pLEASE HELP WITH PHYSICS?

Question

Jean, who likes physics experiments, dangles her watch from a thin piece of string while the jetliner she is in takes off from Dulles Airport (Fig. 4-55). She notices that the string makes an angle of 29° with respect to the vertical while the aircraft accelerates for takeoff, which takes about 20 seconds. Estimate the takeoff speed of the aircraft.

Answer 1

If you draw a diagram of the forces acting on the watch, you'll see that there's just a force of gravity (mg) pulling straight down and the force of the string called tension (T).

Since the string is at an angle, it exerts a horizontal force (Tx) and a vertical force (Ty).

Tx = Tsin29
Ty = Tcos29

Since the plane is accelerating along the runway, the acceleration is only in the x direction, not the y. So the forces in the y direction must add up to zero. That's the same as saying the upward force (Ty) must equal the downward force (mg).

Tcos29 = mg
T = mg / cos29

Newton's 2nd law will give the acceleration along the x direction. The force in this case will be the only force in the x direction which is Tx which equals Tsin29.

F = ma
Tsin29 = ma

rearranging for T gives

T = ma / sin 29

From above, T = mg / cos29, so set them equal to each other.

ma / sin 29 = mg / cos29 (the mass cancels out)

a / sin29 = g / cos 29
a = g (sin 29)/cos 29

a = g tan 29 (because sin/cos = tan)

so a = 9.8 m/s^2 x tan 29 = 5.43 m/s^2

If the plane accelerates for 20 sec, each second it will get 5.43 m/s faster, so its final velocity (vf) will be...

vf = a t
vf = 5.43 m/s^s x 20 s = 108.6 m/s

Answer 2

The angle tells you the amount of acceleration, compared to gravity. since 29º is close to 30º, we can deduce that the acceleration is approximately g/sqrt(3).

Go from there; you should do the exact calculation with the angle though.