Engineering Dynamics: Solving Train Decel on Slope

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Homework Help Overview

The problem involves a train's deceleration while climbing a slope, with specific forces acting on it, including traction and rolling resistance. The context is within the subject area of engineering dynamics.

Discussion Character

  • Exploratory, Conceptual clarification, Mathematical reasoning

Approaches and Questions Raised

  • Participants discuss the forces acting on the train and the equations of motion. There is an attempt to clarify the traction force and its relation to the weight of the train. Some participants question the accuracy of the initial calculations and suggest showing work for better understanding.

Discussion Status

The discussion is ongoing, with participants providing insights into the equations used and questioning assumptions about the setup. There is an emphasis on breaking down the problem into components and understanding the forces involved, but no consensus has been reached on the correct approach or solution.

Contextual Notes

There are indications of confusion regarding the values used for forces and the need for clarity in the equations. Participants are also exploring the necessity of rotating axes for simplification.

JaredPM
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Homework Statement


The 160-Mg train travels with a speed of 80 kmlh when it starts to climb the slope. If the engine exerts a traction force F of 1120 of the weight of the train and the rolling resistance F D is equal to 1/500 of the weight of the train, determine the deceleration of the train. slope is y=1 and x=10
Capture.PNG


****See attached files for picture.

Homework Equations


I summed the forces in the x and y directions because it is going up a slope.

The Attempt at a Solution


the solution I keep getting is in the right direction, but is 1.0m/s^2 off. i.e. -1.5m/s^2 roughly.
 
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The engine exerts a traction force of F(1/20) of the weight of the train. Not 1120
 
Show your work. Simply telling us the wrong answer you got isn't very helpful in seeing where you might have gone wrong.
 
∑F(x)=Fcos(5.7)-F(Resistance)cos(5.7)-F(Normal)cos(84.3)-mg*cos(84.3)=ma
∑F(y)=Fsin(5.7)-F(Resistance)sin(5.7)+F(Normal)sin(84.3)-mg*sin(84.3)=ma

I uploaded a free-body diagram.
 

Attachments

  • f=ma.png
    f=ma.png
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Can you explain your reasoning behind the equations?
 
Well I wanted to prove that the axis didn't need to be rotated. It certainly makes things easier if you do, but I wanted to break everything down into components to get a better understanding of how all of the forces are represented in the problem by using vectors. I want a fool proof way to begin dynamics. I did the above equation way wrong. I didn't give acceleration components for x or y most importantly.

I'm just trying to find a common approach to problems. Pulleys are different.
 

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