MECANIQUE RATIONNELLE

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UMBB Boumerdès, Faculté des sciences, Département de physique Cours exercices, Mécanique Rationnelle : TCT et LMD-ST sem :3 A.KADI On peut vérifier facilement : → → → −→ ⎛ ⎞ ⎜ −15 i + 5 k → → ⎟ ⎛ n • AB = • ⎜− 2 + ⎜ 106 ⎟ i ⎝ ⎝ ⎠ →⎞ j− 6 k ⎟ = 30 − 30 = 0 ⎠ → → → −→ ⎛ ⎞ ⎜ −15 i + 5 k → → → ⎟ ⎛ ⎞ n • AC = • ⎜− − 2 − 3 ⎟ = 15 −15 = 0 ⎜ 106 ⎟ i j k ⎝ ⎠ ⎝ ⎠ → → → −→ ⎛ ⎞ ⎜ −15 i + 5 k → → → ⎟ ⎛ ⎞ n • BC = • ⎜ − 3 + 3 ⎟ = −15 + 15 = 0 ⎜ 106 ⎟ i j k ⎝ ⎠ ⎝ ⎠ La composante, du vecteur, suivant la normale au plan s’écrirait : → → → → → → → → → → → ⎛ ⎞ ⎛⎛ ⎞ 1 ⎛ ⎞⎞ 65 V n = ⎜V • n ⎟ n = ⎜⎜3 i + j− 4 k ⎟ • ⎜−15 i + 5 k ⎟⎟ n = − n ⎝ ⎠ ⎝⎝ ⎠ 106 ⎝ ⎠⎠ 106 → → → ⎛ ⎞ 65 → 65 ⎜ −15 i + 5 k ⎟ 1 → → ⎛ ⎞ V n = − n = − ⎜ ⎟ = ⎜975 i − 325 k ⎟ 106 106 106 106 ⎝ ⎠ ⎝ ⎠ La composante dans le plan (π ) se déduit par : → → → → → → ⎛ ⎞ 1 → → ⎛ ⎞ 1 → → → ⎛ ⎞ Vπ = V −Vn = ⎜3 i + j− 4 k ⎟ − ⎜975 i − 325 k ⎟ = ⎜− 657 i + j− 99 k ⎟ ⎝ ⎠ 106 ⎝ ⎠ 106 ⎝ ⎠ Exercice 15 : Déterminer l’expression générale des vecteurs W orthogonaux aux vecteurs : → → → → → V = − i + 2 j+ 3k 1 → → → → et V = i + 3 j− 5 k . En déduire les vecteurs unitaires porté par W . 2 → Exercice 16 : → → → Soient trois vecteurs libres U , V , W ; montrer qu’il vérifient la relation suivante : → → → → → → → → → → ⎛ ⎞ ⎛ ⎞ ⎛ ⎞ U ∧ ⎜ V ∧W ⎟ + W ∧ ⎜ U ∧ V ⎟ + V ∧ ⎜ W ∧ U ⎟ = 0 ⎝ ⎠ ⎝ ⎠ ⎝ ⎠ Solution : On utilise la formule de développement du double produit vectoriel. → → → ⎛ U ∧ ⎜ V ∧W ⎝ ⎞ ⎟ ⎠ → = V ⎛ ⎜ ⎝ → → → ⎞ U • W ⎟ −W ⎠ ⎛ ⎜ ⎝ → U → ⎞ • V ⎟ ⎠ 43
UMBB Boumerdès, Faculté des sciences, Département de physique Cours exercices, Mécanique Rationnelle : TCT et LMD-ST sem :3 A.KADI → → → → ⎛ ⎞ W ∧ ⎜ U ∧V ⎟ = U ⎝ ⎠ ⎛ ⎜ ⎝ → → → ⎞ W • V ⎟ −V ⎠ ⎛ ⎜ ⎝ → W ∧ → U ⎞ ⎟ ⎠ → → → → ⎛ ⎞ V ∧ ⎜ W ∧U ⎟ = W ⎝ ⎠ ⎛ ⎜ ⎝ → → → ⎞ V • U ⎟ −U ⎠ ⎛ ⎜ ⎝ La somme des trois termes donne : → V → V ⎛ ⎜ ⎝ ⎛ ⎜ ⎝ → → → ⎞ U• W ⎟ −W ⎠ → → → ⎞ U• W ⎟ −V ⎠ ⎛ ⎜ ⎝ ⎛ ⎜ ⎝ → → → ⎞ U• V ⎟ + U ⎠ → → → ⎞ W • U ⎟ −W ⎠ ⎛ ⎜ ⎝ ⎛ ⎜ ⎝ → → V → ⎞ • W ⎟ ⎠ → → → ⎞ W • V ⎟ −V ⎠ U → ⎞ • V ⎟ + ⎠ Comme le produit scalaire est commutatif alors : → W ⎛ ⎜ ⎝ ⎛ ⎜ ⎝ → → → ⎞ W • U ⎟ + W ⎠ → → → ⎞ V • U ⎟ + U ⎠ → → → → → → → → → → → → → ⎛ ⎞ ⎛ ⎞ ⎛ ⎞ ⎛ ⎞ ⎛ ⎞ ⎛ ⎞ ⎜V −V ⎟ ⎜ W • U ⎟ + ⎜W −W ⎟ ⎜ V • U ⎟ + ⎜U −U ⎟ ⎜ V • W ⎟ = 0 ⎝ ⎠ ⎝ ⎠ ⎝ ⎠ ⎝ ⎠ ⎝ ⎠ ⎝ ⎠ Exercice 17 : → F 1 → F 2 ⎛ ⎜ ⎝ ⎛ ⎜ ⎝ → → → ⎞ V • U ⎟ −U ⎠ → → → ⎞ W • V ⎟ −U ⎠ ⎛ ⎜ ⎝ ⎛ ⎜ ⎝ → V → V → ⎞ • W ⎟ ⎠ = → → ⎞ • W ⎟ = 0 ⎠ Soient deux forces et faisant chacune respectivement un angle de 25° et 35° avec la résultante Solution : → R qui a une valeur de 400 N . Déterminer les modules des deux forces. Utilisons la règle des sinus : BC AB AC = = sin 25° sin 35° sinα α = 180° − (25° + 35° ) = 120° or nous avons : AB = F 1 , BC = F2 et AC = R A → F 2 35° 25° → F 1 → R α B 35 ° C D’où : sin 25° sin 35° F2 = R = 195N et F1 = R = 265N sin120° sin120° Exercice 18 : → → → → 2 3 3 2 Soit P = 2t i + 5t j− 7t k , Q = −4t i + 10t j− 2t k → → → → 1) Vérifier les relations suivantes : d dt → → ⎛ ⎞ ⎜ P• Q⎟ = ⎝ ⎠ → → • Q d P dt + → → d Q P• dt d dt → → ⎛ ⎞ ⎜ P∧ Q⎟ = ⎝ ⎠ → d P → ∧ Q dt + → → d Q P∧ dt 44
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MECANIQUE RATIONNELLE

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