The subway train shown travels at a speed of 36 m h ?

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.