Calculos Dinamica

2-

Aa

Determinar la velocidad del bloque A (Va) cuando ΔYa = 10 ft

El tiempo que se requiere para que el bloque A alcance una velocidad de 20 ft/seg

Sistema 1

A: T B: T mb Ab

= =

Wa ma Aa Wb

A: + ∑FY=0 = ma Aa: Wa-T=Wa/G.Aa

T= 200(1- Aa/G)

B: + ∑FY=0 = mbAb: T-Wb=Wb/G.Ab

T= 100(1+ Ab/G)

Aa=Ab

200(1- Aa/G)= 100(1+ Ab/G)

Aa= 1/3 .g= 1/3(32.2 ft/s^2).10.7333 ft/s^2

Aa=10.73 ft/s^2

〖Va〗^2= (Va)^2+2Aa(yf-yi)

ΔYa = 10 ft : 〖Va〗^2=2(10.7333)ft/s^2(10ft) =14.65 ft

C) Va= (va)+ Aa.t

Va= 20 ft : 20ft.(10.7333 ft/s^2).t

T= 1.863 seg

SISTEMA 2

a)

T=100lbs

=

Wa maAa

+ ∑FY=0 = ma Aa: Wa-T=Wa/G.Aa

Aa= (32.2)( 1- 100/200)

Aa= 16.1 ft/s^2

b)

〖Va〗^2= (Va)^2+2Aa(yf-yi)

ΔYa = 10 ft : 〖Va〗^2=2(16.1)ft/s^2(10ft) =17.94 ft/seg

C)

Va= (va)+ Aa.t

Va= 20 ft : 20ft.(16.1ft/s^2).t

T= 1.242 seg

SISTEMA 3

A: T B: T mb Ab

= =

Wa maAa Wb

A:

A: + ∑FY=0 = ma Aa: Wa-T=Wa/G.Aa

T= 2200(1- Aa/G)

B: + ∑FY=0 = mbAb: T-Wb=Wb/G.Ab

T= 2100(1+ Ab/G)

Aa=Ab

2200(1- Aa/G)= 2100(1+ Ab/G)

Aa= 1/43 .g= 1/43(32.2 ft/s^2)

Aa=0.74884 ft/s^2

b)

〖Va〗^2= (Va)^2+2Aa(yf-yi)

ΔYa = 10 ft: 〖Va〗^2=2(0.74884)ft/s^2.(10ft) =3.87 ft/seg

C)

Va= (va)+ Aa.t

Va= 20 ft: 20ft. (0.74884 ft/s^2).t

T= 26.7 seg

3.

a) la desaceleración a2 de la banda

b) el desplazamiento del paquete relativo a la banda cuando esta se detiene

2.2m

X12

A1 A2

V1=0 V2=0 V3=0

1 2: x12 = x1+v1.t+1/2 a1.t_12^2

= 1/2(2 m/s^2 ).〖(1.3s)〗^2= 1.69 m

V2=V1+A1.t_12

= (2 m/s^(2 ) )(1.3 s)= 2.6 m/s

2 3:

V_3^2=V_2^2+2A_2 (X_3-X_2)

0=〖(2.6 m/s)〗^2+2A_2 (2.2-1.69)m

A2= -6.62745 m/s^2

1 2

W

= m(a)

F12

N

+ ∑FY=0 = N-W=0

N=W

Fmax= µs .N

= 0.35 W

F12< Fmax

F12= ma1 m= w/g

= W((2 m/s^2 )/(9.81 m/s^2 ))=0.204 w

F12 (0.204w) < Fmax (0.35)

2 3

W

= m(a)2

F23

N

+ ∑FY=0 = N-W=0

N=W

Fmax= µs .N

= 0.35 W

F<Fmax

-F23=ma2

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