50. (a) The angle made by the cord with the vertical axis is given by θ=cos
−1(18/30) = 53◦. This means
the radius of the plane’s circular path is r= 30 sin θ=24m(wealsocouldhavearrivedatthis
using the Pythagorean theorem). The speed of the plane is
v=4.4(2πr)
1min =8.8π(24 m)
60 s
which yields v= 11 m/s. Eq. 6-17 then gives the acceleration (which at any instant is horizontally
directed from the plane to the center of its circular path)
a=v2
r=112
24 =5.1m/s2.
(b) The only horizontal force on the airplane is that component of tension, so Newton’s second law
gives
Tsin θ=mv2
r=⇒T=(0.75)(11)2
24 sin 53◦
which yields T=4.8N.
(c) The net vertical force on the airplane is zero (since its only acceleration is horizontal), so
Flift =Tcos θ+mg =4.8 cos 53◦+(0.75)(9.8) = 10 N .