What is the acceleration of a plane based on a pendulum's angle?

[tjackson3]

Homework Statement

Granted, I may be thinking too much into this.

You are sitting in a jet airplane as it accelerates at a constant rate down the
runway. Being a good physics student you hold the string of a small pendulum of length
l = 0.75m and mass m = 1 kg. You then measure the angle between the string and a
vertical line and \theta = 37^{\circ}. Assume \sin\ 37 = 3/5; \cos\37 = 4/5; \tan\ 37 = 3/4, and take g = 10 m/s^2

a.) What is the magnitude of the acceleration of the plane?
b.) What is the tension on the string?

Homework Equations

Nonrelativistic coordinate transformation: x' = x - vt

The Attempt at a Solution

There are a couple of ways I can see doing this, but I think both show that I don't completely understand what's going on. My first thought was that the net acceleration is zero, meaning that a = g\cos\theta = 8 m/s^2. This would make the tension \sqrt{m^2g^2 + mg^2\cos^2\theta} = \sqrt{100 + 100.5625} = \sqrt{200.5625} but I don't think that's correct.

Alternatively, it could involve circular motion, although I have no idea what to plug in for the velocity in the centripetal force equation.

Thanks for the help!

 

[ehild]

I suggest you to start every problem-solving with a figure. Here you have to show the forces acting on the moving object, the bob of the pendulum. When the pendulum is stationary with respect to you, it moves together with the aeroplane, with a constant horizontal acceleration. As the bob is connected to the string only, the horizontal component of the tension T provides the force needed to this horizontal acceleration. The other force is gravity, it cancels with the vertical component of T.

ehild

 

[tjackson3]

So is it as simple as a = g\tan 37 = 7.5? Since the diagram looks like

http://img29.imageshack.us/img29/3688/penddia.png

 

[ehild]

Yes, it is so simple...

ehild