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Calculating the Acceleration of a Train Given Angle from Vertical

  • Thread starter teriw
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  • #1
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An engineer is riding The El standing next to a bot holding a yo-yo which is hanging down. As the train starts to move, the boy looks at the engineer strangely as he whips out a protractor and measures a 4.19 degree angle between the string of the yo-yo and the vertical. The engineer quickly calculates the acceleration of the train. What is it?



The sum of the forces = ma = (mv^2)/p



I have drawn separate free body diagrams for the yo-yo and the train car. I used normal and tangent for the yo-yo and x and y for the train car. The train car only has movement on the horizontal axis (x). I summed the forces acting on the yo-yo and set it equal to (mv^2)/p then I solved for a. I am getting stuck when I am trying to find the mathematical relationship between the train car and the yo-yo. Maybe its because of the limited information that was given? I also looked at my equations for constant acceleration but did not find any of them to look helpful in this situation since I was only given the angle.

[tex]\Sigma[/tex]F = Ma
a = (T-mgcos4.19)/m

I am affraid that I might be going about the solution in the wrong way. My homework is due tomorrow so if you give me a little guidance as to what topics I could look up in my book it would be awesome. I cannot find a similar problem anywhere.
 

Answers and Replies

  • #2
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You have two equations:
The horizontal component of the tension in the string is what gives the yo-yo its horizontal acceleration, a. The vertical component of the tension must equal mg, the weight of the yo-yo, as there is no resultant vertical acceleration.
Solve these two for a. (M cancels out; eliminate T)
 

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