A freight train consists of 3 engines at the front, followed by 32 rail cars. Both the engines and rail cars have the same mass of 638,394 kilograms each. Each engine exerts the same force backward on the track to push the train forward. The total force of friction on the entire train is 749,869 Newtons and is evenly distributed among all of the cars and engines. Even with this friction present, the train engines are able to accelerate the train at 0.035 meters/second2. What is the magnitude of the force in Newtons in the coupling between the engine and the first rail car?

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wagonbelleville wagonbelleville · Answer 1 · 2019-11-14T09:31:58+01:00

Answer

consider the second car till the 32nd car as one system .

forces acting on the system:

Force in the direction of motion =T

Friction force=31 x friction force on each car

=[tex]31\times \dfrac{749869}{32+3}[/tex]

= 664169.6857 N

writing force balance equation,

T - f = m a

T- 664169.6857 = 638394 x 31 x 0.035

T = 664169.6857 + 638394 x 31 x 0.035

T = 1356827.1757 N

now consider the first rail car as the single system.

forces acting are:

pulling force in forward direction=T₁

force due to the rest of the 31 cars =T , in backward direction

friction force acting on this car

writing force balance equation:

T₁ - T - friction force = mass ₓ acceleration

T₁ - 1356827.1757 - (749869/35) = 638394 x 0.035

T₁ = 1400595.7943 N

hence, force in the coupling between the engine and the first rail car is 1400595.7943 N.