Question

A fire helicopter carries a 700-kg bucket of water at the end of a 20.0-m long cable. Flying back from a fire at a constant speed of 40.0 m/s, the cable makes an angle of 38.0° with respect to the vertical. Determine the force exerted by air resistance on the bucket.

Answer 1

F = 50636.873 N

Step-by-step explanation:

given,

bucket of water = 700-kg

length of cable = 20 m

Speed = 40 m/s

angle of the cable = 38.0°

let air resistance be = F

tension in rope be = T

T cos 38° = m×g..................(1)

`T sin 38^0= (mv^2)/(l) + F`..........(2)

equation (1)/(2)

`tan 38^0 =((mv^2)/(l) + F)/(mg)`

`0.781=((700* 40^2)/(20) + F)/(700* 9.8)`

F = 50636.873 N

Hence the force exerted on the bucket is equal to F = 50636.873 N