• Author / Uploaded
• Pamela Alegre Campos

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150 N 65°

115° Fr 10°

15°

LEY DE COSENOS 2 2 2 A = B + C - 2 BC Cos A 2

2 2 Fr = ( 100 N) + (150 N) – 2(100 N)(150 N)Cos 115° Fr = 212.55 N

65°

100 N

150 N 65°

115° Fr 10°

15°

El ángulo se va determinar al aplicar la ley de senos 150 N 212.55 N -------- = ----------Sen & Sen 115°

100 N

65°

212.55 N

150 N

Dirección de Fr

Sen & = 0.63959618

= 39.75° + 15°

Sen & = 39.75°

= 54.75°

&

115° 100 N

y

B

u 30° A x

30° 600 LB

Fu

120°

v

C Aplicando ley de senos Fu 600 lb ----------- = ---------Sen 120° Sen 30° Fu = 1039.23 lb

Fv 600 lb ----------- = ---------Sen 30° Sen 30°

Fv = 600 lb

30° Fv

y

45° Fr 45°

75° 60°

F

200 lb 30° x

Aplicando ley de senos

45°

F

Fr 75°

60° 200 lb

F 200 lb ----------- = ---------Sen 60° Sen 45° Fr 200 lb ----------- = ---------Sen 75° Sen 45°

F = 244.96

Fr = 273.2