Simple accelerometer proof (a=tan(theta)) help

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The situation is there is a ball attached to a string in a car, and the angle that the string makes is measured when the car is accelerating. I'm having a bit of trouble with the proof that a=tan(theta)

so far I've gotten that
weight=mg=Tcos(theta)
a=ma=Tsin(theta)

But what I don't understand is why you need to find the ratio between them and how that shows acceleration.
a/g=sin(theta)/cos(theta)
a=gtan(theta)

another one of my questions is what happens to the g in the equation?

Help is much appreciated thanks!
 

Answers and Replies

  • #2
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The situation is there is a ball attached to a string in a car, and the angle that the string makes is measured when the car is accelerating. I'm having a bit of trouble with the proof that a=tan(theta)
It's normal to have trouble. This is not correct. Look at the units. Tan(theta) is dimensionless. How can be equal to the acceleration?

The situation is there is a ball attached to a string in a car, and
so far I've gotten that
weight=mg=Tcos(theta)
a=ma=Tsin(theta)

But what I don't understand is why you need to find the ratio between them and how that shows acceleration.
a/g=sin(theta)/cos(theta)
a=gtan(theta)

another one of my questions is what happens to the g in the equation?

Help is much appreciated thanks!
You have two equations with two unknowns (T and a). As you are interested in a, you can isolate T from the first equation, replace in the other one and solve for a.
a = g tan(theta) makes sense.

PS. It may be that you found the a=tan(theta) in a text where they measure acceleration in units of "g"?
Then what you have there is actually a/g=Tan(theta).
 
  • #3
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AH, ok that makes sense. Thanks for the clarification!
 

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