A helicopter flies to the right (in the +xdirection) at a constant speed of 12m/s, parallel to the surface of the ocean. A 900 kgpackage of supplies is suspended below the helicopter by a cable as shown in Figure the package is also traveling to the right in a straight line, at a constant speed of 12 m/s. The pilot is concerned about whether or not the cable, whose breaking strength is listed at 9300 N , is strong enough to support this package under these circumstances.

(a) Choose the package as the system, and draw a free-body diagram.

(b) What is the magnitude of the tension in the cable supporting the package?

(c) Write the force exerted on the package by the cable as a vector.

(d) What is the magnitude of the force exerted by the air on the package?

(e) Write the force on the package by the air as a vector.

(f) Is the cable in danger of breaking?

Draw a free-diagram for the package. The system here represented by the package, while the surroundings are the gravitational force ${\overrightarrow{F}}_g$, the applied force by the air role="math" localid="1657080450764" ${\overrightarrow{F}}_a$in the opposite direction of the motion and the tension force of the cable $F_T$where $F_T$makes an angle $\theta$ withx-axis. So, it has two components $\left(F_{T,x}\mathrm{and}{F}_{T,y}\right)$. The free-diagram is shown in the figure below, where the direction of the gravitational force ${\overrightarrow{F}}_g$is downward. The force of air to left, while the cable makes an angle $\theta$to the left of the x-axis.

The tension force of the cable $F_T$makes an angle with x-axis. So, it has two components$\left(F_{T,x}\mathrm{and}{F}_{T,y}\right)$ . By using direction cosines we can get $F_{T,y}$as next

$$\begin{gathered} F_{Ty}=F_T\cos \left(90°-\theta \right) \\ {F}_T=\frac{F_{T,y}}{\cos \left(90°-\theta \right)} \end{gathered}$$

Where $\theta$is the angle between $F_T$and the x-axis. Let us find$\theta$ , from the given figure in the textbook and as shown in the drawn free-diagram in part (a) we have a triangle and from this triangle, we could get $\theta$as next

$$\begin{gathered} \tan \theta =\frac{11}{4.5} \\ \theta ={\tan }^{-1}\frac{11}{4.5} \\ \theta =67.75° \end{gathered}$$

From the free body diagram, the net force in the y-axis is zero as the change in the momentum is zero where the net force equals

$$\begin{gathered} \frac{d\overrightarrow{p}}{dt}=F_{net,y} \\ =F_{T2,y}-F_g \end{gathered}$$

The gravitational force is downward and in the negative direction and it is given by,

$F_g=-mg$

Where m is the mass of the package and g is the gravitational acceleration. Now we can plug our values for mand g into above equation to get F_g.

$$\begin{gathered} F_g=-\left(900\mathrm{kg}\right)\left(9.8\mathrm{m}/{\mathrm{s}}^2\right) \\ =-8820\mathrm{N} \end{gathered}$$

The net force is zero, so, let us solve the above equation for $F_{T,y}$and plug the values for$F_g$where,

$$\begin{gathered} F_{T,y}=F_g+F_{net,y} \\ =8820+0 \\ =8820\mathrm{N} \end{gathered}$$

Now we can plug our values for role="math" localid="1657081747754" $F_{T,y}$and $\theta$into the above equation to get $F_T$,

$$\begin{gathered} F_T=\frac{F_{T,y}}{\cos \left(90°-\theta \right)} \\ =\frac{8820}{\cos \left(90°-67.75°\right)} \\ =9530\mathrm{N} \end{gathered}$$

The tension force of the cable $F_T$makes an angle with x-axis. So, it has two components $\left(F_{T,x}\mathrm{and}{F}_{T,y}\right)$. By using direction cosines we can get $F_{T,x}$as next$F_{T,x}=F_T\cos \theta$

Where $\theta$is the angle between the$F_T$and x-axis. Now we can plug our values for $F_T$and $\theta$into the above equation to get$F_{T,x}$,

$$\begin{gathered} F_{T,x}=F_T\cos \theta \\ =\left(9530\right)\cos 67.75° \\ =3608\mathrm{N} \end{gathered}$$

We calculated the y-component of the tension force$F_{T,y}=8820\mathrm{N}$. Also, there is no force component in z-direction, therefore,$F_T,z=0$. The force vector in the three directions is given by,

localid="1657083207944" ${\overrightarrow{F}}_T=F_{T,x}\hat{i}+F_{T,y}\hat{j}+F_{T,z}\hat{k}$

Now we can plug our values for localid="1657082475881" $F_{T,x},F_{T,y}$and $F_{T,z}$into the above equation to get the vector of the tension force,

localid="1657083249607" $$\begin{gathered} {\overrightarrow{F}}_T=F_{T,x}\hat{i}+F_{T,y}\hat{j}+F_{T,z}\hat{k} \\ =\left(3608\hat{i}+8820\hat{j}+0\hat{k}\right) \\ =\left(3608\hat{i}+8820\hat{j}\right)\mathrm{N} \end{gathered}$$

As shown by the free-diagram, the x-component has two forces, the applied force by the air ${\overrightarrow{F}}_a$and the component$F_{T,x}$ . And the net force in x-axis is zero as the change in the momentum is zero where the net force equals

$$\begin{gathered} \frac{d\overrightarrow{p}}{dt}=F_{net,x} \\ =F_{T,x}+F_a \end{gathered}$$

For $F_a$and plug the values for $F_{T,x}$,

$$\begin{gathered} F_a=F_{net,x}-F_{T,x} \\ =0-3608 \\ =-3608\mathrm{N} \end{gathered}$$

We calculated the x-component of the air force

$F_a=-3608\mathrm{N}$

Also, there is no force component in z-direction or in y-direction by the air, therefore,$F_{a,x}=0$and$F_{a,z}=0$

The force vector in the three directions is given by

${\overrightarrow{F}}_T=F_{T,x}\hat{i}+F_{T,y}\hat{j}+F_{T,z}\hat{k}$

Now we can plug our values for$F_{a,x},F_{a,y}$and$F_{a,z}$into above equation to get the vector of the air force,

$$\begin{gathered} {\overrightarrow{F}}_a=F_{a,x}\hat{i}+F_{a,y}\hat{j}+F_{a,z}\hat{k} \\ =\left(3608\hat{i}+0\hat{j}+0\hat{k}\right) \\ =\left(3608\hat{i}\right)\mathrm{N} \end{gathered}$$

As given that the breaking strength of the cable is 9300 N and we calculated the tension force $F_T$acted on the cable by 9530 N. So, breaking strength of the cable is less than the tension force. Hence, the cable is in danger and will be broken.