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Inertial and Non-Inertial Frames of Reference Question

by Kaos_Griever
Tags: frames, inertial, noninertial, reference
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Kaos_Griever
#1
Mar12-07, 01:18 PM
P: 6
1. The problem statement, all variables and given/known data
A rubber stopper of mass 25g is suspended by string from a handrail of a subway car travelling directly eastward. As the subway train nears a station, it begins to slow down, causing the stopper and string to hang at an angle of 13 degrees from the vertical. What is the acceleration of the train? Determine the magnitude of the tension in the string.

The attempt at a solution
Tension of String = (mass)(9.81) / cos 13 degrees
The Horizontal component of Tension = [(mass)(9.81) / cos 13 degrees] sin 13 degrees = (mass)(acceleration)
a = [[(mass)(9.81) / cos 13 degrees] sin 13 degrees] / mass
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Doc Al
#2
Mar12-07, 01:24 PM
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That works. But simplify that answer!
Kaos_Griever
#3
Mar12-07, 01:26 PM
P: 6
I'm not sure how to simplify the answer because I do not have many values to use...

Doc Al
#4
Mar12-07, 01:28 PM
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Inertial and Non-Inertial Frames of Reference Question

You have everything you need.

Cancel what can be canceled; use a single trig expression. Then evaluate to get the numerical answer.
Kaos_Griever
#5
Mar12-07, 01:31 PM
P: 6
Thank you very much! =D It took me a while to understand it.. I thought what I was doing was wrong.
Doc Al
#6
Mar12-07, 01:34 PM
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Your solution is fine!

FYI, here's how I would do it:

Horizontal forces:
[tex]T\sin\theta = ma[/tex]

Vertical forces:
[tex]T\cos\theta = mg[/tex]

Combine (divide one by the other) to get:
[tex]a = g\tan\theta[/tex]
Kaos_Griever
#7
Mar12-07, 01:43 PM
P: 6
Oh, thanks! I really appreciate your help.


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