La grúa

TABLA DE DATOS

| |F1 (N) |Θ1 |F2 (N) |Θ2 |
|Caso 1 |0.9|105 |4.9 |75 |
|Caso 2 |0.8 |4.8 |4.8 |80|
|Caso 3 |1.8 |15 |5.5 |75 |

Distancia de AC= 0.39 m

Distancia de BC=0.65 m

ANALISI DE RESULTADOS

• Caso 1

Coordenadas

A (-0.39, 0)m

B (0.65cos75, 0.65sen75)m

C(O, O)m

Vector posición

rCA = (-0.39, 0) –(0,0) = (0.39) m

|| rCA || =0.39 mUCA = -i

F1 * UCA= 0.9i

rCB = (0.1682 , 0.687)-(0, 0)= (0.1682 , 0.687)m

||rcB||= 0.70 m

UCB = 0.24i + 0.98j

F2 *UCB= 0.24F2i + 0.98F2j

Ecuación 1Ecuación 2

∑ fx =0= F1 – 0.24F2

F1= 1.176 N

∑ fy =0= 0.98F2 – 2.45F2

F2 = 2.5 N

Momento en A no hay porque es paralelo a la fuerza

M B= i jk =

0.168 -0.687 0

1.176 4.802 0

MB = -1.76x10-3 Nm

• Caso 2

Coordenadas

A (0.46cos10, 0.39sen10) m

B (0.65cos80,065sen 80) m

C (0, 0) m

Vector posición

Vector posición

rCA = (-0.453 , 0.067) –(0,0) =

(-0.453, 0.067) m

|| rCA || =0.457 m

UCA = -0.991i + 0.172j

FCA * UCA= -0.991FACi + 0.146FCAjrCB = (0.112 , 0.64)-(0, 0)= (0.112 , 0.64) m

||rcB||= 0.649 m

UCB = 0.172i + 0.986j

FCB *UCB= 0.172FCBi + 0.986FCBj

Ecuación 1Ecuación 2

∑ fy =0= 0.172FAC – 0.986FCB - W

FCB = -0.172FCA +2.45

0.986

∑ fx =0= -0.987FAC – 0.172FCB

∑ fx =0= -0.987FAC – 0.172(-0174 +2.48)

0= -1.016FCA + 0.652...