Physics: Principles with Applications (7th Edition)

Published by Pearson
ISBN 10: 0-32162-592-7
ISBN 13: 978-0-32162-592-2

Chapter 4 - Dynamics: Newton's Laws of Motion - General Problems - Page 106: 79

Answer

(a) The air exerts a lift force of 87,600 N upward on the helicopter rotors. (b) The cable exerts a force of 11,400 N upward on the frame. (c) The cable exerts a force of 11,400 N downward on the helicopter.

Work Step by Step

(a) The lift force $F_L$ provides the force to accelerate both the helicopter and the frame at a rate of $0.80 ~m/s^2$: $\sum F = ma$ $F_L -mg = ma$ $F_L = m(a + g)$ $F_L = (7180 ~kg + 1080 ~kg)(0.80 ~m/s^2 + 9.80 ~m/s^2)$ $F_L = 87,600 ~N$ The air exerts a lift force of 87,600 N upward on the helicopter rotors. (b) The tension $F_T$ provides the force to accelerate the frame at a rate of $0.80 ~m/s^2$: $\sum F = ma$ $F_T -mg = ma$ $F_T = m(a + g)$ $F_T = (1080 ~kg)(0.80 ~m/s^2 + 9.80 ~m/s^2)$ $F_T = 11,400 ~N$ The cable exerts a force of 11,400 N upward on the frame. (c) The tension in the cable exerts the same magnitude of force on the helicopter as it does on the frame. Therefore, the cable exerts a force of 11,400 N downward on the helicopter.
Update this answer!

You can help us out by revising, improving and updating this answer.

Update this answer

After you claim an answer you’ll have 24 hours to send in a draft. An editor will review the submission and either publish your submission or provide feedback.