Question

A helicopter carries a 1000-kg-car suspended from a rope below it. The helicopter flies horizontally at a constant speed of 25 m/s. The rope carrying the car makes an angle of 20° with respect to the vertical.(a) What is the tension in the rope?(b) What is the force of the wind resistance on the car?(c) Use the Special Cases sense-making technique on your answer to part (a) and (b) for a situation where the angle of the rope is at both of the two extrema (0º and 90º)?

Answer 1

Answer:

(a) Tension, T = 28.653 kN

(b) Wind resistance force, $26.925\ kN$

Solution:

As per the question:

Mass of the car, m = 1000 kg

Speed of the helicopter, v = 25 m/s

Angle made by the rope, $theta = 20^{\circ}$

Now,

(a) To calculate the tension, T in the car:

Tension along the direction of motion, $T_{h} = Tcos20^{\circ}$

Tension along the vertical direction, $T_{v} = Tsin20^{\circ}$

Now, let the force due to the wind directed in the opposite direction of the motion be $F_{W}$ and it balances the horizontal component of the tension, T.

The vertical component is balance by the weight of the car, i.e., mg that acts vertically downwards.

Now,

$T_{v} = mg$

$Tsin20^{\circ} = 1000\times 9.8$

T = 28653 N = 28.653 kN

(b) The force of the wind resistance:

$F_{W} = T_{h}$

$F_{W} = 2cos20^{\circ} = 26925\ N = 26.925\ kN$

(c) Now,

• If the angle made by the rope with the vertical is $0^{\circ}$:

$mg = Tsin(90^{\circ} - 0^{\circ})$

$Tsin90^{\circ} = mg = 9800\ N$

The tension in the rope will be equal to the weight the car.

Wind resistance force, $F_{W} = Tcos90^{\circ} = 0\ N$

• If the angle made by the rope with the vertical is $90^{\circ}$:

$mg = Tsin(90^{\circ} - 90^{\circ})$

T = 0 N

Wind resistance force, $F_{W} = Tsin0^{\circ}$

$Tsin0^{\circ} = mg$

$F_{W} = \infty$

There will be no tension in the rope and wind resistance will be infinite.