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A simple accelerometer

You tape one end of a piece of string to the ceiling light of your car and hang a key with mass m to the other end (Figure 5.7). A protractor taped to the light allows you to measure the angle the string makes with the vertical. Your friend drives the car while you make measurements. When the car has a constant acceleration with magnitude a toward the right, the string hangs at rest (relative to the car), making an angle $B$ with the vertical.

  • (a) Derive an expression for the acceleration $a$ in terms of the mass m and the measured angle $B$.

  • (b) In particular, what is a when $B$ = 45? When $B$ = 0?

I don't care about the answers, the important thing is the following:-

The book says The string and the key are at rest with respect to the car, but car, string, and key are all accelerating in the +x direction. Thus, there must be a horizontal component of force acting on the key.

That's the reason the book decided to consider a force in the $+x$ direction, but I'm looking for a better explanation: how would I find detect the force in the $+x$ direction in another way? To me, when I draw the free body diagram of the string, there looks to be no force acting on the $+x$ direction! I understand it starts with noticing that the string is attached to the ceiling of the car, and that the car has force causing acceleration in one direction, but I don't know how to go further than that.

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There is a force, remember f = ma ? –  Martin Beckett Feb 27 '12 at 18:16
    
What type of force is it? –  w4j3d Feb 27 '12 at 18:40
    
I understand there is force, but there could be multiple forces, and the force of the string (horizontally) doesn't have to be equal to the force of the car (horizontally), since they are not an action-reaction pair. –  w4j3d Feb 27 '12 at 18:45
    
There is only the string tension and gravity acting on the key (no fictitious "force of the car"). See my answer for more details. –  kleingordon Apr 28 '12 at 9:19

4 Answers 4

It is a fictitious force owing to the fact that the key is not in an inertial reference frame, but is in fact in circular motion (it is under tension by string to the car which is in circular motion relative to the ground)

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Since the net acceleration is in the +x direction, the x component of the tension in the string must be providing the net force, since the only other force that applies, gravity, has no x component. Meanwhile, the y component of the string tension cancels the y acceleration from gravity. There is no other force acting on the key.

Finally, I believe there is a typo in the statement of the question. Your expression for the acceleration will include the gravitational acceleration g but not the mass of the key.

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Did you forget to include the D'Alembert forces in the Free Body Diagram?

http://en.wikipedia.org/wiki/D'Alembert's_principle

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Acceleration and force are not the same thing.

So first simplify the problem by just talking about acceleration.

For example, suppose the car is accelerating forward at a rate of 1.0 meter per second per second. It also feels an "acceleration" due to gravity, which is 9.8 m/(s^2). If you want to know the angle at which the key hangs, it is the angle whose tangent is 1.0/9.8. Hopefully that is enough for you to run with.

Then if you want to know force, make use of f=ma. i.e. multiply the acceleration by the vehicle's mass.

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