Question

A fire helicopter carries a 580-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 = 41,954 N

Explanation:

given,

mass of bucket = 580 Kg

length of the cable = 20 m

velocity = 40 m/s

angle made = 38.0°

T cos 38° = m g..............(1)

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

dividing equation (2) by (1)

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

$tan 38^0 = \dfrac{\dfrac{580\times 40^2}{20} + F}{580 \times 9.81}$

$4445.36 = \dfrac{580\times 40^2}{20} + F$

F = -46400 + 4445.36

F = -41,954 N

hence, the force is acting in the opposite direction as assumed.

F = 41,954 N