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

(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* 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.