решения курсовых (семестровых) работ из задачника / Динамика / d01_17

G,

N

F .

& ' ,

m × x1 = G ×sinα - f ×G × cosα x1 = g ×sinα - f × g × cosα .

#, :

x1 = g × (sinα - f × cosα )× t + C1 ;

x1

= [g × (sin α - f × cos α ) 2]× t 2 + C1 × t + C2 .

10 = 1 ; 10 = 2 .

) % # : 1 = υ ; 2 = 0 . &:

x1

= g × (sinα - f × cosα )× t +υ

=

g × (sinα - f × cosα )

× t2 +υ

× t

1

2

http://www.termeh.ru/

x1 = , . .

( τ , , x1 = υB ;

υB = g × (sinα - f × cosα )×τ +υA (1)

=

g × (sinα - f × cosα )

×τ 2 +υA ×τ (2)

2

*

(2)

: 2 = (υB +υA )×τ

,

υA = 0,5υB ,

2 = (υB + 0,5υB )×τ ;

2 = 1,5υB ×τ . +υB =

2

=

2 ×6

= 8 ;

1,5×τ

1,5×1

*

(1)

,

:

υB = g × (sinα - f × cosα )×1+ 0,5υ

0,5υB = 9,81× (sin 45

- f × cos 45 )= 9,81× (0,707 - f ×0,707) = 6,94 - 6,94 f

;

0,5×8 = 6,94 - 6,94 f

; 4 = 6,94 − 6,94 f ;

6,94 f = 2,94; f = 0,424

2.

$ .

m× = G

:

m × x = 0 ;

= g .

) # t=0:

= 0 ;

x = 0 ;

= υB × cosα ;

= υ ×sinα

x

# #:

= C3 ;

= g × t + C4 ;

x

x = C3 × t + C5 ;

= g ×t2 2 + C4 × t + C6

) , # t=0:

= 4 ;

x = 3 ;

x = 5 ;

= 6

+%, 3 = υ × cosα ; 4 = υ × sinα ;

C5 = 0 ;

C6 = 0 .

! %% :

:

x = υB × cosα ; = g × t +υ × sinα

x = υB × cosα ×t ;

=

g × t2

+υB ×sinα × t .

2

%,

t

*

%

g × x2

=

2×υB2 × cosα + x × tgα (3)

6 =

9,81× d 2

+ d × tg45

;

6 =

9,81× d 2

+ d ×1

; 6 = 0,153× d 2

+ d ; 0,153× d 2

+ d - 6 .

2

×82 × cos2

α

2

× 64 × 0,50

d =

-1± 12 + 4× 0,153× 6

=

-1±

1+ 3,67

=

-1± 2,16

; d=3,79

2 ×0,153

0,306

0,306