Physics for Scientists and Engineers: A Strategic Approach with Modern Physics (4th Edition)

Published by Pearson
ISBN 10: 0133942651
ISBN 13: 978-0-13394-265-1

Chapter 7 - Newton's Third Law - Exercises and Problems: 44

Answer

$F_{prop} = (12306~\hat i+33810~\hat j)~N$

Work Step by Step

The helicopter is flying horizontally. Therefore the vertical component of $F_{prop}$ is equal in magnitude to the weight of the helicopter-and-crate system. $F_{prop,y} = (3200~kg+250~kg)(9.80~m/s^2)$ $F_{prop,y} = 33810~N$ Note that the vertical component of the tension in the cable is equal in magnitude to the weight of the crate. We can find the horizontal component of the tension in the cable. Let $m_c$ be the mass of the crate. So; $\frac{T_x}{T_y} = tan(\theta)$ $T_x = T_y~tan(\theta)$ $T_x = (m_c~g)~tan(\theta)$ We can find the horizontal acceleration of the crate. The horizontal component of the tension provides the force to accelerate the crate in the direction of motion. $T_x = m_c~a_x$ $a_x = \frac{T_x}{m_c}$ $a_x = \frac{(m_c~g)~tan(\theta)}{m_c}$ $a_x = g~tan(\theta)$ $a_x = (9.80~m/s^2)~tan(20^{\circ})$ $a_x = 3.567~m/s^2$ Since the crate moves with the helicopter, the system of the helicopter and crate has this acceleration. The horizontal component of $F_{prop}$ provides the force to accelerate the system. $F_{prop,x} = (3200~kg+250~kg)~a_x$ $F_{prop,x} = (3200~kg+250~kg)(3.567~m/s^2)$ $F_{prop,x} = 12306~N$ We can write an expression for $F_{prop}$ in component form. $F_{prop} = (12306~\hat i+33810~\hat j)~N$
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.