Question

A $3200kg$helicopter is flying horizontally. A $250kg$crate is suspended from the helicopter by a massless cable that is constantly $20°$from vertical. What propulsion force F u prop is being provided by the helicopter’s rotor? Air resistance can be ignored. Give your answer in component form in a coordinate system where $\hat{i}$points in the direction of motion and $\hat{j}$points upward.

Short answer

Propulsion force,$\overrightarrow{F}=\left(12.3kN\right)\hat{i}+\left(33.8kN\right)\hat{j}$

Step-by-step solution

Given information 

The mass of the helicopter $M=3200kg$

The mass of the crate$m=250kg$

Explanation 

The expression to calculate the vertical component of force is,

$f_y=mg$

The cable makes an angle with the vertical component of the force.

$$\begin{gathered} \tan \left(\theta \right)=\frac{f_x}{f_y} \\ {f}_x=mg\tan \left(\theta \right) \end{gathered}$$

The acceleration of both the systems remains the same so, the horizontal component of the force on the system is given as,

localid="1649168708556" $$\begin{gathered} \frac{f_x}{m}=\frac{f_y}{M+m} \\ \frac{mg\tan \left(\theta \right)}{m}=\frac{F_x}{M+m} \\ {F}_x=\left(M+m\right)g\tan \left(\theta \right) \\ {F}_x=\left(3200kg+250kg\right)\left(9.8m/s^2\right)\tan \left(20°\right) \\ =12305.8N \end{gathered}$$

Now calculate vertical force using

$$\begin{gathered} F_y=\left(M+m\right)g \\ =\left(3200kg+250kg\right)\left(9.8m/s^2\right) \\ =33.8kN \end{gathered}$$

Therefore the propulsion force provided by the helicopter's rotor is

localid="1649169065712" $\overrightarrow{F}=\left(12.3kN\right)\hat{i}+\left(33.8kN\right)\hat{j}$