Lösungen für Aufgaben zur Technischen Mechanik – Statik · Lösungen für Aufgaben zur Technischen Mechanik – Statik - Lösung 2.1.8 Lösung 2.2.3 Lösung 2.2.7 S1 FS1 FS2 FG

Lösungen für Aufgaben zur Technischen Mechanik – Statik - Lösung 2.1.8 

 Lösung 2.2.3 

   Lösung 2.2.7 

FS1FS2

FG = mgK

α1 α2

x

yF1

F2

F3F4

0A

B

r

S

0 F

FG

FS1 FS2

FS3

→: -FS1cosα1 + FS2cosα2 = 0 F FS

S2

1=cos

cos1

2

αα

↑ : FS1sinα1 + FS2sinα2 - mg = 0 FS1(sinα1 + tanα2 cosα1) = mg

°===−−+

=

°====

1,34677,06062,4tan

3,278889,05,4

4cos

2

22

2

11

αα

αα

bcbla

lb

Ergebnis: FS1 = 277N, FS2 = 297N

°=°−==

=+=+=

30060tantan

23 2222

RRx

RyR

RyRxR

FF

FFFFFF

αα

FRx = Σ Fix = F1cos60° + F2cos60° - F3cos60° + F4cos60° = 2Fcos60° = F FRy = Σ Fiy = F1sin60° - F2sin60° - F3sin60° - F4sin60° = -2Fsin60° =−F 3

← : F + FS1 cos 60° - FS3 cos 45° = 0 ↑ : - FS2 - FS1 sin 60° - FS3 sin 45° - FG = 0 ↶0 : 2FS1a sin60° - FS1a cos60° + FG a = 0

Ergebnisse: FS1 = - 812 N....Druckstab FS2 = - 891 N....Druckstab FS3 = 840 N....Zugstab

Lösung 3.1.2 

Ergebnis:   Lösung 3.1.12 

Ergebnis: FB = 3,82F; FAH = - 1,09F; FAV = 2,7F Lösung 3.2.3 

Teil I: → : FAH - FGH = 0 ↑ : FAV - FGV - F1 - 5qa = 0 A : MA - F1 . 3a - 5qa . 2,5a - FGV

. 5a = 0 Teil II: → : FGH = 0

↑ : FB + FGV - F2= 0 G : -F2 . 3a + FB . 5a = 0

FC

FAH FAV F sinα

F cosα

3F FBcos30°

FAHFAV

4F

2F

FBsin30°

FAH FAV

2,5a 5qaF1

FGV

FGH

FGV

FGHF2

2a FB

I

II

MA

kNFFF

kNFosFF

kNFFF

AV

AH

C

1343sin

21

1521c

39343sin

23

===

===

===

α

α

α

→ : FAH - F cosα = 0 ↑ : -FAV - F sinα + FC = 0 ↶A : -F sinα . 3a + FC

. 2a = 0

→ : FAH + 3F - FB sin30° = 0 ↑ : FAV + FB cos30° - 2F - 4F = 0 ↶A : FB cos30° . 4a + FBsin30° . 2a - 2F . a - 3F . a - 4F . 3a = 0

Ergebnis:

F F N F F F N F F F F qa N

M F a F a qa Nm

B GH GV AH AV

A

= = = = = = = + + =

= + + =

35

960 0 25

640 0 25

5 2440

3 2 252

8100

2 2 1 2

1 22

Lösung 3.2.9 

→ : - FAH + FGHl = 0 ↑ : FAV - FGVl = 0 A : -FGVl a + MA = 0 → : - FGHl + FGHr = 0 ↑ : -FGVr + FGVl - F1 = 0 → : -FGHr + F2 = 0 ↑ : FB + FGVr = 0 G : F2 c + FB a = 0

Ergebnis: Lösung 3.2.22 

Symmetrie: FAH = 0; FAV = FB = 1,5F = 30kN

Zur Geometrie:

l a atan aa

l a

l a l a

1 1

2 1

3 46

23

113

6 73

23

33 7 311

15 25

97

52 15

= + = = =

= − =

= → = ° = → = °

= → = °

α α

α α β β

γ γ

; ;

, ,

,

tan

tan tan = al

tan = 3al

1

2

→: FS1cosα + FS2cosβ = 0 ↑ : FAV + FS1sinα + FS2sinβ = 0 FS1= -91,4kN FS2 = 78,8kN

FAV

FAH

MA FGVl FGHl

F1

F2

FB

FGHl

FGVl FGVr

FGHr

FGVrFGHr

FF

F

FB

FAH

FAV

III

III

3a 3a

2a

a a

l1 l2

γ α

α β

atanα

I

FAV

FS1FS2

α β

;;

;;

;;

221

2121

222

acFFcFaFM

acFFF

acFFF

acFFFFFFF

GVrA

GVlAV

BGHrGHlAH

⋅=⋅+⋅=

⋅+=⋅+=

⋅−====

: -FS1 + FS4 - Fsinα = 0 : -FS3 - Fcosα = 0 FS3 = -16,6kN FS4 = -80,4kN

→: FS6 -FS3sinα + FS5 cosγ - FS2cosβ = 0 ↑ : FS5sinγ + FS3cosα - FS2sinβ = 0 FS5 = 43,9kN FS6 = 40kN

Lösung 3.2.25 

→: F - FAH = 0 FAH = F ↑ : FB - F + FAV = 0 FAV = 0,25F A : -F . a - F . 2a + FB . 4a = 0 FB = 0,75 F

I F a F a F a F F

II F a F a F F

F F F F

aus F F

S B S

S B S

S B S

S

:

:

:

,

3 3

1 1

2 2

2

32

0 18

2 0 32

0 34sin

2 0 84

⋅ − ⋅ + ⋅ = = −

− ⋅ + ⋅ = =

↑ + = = −

= ° = −

sin

tan folgt = 63,4

αα

α α

Lösung 3.3.2 

→ : - FBH + F2 cos60° + F3 cos60° = 0 ↑ : FA + F2 sin60° - F1 - F3 sin60° + FBV = 0 A : - F1

. a + F2 sin60° . 2a - F3 sin60° . 4a + FBV . 3a = 0 Ergebnis: FA = 756 N; FBH = 1000 N; FBV = 1244 N

0 1≤ ≤z a → : FL(z1) = 0; ↑ : FQ(z1) = FA = 756 N S1 : M(z1) = FA z1;

z1 = 0: M(0) = 0 z1 = a: M(a) = 756 Nm

II

F

FS1

FS4

FS3α

IIIFS6

FS5FS3

FS2

α γβ

FF

FB

FAH

FAV

FS3F

FB

FS1

FS2

αI

II

FBHFA

F1

F2

F3

FBV

FAz1

FL(z1)

FQ(z1)

M(z1)

0 2≤ ≤z a

→ : FL(z2) = 0; ↑ : FQ(z2) = FA - F1 = -1244 N S2 : M(z2) = FA (a + z2) - F1 z2; z2 = 0: M2(0) = 756 Nm z2 = a: M2(a) = -488 Nm

0 3≤ ≤z a

→ : FL(z3) = - FBH + F3 cos 60° = - 500 N; ↑ : FQ(z3) = - FBV + F3 sin 60° = -378 N; S3 : M(z3) = FBV (a - z3) - F3 sin 60° . (2a-z3); z3 = 0: M3(0) = - 488 Nm z3 = a: M3(a) = -866 Nm

0 4≤ ≤z a

→ : FL(z4) = F3 cos 60° = 500 N; ↑ : FQ(z4) = F3 sin 60° = 866 N; S4 : M(z4) = - F3 sin 60° . (a-z4); z4 = 0: M4(0) = - 866 Nm z4 = a: M4(a) = 0

M Nmmax = 866

FL(z2)a

F1 M(z2)

FQ(z2)FA

z2

FL(z3)a

F3M(z3)

FQ(z3)

FBVz3 a-z3FBH

FL(z4)

F3M(z4) FQ(z4)

z4 a-z4

FBHFA

F1

F2

F3

FBV

FL in N+-

FQ in N

500

500

+ +

-

866

378

1244

756

M in Nm866

488

756

-+

Lösung 3.3.6 

A: M Fc ql ql c lA = − = −⎛

⎝⎜⎞⎠⎟

12 2

2

M ist

positiv für c l

für c l

negativ für c lA

>

=

<

⎧

⎨

⎪⎪

⎩

⎪⎪

20

2

2

B: M Fc qlcB = =

MB ist positiv für beliebige c, bei c l≥

2 ist MB > MA, ⏐MA⏐=⏐MB⏐ist nur für

negatives MA möglich.

- MA = MB oder − −⎛⎝⎜

⎞⎠⎟ = =ql c l qlc c l

2 4

Lösung 3.3.7 

→ : FAH = 0 ↑ : - 4qa + FAV + FB = 0 A : -4qa . a + FB . 3a = 0

Ergebnis: ;3,1334;7,26

38;0 kNqaFkNqaFF BAVAH =====

0 1≤ ≤z a → : FL(z1) = 0 ↑ : FQ(z1) = - qz1 z1 = 0: FQ(0) = 0 z1 = a: FQ(a) = -qa = -10kN

S1 : M(z1) = - 12 1

2qz ;

z1 = 0: M(0) = 0 z1 = a: M(a) = - 0,5 qa2 = - 5kNm

0 32≤ ≤z a → : FL(z2) = 0 ↑ : FQ(z2) = - FB + q(3a - z2)

z2 = 0: FQ(0) = 53

qa = 16,7kN;

z2 = 3a: FQ(a) = − 43

qa = - 13,3kN

S2 : M(z2) = FB . (3a-z2) - ( )12

3 22q a z− ;

Fql

l

B

FAVFAH

MA c

BMB

c

F

FAH FB3a

4qa2a

FAV

a

z1

FL(z1)

FQ(z1)

M(z1)qz1

FL(z2) FB

M(z2) FQ(z2)

z2 3a-z2

q(3a-z2)

z2 = 0: M2(0) = − 12

2qa = -5kNm z2 = 3a: M2(3a) = 0

Besondere Werte:

- Querkraft-Nullstelle: mazzFz Q 66,1350)(: 222 ===

- Extremwert für M: ( )M z qa kNm228

98 9= = , = Mmax

Lösung 3.3.13 

→ : FAH + F = 0 ↑ : FAV - qa = 0 A : MA - F . a - qa . 0,5a = 0 Ergebnis: FAV = qa; FAH = -qa; MA = 1,5qa2

0 1≤ ≤z a ← : FL(z1) = 0 ↑ : FQ(z1) = -qz1 FQ(0) = 0 FQ(a) = -qa S1 : M(z1) = - 0,5q z1

2 z1 = 0: M(0) = 0 z1 = a: M(a) = - 0,5qa2

0 2≤ ≤z a ← : FQ(z2) = 0 ↓ : FL(z2)= - qa S2 : M(z2) = - 0,5qa2

-

--

+

+

FQ

M

10kN

16,7kN

13,3kN

5kNm

8,9kNm

FAV

F

MA

FAH

qa

FL(z1)

M(z1) FQ(z1)

z1

qz1

FL(z2)

z2

M(z2)

FQ(z2)

qa

0 3≤ ≤z a ← : FQ(z3) =- F = -qa ↓ : FL(z2)= - qa S2 : M(z2) = - 0,5qa2 - Fz3 z3 = 0: M(0) = - 0,5qa2 z3 = a: M(a) = - 1,5qa2

⏐ Mmax ⏐ = 1,5 qa2

Lösung 3.3.23 

Aus Symmetriegründen im Teil I folgt F F qaA GV= =12

0 ≤ z ≤ l ↑ : FQ(z) = FA -qz

S : M(z) = FA . z - 12

2qz

M(0,5a) = 0,125qa2; M(a) = 0 M(l) = - 0,5ql(l-a) ⏐Mmax⏐I = ⏐Mmax⏐II

( )18

12

4 4 4 4 0 2 2 1 0 8282 2 2 2 2qa ql l a a l la a la l a l l= − = − + − = = − =( ) ,

Lösung 4.1.3 

x: -FS2 sin45° + FS1 sin45° = 0 FS1 = FS2 y: -FS2 cos45° - FS1 cos45° - FS3 sin45° = 0 z: -F - FS3 cos45° = 0

Ergebnis: F F F N F F NS S S1 2 32

2707 2 1414= = = = − = −;

FL(z3)

z3

M(z3)

FQ(z3)

qa

F

FL in qa FQ in M in qa2

- -

-

-- 1

1 1

1,5

0,5

zFL(z)

FQ(z)

M(z)

FA

qz

45°

45°45°

Fx

y

z

FS1

FS2

FS3

Lösung 4.2.5 Aus Symmetriegründen folgt: FS4 = FS5 und FS2 = FS3

Knoten I:

↓ + = = = − = −

+ ⋅ = = −

=

: ,

:

F F F F F F

F F F F

F F

S S S

S S S S

S

2 16

0 62

1 2

2 56

25

0 2 63

2

5 5 4

6 5 6 5

6

Knoten II:

: ,

: ,

− + = = = =

↓ + = = = − = −

F F F F F F

F F F F F F

S S S S

S S S S S

6 1 1 6

1 2 2 3 1

12

0 2 2 2 2 8

22

2 22

0 12

1 4

Lösung 4.3.1 

Nach dem Erstarrungsprinzip gilt:

Fu1 . r01 - Fu2 . r02 = 0 oder F F rru u2 1

01

02

= ⋅

Fu1 = 4500N Fa1 = 1640N Fr1 = 1740N Fu2 = 10800N Fa2 = 3930N Fr2 = 4180N

Lagerreaktionen: FAx - Fr1 - Fu2 + FBx = 0 A: -Fu1l1 - Fr2(l1 + l2) + Fa2r02 + FBy(l1 + l2 + l3) = 0 ↑: FAy - Fr2 - Fu1 + FBy = 0 A: -Fr1l1 - Fu2(l1 + l2) - Fa1r01 + FBx(l1 + l2 + l3) = 0 →: FAz + Fa1 - Fa2 = 0 Ergebnis: FAx = 3660N FAy = 5580N FAz = 2290N FBx = 8880N FBy = 3100N 0 ≤ z1 ≤ l1

FL1 = - FAz = - 2290N FQx1 = - FAx = -3360N FQy1 = FAy = 5580N Mx1 = FAy . z1 Mx1(0) = 0 Mx1(l1) = 307Nm My1 = - FAx . z1 My1(0) = 0 My1(l1) = -201Nm Mz1 = 0

a 5a 6

I

II F

a

2a

a

FS6

FS4

FS5

FS1 FS2FS3

FAz

FAxFAy

y Fu1

Fu2

Fa1

Fa2

Fr1

Fr2

FBx

zFBy

z1

Mx1My1

Mz1FL1FQx1FQy1

FAx FAy

FAz

0 ≤ z2 ≤ l2 FL2 = - FAz - Fa1 = - 3930N FQx2 = - FAx + Fr1= -1920N FQy2 = FAy - Fu1 = 1080N Mx2 = FAy . (l1 + z2) -Fu1 . z2 Mx2(0) = 307Nm Mx2(l2) = 383Nm My2 = - FAx . (l1 + z2) + Fr1 . z2 - Fa1 . r01 My2(0) = -398Nm My2(l2) = -533Nm Mz2 = -Fu1 . r01 = -540Nm

0 ≤ z3 ≤ l3

FL3 = 0 FQx3 = FBx = 8880N FQy3 = - FBy = -3100N Mx3 = FBy . (l3 - z3) Mx3(0) = 186Nm Mx3(l3) = 0 My3 = - FBx . (l3 - z3) My3(0) = -533Nm My3(l3) = 0 Mz3 = 0

2290N

3930N

-FL

y-z-Ebene x-z-Ebene

+

-FQy

5580N1080N

3100N

Mx

+

-

307Nm 383Nm186Nm

540NmMz

-+

FQx

3660N 1920N

8880N

-201Nm398Nm

533Nm

My

Mbmax bei z2 = l2 = 70mm M M z l M z l Nm Nmx ybmax = = + = = + =2

22 2 2

22 2

2 2383 533 656( ) ( )

z2

Mx2

My2

Mz2FL2FQx2

FQy2FAx FAy

FAz

l1

Fr1

Fa1

Fu1

z3

Mx3 My3

Mz3 FL3

FQx3FQy3 FBx

FBy

l3-z3

Lösung 5.4 

[ ] [ ] [ ] [ ] [ ]i A mm x mm y mm x A mm y A mmi Si Si Si i Si i

2 3 3 1 600 10 15 6000 9000 2 900 50 7,5 45000 6750 3 1100 90 27,5 99000 30250 Σ 2600 150000 46000

Damit x mm mm y mm mmSF SF= = = =150000

260057 5 46000

260017 7, ; , ;

Lösung 6.2 

[ ] [ ] [ ] [ ] [ ] [ ]i A h y h y A h I h I h y A hi Si Si i xxi yyi Si i2 3 4 4 2 4

1 40 0,5 20 83,33 213,33 10 2 -24 0 0 -32 -72 0

Σ 16 20 51,33 141,33 10

y h h I I h IS xx xx yy= = = = − ⋅⎛⎝⎜

⎞⎠⎟ = =

2016

1 25 61 33h 61 33 2516

16 36 33h 141 33h4 4 4 4, ; , ; , , ; ,

Ixy = 0; I1 = Iyy; I2 = Ixx

S3A2

A1

A3

S2S1

x

y

xS

SyS

S1

S2

y y,

x

Lösung  6.6 

[ ] [ ] [ ] [ ] [ ]i A a x a y a x A a y A ai Si Si Si i Si i2 3 3

1 15 -1,5 2,5 -22,5 37,5 2 7,5 1 1,667 7,5 12,5 Σ 22,5 -15 50

x a a y a aS S= − = − = =23

0 667 209

2 222, ; , ;

Fortsetzung der Tabelle: [ ] [ ] [ ] [ ] [ ] [ ]I a I a I a y A a x A a x y A axxi yyi xyi Si i Si i Si Si i

4 4 4 2 4 2 4 4 31,25 11,25 0 93,75 33,75 -56,25 10,42 3,75 3,125 20,84 7,5 12,5 41,67 15 3,125 114,59 41,25 -43,75 Ixx = (41,67 + 114,59)a4 = 156,26a4; Iyy = (15 + 41,25)a4 = 56,25a4 Ixy = (3,125 + 43,75) = 46,875a4 I I y Axx xx S= − =2 (156,26 - 111,09)a4 = 45,17a4 I I x Ayy yy S= − =2 (56,25 - 10)a4 = 46,25a4

I I x y Axy xy S S= + = (46,875 - 33,317)a4 = 13,56a4 Hauptträgheitsmomente und Hauptachsen:

( ) ( ) ( )I a a1 22 2 4 445 71 0 54 13 56 45 71 13 57, ( , , , ) , ,= ± − + = ±

I1 = 59,28a4; I2 = 32,14a4

°==−

=−

= 1,4604056,156,13

17,4528,59tan 011

01 ϕϕxy

xx

III

Lösung 7.3 

C F a F a F F

B F x F a x

F F F

F x F a x x a x

G N N G

H G

Hmax N G

G G

:

:

⋅ − ⋅ = =

± ⋅ − −⎛⎝⎜

⎞⎠⎟ =

= =

± ⋅ − −⎛⎝⎜

⎞⎠⎟ = ± ⋅ − + =

20 1

2

20

12

12 2

0 12 2

0

0 0

0 0

µ µ

µ µ

Lösung 7.5 

( )

( )2;5,00123:45

0cotcot143

0scosls87cos

8:

0:

0:

0201020

020

0max0max

−===−+°=

=−++

==

=⋅+⋅+⋅−⋅

=+−↑

=−−→

µµµµα

αµαµ

µµ

αααα

NBHBNAHA

HBNB

HBHA

NBNA

FFFF

inlFlFinFlFA

FFF

FFF

xFG

FH

FN

FS

A B

C

F

F

FNB

FNA

FHB

FHA

α

Lösungen für Aufgaben zur Technischen Mechanik – Festigkeitslehre - Lösung 2.16 

→ =

↑ = − +

− − =

= −

: ( )

: ( )

: ( )( )

F

F F F qa

B F a qa F al f

BH

BV C S

S C

0 1

2 2

2 2 0 32 4

2

∆

Für die lineare Feder gilt: f FcC= und damit wird aus Gleichungen (3) und (4)

2 2

2 0

14

24

14

14

F F qaF lEA

Fc

F qaclEA

Fqa cl

EAclEA

F qaclEA

S C

S C

S C BV

− =

+ =

=+

= −+

=+

; ;

Lösung 2.22 

↓ : FS1 + FS2 + FS3 + F = 0 ( 1 ) A : FS2 a + 2 FS3 a + F x = 0 ( 2 )

w F lEAi

Si i= ( 3 ), ( 4 ), ( 5 )

1.) w1 = w2 = w3 F aEA

F aEA

F aEA

F F F FS S SS S S S

1 2 32 1 3 1

32

2 23

12

= = = = ( 6 ), ( 7 )

( 6 ) und ( 7 ) in ( 1 ) : F F F F F FS S S S1 1 1 123

12

613

+ + = − = −

( 6 ) und ( 7 ) in ( 2 ) : 23

53

10131 1

1F a F a Fx x FF

a aS SS+ = − = − =

2.)

σ σ σ11

22

336

134

133

13= = − = = − = = −

FA

FA

FA

FA

FA

FA

S S S

B2qa FS

FC

f ∆l

x FA

FS1 FS2 FS3

3.)

( )w w w2 1 312

= + ( 6’ )

( 1 ) bis ( 5 ) und ( 6’) liefern FS1 + FS2 + FS3 = - F ( 1’ )

F a F a FaS S2 32 23

+ = − ( 2’ )

32

122 1 3F F FS S S= + ( 6’ )

FS1 + FS2 + FS3 = - F (I) FS1 -3 FS2 + 2FS3 = 0 (II) F F FS S2 32 23

+ = − (III)

I - II : 4 FS2 - FS3 = -F III + 2.(I - II) : 9 832F FS = −

F F F F F F

w FaEA

w FaEA

w FaEA

S S S2 3 1

2 3 1

827

527

1427

49

1027

1427

= − = − = −

= − = − = −

Lösung 4.1

→ =

↑ + − =

− − =

− =

= = = = −

= =

:

:

:

:

; ; ;;

F

F F ql

C F l ql M

G F l ql

F ql F F ql M qlF F ql

CH

B CV

B C

B

B CH CV C

GH GV

0

2 0

3 4 0

2 2 0

00

2

2

2

( )

0 2 00 0

12

1 2

1 2

1 1 2

1 1 12

2 2

≤ ≤ ≤ ≤

= =

= − = −

= − = −

z l z lF z F z

F z q l z F z ql

M z qlz qz M z qlz

L L

Q Q

( ) ( )

( ) ( )

( ) ( )

Aufgabe 6.2 (Statik): Ixx = 36,33h4

w1 w2 w3

2ql

FB

FB

2ql

FGV

FGH

MC

FCV

FCHz1 z2

12

2ql

ql

ql

ql2

-

-

+

+

FQ-Verlauf

M-Verlauf

Maximalwerte für y: Unterseite y = e1 = 3,25h Oberseite y = -e2 = -1,75h Spannungsverteilung an der Einspannstelle (Maximales Moment)

σ

σ

σ

zxx

z

z

y qlI

y

y e qlh

Druckspannung

y e qlh

Zugspannung

( )

( ) ,

( ) ,

= −

= = − ⇒

= − = ⇒

2

1

2

3

2

2

3

0 0894

0 0482

Lösung 4.8

σ

σ

( ) ( )( )

( ) ( ) ( )

( ) ( )

( )

z M zW z

M z F z W z bh z

h z h zl

W z bh zl

z F z

bh zl

xx

x

max

max

= = ⋅ =

= +⎛⎝⎜

⎞⎠⎟ = +⎛

⎝⎜⎞⎠⎟

=⋅

+⎛⎝⎜

⎞⎠⎟

2

002 2

02

2

6

1 26

1 2

6

1 2

Ort der max. Biegespannung entweder an der Einspannstelle (max. Moment) oder an der

Stelle, an der d zdz

σmax ( )= 0 ist.

d zdz

Fbh

zl

zl

zl

zl

zl

zl

zl

σmax ( )=

+⎛⎝⎜

⎞⎠⎟ − +⎛

⎝⎜⎞⎠⎟

+⎛⎝⎜

⎞⎠⎟

=

+⎛⎝⎜

⎞⎠⎟ − = =

6 1 2 4 1 2

1 20

1 2 4 0 12

02

2

4

( ) 20

max20max 3

243

2 bhFll

bhFll

==⎟⎠⎞

⎜⎝⎛ σσ

Die absolut größte Spannung beträgt σmax =34 0

2

Flbh

und tritt an der Stelle z l=

2 auf.

Lösung 4.18  Größte Beanspruchung an der Einspannstelle! Max. Spannung an einem Eckpunkt des Querschnittes, d.h.:

σzy

Zugsp.

Drucksp.

h(z)

z

h03h0

σmax(z)

z 0,5ll

σmax

mmbmmbFlh

hFlb

hFlb

hFlb

hbFl

bhFl

hblFM

bhlFM

MMx

My

M

gewzul

zul

zulzulzul

607,5824

1112

06246246

W

6W22

WWII

3

2

22

22max

2

yymax

2

xxmax

y

ymax

x

xmaxmax

yy

ymaxmax

xx

xmaxmax

==⎟⎟⎠

⎞⎜⎜⎝

⎛ ⋅+±=

=−⋅−⇒=+=

=⋅=

=⋅=

+=+=

σσ

σσσσ

σ

Nur das positive Vorzeichen ist physikalisch sinnvoll. Lösung 4.33 

( ) ( )

F F F F

z a z a

M z F z Fz M z F a z F a z

B C

B C

= =

≤ ≤ ≤ ≤

= = = − = −

23

13

0 0 223

2 13

2

1 2

1 1 1 2 2 2

;

( ) ( )

EIv z Fz EIv z F a z

EIv z Fz C EIv z F a z C

EIv z Fz C z C EIv z F a z C z C

RB ÜB v z Cv z a C a C

v z a v z Fa C

′′ = − ′′ = − −

′ = − + ′ = − +

= − + + = − − + +

= = ⇒ =

= = ⇒ + =

= = = ⇒ − +

( ) ( ) ( )

( ) ( ) ( )

( ) ( ) ( )

/ : ( )( )

( ) ( )

1 1 2 2

1 12

1 2 22

3

1 13

1 1 2 2 23

3 2 4

1 2

2 3 4

1 23

23

13

2

13

16

2

19

118

2

0 0 02 0 2 0

0 19 1

34

1 22

12

3

12

2 32

43

49

0 13

23

59

0 49

89

a Fa C

v z a v z Fa C Fa C

C Fa C C Fa C Fa

= − +

′ = = ′ = ⇒ − + = +

= = = − =

( ) ( )

v z FaEI

za

za

v z FaEI

za

za

za

v v z a v z FaEIF

( ) ( )

( ) ( )

1

31 1

3

2

32 2

22

3

1 2

3

95

188 4 6

0 49

= − ⎛⎝⎜⎞⎠⎟

⎡

⎣⎢

⎤

⎦⎥ = + − ⎛

⎝⎜⎞⎠⎟ + ⎛⎝⎜

⎞⎠⎟

⎡

⎣⎢

⎤

⎦⎥

= = = = =

z1 z2

F

FB FC

Lösung 4.45 

Die Aufgabe ist einfach statisch unbestimmt. Gleichgewichtsbedingungen: ← − =

↑ + − =

⋅ − ⋅ =

:

::

F F

F F qaB F a F a

AH BH

AV BV

AV AH

0

2 00

( )

( )

( )

( )

M z F z qz M z q a z

EIv z qz F z EIv z q a z

EIv z qz F z C EIv z q a z C

EIv z qz F z C z C EIv z q a z C z C

AV

AV

AV

AV

( ) ( )

( ) ( )

( ) ( )

( ) ( )

1 1 12

2 22

1 12

1 2 22

1 13

12

1 2 23

3

1 14

13

1 1 2 2 24

3 2 4

12

12

12

12

16

12

16

124

16

124

= ⋅ − = − −

′′ = − ⋅ ′′ = −

′ = − ⋅ + ′ = − − +

= − ⋅ + + = − + +

M z F zEIv z F z

EIv z F z C

EIv z F z C z C

BH

BH

BH

BH

( )( )

( )

( )

3 3

3 3

3 32

5

3 33

5 3 6

12

16

= − ⋅

′′ = ⋅

′ = ⋅ +

= ⋅ + +

RB/ÜB: 1 0 0 02 0 0 0

3 0 124

16

0 16

124

4 0 0 124

0 124

5 0 16

0 16

6 0 16

12

16

1 2

3 6

14 3

1 12 3

24

4 44

33

5 52

1 23 2

13

. ( ). ( )

. ( )

. ( )

. ( )

. ( ) ( )

v z Cv z C

v z a qa F a C a C F a qa

v z qa C C qa

v z a F a C a C F a

v z a v z qa F a C qa C

AV AV

BH BH

AV

= = ⇒ =

= = ⇒ =

= = ⇒ − + = = −

= = ⇒ + = = −

= = ⇒ + = = −

′ = = ′ = ⇒ − + = − + 3

3 22

53

3 32 37 0 1

216

13

16

. ( ) ( )′ = = ′ = ⇒ + = − + = +v z a v z F a C qa C C F a qaBH BH

Ergebnisse aus GGW und RB/ÜB:

aus 6. folgt mit den GGW F qa F F qaAV AH BH= ⇒ = =3

163

16 und F qaBV =

2916

FAH

FAV

FBH

FBV

z1 z2

z3

C

q

Momentenverlauf:

( )

( )

M z qz a z F z qa qz z a M z qa

M z q a z

M z qaz

Q( ) ( )

( )

( )

* * * *1 1 1 1 1 1 1

2

2 22

3 3

12

38

316

0 316

9512

123

16

= −⎛⎝⎜

⎞⎠⎟ = − = = =

= − −

= −

Verschiebung des Punktes C:

( )v v z aEI

C a C qaEIC = = = + =( )2 3 4

41 316

Lösung 4.50 

FC = v(z2 = 0) . c

( )

( )

( )

( )

M z qz M z q a z F z

EIv z qz EIv z q a z F z

EIv z qz C EIv z q a z F z C

EIv z qz C z C EIv z q a z F z C z C

C

C

C

C

( ) ( )

( ) ( )

( ) ( )

( ) ( )

1 12

2 22

2

1 12

2 22

2

1 13

1 2 23

22

3

1 14

1 1 2 2 24

23

3 2 4

12

12

12

12

16

16

12

124

124

16

= − = − + +

′′ = ′′ = + −

′ = + ′ = + − +

= + + = + − + +

RB/ÜB:

1 0 12

43

2 0 23

13

2 32 3

2 44 3

. ( )

. ( )

′ = = ⇒ = −

= = ⇒ = −

v z a C F a qa

v z a C qa F a

C

C

3 0 12

43

4 0 2 56

1 2 1 32 3

1 2 2 4 14 3

. ( ) ( )

. ( ) ( )

′ = = ′ = ⇒ = = −

= = = ⇒ = − = −

v z a v z C C F a qa

v z a v z C C C a qa F a

C

C

516

2qa0,5qa2

316

2qa

M(z)

-9

5122qa

316 a

-

-

FC z1z2

MA

FA

D

F v z c qaEI

F aEI

c F qacEIa

c

F c qa

v v z CEI

qaEI

F aEI

v c qaEI

v c qaEI

qaEI

qaEI

CC

C

C

DC

D D

= = ⋅ = −⎛⎝⎜

⎞⎠⎟ ⋅ ⇒ =

+⎛⎝⎜

⎞⎠⎟

→ ∞ =

= = = = −

→ = →∞ = − =

( )

( )

( )

( ) ( )

2

4 3

3

12

4 3

4 4 4 4

0 1724 3

17

8 3

178

0 2 56

0 2 2 8548

1148

Lösung 5.6 

( )14 4

100 100 25

216

1

100 1 100 1

11 100 16

151

22 2

2

3 4

4

.

% % %

.

% % %

m A l D l m A l D d l

m mm

dD

MW

MW

W D W W dD

WW d

D

v v h h

v h

v

vt

tvh

t

thtv th tv

h v

v

tv

th

= ⋅ ⋅ = ⋅ ⋅ = ⋅ ⋅ = − ⋅ ⋅

−⋅ = ⎛⎝⎜

⎞⎠⎟ ⋅ =

= = = = − ⎛⎝⎜⎞⎠⎟

⎛⎝⎜

⎞⎠⎟

−⎛⎝⎜

⎞⎠⎟ ⋅ = −

⎛

⎝⎜

⎞

⎠⎟ ⋅ =

− ⎛⎝⎜⎞⎠⎟

−

⎛

⎝

⎜⎜⎜⎜

⎞

⎠

⎟⎟⎟⎟⋅ = −

ρπ

ρ ρπ

ρ

τ τπ

τ ττ

⎛⎝⎜

⎞⎠⎟ ⋅ =

−⎛⎝⎜

⎞⎠⎟ ⋅ = −

⎛

⎝⎜

⎞

⎠⎟ ⋅ =

− ⎛⎝⎜⎞⎠⎟

−

⎛

⎝

⎜⎜⎜⎜

⎞

⎠

⎟⎟⎟⎟⋅ = −⎛

⎝⎜⎞⎠⎟ ⋅ =

100 6 7

3 100 1 100 1

11 100 16

151 100 6 74

% , %

. % % % % , %ϑ ϑϑ

h v

v

tv

th

II d

D

Lösung 7.11  Momente an der Einspannstelle: Biege- und Torsionsmoment erreichen an der Einspannstelle ihr Maximum.

( )( )

M F F a Nmm

M F F b Nmm

bmax

tmax

= − ⋅ = ⋅

= − ⋅ = ⋅

2 80 10

60 10

1 24

1 24

Spannungen: ( )σπmax =

⋅ ⋅−

M DD D

bmax a

a i

324 4 ( )τ

πmax =⋅ ⋅−

M DD D

tmax a

a i

164 4

Der erforderliche Innendurchmesser folgt aus der Beziehung σ στσ

σV zul4

2

1 3= +⎛⎝⎜

⎞⎠⎟ ≤max

max

max

( )σπ

πσ

zula bmax

i

tmax

bmax

i aa bmax

zul

tmax

bmax

D MD

MM

D DD M M

Mmm

≥−

+⎛

⎝⎜

⎞

⎠⎟

≤ − +⎛

⎝⎜

⎞

⎠⎟ =

321 3

2

321 3

242 3

4

2

4

2

4

Da4

,

Gewählt: Di = 42 mm

Spannungsnachweis: σ σVvorh zulN

mmN

mm= < =154 8 1602 2,

Lösung 7.13  Biege- und Torsionsmomente an der Einspannstelle:

FN = 0 M Fa M Fa M Fax y t= = =2 3 3

M M M Fa Fa Nmmres x y= + = + = = ⋅2 2 64 9 13 1 8028 10,

Spannungen: ( ) ( )

( )

σπ

τπ

σ σ τπ

max max

max max

=⋅

−=

⋅−

= + =−

+ ⎛⎝⎜

⎞⎠⎟

32 16

3 32 32

4 4 4 4

2 24 4

22

M dd d

M dd d

dd d

M M

res a

a i

t a

a i

Vmaxa

a ires

t

Erforderlicher Innendurchmesser: Aus σ σVmax zul≤ folgt

d d d M M mmi aa

zulres

t≤ − + ⎛⎝⎜

⎞⎠⎟ =4 2

2

432 3

294 9

π σ, digew = 94mm

σ σvorh zulN

mmN

mm= < =103 24 1202 2,

FF

3Fa3Fa

2Fa

Lösungen für Aufgaben zur Technischen Mechanik – Dynamik - Lösung 1.6 

Bereich I x a x a t C x a t C t C

AB t x x C C

t t x t v x t s t va

s s a t m

Bereich II x x C x C t C

ÜB t t x v x s C v ms

C s vt m

t t x t v x t

A A A

AA

: && &

: & , ,

: &( ) , ( )

: && &

: & , ,

: &( ) , ( )

= = + = + +

= = = ⇒ = =

= = = ⇒ = = = =

= = = +

= = = ⇒ = = = − = −

= = =

12

1 2

1 2

1 1 1 1 1 1 12

3 3 4

1 1 3 4 1 1

2 2 2

12

0 0 0 0 0

20 12

160

0

16 160

( )

( )

( ) ( ) ( )

s vt s vt v t t s

Bereich III x a x a t C x a t C t C

ÜB t t x v x s C v a t C s vt a t

Endbedingung t t x x s S a t t v und

S a t v a t t s vt a t a t t v t t

B B B

B B

B

B B B B

2 2 1 1 2 1 1

52

5 6

2 2 5 2 6 2 2 22

3 3 3 2

32

2 3 2 2 22

3 22

3 2

12

12

0 0 1

12

12

12

= + − = − +

= = + = + +

= = = ⇒ = − = − +

= = = = ⇒ − + =

= + − + − + = − + −

: && &

: & ,

: & , ( )

( )

( ) ( ) ( )

+

= + ⇒ = +−

− = + + = =

= − = ⇒ = − − − − =

s

t t va

in t t S sv

va

s s t

t s s s S a t t v t t m

B Bges

B

2

2 3 3 11

2 2 3 22

3 2

2

2 12

20 23 8 51

51 16 35 12

400

( )

( )

Lösung 1.7  1 0

13

112

0 0 0 0 0 0 0

1

13

13

112

112

1

12

13

1 14

1 2

1 2

1 1 1 1 1 12

11

12 4

1 1 13

1 1 14

. :

( ) ( ) ( )

: ( ) ( )

: ( )

( ) ( )

Bereich t t

a t K t v t K t C s t K t C t C

t v s C C

t t a t a a K t K at

ms

v t K t ms

s t K t m

≤ ≤

= = + = + +

= = = ⇒ = =

= = ⇒ = = =

= = = =

212

13

13

1 23

112

112

12

23

14

2 23

43

12

1 2

1 1 3 12

3 4

1 2 1 1 1 1 1 3 3 3

2 1 1 1 4 4

2 2 2 1 2 3

2 2 1 2

.

( ) ( ) ( )

: ( ) ( )

( ) ( )

( )

( )

Bereich t t t

a t a v t a t C s t a t C t C

ÜB t t v t v t ms

ms

a t C ms

C C ms

s t s t m m m m C C m

t t v t a t C ms

ms

ms

s t a t

≤ ≤

= = + = + +

= = = ⇒ = + = + = −

= = ⇒ = − + =

= = + = − =

= 23 2 4 2 4

314

1112

+ + = − + =C t C m m m m

( )

mCsmC

CtCmmmmtsts

Csm

smCtKtK

sm

smtvtvttÜB

smtKK

sm

ttaKttKa

tKKtatt

tKKaatatt

CtCtKtKtsCtKtKtvtKKta

tttBereich

1219

38

634

1211

1211)()(

2621

34

34)()(:

31

00)(:

)(:61

21)(

21)()(

.3

65

6252322

552232223222

2332323

132331

33233

2321122

653

32

252

3232

32

=−=

+++−=⇒==

+−=++=⇒===

=−=−=−−

=⇒−=−

++===

−+===

+++=++=+=

≤≤

Lösung 2.1 

ω ω

ω ω ω ω

ω ω ππ

π

W W

W G ZW K ZWG

KW

G

K

ZW G M KG

K

K

GM

K

G M

K

G M

K

G

D v vD

r r rr

rr

vD

r r rr

vD

rr

n rr

vn D

i

i rr

vn D

zz

⋅ = ⇒ =

⋅ = ⋅ ⇒ = ⋅ = ⋅

⋅ = ⋅ ⇒ ⋅ = ⋅⎛

⎝⎜

⎞

⎠⎟ = =

= = = =

22

2

2 2

0 18

22

,

Lösung 3.5 

542422

322

2222

11111101

)(021)(

0)(

CtCtxCxx

CtCgttyCgtygy

CCtvtxvxHy

+===

++−=+−=−=

=+===

&&&

&&&

&

gHvvt

gHtvt

tvgtHtyty

vvtxtxttEB

CyCx

invCinvyvCvxtAB

EEE

EEEE

EEE

0

2

112,1

012

12

021

1221

32

52

2222

2422

2g

tang

tan02g

tan2

tan21)()(

cos)()(:

0000

sscoscos0:

−⎟⎟⎠

⎞⎜⎜⎝

⎛±=⇒=+−

⋅+−=⇒=

=⇒==

=⇒==⇒==⇒==⇒==

ααα

α

α

αααα

&

&

Zahlenwerte: v kmh

t s t sE E2 1 21200 3 67 55 1= = =, , ( , )

Lösung 3.13 

( ) ( )J M J M t tt

t t J M t M Jt

J m D m d m e

&&

:

ϕ ω ω

ω

ωω

= − = −

= =

= = ⇒ =

= ⎛⎝⎜

⎞⎠⎟ − ⎛

⎝⎜⎞⎠⎟ +

⎧⎨⎩

⎫⎬⎭

0 0 0 0

0 0

1 1 0 1 01

1

1

2

2

2

22

0 0

12 2

6 12 2

( )[ ]( )[ ]

m a D m b d J aD bd d e

Mt

aD bd d e n

Zahlenwerte n s M Nm

1

2

2

24 2 2 2

01

1

4 2 2 21

1

11 1

0

4 4 326 8

326 8

2

4 77 286 5 0 3933

= = ⇒ = − +

= − + =

= ≡ =− −

ρπ

ρπ ρπ

ω ρπ ωπ

: , , ,min

m y2&&

m1

m2

m2g

x

y

m x2&&

J&&ϕ

ϕ

M0

J

Lösung 3.20 

ZB x x x xr

J m r

F m g m x m g m xF m g m x m g m x

J F r F r J xr

m gr m xr m gr m xr

x g m m

m m m

S

S

S S

:

&& &&

&& &&

&&&&

&& &&

&&

3 1 2

22

1 1 1 1 1 1

2 3 3 3 3 3

2 2 1 3 3 1 1

1 3

1 2 3

12

0

12

= = = =

=

= − = −

= + = +

+ − = = + + − +

=−

+ +

ϕ ϕ

ϕ

( )

( )

( )

T m x m x J x m m m

U m gx m gx gx m m

L T U x m m m gx m m

ddt

Lx

Lx

Lx

g m m Lx

x m m m ddt

Lx

x m m

= + + = + +⎛⎝⎜

⎞⎠⎟

= − + = − −

= − = + +⎛⎝⎜

⎞⎠⎟ + −

⎛⎝⎜

⎞⎠⎟− =

= − = + +⎛⎝⎜

⎞⎠⎟

⎛⎝⎜

⎞⎠⎟= +

12

12

12

12

12

12

12

0

12

12

1 12

3 32

22 2

1 2 3

1 1 3 3 1 3

21 2 3 1 3

1 3 1 2 3 1

& & & &

&

&

&&

&&&

ϕ

∂∂

∂∂

∂∂

∂∂

∂∂

( )

( )

( )

2 3

21 2 3 1 3

1 2 3 1 3

0

12

12

12

0

+⎛⎝⎜

⎞⎠⎟

+ = ⇒ + =

+ = + +⎛⎝⎜

⎞⎠⎟ − −

+ +⎛⎝⎜

⎞⎠⎟ − − =

m

T U konst ddt

T U

T U x m m m gx m m

xx m m m gx m m

.

&

&&& & Lösung 3.26 

( )

( )

ϕ ϕ= = − = −⎛⎝⎜

⎞⎠⎟

= − −⎛⎝⎜

⎞⎠⎟

− − − = ⋅

+ + −⎛⎝⎜

⎞⎠⎟

⎧⎨⎩

⎫⎬⎭= −⎛

⎝⎜⎞⎠⎟

= + = =

xr

x R r Rr

x

F m g m Rr

x

F R r J xr

m x rr

x m Jr

m Rr

m g Rr

J J J J m r J m R

S

S

A B A A B B

21 2

1 1 2

22

2 2

2 222 1

2

1

22 2

1

1

0 1

1 1

2 12

12

&&

&&&&

&&

J&&ϕ2

x1 x3

ϕ2

FS1 FS2

m1g m3gm x3 3&&m x1 1&&

J2&&ϕ

m x2 2&&

x2

x1

FS

FS

m1g

11xm &&

FT

ϕ

FN

m2g

{ } { } { }{ }{ }

22

22

21

121

22

44

22

22222

244

2

222

11

1

2121

22

⎟⎠⎞

⎜⎝⎛ −

+

⎟⎠⎞

⎜⎝⎛ −

+=⎟

⎠⎞

⎜⎝⎛ −=

++

=

+=⇒+=+=

⋅=⋅=+=

rRr

J

rR

mm

gmxrRx

bRdrbRdrmJ

bRdrmbRdrmbRdrJ

bRmdrmmmm BABA

&&&&

πρπρπρ

ρπρπ

( ) ( )T m x m x J U m gx z B ddt

T U= + + = − + =12

02 22

1 12

22

1 1& & & . .ϕ

Lösung 3.28 

( ) αα

αα

αα

αϕαϕ

ϕϕϕ

sin71sin

sin74sin2

21

21

0sinsin

0sin:0sin:

22

2121

11212222

2222

1111

2121

mggxxmF

gxmgmmmmx

mrJmrJmmm

gmxmrxJgmxm

rxJ

rgmrxmrFJmrgmrxmrFJm

rxxxx

S

S

S

=−+=

==⎟⎠⎞

⎜⎝⎛ +++

====

=−++−+

=⋅−+−=⋅−++

=====

&&&&

&&&&

&&&&

&&&&

&&&&

&&&&

Lösung 3.31 

( )

( )

ZB sr

r r rr

sr

Masse m ms mg FScheibe J M F r

Scheibe J J F r F r

F ms mg F Mr

J rr

sr

J J sr

rr

M J rr

sr

msr mgr

S

A S

S S

S SA

A

:

: &&

: &&

/ : &&

&&&&

&& &&&&

ϕ ϕ ϕ ϕ

ϕ

ϕ

23

2 2 1 1 12

1 3

2

1 1 1 1

2 3 2 1 2 2 3

2 11

12

12

3

2 33

2

11

22

12

33 3

01 0

2 3 0

0

= = ⇒ =

+ − =

− + =

+ − + =

= + = −

+ − + + + =

( FS = 0 , da Riemenabtriebsseite ohne Last)

m x1 1&&

J1 1&&ϕJ2 2&&ϕx1

x2

FS

FN1

FT1m1g

FT2

FN2m2g

rr

ϕ2

ϕ1 m x2 2&&

( )J J2 3 2+ &&ϕ

J1 1&&ϕ FS1

FS=0

FS2

mg s

ms&&

ϕ2ϕ1

r1

r2

r3

MA

( )

( )

KssKdssdssdssdKsoder

Kss

Kst

KtsKtsK

mr

JJrJ

rr

mgsMM

mgrrrMMsfürM

mr

JJrJ

rr

mgMrr

r

s

AA

AAA

A

221.3

212.2

0.1

2

2

23

322

1

1

2

3

2

min

2

31minmin

23

322

1

1

2

3

2

31

2

=====⇒=

===

++

+⎟⎟⎠

⎞⎜⎜⎝

⎛=⇒=

==⇒=

++

+⎟⎟⎠

⎞⎜⎜⎝

⎛

−=

&&&&&

&&&&

&&&

&&&&

Lösung 3.39 

K

rrJ

rrJ

MMrrJ

rrJ

rJFrFJ

MrFJrrrr

UU

U

=

⎟⎟⎠

⎞⎜⎜⎝

⎛+

==−⎟⎟⎠

⎞⎜⎜⎝

⎛+

=⇒=−

=−+

=⇒=

2

12

1

21

22

12

1

212

22

2222

111

21

212211

0

0

0

ϕϕ

ϕϕ

ϕ

ϕϕϕϕ

&&&&

&&&&

&&

&&&&&&

( ) Nmbrbrt

rrnMbrrmJbrrmJ

nrrJ

rrJ

tMKttt

KtCCKtK

6,42222

2:

00)0(

22

212

11

2122

2

422

22

21

412

11

1

222

12

1

21

1

212221

2222

=+=====

=⎟⎟⎠

⎞⎜⎜⎝

⎛+=⇒===

==⇒=+==

ρππρπρ

πωωωωϕ

ϕϕϕϕ

&

&&&&&

Lösung 4.2 

( )ϕϕϕ

ϕϕϕ

ϕϕ

ϕ

coscos2)(

cos)(21cos

)(21cos

0cos

0

20

222

101

2211

−=

+=

==

==+=+

grv

mgrmvmgr

mvTmgrU

TmgrUTUTU

( )

( ) ( ) NmgFmgmgvrmF

smgrv

SS 4,92833)(cos3cos2cos)()(

05,61cos2)(

02

0

=+=−=−=

=+=

πϕϕϕϕϕ

ϕπ

22ϕ&&J

11ϕ&&J

ϕ2 ϕ1

FU

M FN

rvm

2

U=0

(2)

mg

FS ϕ ϕ0

(1)

ϕ&&mr

Lösung 4.10 

Energiebilanz im Rohr: mclvmvcl =⇒= 0

20

2

21

21

00000:

21

0

34

0102

432

3

211

=⇒==⇒==⇒==⇒==

++−=+−=−=

+===

CyhChyvCvxCxtAB

CtCgtyCgtygy

CtCxCxx

&

&

&&&

&&&

EB: chmgwlh

vwg

vwttvwywxtt EEE 2

0210

22

000 ==−⎟⎟

⎠

⎞⎜⎜⎝

⎛==⇒===

Zahlenwerte: l = 9,9 cm; v0 = 8,09 m/s Lösung 4.13 

Bewegung auf der schiefen Ebene: αα sin221sin 1

21 glvmvmgl =⇒=

ββββ

ββω

ωϕ

cossin0cossin:

0sincos: 2

1

SNNS

NS

FmgFFFmg

FFmrrv

+==−+↓

=−−→

==&

cm

lr

wh

wwhÜberhöhungg

rfürF

FmgFmr

S

SS

6,20

sin21

tan1tansin:tan0

0sincos

sincos

2

*2

**

2**

2

=

⎟⎠⎞

⎜⎝⎛+

=

+=====

=⋅+

−−

α

βββωβββ

ββ

ββω

Lösung 4.14 

21313

233

2113311

2112

22

2112211

)(221

21

221

21

vhhgvmvmghmvmghUTUT

vghvmvmvmghUTUT

+−=+=+⇒+=+

+==+⇒+=+

mg

x

y

mg β FS FN

mrω2 r

rvm

23 3

mg

FN

v3

( ) mrsmv

gvhhrr

gvrrrfürF

rvmmgF

rvmmgF

N

NN

917,295,62

0

0:

2

21

31*

23**

23

23

>=+−=>

=⇒==

−==+−↑