Two 440,000-lb locomotives pull 44 250,000-lb coal hoppers. The train starts fro

Question

Two 440,000-lb locomotives pull 44 250,000-lb coal hoppers. The train starts from rest and accelerates uniformly to a speed of 38 mi/hr over a distance of 9900 ft on a level track. The constant rolling resistance of each car is 0.003 times its weight. Neglect all other retarding forces and assume that each locomotive contributes equally to the tractive force. Determine (a) the tractive force exerted by each locomotive at 19 mi/hr, (b) the power required from each locomotive at 19 mi/hr, (c) the power required from each locomotive as the train speed approaches 38 mi/hr, and (d) the power required from each locomotive if the train cruises at a steady 38 mi/hr.

Explanation / Answer

rolling resistance = 0.003 *(2*440000 + 44250000)*.4536 *9.8 (.4536 is conversion from lb to kg) = 591284.47 N a) TRACTION FORCE BY ONE LOCOMOTIVE = TOTAL RESISTANCE/2 + mass*accelaration TRACTION FORCE BY ONE LOCOMOTIVE = 295642.23 N +m*a v^2=u^2 +2*a*s 38/^2 =0+2*a*9900*0.000189394 ( here we can find a now so we can find traction force b) power required =force *velocity =traction force* velocity c) when train speed approaches 38 then train stops accelarating so there will no m*a force in traction traction =rolling resistance =591284.47 N d) power=FORCE*VELOCITY

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