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

Answer:

F = 50636.873 N

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)

$T sin 38^0= \dfrac{mv^2}{l} + F$..........(2)

equation (1)/(2)

$tan 38^0 =\dfrac{\dfrac{mv^2}{l} + F}{mg}$

       $0.781=\dfrac{\dfrac{700\times 40^2}{20} + F}{700\times 9.8}$

           F = 50636.873 N

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