1 esercizio 1 esonero Meccanica applicata alle macchine

VWED

N ( ED)

+

CORIZZ )

. VE

diz

4g/

* .

VD TRIANGOLO

~ IVEI m/5

IVD) 5

'SOSCELE 66

=

= .

RETTAN GOLO

diz 41 (E D)

-

. ..

D

· cost

0

-ini-a-we a

(B - B

a

-

2

wi 500 m/s

B 08

EB = 200

=

= &

.

Be relativa

Corpoz cinematica

-

AB azeeB IB +

+

=

- aTBc

N -w

EB= + N

VzelB

= 23 1

-N -

Wis

diz C

1/B-

. 131(B 2) =

-

12w3nVzeelaBl 45

Sim

aB = .

-

wa ↳

C

(B 2w3/vzeel

- leblaim

die

- is

& area

45) +

w3

↓ =

,

A BC

IVBicos45

IVzeel m/s

93

2

= = .

zwanvece ↳ Orario

(verso

zade

l3 1654

=

DE 3

Corpo wE(D c)

c)

w3r(D

ED -

- -

= ap + 3 m/s

lie(d-c)) 1654 3 496

= =

0

. . .

2

I-w(D-1) 7 m/s

86 3

18 106

= 0 =

. .

. . m/s

32 72

10 Sot

1 796 6

=

106

= + .

.

.

e ar t

Wb(B-) =

↳ ↑

2 c)

(B

wz

= -

VE 2

↑

si C G &

O · H

4 B 3

&

90XF

1 .

se

! BC 15cm 15

B DH

BE :

30cm

AC CH

A CG

= GH

= :

= =

= cm

E =

:

4 m/s (costante M Kg

=

VE 100

=

vor

· (B-

w4 &

E)

w4n(B

VB = VE + - ·4

YA P

A)

VB win(B

+

= E se ve

- ,

-

Es

= e

18 86

W4 .

ORARIA

Zad/S ANTIORARIA

18 86

WH W4

= = .

fa W4C

Ival

a)

w4n(c =

va +

= - D)(VH) +

YD w3n(H

Vi w3

+ = D

= - improprio

Al / DH-

+ 4

R punto

corpo

C . .

↳ V VG

= = - m/s

(VE) 5 66

56

18

3 =

= 0 . .

.

.

ag

· wi(B

da A)

win(B 1)

ab +

= - -

- wi(B

a E) E)

w41/B

aB = =

+ - - wi(B

wY(B E)

w1n(B w4n(B

A) E)

A) =

- -

-

-

-

- N

d

d N wi(B 36m/

))

1 53

=

& wülB-El -

-

E)

w4n(B .

-

- 45m/b

Va 75

53 36

=B =

< = . . .

wi(B A)

- -

A) =I=

w11(B 7

- w4

Wi 355

= .

WI

4 ANTIORARIA

ANTIORARIO 2m/

wi(c

= a)

wan(c 150

a) 2qB

= .

2 + =

-

-

- wz(H D

b)

ab w3r(H

+

an -

= - -

w((H c

c)

w2r(H

+

alt = ac - -

-

D)

wSr(H

1

Vi = -

+ S

a)

w2r(c 86

18

w3

Vi Vc = w2

= = =

=

- .

HD

LA =

CA/ID

(4 0

= 9 m/s

150

=

ab ac

= .