The subway train shown travels at a speed of 36 m h . determine the force in each coupling when the brakes are applied knowing that the braking force is 5000 l b on each car.
Respuestas a la pregunta
Respuesta:
El tren subterráneo que se muestra viaja a una velocidad de 36 m h. determine la fuerza en cada acoplamiento cuando se aplican los frenos sabiendo que la fuerza de frenado es de 5000 l b en cada automóvil.
Explicación:
The force in each coupling when the brakes are applied are :
FAB = 3 kips ; FBC = 2 kips .
The force in each coupling when the brakes are applied is calculated as follows :
WA = 30 tons = 60 kips
WB = 40 tons = 80 kips
WC= 30 tons = 60 kips
Fr = 5000 lbf on each car
V1 = 36 mph = 36 mi/h * 1h/3600s* 1609m/1 mi *1ft/0.3048 m = 52.78 ft/s
K1 = 1/2 * ( WA + WB+ WC)/g *V1²
K1 = 1/2 * ( 200000 lbf)/32.2 ft/s2 * ( 52.78 ft/s)²
K1 = 8651330.43 Lbf* ft
U1→2 = -Fr *d = -10000 lbf * x
K2 =0
K1 + U1→2 = K2
8651330.43 Lbf* ft - 10000 lbf* x = 0
x = 865.13 ft
Force in coupling AB:
Replace wagon A with a coupling. Assume a tensile force on the coupling. Force FAB is acting on the coupling :
K1 = 1/2 * ( WB + WC)/g * V1²
K1 = 1/2 * ( 140000 lbf )/32.2 ft/s2* ( 52.78 ft/s)² = 6055931.3 lbf*ft
U1→2 = FAB *865.13 ft -10000 lbf* 865.13 ft
K2 =0
K1 + U1→2 = K2
6055931.3 lbf*ft + FAB *865.13 ft -10000 lbf* 865.13 ft =0
FAB = 3000 lbf = 3 kips ( tension)
Force in coupling BC :
Replace wagons A and B with a coupling. Assume a tensile force on the coupling . Force BC is acting on the coupling :
K1 = 1/2 * ( 60000 lbf )/32.2 ft/s2 * ( 52.78 ft/s)² = 2595399.13 lbf*ft
U1→2 = 865.13 ft * FBC - 5000 Lbf* 865.13 ft
K2 =0
K1 + U1→2 = K2
2595399.13 lbf*ft + 865.13 ft * FBC - 5000 Lbf* 865.13 ft = 0
FBC = 2000 lbf = 2 kips ( tension)
The statement with its respective figure is attached.
