Question

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.

Answer 1

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:

Answer 2

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.