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3.126. El cabezal del t aladro radial originalmente estaba colocado con el brazo AB paralelo del eje z. mientras que la broca y la porta brocas estaban colocados paralelos al eje y. El sistema se roto 25° respecto al eje y 20° alrededor de la línea de centros del brazo horizontal AB, hasta que quedo en la posición mostrada. El proceso de taladro comienza al ence nder el motor y rotar la manivela hasta que la broca entre en contacto con la pieza de trabajo. Reemplace la fuerza y el par de ejercidos por el taladro por un sistema equivalente fuerza –par en el centro o de la base de la columna vertical.
SOLUCION:
tenemos
o
tenemos
donde
F 11sin20°cos25° − cos20° − sin20°sin25° 3. 4 097 − 10. 3 366 − 1. 5 8998 3.41 − 10. 3 4 − 1. 5 90 × × × 5.14.9167. si n 25° + 15. 15 . + 14. cos25° 9 16 7. + 15 . . +12. + 12. 6883 6 88 3 90′.20°cos25° − cos20° − sin20°sin25
14.si 14. sin25° 25° + 15 + 14cos25° 14cos25° 5.9167 167 + 15. + 12.6883 883 90. 20°cos25° − cos20° − si n 20°si n 25° 27.898. − 84.572. − 13.0090. 3.5.94109767 −10.135366 1.125.8998 6883 + 27. 8 98 − 84. 5 72 − 13. 0 090 090 . 135.202. 02 . − 31.901. 01 . − 125.313. 13 . 135.2 − −31.9. − 125.3.
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or
3.127. Tres niños se encuentran parados en la balsa de 5 x 5m. Si el peso de los niños que están parados en A, B Y C es de 375 N, 260 N Y 400 N, respectivamente, determinar la magnitud y el punto de aplicación de la resultante de los tres pesos.
SOLUCION:
tenemos
tenemos
tenemos
∑: + + −375 − 260 − 400 −1035 1035 ∑ : + + 3753 + 2600.5 +3.04004. 7 5 1035 483 3.05 375. + 2601.5+ 4004.75 1035 2.57 2.5749 M Z
: FA ( X B ) FB ( X B ) FC ( X C ) R( X D )
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3.128. tres niños se encuentran parados en la balsa de 5x5m. Los pesos de los niños que están parados en A, B Y C son de 375 N, 260 N y 400 N, respectivamente. Si un cuarto niño que pesa 425 N se sube a la balsa, determine donde debe estar parado si los otros niños permanecen en la posición mostrada y si la línea de acción de la resultante del peso de los cuatro niños debe pasar por el centro de la balsa.
SOLUCION
tenemos
tenemos
∑: + + −375 − 260 − 400 − 425 −1460 ∑ : + + +
375 4. 3753 3 + 260 2600.0.5 + 400 400 4. 7 5 5 + 425N 425N 14602. 5 1.16471
1.165
tenemos
M
Z
: FA ( X B ) FB ( X B ) FC ( X C ) FD ( X D ) R( X H )
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3.129 Cuatro señalamientos se montan en un marco que esta sobre la carretera y las magnitudes de las fuerzas horizontales del viento que actúan sobre las señales son la que muestran en la figura. Determine la magnitud y el punto de aplicación de la resultante de las cuatro fuerzas del viento cuando a = 1 ft y b = 12 ft.
SOLUCIÓN: Tenemos
Suponga que la R resultante se aplica en el punto P cuyas coordenadas son (x, y, 0). La equivalencia luego requiere
∑ :−105 − 90 − 160 − 50 − 405 ∑ :5105− 190 + 3160 +5.550 −405 −405 −2.94
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R actúa 1 2,60 pies a la derecha del miembro AB y 2,94 pies debajo del miembro BC.
3.130. cuatro señalamientos se montan en un marco que esta sobre la carreta y las magnitudes de las fuerzas horizontales del viento que actúan sobre las señales son las que se muestran en la figura. Determine a y b tales que el punto de aplicación de la resultante de las cuatro fuerzas se encuentre en G.
SOLUCION:
Dado que R actúa en G, equivalencia, entonces yo necesito que 1M del sistema de fuerzas aplicado también sea cero. Entonces un
∑:−224 − 392 − 176 0.722 ∑
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131.1 un grupo de estudiantes carga la plataforma de un carácter de 2x3.3 con las cajas de o.66x 0.66m y con caja de 0.66x 0.66x1,2m, cada una de las cajas se s e coloca en la parte posterior del tráiler, de tal forma que queden alinearse con la parte trasera y con los costados del tráiler. Determine la carga mínima que los estudiantes deben deben colocar col ocar en una caja adicional de 0.66x 0.66x1.2m y el sitio en el tráiler donde deben asegurarla si ninguna parte de las cajas debe salirse de los costados. Además, suponga que cada caja está cargando uniformemente y que la línea de acción de la resultante r esultante del peso de las cuatro cajas pasa por el punto de intersección de las líneas centrales y el eje del tráiler. (sugerencia: tome en cuenta que las cajas pueden colocarse sobre sus extremos o sobre sus costados.)
SOLUCION:
Para el peso más pequeño en el remolque de modo que la fuerza resultante de los cuatro pesos actúe sobre el eje en la intersección con la línea central del remolque, la caja agregada de 0.66x0.66x1.2 m debe colocarse adyacente a una de las bordes del remolque con el lado 0.66 x 0.66-m en la parte inferior. Los bordes a considerar se basan en la ubicación de la resultante para los tres pesos dados.
tenemos
tenemos
∑ :−2 : −224 24 − 392 392 − 176 176 −792
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A partir de la declaración declaración del problema, problema, se sabe que que la resultante resultante de R de la carga carga original original y la carga W más ligera pasa a través del punto de intersección de las dos líneas centrales. Por lo tanto, IMa - 0. Además, dado que la carga más ligera W debe ser lo más pequeña posible, la cuarta caja debe colocarse lo más lejos posible de G sin que la caja sobresalga del remolque. Estos dos requisitos implican
=0.33m
:∑ : 1− 1 − 0.33 33 × − 1.09101 − 1 × 792 0 344.00 :∑ : − 1.5 × 3.44 − 1.5− 0.83475 ×792 0 3.032 2.97 :∑ : 2.2.97 − 1.1.5 × − 1.5− 0.83475 × 792 0 358.42 :∑ : 1 − × 358.42 − 1.29101 − 1 × 792 0 0.357 L 358
Ahora debe verificar si esto es físicamente posible
que no es aceptable
El peso mínimo, de la cuarta caja es
Y se coloca en el extremo (A 0.66x0.66-m del lado AD. de lado hacia abajo) a lo largo del lado j42? con el centro del campo a 0.357 m del lado AD
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SOLUCION: Primero reemplace las tres cargas conocidas con una sola fuerza equivalente R aplicada a la coordenada (XR, 0, Z / () La equivalencia requiere
O
∑ :−224 − 392 − 176 − 792 M
x
)(224 N ) (0.6m)( )(392 N) (2m) (2m)(176 N) z k (792) (0.33m)(2
576.20
Verificando si esto es físicamente posible
:∑: − 1.5 × 576.20− 1.5− 0.83475 × 192 0 2.414
A o lo cual es aceptable Con zn = 2,7 m a
O
:∑: 2.2.7 − 1.5 5 −1.5− 0.83475 × 792 0 439
Como este es menor que el primer caso, el peso máximo de la cuarta caja es
576 N W = 576 N y se coloca con un lado de 0.66x1 .2 m hacia abajo, un borde de 0.66 m junto con el lado AD, y el centro a 2.41 m del lado de DC.
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SOLUCION: Sistema de par de fuerza en O: R=Pi+ pj + Pk = P(i + j + k)
M = aj x Pi + ak x Pj + ci x Pk
M
Ro
= -Pak-Pai-Paj
− + + Ro
Como R y M tienen la misma dirección, forman una llave con M, = M Por lo tanto, el eje de la llave es la diagonal OA. Notamos eso.
cos cos cos √ 3 √ 13 √ 3 54.7° −√ −√ 3 −√ 3
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3.134 sometida a tres fuerzas de metal laminado se dobla en la forma que se muestra en la figura. Si las fuerzas tienen la misma magnitud P. reemplazando por una llave de torsión equivalente y determine a) la magnitud y la dirección de la fuerza r esultante R, b) el paso de la llave de torsión y e c) el punto donde el eje de la llave de torsión intersecta al plano xz .
SOLUCION:
Primero reduzca las fuerzas dadas a un sistema equivalente de fuerza-pareja (R, Mj) en el origen Tenemos
∑ : − + + + R=Pk
o
:− + [− + 52 ] − − − + 52
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entonces
52
o
52
Los componentes de la llave son (R, M (), donde M, = Mx X Eje> y se supone que el eje de la llave se cruza con el plano xy en e n el punto Q cuyas coordenadas son (x, y, 0 ).
× × − − − + 25 − 52 + + Coeficiente de igualación I:
-aP=yP
o
y= -a
j:
-ap =-xP
o
X=a
El eje de la llave es paralelo al eje z e interseca el plano xy en
x= a, y=-a
3.135 Las fuerzas y los l os pares mostrados se aplican sobre dos tornillos mediante los que se sujeta una placa de metal a un bloque de madera. Rechaza las fuerzas y l os pares a una llave de torsión equivalente y determine a) la fuerza resultante y los pares a una llave de torsión c) el punto donde el eje de la llave de torsión intersecta el plano xz.
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Solución Primero, reduzca el sistema de fuerza dado a un sistema de fuerza-pareja Tenemos
Tenemos
(a) (b) tenemos
Tono
∑:−20 − 15
R=25N
∑:∑ + ∑ ∑ − 4 . – 3.
R= -(20.0N)i-(15.0N)j
. −0.8 − 0.6−4. 5. 5. 5.. 0.200
O P= 0.200m
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3.136 Las fuerzas y los pares mostrados se aplican sobre dos tornillos mediante los que
se sujeta una placa de metal a un bloque de madera. Rechaza las fuerzas y los pares a una llave de torsión equivalente y determine a) la fuerza resultante y los pares a una
llave de torsión c) el punto donde el eje de la llave de torsión intersecta el plano xz. Solución:
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(a)
(b) tenemos
(c) tenemos
Requiere
−21 . −.35. − 12. 12. . −12. . 0.57143 1221. + − 35. × 35. xi + zK ×21lb 35i −21 −21k.+21 21
o R = -(21.0 lb)
y
O p= 0.571 in
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3.137 dos pernos A y B se aprietan aplicando las fuerzas y el par mostrados en la figura. Reemplace las dos llaves de torsión por una sola llave de torsión equivalente y determine a) la resultante R. b) el paso de la llave de torsión equivalente c) el punto donde l eje de la llave de torsión intersecta al plano xz.
Solución:
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− −84116− 80 −15. −15.6. 6. + 2. 2. − 82. 82.4.] . ] 55.379. −40.102. 102. − 38. 38.102. 02. 379. 0.47741 55.116 0.477 + − −15. −15.6 +2 + 2 − 82.44 − 40.102 − 38.192. . −15.6. + 42.102. 02. − 44. 44.208. 08. −15.6 +42.102 − 44.208 + 84 − 80 84 84 + + 80 80 − − 84 84
(b) tenemos
y
tenemos
requiere
de i
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3.138 dos pernos A y B se aprietan aplicando las fuerzas y el par mostrados en la figura. Reemplace las dos llaves de torsión por una sola llave de torsión equivalente y determine a) la resultante R. b) el paso de la llave de torsión equivalente c) el punto donde l eje de la llave de torsión intersecta al plano xz
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(b)tenemos
y
el tono
tenemos
. −8.00 − 31.15.400 − 26 . 152 152. . − 210. 210. − 220. 220. − 62.62.818 18.. − 117. 117.783 783.. − 207.07.30.. 2.46.56. 6. 7.85 246.31.546 1.8522 . + − 152 − 210 − 220 − −62.818 − 117.783 − 207.30 o
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3.139dos cuerdas atadas en A y B se usan para mover el tronco de un árbol caído se muestra en la figura caído como se muestra en la figura. reemplace las fuerzas ejercidas por las cuerdas por una sola llave de torsión equivalente y determine a) la fuerza resultante b) él puso de la llave de torsión y c) el punto donde el eje de la llave de torsión intersecta el plano yz
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(c)
O
tenemos
−601.26.× 29.1937 23 23 + 5 +18. + 18.5 −461. −461.93. 3. − 100. 100.42.2. − 371. 371.56. 6. − −3600 −3600 +6300 + 6300 + 1800 1800 − −461.3 − 100.42121 − 371.56 −3138 −3138..1. . + 6400. 6400.4. . + 2171. 2171.6. .
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solución:
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donde
8 325 2 − 20 − . 245 − − −15√ 15√ 5
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3.141. determine si el sistema sistema fuerza-par mostrado en la figura puede puede reducirse a una sola fuerza equivalente R. si esto es posible, determine R y el punto donde la línea de acción de R intersecta al plano yz. Si la reducción no es posible, reemplace y determine su resultante, su paso y el punto donde su eje interesa al plano yz.
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Para poder reducir las fuerzas originales y las parejas a una sola fuerza equivalente, R y M deben ser perpendiculares. Por lo tanto, R • M = 0. Sustituimos
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acción de R intersecta al plano yz. Si la reducción no es posible, reemplace y determine determi ne su resultante, su paso y el punto donde su eje interesa al plano yz.
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Entonces
−6 − 6 + 18 160. 6160√ 1111
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