din

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Solución N°1 Solución N°2 VB= VA+ X r B/A VA =0 VB = X (173.21 i + 100 j) VB = 173.21 j – 100 i……… (1) VB = VC + X r B/C VB = 100 i + X ( -229.13 i + 100 j ) VB = (100 -100 ) i – 229.13 j ……(2) De 1 y 2 -100 = 100 -100 -0.76 = = -0.568 rad / s = 0.431 rad / s

Transcript of din

Solución N°1

Solución N°2

VB= VA+ X r B/A

VA =0

VB = X (173.21 i + 100 j)

VB = 173.21 j – 100 i……… (1)

VB = VC + X r B/C

VB = 100 i + X ( -229.13 i + 100 j )

VB = (100 -100 ) i – 229.13 j ……(2)

De 1 y 2

-100 = 100 -100

-0.76 =

= -0.568 rad / s

= 0.431 rad / s

āB = āA + α AB X r B/A + X ( X r B/A)

āB = α AB k X ( 173.21 i + 100 j ) + (-0.568k)x (-0.568k x ( 173 i + 100 j))

āB = (-100 α AB – 55.88 ) i + ( 173.21 α AB – 32.26 ) j ………(3)

āB = āC + α BC X r B /C + X ( X r B/C)

āB= 10i + α BC k x (-229.13 i + 100j ) + 0.431 k x ( 0.431 k x (-229.13 i + 100j ))

āB= (10 + 100 α BC + 42.57 ) i – ( 229.13 α BC + 18.58 ) j ………(4)

D e 3 y 4

α AB = -4.76 rad / s2

α BC = 3.68 rad / s2

Solución N°3

Solución N° 4

PARA I

…. (1)

……(2)

1 y 2

REEMPLAZANDO :

M1 = ∫

M1 =

PARA II

Y = k x^2

Para y=q x = 2a

K = q / 4 a^2

…(3)

..

En (3)

..

M2 = ∫

M2 =

TOMANDO MOMENTO EN EL PUNTO A :

M1 – Kya + kya/ 2 + 4kya + M2 = 0

-

- kya + kya / 2 + 4kya +

= 0

a) +

y = 0

b)

T = 2π /

T = 2 π (

)