Chapter 11.5, Problem 11.163P

During a parasailing ride, the boat is traveling at a constant 30 km/hr with a 200-m long tow line. At the instant shown, the angle between the line and the water is 30° and is increasing at a constant rate of 2°/s. Determine the velocity and acceleration of the parasailer at this instant.

Fig. P11.163 and P11.164

To determine

The velocity $\left(v\right)$ and acceleration $\left(a\right)$ of the parasailer at this instant.

Answer to Problem 11.163P

The velocity $\left(v\right)$ and acceleration $\left(a\right)$ of the parasailer at this instant are $\underset{\_}{13.280\text{m/s}\left(27.08°\right)}$ and $\underset{\_}{0.2437\text{m/s}\left(30.00°\right)}$ respectively.

Explanation of Solution

Given Information:

The boat is traveling at a constant speed $\left(v_B\right)$ of $30\text{km/hr}$.

The radius (r) of tow line is $200\text{m}$.

The angle $\left(\theta \right)$ between the line and the water is $30°$ and increasing at a constant $\left(\dot{\theta }\right)$ rate of $\text{2°/s}$.

Calculation:

Convert the kilometer per hour to meter per second.

Consider $\left(v_B\right)$:

$\begin{array}{c}\left(v_B\right)=30\text{km/hr}\times \frac{1000\text{m}}{1\text{km}}\times \frac{1\text{hr}}{3600\sec }\\ =8.333\text{m/s}\end{array}$

Show the Free body diagram of parasailer and boat as in Figure (1).

Write the velocity $\left(v_B\right)$ of the boat in term of vector:

$v_B=8.333i\text{m/s}$

The acceleration vector of the boat is as follows:

$a_B=0$

Differentiate angle $\left(\dot{\theta }\right)$ with respective to time (t).

$\ddot{\theta }=0$

Differentiate radius (r) with respective to time (t).

$\dot{r}=0$

Differentiate $\left(\dot{r}\right)$ with respective to time (t).

$\ddot{r}=0$

Write the expression for velocity vector $\left(v_P\right)$ of parasailer:

$v_P=v_B+v_{P/B}$ . (1)

Here, $v_{P/B}$ is relative velocity vector of parasailer with respect to boat.

Write the expression for acceleration vector $\left(a_P\right)$ of parasailer:

$a_P=a_B+a_{P/B}$ (2)

Here, $a_{P/B}$ is relative acceleration vector of parasailer with respect to boat.

Calculate the velocity vector ${\left(v_{P/B}\right)}_{\text{radial}}$ of parasailer with respect boat using radial and transverse component:

${\left(v_{P/B}\right)}_{\text{radial}}=\dot{r}{e}_r+r\dot{\theta }{e}_{\theta }$

Substitute 0 for $\dot{r}$, $200\text{m}$ for r and $\text{2°/s}$ for $\dot{\theta }$.

$\begin{array}{c}{\left(v_{P/B}\right)}_{\text{radial}}=0e_r+\left(200\right)\left(2°/s\times \frac{\pi \text{rad}}{180°}\right)e_{\theta }\\ =\left(6.981\text{m/s}\right)e_{\theta }\end{array}$

Calculate the acceleration vector ${\left(a_{P/B}\right)}_{\text{radial}}$ of parasailer with respect boat using radial and transverse component:

${\left(a_{P/B}\right)}_{\text{radial}}=\left(\ddot{r}+r{\dot{\theta }}^2\right)e_r+\left(r\ddot{\theta }+2\dot{r}\dot{\theta }\right)e_{\theta }$

Substitute 0 for $\dot{r}$, $200\text{m}$ for r, 0 for $\ddot{r}$, $\text{2°/s}$ for $\dot{\theta }$, $30°$ for $\theta$, and 0 for $\ddot{\theta }$.

$\begin{array}{c}{\left(a_{P/B}\right)}_{\text{radial}}=\left(0+\left(200\right){\left(2°/s\times \frac{\pi \text{rad}}{180°}\right)}^2\right)e_r+\left(\left(200\right)\left(0\right)+2\times \left(0\right)\times \left(0\right)\right)e_{\theta }\\ =\left(-0.2437{\text{m/s}}^{\text{2}}\right)e_r\end{array}$

Write the velocity vector $\left(v_{P/B}\right)$ of parasailer with respect boat in rectangular coordinates using Equation (1):

$v_{P/B}={\left(v_{P/B}\right)}_{\text{radial}}\sin \theta i+{\left(v_{P/B}\right)}_{\text{radial}}\cos \theta j$

Substitute $6.981\text{m/s}$ for ${\left(v_{P/B}\right)}_{\text{radial}}$ and $30°$ for $\theta$.

$\begin{array}{c}{v}_{P/B}=6.981\times \sin \left(30°\right)i+6.981\times \cos \left(30°\right)j\\ =3.491i+6.046j\end{array}$

Write the acceleration vector $\left(a_{P/B}\right)$ of parasailer with respect boat in rectangular coordinates using the Equation (2):

$a_{P/B}={\left(a_{P/B}\right)}_{\text{radial}}\left(-\cos \theta \right)i-{\left(a_{P/B}\right)}_{\text{radial}}\sin \theta j$

Substitute $-0.2437{\text{m/s}}^{\text{2}}$ for ${\left(a_{P/B}\right)}_{\text{radial}}$ and $30°$ for $\theta$.

$\begin{array}{c}{a}_{P/B}=-0.2437\times \left(-\cos 30°\right)i+0.2437\times \sin \left(30°\right)j\\ =0.2111i-0.1219j\end{array}$

Calculate the velocity vector $\left(v_P\right)$ of parasailer:

Substitute $3.491i+6.046j$ for $v_{P/B}$ and $8.333i\text{m/s}$ for $v_B$ in Equation (1).

$\begin{array}{c}{v}_P=8.333i+3.491i+6.046j\\ =11.824i+6.046j\end{array}$

Here, $11.824\text{m/s}$ for ${\left(v_p\right)}_x$ and $6.046\text{m/s}$ for ${\left(v_p\right)}_y$.

Calculate the velocity $\left(v_P\right)$ of parasailer using the relation:

$\left(v_p\right)=\sqrt{{\left(v_P\right)}_x^2+{\left(v_P\right)}_y^2}$

Substitute $11.824\text{m/s}$ for ${\left(v_p\right)}_x$ and $6.046\text{m/s}$ for ${\left(v_p\right)}_y$.

$\begin{array}{c}\left(v_p\right)=\sqrt{{\left(11.824\right)}^2+{\left(6.046\right)}^2}\\ =\sqrt{176.361}\\ =13.280\text{m/s}\end{array}$

Calculate the angle $\left(\alpha \right)$:

$\alpha ={\tan }^{-1}\left(\frac{{\left(v_p\right)}_y}{{\left(v_p\right)}_x}\right)$

Substitute $11.824\text{m/s}$ for ${\left(v_p\right)}_x$ and $6.046\text{m/s}$ for ${\left(v_p\right)}_y$.

$\begin{array}{c}\alpha ={\tan }^{-1}\left(\frac{6.046}{11.824}\right)\\ =27.08°\end{array}$

Calculate the acceleration vector $\left(a_P\right)$ of parasailer:

Substitute $0.2111i-0.1219j$ for $a_{P/B}$ and 0 for $a_B$ in Equation (2).

$\begin{array}{c}{a}_P=0+\left(0.2111i-0.1219j\right)\\ =0.2111i-0.1219j\end{array}$

Here, $0.2111{\text{m/s}}^2$ for ${\left(a_p\right)}_x$ and $-0.1219{\text{m/s}}^{\text{2}}$ for ${\left(a_p\right)}_y$.

Calculate the acceleration $\left(a_P\right)$ of parasailer using the relation:

$\left(a_p\right)=\sqrt{{\left(a_P\right)}_x^2+{\left(a_P\right)}_y^2}$

Substitute $0.2111{\text{m/s}}^2$ for ${\left(v_p\right)}_x$ and $-0.1219{\text{m/s}}^{\text{2}}$ for ${\left(v_p\right)}_y$.

$\begin{array}{c}\left(a_p\right)=\sqrt{{\left(0.2111\right)}^2+{\left(-0.1219\right)}^2}\\ =\sqrt{0.0594}\\ =0.2437\text{m/s}\end{array}$

Calculate the angle $\left(\varphi \right)$:

$\varphi ={\tan }^{-1}\left(\frac{{\left(a_p\right)}_y}{{\left(a_p\right)}_x}\right)$

Substitute $0.2111{\text{m/s}}^2$ for ${\left(v_p\right)}_x$ and $-0.1219{\text{m/s}}^{\text{2}}$ for ${\left(v_p\right)}_y$.

$\begin{array}{c}\alpha ={\tan }^{-1}\left(\frac{-0.1219}{0.2111}\right)\\ =-30.00°\end{array}$

Therefore, the velocity $\left(v\right)$ and acceleration $\left(a\right)$ of the parasailer at this instant are $\underset{\_}{13.280\text{m/s}\left(27.08°\right)}$ and $\underset{\_}{0.2437\text{m/s}\left(30.00°\right)}$ respectively.