Problème des deux poulies ajouté.

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Guyot 2019-01-25 22:24:47 +01:00
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166 changed files with 4626 additions and 1680 deletions

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@ -1088,7 +1088,108 @@ On lâche la première à vitesse initiale nulle. Calculez la vitesse de la seco
\end{exos}
\begin{exos}
Un exercice de test.
Une poulie est accrochée au plafond par son axe de rotation. D'un côté elle soutient une masse m de \unit{5}{\kilo\gram} et de l'autre une autre poulie par l'intermédiaire d'une corde accrochée au plafond. À l'axe de rotation de cette dernière est suspendu une autre masse M de \unit{5}{\kilo\gram}. La figure \ref{deuxpoulies} présente la situation.
\begin{figure}[h]
\centering
\caption[Deux poulies]{Deux poulies}\label{deuxpoulies}
\medskip
\def\svgwidth{6cm}
\input{Annexe-Exercices/Images/deuxpoulies.eps_tex}
\end{figure}
Déterminez l'accélération de chaque masse.
\begin{solos}
On verra ci-dessous que l'utilisation de la seconde loi de Newton permet d'obtenir une relation entre l'accélération de chaque masse, constituant une équation à deux inconnues. Pour déterminer la valeur des deux accélérations, il est donc nécessaire de trouver une seconde équation entre ces deux équations.
\medskip
Imaginons une poulie suspendue à une corde qui dépasse de celle-ci des deux côtés d'une longueur L, comme présenté sur la figure \ref{cordepoulie}. La demi-circonférence de la poulie vaut aussi L.
\begin{figure}[h]
\centering
\caption[Corde poulie]{Corde et poulie}\label{cordepoulie}
\medskip
\def\svgwidth{2cm}
\input{Annexe-Exercices/Images/cordepoulie.eps_tex}
\end{figure}
On tire sur la corde à gauche pour la faire monter d'une hauteur L.
\smallskip
La question est : \emph{de quelle hauteur monte la poulie ?}
\smallskip
Pour le déterminer, il faut considérer qu'en tirant sur la corde pour la faire monter d'une hauteur L, on amène le point A qui est à l'origine au contact de la poulie à la place du point B du haut de la corde (voir figure \ref{cordepoulie}). Si la poulie restait à sa place, on aurait la situation de la figure \ref{cordepoulietiree}). Mais, elle est en réalité libre de monter. Ce qui reste fixe est le point C de la figure \ref{cordepoulie}.
\begin{figure}[h]
\centering
\caption[Corde tirée poulie]{Corde tirée et poulie}\label{cordepoulietiree}
\medskip
\def\svgwidth{2cm}
\input{Annexe-Exercices/Images/cordepoulietiree.eps_tex}
\end{figure}
On a donc a répartir une longueur 2L de corde entre le point A de la figure \ref{cordepoulietiree} et le point C de la figure \ref{cordepoulie}. Comme la demi-circonférence de la poulie vaut L, il reste une longueur L à répartir des deux côtés de la poulie, soit L/2 de chaque côté, comme le montre la figure \ref{cordepoulietireejuste}.
\begin{figure}[h]
\centering
\caption[Corde tirée poulie juste]{Corde tirée et poulie montée}\label{cordepoulietireejuste}
\medskip
\def\svgwidth{3.1cm}
\input{Annexe-Exercices/Images/cordepoulietireejuste.eps_tex}
\end{figure}
Ainsi, quand on tire la corde d'une longueur L, la poulie monte d'une longueur L/2.
\bigskip
Pour revenir au système des deux poulies du problème, la remarque précédente se traduit par le fait que quand la masse m descent d'une longueur L, la masse M monte d'une longueur L/2.
L'accélération étant une distance divisée par un temps au carré, on comprends facilement que cela signifie que l'accélération de la masse m vaut simplement le double de celle de la masse M, soit :
\[a_m=2\cdot a_M\]
Cette équation constitue une première relation entre les deux inconnues que sont les accélérations de chaque masse.
\bigskip
Les masses étant égales, on pourrait aussi croire que le système est en équilibre. Ce n'est pas le cas. Pour le comprendre, considérons la poulie qui n'est pas accrochée au plafond.
\smallskip
À l'instar d'une poulie suspendue au plafond à laquelle on accroche deux masses identiques pendantes par l'intermédiaire d'une corde, la force totale qu'exerce sur elle le plafond vaut évidemment le poids total des deux masses, puisque la poulie ne bouge pas. Or, pour que chaque masse individuellement ne bouge pas, il faut que la corde qui passe dans la poulie exerce sur chacune d'elle une force égale à son poids. De chaque côté de la poulie, la corde exerce donc une même force et la poulie est tirée vers le bas par l'ensemble de ces deux forces.
\smallskip
Ainsi, sur la poulie qui n'est pas accrochée au plafond, la force totale exercée vers le haut vaut deux fois la tension dans la corde. Vers le bas, seule le poids de la masse M qui lui est suspendue est présent.
\smallskip
De l'autre côté, la masse m retenue par la corde qui passe sur la poulie suspendue au plafond est soumise à un poids identique vers le bas et à une seule force vers le haut exercée par la corde. Comme la masse de la corde est nulle, la tension dans la corde est la même partout.
\smallskip
Finalement, la masse M est tirée vers le haut par deux fois la tension dans la corde et la masse m est retenue par une fois la tension dans la corde. Clairement donc, la première monte et la seconde descend.
\medskip
Le choix du système d'axe est donc clair : vers le haut pour la masse M qui monte et vers le bas pour m qui descend, le mouvement se faisant dans cette direction avec une accélération a identique pour les deux masses.
\smallskip
On peut alors écrire les équations du mouvement de chaque masse ainsi :
\begin{align*}
m\cdot g-T&=m\cdot a_m\\
2\cdot T-M\cdot g&=M\cdot a_M
\end{align*}
En multipliant par deux la première équation, on peut les additionner pour en tirer l'accélération.
\begin{align*}
2\cdot m\cdot g-2\cdot T&=2\cdot m\cdot a_m\\
+\;\;\;\;\;\;\;\;\;\;2\cdot T-M\cdot g&=M\cdot a_M\\
\hline\\
2\cdot m\cdot g-M\cdot g&=2\cdot m\cdot a_m+M\cdot a_M
\end{align*}
Cela constitue la seconde relation entre les deux inconnues que sont les accélérations de chaque masse.
\medskip
Avec la première relation établie ci-dessus entre les accélérations des deux masses (\(a_m=2\cdot a_M\)), on a alors~:
\begin{align*}
2\cdot m\cdot g-M\cdot g&=2\cdot m\cdot a_m+M\cdot a_M\;\Rightarrow\\
2\cdot m\cdot g-M\cdot g&=2\cdot m\cdot 2\cdot a_M+M\cdot a_M\;\Rightarrow\\
2\cdot m\cdot g-M\cdot g&=4\cdot m\cdot a_M+M\cdot a_M\;\Rightarrow\\
a_M&=\frac{2\cdot m-M}{4\cdot m+M}\cdot g
\end{align*}
Dans le cas où les deux masse sont identiques (M=m), on a alors :
\[a_M=\frac{2\cdot m-m}{4\cdot m+m}\cdot g=\frac{m}{5\cdot m}\cdot g=\frac{1}{5}\cdot g\]
et pour l'accélération de m~:\[a_m=2\cdot a_M=\frac{2}{5}\cdot g\]
\end{solos}
\end{exos}
\begin{exos}
Un énoncé de test.
\begin{solos}
Un corrigé de test.
\end{solos}

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@ -1088,7 +1088,105 @@ On lâche la première à vitesse initiale nulle. Calculez la vitesse de la seco
\end{exos}
\begin{exos}
Un exercice de test.
Une poulie est accrochée au plafond par son axe de rotation. D'un côté elle soutient une masse m de \unit{5}{\kilo\gram} et de l'autre une autre poulie par l'intermédiaire d'une corde accrochée au plafond. À l'axe de rotation de cette dernière est suspendu une autre masse M de \unit{5}{\kilo\gram}. La figure \ref{deuxpoulies} présente la situation.
\begin{figure}[h]
\centering
\caption[Deux poulies]{Deux poulies}\label{deuxpoulies}
\medskip
\def\svgwidth{6cm}
\input{Annexe-Exercices/Images/deuxpoulies.eps_tex}
\end{figure}
Déterminez l'accélération de chaque masse.
\begin{solos}
On verra ci-dessous que l'utilisation de la seconde loi de Newton permet d'obtenir une relation entre l'accélération de chaque masse, constituant une équation à deux inconnues. Pour déterminer la valeur des deux accélérations, il est donc nécessaire de trouver une seconde équation entre ces deux équations.
\medskip
Imaginons une poulie suspendue à une corde qui dépasse de celle-ci des deux côtés d'une longueur L, comme présenté sur la figure \ref{cordepoulie}. La demi-circonférence de la poulie vaut aussi L.
\begin{figure}[h]
\centering
\caption[Corde poulie]{Corde et poulie}\label{cordepoulie}
\medskip
\def\svgwidth{2cm}
\input{Annexe-Exercices/Images/cordepoulie.eps_tex}
\end{figure}
On tire sur la corde à gauche pour la faire monter d'une hauteur L.
\smallskip
La question est : \emph{de quelle hauteur monte la poulie ?}
\smallskip
Pour le déterminer, il faut considérer qu'en tirant sur la corde pour la faire monter d'une hauteur L, on amène le point A qui est à l'origine au contact de la poulie à la place du point B du haut de la corde (voir figure \ref{cordepoulie}). Si la poulie restait à sa place, on aurait la situation de la figure \ref{cordepoulietiree}). Mais, elle est en réalité libre de monter. Ce qui reste fixe est le point C de la figure \ref{cordepoulie}.
\begin{figure}[h]
\centering
\caption[Corde tirée poulie]{Corde tirée et poulie}\label{cordepoulietiree}
\medskip
\def\svgwidth{2cm}
\input{Annexe-Exercices/Images/cordepoulietiree.eps_tex}
\end{figure}
On a donc a répartir une longueur 2L de corde entre le point A de la figure \ref{cordepoulietiree} et le point C de la figure \ref{cordepoulie}. Comme la demi-circonférence de la poulie vaut L, il reste une longueur L à répartir des deux côtés de la poulie, soit L/2 de chaque côté, comme le montre la figure \ref{cordepoulietireejuste}.
\begin{figure}[h]
\centering
\caption[Corde tirée poulie juste]{Corde tirée et poulie montée}\label{cordepoulietireejuste}
\medskip
\def\svgwidth{3.1cm}
\input{Annexe-Exercices/Images/cordepoulietireejuste.eps_tex}
\end{figure}
Ainsi, quand on tire la corde d'une longueur L, la poulie monte d'une longueur L/2.
\bigskip
Pour revenir au système des deux poulies du problème, la remarque précédente se traduit par le fait que quand la masse m descent d'une longueur L, la masse M monte d'une longueur L/2.
L'accélération étant une distance divisée par un temps au carré, on comprends facilement que cela signifie que l'accélération de la masse m vaut simplement le double de celle de la masse M, soit :
\[a_m=2\cdot a_M\]
Cette équation constitue une première relation entre les deux inconnues que sont les accélérations de chaque masse.
\bigskip
Les masses étant égales, on pourrait croire que le système est en équilibre. Ce n'est pas le cas. Pour le comprendre, considérons la poulie qui n'est pas accrochée au plafond.
\smallskip
À l'instar d'une poulie suspendue au plafond à laquelle on accroche deux masses identiques pendantes par l'intermédiaire d'une corde, la force totale qu'exerce sur elle le plafond vaut évidemment le poids total des deux masses, puisque la poulie ne bouge pas. Or, pour que chaque masse individuellement ne bouge pas, il faut que la corde qui passe dans la poulie exerce sur chacune d'elle une force égale à son poids. De chaque côté de la poulie, la corde exerce donc une même force et la poulie est tirée vers le bas par l'ensemble de ces deux forces.
\smallskip
Ainsi, sur la poulie qui n'est pas accrochée au plafond, la force totale exercée vers le haut vaut deux fois la tension dans la corde. Vers le bas, seule le poids de la masse M qui lui est suspendue est présent.
\smallskip
De l'autre côté, la masse m retenue par la corde qui passe sur la poulie suspendue au plafond est soumise à un poids identique vers le bas et à une seule force vers le haut exercée par la corde. Comme la masse de la corde est nulle, la tension dans la corde est la même partout.
\smallskip
Finalement, la masse M est tirée vers le haut par deux fois la tension dans la corde et la masse m est retenue par une fois la tension dans la corde. Clairement donc, la première monte et la seconde descend.
\medskip
Le choix du système d'axe est donc clair : vers le haut pour la masse M qui monte et vers le bas pour m qui descend, le mouvement se faisant dans cette direction avec une accélération a identique pour les deux masses.
\smallskip
On peut alors écrire les équations du mouvement de chaque masse ainsi :
\begin{align*}
m\cdot g-T&=m\cdot a_m\\
2\cdot T-M\cdot g&=M\cdot a_M
\end{align*}
En multipliant par deux la première équation, on peut les additionner pour en tirer l'accélération.
\begin{align*}
2\cdot m\cdot g-2\cdot T&=2\cdot m\cdot a_m\\
+\;\;\;\;\;\;\;\;\;\;2\cdot T-M\cdot g&=M\cdot a_M\\
\hline\\
2\cdot m\cdot g-M\cdot g&=2\cdot m\cdot a_m+M\cdot a_M
\end{align*}
Avec la relation établie ci-dessus entre les accélérations des deux masses (\(a_m=2\cdot a_M\)), on a~:
\begin{align*}
2\cdot m\cdot g-M\cdot g&=2\cdot m\cdot a_m+M\cdot a_M\;\Rightarrow\\
2\cdot m\cdot g-M\cdot g&=2\cdot m\cdot 2\cdot a_M+M\cdot a_M\;\Rightarrow\\
2\cdot m\cdot g-M\cdot g&=4\cdot m\cdot a_M+M\cdot a_M\;\Rightarrow\\
a_M&=\frac{2\cdot m-M}{4\cdot m+M}\cdot g
\end{align*}
Dans le cas où les deux masse sont identiques (M=m), on a alors :
\[a_M=\frac{2\cdot m-m}{4\cdot m+m}\cdot g=\frac{m}{5\cdot m}\cdot g=\frac{1}{5}\cdot g\]
et pour l'accélération de m~:\[a_m=2\cdot a_M=\frac{2}{5}\cdot g\]
\end{solos}
\end{exos}
\begin{exos}
Un énoncé de test.
\begin{solos}
Un corrigé de test.
\end{solos}

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