# Chapter 7 - Newton's Third Law - Exercises and Problems - Page 180: 44

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

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

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