Question

You are sitting in a commercial jetliner. It is on the runway and ready to take off. Before it does, clever soul that you are, you hang a .5 meter long string from the ceiling just overhead and attach to its free end a weight whose mass is m = .05 kg. The jet begins its acceleration. As it does, the string and mass swing toward you until the string comes into equilibrium at a constant angle of θ = 26o with the vertical.

a.) What is the jet's acceleration?

b.) Once the jet gains altitude and proceeds with a constant velocity, what will the string and mass be doing (that is, will the 26o angle have changed)? Explain

Answer 1

a.=4.78 m/s2

b. the thread will remain vertical

Step-by-step explanation:

a.

For jet acceleration as you know,

Weight due to gravity given as

F= w =mg

and body movement with acceleration as

F= ma

`tan\theta=(ma)/(mg)`

`tan\theta=(a)/(g)`

`a=gtan\theta`

`a=9.81* tan 26^(0)`

`a=4.78(m)/(s^(2) )`

Its acceleration of plane

b.

For planes attaining constant velocity

a = 0 and tan∅ = 0

so Ф = 0

hence the thread will be remain vertical