ejercicio de centros de gravedad

UNIVERSIDAD NACIONAL DE CAJAMARCA

FACULTAD DE INGENIERA

ESCUELA ACADMICO PROFESIONAL DE INGENIERA CIVIL

ESTTICA

TEMA :

DESARROLLO DE LA SEGUNDA PRCTICA CALIFICADA

DOCENTE:

ING. TARSICIO VALDERRAMA SORIANO

ALUMNOS :

ALTAMIRANO SEGURA, Roiser

CARRERA TERRONES, Jos Wilson

VASQUEZ AGIP, Jos Kevins

CICLO : VACACIONAL

FECHA : Cajamarca febrero de 2015


2 CENTRO DE GRAVEDAD

PROBLEMA UNICO: Hallar el centro de gravedad del rea que se muestra en la figura.

160

4

314899953.16

5

3

3

1

VSVE

mVQ

FG

BACL

CBKB

VA Es altura del AFD

Tmpano circula de radio 4 m DESARROLLO DEL EJERCICIO.

1. HALLAMOS LAS COORDENADAS DE TODOS LOS PUNTOS DE LA FIGURA



a. Hallamos el punto L

|AL| = 3

5|BA|

= (-11, 3, -3)

|AB| = 11.78982612

|AB| = (0.93300782, 0.25445668, 0.25445668)

|AL| = 7.07389567

= (-6.6, 1.8, -1.8)

L = A +|AL|

b. Hallamos el punto C, en el rectngulo ABCD

= (3.75,3.55,17.3)

|AD| = 18.05422388 |AD| = (0.207707, -0.196630, -0.958224)

|BC| = 18.05422388

= (3.75, -3.55, -17.3)

C = + B

c. Hallamos el punto K

|KB| = 1

3|CB|

= (3.75, 3.55, 17.3)

|BC| = 18.05422388 |BC| = (0.207707, -0.196630, -0.958224)

|KB| = 6.01807463

= |KB| . |BC|

= (1.25,1.18333333, 5.76666667)

K = +

d. Hallamos el punto I

|CI| = 4

L = (-9.1, 25.3, -11.8)

C = (-9.75, 22.95, -30.3)

K = (-12.25, 25.31666667, -18.76666667)



|CB| = (-0.207707, 0.196630, 0.958224)

= |CI| . |CB|

= (0.83083051, 0.78651955, 3.83284808)

= +

e. Hallamos el punto H

|CH| = 4 |CD| = (0.933008, -0.254457, 0.254457)

= |CH| . |CD|

= (3.73203129,1.01782672, 1.01782671)

= +

f. Hallamos el punto J

|HJ| = 4

|CI| = (-0.20770761, 0.19662989, 0.95812452)

= |HJ| . |CI|

= (-0.83083051, 0.78651955, 3.83284808)

= +

g. Hallamos el punto F

|FG| = 16.31489995

F = (X, Y, Z)

G = (-4, 26.4067345724, 21.1765433395)

Se tiene la siguiente ecuacin:

( + 4)2 + ( 26.40673457)2 + ( 21.17654334)2 = (16.31489995)2.

(1)

En el plano GFA, se tiene:

. = |FA| . |FG|

. cos(90) = 0

2 + 6.3 + 2 49.90673457 + 2 11.17654334 + 418.7928290= 0 (2)

En el plano FAGD, se tiene:

. ( ) = 0

I = (-10.58083051, 23.73651955, -26.46710192)

H = (-6.01796871, 21.93217328, -29.28217328)

J = (-6.8487992274, 22.71869283, -25.44927521)



60.39022075X + 90.96203752Y 5.57525465Z = 2042.38487638 (3)

De las ecuaciones (1), (2) y (3) se tiene:

X = 8

Y = 18

Z = 14

h. Hallamos el punto V

. = 0 2 5.5 + 2 41.5 + 2 4 + 263 = 0 . . (4)

. = 0 2 1.25 + 2 43.45 + 2 + 37.3 + 738.7 = 0 . . (5)

En el plano AFVD se tiene:

. ( ) = 0

180.35X + 271.65Y - 16.65Z 6099.4 = 0 (6)

De las ecuaciones (4), (5) y (6) se tiene:

X = 3.87397546

Y = 19.19196264

Z = -11.24515756

i. Hallamos el punto Q

|VQ| = 4 |VA| = (-0.81786782, 0.55277984, 0.15977066)

= |VQ| . |VA|

= (3.27147130, 2.21111937, 0.63908266)

= +

j. Hallamos los puntos E y S

|VE| = |VS| = 160

|FD| = (0.16112332, 0.04654674,0.98583603

= |VE| . |FD|

F = (8, 18, 14)

V = (3.87397546, 19.19196264, -11.24515756)

Q = (0.60250416, 21.40308201, -10.60607489)



= (2.03806673, 0.58877483, 12.46994801)

= +

= |SV| . |FD|

= (2.03806673, 0.58877483, 12.46994901)

=

k. Hallamos el punto T

(|TV| )2 + (|VS| )2 = (|TS| )2

|TS| = |TV| + 4

|VS| = 160

|TV| = 18 |AV| = (0.81786782, -0.55277984, -0.15977067)

= |TV| . |AV|

= (14.72162087, -9.95003715, -2.87587199)

= +

l. Hallamos el ngulo

= arctan (

)

HALLANDO EL CENTRO DE GRAVEDAD DE LA ENJUTA PARABOLICA

E = (1.83590873, 19.78073748, -23.71510656)

S = (5.91204218, 18.60318781, 1.22479145)

T = (18.59559633, 9.24192549, -14.12102955)

= 35.0968012276



=

AREA=145.5697907

| | =3

4||

| | = 20.0756

= | |

| | =3

10| |

| | = 4.89447

= | |

C (3.375, 19.375, 8)

G (1.775, 21.8970203717, 10.1529630018)



HALLANDO EL CENTRO DE GRAVEDAD DEL RECTANGULO

=

AREA= 212.8561603524

| | =1

2| |

| | = 5.89491

= | |

| | =1

2| |

| | = 9.02711

= | |

C (-4.25, 21.45, -28.8)

G (-6.125, 23.225, -20.15)



HALLANDO EL CENTRO DE GRAVEDAD DEL TIMPANO CIRCULAR

=

AREA = 3.4336293856

| | =2| |

3(4)

| | = 3.106528244

= | |

| | = 4 2| |

3(4)

| | = 0.893471756

= | |

C (-8.9163838625, 22.722650144, -30.072650144)

G (-9.10196476, 22.898333395, 29.21650360)



HALLANDO EL CENTRO DE GRAVEDAD DE LA SEMIPARABOLA

=

AREA = 30.675058328

| | =5

8| |

| | = 2.94746

= | |

| | =3

5||

| | = 3.61084

= | |

C (-10.75, 25.75, -12.25)

G (-10, 25.04, -15.71)



HALLANDO EL CENTRO DE GRAVEDAD DEL TRIANGULO

=

AREA = 163.2460317282

|| =1

3(| | + | |)

| | = 22.50041440

= | |

| | =1

3||

| | = 2.59780177

= | |

C (4.3746584865, 19.0473208817, -8.1817191864)

G (2.250, 20.483333334, -7.766666667)



HALLANDO EL CENTRO DE GRAVEDAD DEL SEGMENTO CIRCULAR

=

AREA = 68.792501824

EL ANGULO DE = . , = = =

= .

= .

=()^

()

=(.)^

(.(.)(.)= . ,

=

| | = 19.613161791788

= | |

G (2.554622331822, 20.08368596503, -10.9874216248889)



FIGURA AREA A *A *A *A

ENJUTA PARABOLICA 145.569791 1.775000 21.897020 10.152963 258.3863785 3187.54467 1477.9647

RECTANGULO 212.856160 -6.125000 23.225000 -20.150000 -1303.74398 4943.58432 -4289.05163

TIMPANO CIRCULAR -3.433629 -9.101965 22.898333 29.216504 31.25277367 -78.6243904 -100.318645

SEMIPARABOLA -30.675058 -10.000000 25.040000 -15.710000 306.7505833 -768.103461 481.905166

TRIANGULO 163.246032 2.250000 20.483333 -7.766667 367.3035714 3343.82288 -1267.87751

SEGMENTO CIRCULAR -68.792502 2.554622 20.083686 -10.987422 -175.738861 -1381.60700 755.852222

SUMA 418.770793 -515.789537 9246.61703 -2941.5257

Entonces el centro de gravedad de la figura es:

5-1.2316750

*

6

1

6

1

t

t

t

t

Ai

XAi

X

822.0803770

*

6

1

6

1

t

t

t

t

Ai

YAi

Y

0-7.0241902

*

6

1

6

1

t

t

t

t

Ai

ZAi

Z

ejercicio de centros de gravedad

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