Chapter 6, Problem 44PQ

(a)

To determine

The angle of the rope with the vertical.

Answer to Problem 44PQ

The angle of the rope with the vertical is $17.0°$.

Explanation of Solution

Assume the man as a particle of mass $M$ and diameter equal to $0.500\text{m}$.

The following figure gives the direction of all forces on the man.

Apply equilibrium condition of forces along $y$ direction.

  $\Sigma F_y=0$ (I)

Here, $\Sigma F_y$ is the net force along $y$ direction.

Write the expression for net forces along $x$ direction.

  $\Sigma F_x=Ma$ (II)

Here, $\Sigma F_x$ is the net force along $x$ direction, $M$ is the mass of man and $a$ is the acceleration of the man.

From figure1, expand equation (I) using all forces along $y$ direction.

  $T\cos \theta -Mg=0$

Here, $T$ is the tension in the rope, $\theta$ is the angle of the rope with the vertical and $g$ is the acceleration due to gravity.

Rearrange above equation to get $T\cos \theta$.

  $T\cos \theta =Mg$ (III)

From figure1, expand equation (II) using all forces along $x$ direction.

  $T\sin \theta -F_D=Ma$

Here, $F_D$ is the drag force.

Rearrange above equation to get $T\sin \theta$.

  $T\sin \theta =F_D+Ma$ (IV)

Divide equation (IV) by (III) to get $\tan \theta$.

  $\tan \theta =\frac{F_D+Ma}{Mg}$ (V)

In problem it is allowed to ignore effect of air drag force.

Substitute $0$ for $F_D$ in above equation to get $\tan \theta$.

  $\begin{array}{c}\tan \theta =\frac{0+Ma}{Mg}\\ \tan \theta =\frac{a}{g}\end{array}$ (VI)

Conclusion:

Substitute $3.00\text{m}/{\text{s}}^2$ for $a$ and $9.81\text{m}/{\text{s}}^2$ for $g$ in equation (VI) to get $\theta$.

  $\begin{array}{c}\tan \theta =\frac{3.00\text{m}/{\text{s}}^2}{9.81\text{m}/{\text{s}}^2}\\ \theta ={\tan }^{-1}\left(\frac{3.00\text{m}/{\text{s}}^2}{9.81\text{m}/{\text{s}}^2}\right)\\ =17.0°\end{array}$

Therefore, the angle of the rope with the vertical is $17.0°$.

(b)

To determine

The expression for the angle made by the rope with the vertical as a function of time by considering drag force due to air.

Answer to Problem 44PQ

The expression for the angle made by the rope with the vertical as a function of time by considering drag force due to air is $\tan \theta =0.449+\left(2.45\times {10}^{-2}\right)t+\left(1.05\times {10}^{-3}\right)t^2$ .

Explanation of Solution

Write the expression for the drag force.

  $F_D=\frac{1}{2}C\rho A{v}^2$ (VII)

Here, $F_D$ is the drag force, $C$ is the drag coefficient between man and air, $\rho$ is the density of air, $A$ is the area of surface in contact with air and $v$ is the speed of man.

Since man is modeled as a particle of dimeter $0.500\text{m}$, write expression for the area of particle responsible for drag force.

  $A=\pi {\left(\frac{d}{2}\right)}^2$ (VIII)

Write the expression for the velocity of the particle at each instant of time.

  $v=v_0+at$

Here, $v$ is the speed of particle at any instant of time, $v_0$ is the initial speed, $a$ is the acceleration and $t$ is the time.

Conclusion:

Substitute $0.500\text{m}$ for $d$ in equation (VIII) to get $A$.

  $\begin{array}{c}A=\pi {\left(\frac{0.500\text{m}}{2}\right)}^2\\ =0.196{\text{m}}^2\end{array}$

Substitute $0.500$ for $C$, $1.29\text{kg}/{\text{m}}^3$ for $\rho$ , $0.196{\text{m}}^2$ for $A$ and $v_0+at$ for $v$ in (VII) to get $F_D$.

  $\begin{array}{c}{F}_D=\frac{1}{2}\left(0.500\right)\left(1.29\text{kg}/{\text{m}}^3\right)\left(0.196{\text{m}}^2\right){\left(v_0+at\right)}^2\\ =0.0632{\left(v_0+at\right)}^2\end{array}$

Substitute $35.0\text{m}/\text{s}$ for $v_0$ and $3.00\text{m}/{\text{s}}^2$ for $a$ in above equation to $F_D$.

  $\begin{array}{c}{F}_D=0.0632{\left(35.0\text{m}/\text{s}+3.00\text{m}/{\text{s}}^{2}t\right)}^2\\ =77.42+13.27t+0.569t^2\end{array}$

Substitute $77.42+13.27t+0.569t^2$ for $F_D$, $55.0\text{kg}$ for $M$, $3.00\text{m}/{\text{s}}^2$ for $a$ and $9.81\text{m}/{\text{s}}^2$ for $g$ in equation (V) to get $\tan \theta$.

  $\begin{array}{c}\tan \theta =\frac{77.42+13.27t+0.569t^2+\left(55.0\text{kg}\right)\left(3.00\text{m}/{\text{s}}^2\right)}{\left(55.0\text{kg}\right)\left(9.81\text{m}/{\text{s}}^2\right)}\\ =0.449+\left(2.45\times {10}^{-2}\right)t+\left(1.05\times {10}^{-3}\right)t^2\end{array}$

Therefore, the expression for the angle made by the rope with the vertical as a function of time by considering drag force due to air is $\tan \theta =0.449+\left(2.45\times {10}^{-2}\right)t+\left(1.05\times {10}^{-3}\right)t^2$ .

(c)

To determine

Whether terminal speed is a meaningful concept for this situation and explain the reason.

Answer to Problem 44PQ

Terminal speed is not a meaningful concept for this situation since helicopter provides net acceleration at every time and never reaches a zero net force situation.

Explanation of Solution

Terminal speed is constant speed attained by a body moving in a fluid so that drag force is proportional to velocity of the body. Consider situation of a body moving through a fluid. The forces acting are gravitational force and drag force. Gravitational force is a constant force whereas drag force depends on velocity of the body at each instant of time. At particular point when drag force equal to gravitational force the body takes constant velocity.

In this case drag force depends on square of velocity of the body. As the drag force increases the helicopter exerts more force to provide constant acceleration $a$. Therefore, the body cannot achieve terminal speed.

(d)

To determine

The effect on tension if the helicopter continues to accelerate and the result for a real rope in this situation.

Answer to Problem 44PQ

The tension in the helicopter is inversely proportional to cosine of angle made by the rope with the vertical. As helicopter accelerates, angle made by the rope with the vertical increases. Thus, tension in the rope increases as the helicopter accelerates.

Explanation of Solution

Rearrange equation (III) to get $T$.

  $T=\frac{Mg}{\cos \theta }$

$M$ and $g$ are constants. Tension in the rope is inversely proportional to cosine of the angle made by the rope with the vertical. From figure 1, as the helicopter accelerates the angle made by the rope with vertical increases. So that the value of $\cos \theta$ decreases which in turn increases the tension in the rope.

Therefore, tension in the rope continues to increase as the helicopter accelerates. A real rope has certain limit to withstand tension. After a particular value it will break.