Enter your answer to Part (b), in m/s, in the box below.

Question

A toy airplane, with a motor-driven propeller, flies in a horizontal circle due to a wire of length 18.5 meters attached to the tip of the left wing (as shown in the picture.) The other end of the wire is attached to the ground, and the wire makes an angle of 27.0^O with the horizontal. The wings are tilted so that they are parallel to the wire. The tension in the wire is 140 N and the mass of the plane is 3.26 kg. (Note: the plane experiences a lift force due to the air flow across the wing surface that is perpendicular to the wing.)

Find:

  (a) the lift force acting on the plane;

  (b) the speed of the plane. 

Enter your answer to Part (b), in m/s, in the box below.  

Answer 1

This problem utilizes the principles of vertical force equilibrium and centripetal force in uniform circular motion.

We first take note of the given quantities:

Length of wire l = 18.5 m

Angle with horizontal θ = 27.0°

Tension in wire T = 140 N

Mass of airplane m = 3.26 kg

Acceleration due to gravity g = 9.8 m/s²

We then identify the unknown quantities:

(a) Lift force acting on the plane L = ?

(b) Speed of the plane v = ?

For part (a), we balance the forces in the vertical direction where acceleration is zero:

L cosθ - T sinθ - mg = 0

We rearrange to isolate and solve for L:

L = (T sinθ + mg) / cosθ

L = (140 N * sin(27.0°) + 3.26 kg * 9.8 m/s²) / cos(27.0°)

L = 175.9 N

For part (b), we apply Newton's second law in the horizontal direction. The radius of the horizontal circular motion is:

r = l cosθ

The horizontal components of the lift and tension forces provide the net centripetal force:

m * a = m(v² / r) = L sinθ + T cosθ

Substituting r = l cosθ into the equation:

m(v² / (l cosθ)) = L sinθ + T cosθ

We rearrange to isolate and solve for v:

v = sqrt( l cosθ * (L sinθ + T cosθ) / m )

v = sqrt( 18.5 m * cos(27.0°) * (175.9 N * sin(27.0°) + 140 N * cos(27.0°)) / 3.26 kg )

v = 32.16 m/s