Examen 8 Novembre 2017, réponses PDF

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Université Lille 1  Sciences et Technologies Année 20172018 Licences SESI & SESI/PEIP  Semestre 1 Bases de la Mécanique Devoir surveillé  Novembre 2017 / Corrigé Aucun document ni appareil électronique autorisé Durée : 2 heures Questions de cours Question 1 Il existe un référenti...

Description

S { ( )} F S → S = {0} S1 S2

S2 S1

{F (S1 → S2 )} = − {F (S2 → S1 )}

A C − → n A = +x ,

− → n B = +y , S 2

A

n A O 2

− → n C = −x S 1

D n C O 1

y n B

z

B x

C

B

S1 } { {− → } −m1 g y P1 {F1 } = = − → − → 0 0 O O1 1

S2 } { {− → } −m2 g y P2 {F2 } = = − → − → 0 0 O O2 2

A {− { } → } → nA NA − RA {F3 } = = − → − → 0 0 A A B { } {− → } → nB NB − RB = {F4 } = − → − → 0 0 B B C { } {− → } → nC NC − RC = {F5 } = − → − → 0 0 C C A

B

C

NA > 0 ,

R a y o n r

NB > 0 ,

a

R a y o n 2 r S 2

A

NC > 0

S 1

R A O 2

P 2

b

D

P 1 g O 1

y

C R C

R B

z

B x

O1 − → − → − → − → − → − → RA + RB + RC + P 1 + P 2 = 0 → −−−→ − → −−→ − − → O1 A ∧ R A + O1 O2 ∧ P 2 = 0



−−→ − → O1 A ∧ R A = 

  −(a + r)  +NA  ∧  0  = b

0 0

0  0  −N  A b  0 −a 0 → −−−→ − O1 O2 ∧ P 2 =  b  ∧  −m2 g  =  0  0 0 m2 ga

 

NA − NC = 0 NB − (m1 + m2 )g = 0 − NA b + m2 ga = 0 NA NB

NC

NB = (m1 + m2 )g > 0 m2 ga >0 b m2 ga >0 NC = NA = b

NA =

D → −−→ − − → DE ∧ R 42 = 0 −−→ DE

− → R 42

− → R 42 − → R 42 = X42 x + Y42 y Y42 = 0

S1 S3 S1 } } { {− → Fx x + Fy y F {F31} = − = − → → 0 0 A A

S2

(E, x)

S4 S1 } { {− → } X41 x + Y41 y R 41 {F41} = = − → − → 0 0 C C S4 S2 } } { {− → X42 x R 42 {F42} = = − → − → 0 0 E E

a

a S 2 a

R 42 E

D

y

S 1 d

R 41 C

z

F B

A

x

C − → → − → − → − R 41 + R 42 + F = 0 → −→ − → − −−→ − → CE ∧ R 42 + CA ∧ F = 0 −−→ − → CE ∧ R 42



      X42 0 0 = d ∧  0  =  0  , −X42d 0 0

     F 0 2a −→ − →    x  0  0 ∧ Fy = CA ∧ F = 2Fy a 0 0

X41 + X42 + Fx = 0 Y41 + Fy = 0 − X42d + 2Fy a = 0 X41 Y41

X42

Y41 = −Fy 2Fy a X42 = d X41 = −X42 − Fx = −Fx −

2Fy a d

Fx = −F sin α ,

Fy = +F cos α

− → F

− → P

} { {− → } −mg z P {F1 } = − = − → → 0 0 G G } { {− → } −F y F {F2 } = − = − → → 0 0 C C } {− { → } ROx x + ROy y + ROz z RO {F3 } = − = → MOx x + MOy y + MOz z MO O O

2 a

S 2 S 1 A C

2 b F

c H G g

z P

x

d M O y

R O

L

O

O − → − → − − → → RO + F + P = 0 − → → −−→ − → −−→ − → − M O + OC ∧ F + OG ∧ P = 0

      L−a 0 F (H − b) →  −−→ −  , OC ∧ F = 0  ∧  −F  =  0 H −b 0 −F (L − a)

      0 0 c →  −−→ − OG ∧ P = 0  ∧  0  =  mgc  0 d −mg

ROx = 0 ROy − F = 0 ROz − mg = 0 MOx + F (H − b) = 0 MOy + mgc = 0 MOz − F (L − a) = 0 ROx ROy ROz MOx MOy

− → RO

 0 = F  , mg 

 −F (H − b) − → M O =  −mgc  F (L − a) 

MOz...