eso-pic} % pour mettre une image en fond de première page
|
||||
\newcommand\BackgroundPic{
|
||||
\put(0,0){
|
||||
\parbox[b][\paperheight]{\paperwidth}{%
|
||||
\vfill
|
||||
\centering
|
||||
%\includegraphics[width=\paperwidth,height=\paperheight,keepaspectratio]{TrouNoir.eps}%
|
||||
\includegraphics[width=\paperwidth,height=\paperheight]{TrouNoir.eps}%
|
||||
\vfill
|
||||
}}}
|
||||
%--------------------------------------------------------------
|
||||
% inclusion des macros
|
||||
\include{Macros}
|
||||
|
||||
\begin{document}
|
||||
|
||||
%--------------------------------------------------------------
|
||||
% inclusion des Préfaces
|
||||
\include{Prefaces/Prefaces}
|
||||
%--------------------------------------------------------------
|
||||
|
||||
\twocolumn
|
||||
|
||||
%--------------------------------------------------------------
|
||||
% table des matières, liste des figures et liste des tables
|
||||
\tableofcontents{}
|
||||
|
||||
\renewcommand{\listfigurename}{Liste des figures}
|
||||
\listoffigures
|
||||
|
||||
\myclearpage
|
||||
|
||||
\listoftables
|
||||
%--------------------------------------------------------------
|
||||
|
||||
\myclearpage
|
||||
|
||||
%--------------------------------------------------------------
|
||||
% inclusion du chapitre d'introduction
|
||||
\include{Introduction/Introduction}
|
||||
|
||||
% inclusion du chapitre de cinématique
|
||||
\include{Cinematique/Cinematique}
|
||||
|
||||
% inclusion du chapitre de dynamique
|
||||
\include{Dynamique/Dynamique}
|
||||
|
||||
% inclusion de la Mécanique en plusieurs dimensions (OS)
|
||||
\opt{OS}{\include{MecaniqueDim/MecaniqueDim}}
|
||||
|
||||
% inclusion de la Mécanique différentielle (OS)
|
||||
\opt{OS}{\include{MecaniqueDifferentielle/MecaniqueDifferentielle}}
|
||||
|
||||
% inclusion de la Quantité de mouvement (OS)
|
||||
\opt{OS}{\include{QtiteMvt/QtiteMvt}}
|
||||
|
||||
% inclusion du chapitre d'énergie
|
||||
\include{Energie/Energie}
|
||||
|
||||
% inclusion de lénergie (OS)
|
||||
\opt{OS}{\include{EnergieOS/EnergieOS}}
|
||||
|
||||
% inclusion du chapitre de thermodynamique
|
||||
\include{Thermodynamique/Thermodynamique}
|
||||
|
||||
% inclusion de la Thermodynamique (OS)
|
||||
\opt{OS}{\include{ThermodynamiqueOS/ThermodynamiqueOS}}
|
||||
%--------------------------------------------------------------
|
||||
|
||||
\appendix
|
||||
|
||||
\onecolumn
|
||||
|
||||
%--------------------------------------------------------------
|
||||
% inclusion de l'annexe Unités du système international
|
||||
\include{Annexe-UnitesInternationales/Annexe-UnitesInternationales}
|
||||
|
||||
% inclusion de l'annexe Deux systèmes de coordonnées
|
||||
\include{Annexe-SystemeCoordonnees/Annexe-SystemeCoordonnees}
|
||||
|
||||
% inclusion de l'annexe Mesures de distances
|
||||
\include{Annexe-MesuresDistances/Annexe-MesuresDistances}
|
||||
|
||||
% inclusion de l'annexe Travaux pratiques
|
||||
\include{Annexe-TravauxPratiques/Annexe-TravauxPratiques}
|
||||
|
||||
% inclusion de l'annexe Rotations
|
||||
\include{Annexe-Rotations/Annexe-Rotations}
|
||||
|
||||
% inclusion de l'annexe MRUA développements
|
||||
\include{Annexe-MRUA/Annexe-MRUA}
|
||||
|
||||
% inclusion de l'annexe Chute de la lune
|
||||
\include{Annexe-ChuteLune/Annexe-ChuteLune}
|
||||
|
||||
% inclusion de l'annexe Satellite en orbite géostationnaire
|
||||
\include{Annexe-Satellites/Annexe-Satellites}
|
||||
|
||||
% inclusion de l'annexe Relativité
|
||||
\include{Annexe-Relativite/Annexe-Relativite}
|
||||
|
||||
% inclusion de l'annexe Marées
|
||||
\include{Annexe-Maree/Annexe-Maree}
|
||||
|
||||
% inclusion de l'annexe Énergies
|
||||
\include{Annexe-Energies/Annexe-Energies}
|
||||
|
||||
% inclusion de l'annexe Exercices
|
||||
\include{Annexe-Exercices/Annexe-Exercices}
|
||||
|
||||
% inclusion de l'annexe Incertitudes (OS)
|
||||
\opt{OS}{\include{Annexe-Incertitudes/Annexe-Incertitudes}}
|
||||
%--------------------------------------------------------------
|
||||
|
||||
\theendnotes
|
||||
|
||||
\myclearpage
|
||||
|
||||
%\nocite{*} % pour mettre toutes les entrées dans la biblio, y compris celles non citées ds le texte
|
||||
\bibliographystyle{apalike-fr} % le style de bibliographie apalike francisé. Attention, il ne faut pas le module apalike, mais il faut le module natbib.
|
||||
%\bibliographystyle{apalike} % le style de bibliographie
|
||||
\bibliography{../Bibliographies/BiblioCours} % définit le fichier des réf. biblio. nom.bib
|
||||
% le logiciel utilisé pour faire la biblio est pybliographic (pybliographer en ligne de commande) !!! suite à la suppression de
|
||||
% pybliographic (pybliographer a maintenant une ui) de Lenny, j'utilise JABREF <-- très très bien.
|
||||
\printindex{} % attention, pour que l'index soit mis à jour il faut lancer la commande de konsole : makeindex -s ../Perso.ist nom_du_cours.idx
|
||||
% le Perso.ist est un fichier disant qu'il faut mettre des lettres entre les parties de l'index
|
||||
|
||||
%\renewcommand{\listfigurename}{Liste des figures}
|
||||
%\listoffigures % ai tout basculé après la table des matières car je n'arrivais pas à éliminer le petit bug ci-dessous. Mais le compagnon latex fait idem.
|
||||
%\listoftables % un petit problème (bug ?) d'alignement m'a fait retirer provisoirement cette liste
|
||||
|
||||
\end{document}
|
||||
\endinput % termine l'importation des solutions
|
||||
@ -389,14 +389,14 @@
|
||||
\contentsline {subsection}{\numberline {L.1.7}Relatifs \IeC {\`a} la physique aristot\IeC {\'e}licienne}{210}
|
||||
\contentsline {subsection}{\numberline {L.1.8}Relatifs \IeC {\`a} la physique newtonienne}{210}
|
||||
\contentsline {subsection}{\numberline {L.1.9}Relatifs aux forces}{212}
|
||||
\contentsline {subsection}{\numberline {L.1.10}Relatifs \IeC {\`a} l'\IeC {\'e}nergie}{212}
|
||||
\contentsline {subsection}{\numberline {L.1.10}Relatifs \IeC {\`a} l'\IeC {\'e}nergie}{213}
|
||||
\contentsline {subsection}{\numberline {L.1.11}Relatifs \IeC {\`a} la conservation de l'\IeC {\'e}nergie}{213}
|
||||
\contentsline {subsection}{\numberline {L.1.12}Relatifs \IeC {\`a} l'\IeC {\'e}nergie hydraulique}{213}
|
||||
\contentsline {subsection}{\numberline {L.1.13}Relatifs \IeC {\`a} l'\IeC {\'e}nergie \IeC {\'e}olienne}{213}
|
||||
\contentsline {subsection}{\numberline {L.1.14}Relatifs \IeC {\`a} l'\IeC {\'e}nergie solaire}{214}
|
||||
\contentsline {section}{\numberline {L.2}Solutions}{214}
|
||||
\contentsline {section}{\numberline {L.3}Solutions OS}{220}
|
||||
\contentsline {chapter}{\numberline {M}Ordre de grandeur, erreur et incertitudes}{221}
|
||||
\contentsline {section}{\numberline {M.1}Ordre de grandeur}{221}
|
||||
\contentsline {section}{\numberline {M.2}\IeC {\'E}cart et erreur}{221}
|
||||
\contentsline {section}{\numberline {M.3}Incertitude}{222}
|
||||
\contentsline {chapter}{\numberline {M}Ordre de grandeur, erreur et incertitudes}{223}
|
||||
\contentsline {section}{\numberline {M.1}Ordre de grandeur}{223}
|
||||
\contentsline {section}{\numberline {M.2}\IeC {\'E}cart et erreur}{223}
|
||||
\contentsline {section}{\numberline {M.3}Incertitude}{224}
|
||||
|
||||
@ -20,6 +20,71 @@
|
||||
|
||||
\end{Solution OS}
|
||||
\begin{Solution OS}{2}
|
||||
Deux forces agissent ici : la réaction du plan, qui lui est normale (c'est-à-dire perpendiculaire), et le poids de la masse. Comme la réaction du plan n'a aucune composante parallèlement au plan, elle ne peut être responsable de l'accélération de la masse le long de celui-ci.
|
||||
|
||||
Il faut donc trouver la composante du poids qui est parallèle au plan incliné. L'angle entre le poids et un plan horizontal est de \unit{90}{\degree}. Quand le plan est incliné, cet angle diminue de la valeur de l'inclinaison. L'angle \(\beta\) entre le plan incliné et le poids et donc \(\beta=90-\alpha\).
|
||||
|
||||
Comme la projection du poids selon l'angle \(\beta\) correspond à sa composante parallèle au plan, dans le triangle rectangle composé du poids comme hypoténuse et de ses composantes parallèle et perpendiculaire au plan, la composante parallèle au plan correspond au côté adjacent. Ainsi, on peut écrire~:
|
||||
\begin{align*}
|
||||
P_{//}&=P\cdot\cos(\beta)=P\cdot\cos(90-\alpha)\\
|
||||
&=P\cdot\sin(\alpha)
|
||||
\end{align*}
|
||||
La seconde loi de Newton s'écrit donc le long du plan incliné~:
|
||||
\begin{align*}
|
||||
\sum F^{ext}=P\cdot\sin(\alpha)&=m\cdot a\;\Rightarrow\\
|
||||
m\cdot g\cdot\sin(\alpha)&=m\cdot a\;\Rightarrow\\
|
||||
a&=g\cdot\sin(\alpha)\;\Rightarrow\\
|
||||
a&=9,81\cdot\sin(20)=\unit{3,36}{\metre\per\second\squared}
|
||||
\end{align*}
|
||||
Avec une vitesse initiale nulle, pour un MRUA d'accélération calculée ci-dessus, la vitesse au bout d'un temps t=\unit{2}{\second} s'obtient par~:
|
||||
\[v=a\cdot t+v_0=3,36\cdot 2=\unit{9,72}{\metre\per\second}\]
|
||||
|
||||
\medskip
|
||||
Une autre manière de résoudre le problème est de procéder avec méthode. Le système qu'on doit choisir est bien évidemment la masse m, puisque c'est de celle-ci qu'on cherche l'accélération pour en trouver la vitesse au bout de \unit{2}{\second}.
|
||||
\begin{figure}[h]
|
||||
\centering
|
||||
\caption[Plan incliné]{Le plan incliné}\label{incline}
|
||||
\medskip
|
||||
\def\svgwidth{6cm}
|
||||
\input{Annexe-Exercices/Images/incline.eps_tex}
|
||||
\end{figure}
|
||||
|
||||
La figure \ref{incline} présente ensuite le dessin des forces extérieures et le choix du système d'axes. Remarquez que ce dernier l'a été selon l'inclinaison du plan. Il aurait pu ne pas en être ainsi, mais ce choix simplifie les calculs, car la masse étant contrainte à se déplacer le long du plan, son accélération perpendiculairement est nulle. Les équations de la seconde loi de Newton, obtenues par projection des forces extérieures et de l'accélération selon les axes, peuvent alors s'écrire :
|
||||
\begin{align*}
|
||||
\sum F^{ext}_x&=P_x=m\cdot a_x &\text{sur l'axe x}\\
|
||||
\sum F^{ext}_y&=R-P_y=m\cdot a_y=0 &\text{sur l'axe y}
|
||||
\end{align*}
|
||||
car l'accélération perpendiculairement au plan est nulle, comme déjà mentionné.
|
||||
|
||||
\smallskip
|
||||
Si on considére l'angle \(\alpha\), en s'imaginant le plan incliné horizontal, on comprends qu'il se reporte entre le vecteur poids \(\overleftarrow{P}\) et sa compostante selon y \(P_y\).
|
||||
|
||||
Avec le triangle rectangle formé par le poids et ses composantes et un peu de trigonométrie, on peut en déduire~:
|
||||
\begin{align*}
|
||||
P_x &= P\cdot \sin(\alpha)\\
|
||||
P_y &= P\cdot \cos(\alpha)
|
||||
\end{align*}
|
||||
Comme par ailleurs on sait que \(P=m\cdot g\), on peut réécrire les équations de Newton sur les axes comme~:
|
||||
\begin{align*}
|
||||
\sum F^{ext}_x&=m\cdot g\cdot \sin(\alpha)=m\cdot a_x\\
|
||||
\sum F^{ext}_y&=R-m\cdot g\cdot \cos(\alpha)=m\cdot a_y=0
|
||||
\end{align*}
|
||||
La première de ces équations permet de trouver l'accélération du bloc selon le plan incliné~:
|
||||
\begin{align*}
|
||||
a=a_x&=g\cdot \sin(\alpha)\\
|
||||
&=9,81\cdot\sin(20)=\unit{3,36}{\metre\per\second\squared}
|
||||
\end{align*}
|
||||
|
||||
\smallskip
|
||||
Mais une information supplémentaire nous est donnée par la seconde équation, c'est la valeur de la réaction R au plan~:
|
||||
\begin{align*}
|
||||
R&=m\cdot g\cdot \cos(\alpha)\\
|
||||
&=3\cdot 9,81\cdot \cos(20)=\unit{27,67}{\newton}
|
||||
\end{align*}
|
||||
|
||||
|
||||
\end{Solution OS}
|
||||
\begin{Solution OS}{3}
|
||||
Un corrigé de test.
|
||||
|
||||
\end{Solution OS}
|
||||
|
||||
Loading…
Reference in New Issue
Block a user