Question

Figure 4-57 shows the path taken by a drunk skunk over level ground, from initial point i to final point f .The angles are ${\mathit{\theta }}_{\mathbf{1}}\mathbf{=}\mathbf{30}\mathbf{.}{\mathbf{0}}^{\mathbf{°}}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}{\mathit{\theta }}_{\mathbf{2}}\mathbf{=}\mathbf{50}\mathbf{.}{\mathbf{0}}^{\mathbf{°}}$, and ${\mathit{\theta }}_3\mathbf{=}\mathbf{80}\mathbf{.}{\mathbf{0}}^{\mathbf{°}}\mathbf{}$and the distances are ${\mathit{d}}_{\mathbf{1}}\mathbf{=}\mathbf{5}\mathbf{.}\mathbf{00}\mathbf{}\mathit{m}\mathbf{}\mathbf{,}\mathbf{}\mathbf{}\mathbf{}{\mathit{d}}_{\mathbf{2}}\mathbf{=}\mathbf{8}\mathbf{.}\mathbf{00}\mathbf{}\mathit{m}$, and ${\mathit{d}}_{\mathbf{3}}\mathbf{=}\mathbf{12}\mathbf{.}\mathbf{0}\mathbf{}\mathit{m}$. What are the (a)magnitude and (b) angle of the skunk’s displacement from i to f ?

Short answer

Answer

1. Magnitude of the skunk’s displacement from i to f is

2. Angle of skunk’s displacement from i to f is ${129}^{°}$measured in clockwise direction.

Step-by-step solution

Given information

It is given that,

$$\begin{gathered} d_1=5.0m \\ {\theta }_1={30}^{°} \\ {d}_2=8.0m \\ {\theta }_2={50}^{°} \\ {d}_3=12.0m \\ {\theta }_3={80}^{°} \end{gathered}$$

Determining the concept

This problem is based on the simple concept of magnitude and direction of resultant vector which is the sum of two or more vectors.Thus, first resolve all the displacements along x and y directions. Then, the sum of components along x and y direction can be calculated separately. Then, by using the formulas the magnitude and angle made by resultant with +x axis will be calculated.

Formula:

Magnitude of resultant vector is given by,

$R=\sqrt{R_x^2+R_y^2}$ …(i)

And the Direction of the resultant vector is given by,

$\theta ={\tan }^{-1}\left(\frac{R_y}{R_x}\right)$ …(ii)

Where, R is the resultant position vector, $R_x$is horizontal component of position vector, $R_y$is vertical component of position vector and$\theta$is direction of resultant vector.

(a) Determining the magnitude of the skunk’s displacement

Resultant displacement can be calculated by finding its components,

$$\begin{gathered} d_x=d_{1x}+d_{2x}+d_{3x} \\ {d}_y=d_{1y}+d_{2y}+d_{3y} \\ {\theta }_{1x}=d_1\times \cos \left({}_1\right) \\ =5.0m\times \cos \left({30}^{°}\right) \\ =4.33m \\ {\theta }_{2x}=\theta d_2\times \cos \left({180}^{°}+{}_1-{}_2\right) \\ =8.0m\times \cos \left({160}^{°}\right) \\ =-7.5m \\ {\theta }_{3x}=\theta d_3\times \theta \cos \left(360-{}_3-{}_2+{}_1\right) \\ =12.0m\times \cos \left({260}^{°}\right) \\ =-2.1m \end{gathered}$$

Hence,

$$\begin{gathered} d_x=4.33m-7.5m-2.1m \\ =-5.27m \\ {\theta }_{1y}=d_1\times \sin \left({}_1\right) \\ =5.0m\times \sin \left({30}^{°}\right) \\ =2.5m \\ {\theta }_{2y}=\theta d_2\times \sin \left({180}^{°}+{}_1-{}_2\right) \\ =8.0m\times \sin \left({160}^{°}\right) \\ =2.74m \\ {\theta }_{3y}=\theta d_3\times \theta \sin \left({360}^{°}-{}_3-{}_2+{}_1\right) \\ =12.0m\times \sin \left({260}^{°}\right) \\ =-11.82m \end{gathered}$$

Hence,

$$\begin{gathered} d_y=2.5m+2.74m-11.82m \\ =-6.58m \end{gathered}$$

Therefore, using the notation given in equation (i), the magnitude of displacement is,

$$\begin{gathered} d=\sqrt{d_x^2+d_y^2} \\ =\sqrt{\left({\left(-5.27m\right)}^2\right)+\left({\left(-6.58m\right)}^2\right)} \\ =8.43m \end{gathered}$$

Therefore, the magnitude of displacement is 8.43 m.

(b) determining the angle of skunk’s displacement

From the equation (ii), the direction of resultant displacement is given by

$$\begin{gathered} \theta =\left(\frac{d_y}{d_x}\right) \\ =\left(\frac{-6.58m}{-5.27m}\right) \\ ={51}^{°} \end{gathered}$$

Components of displacement are in third quadrant.

Therefore, angle from role="math" localid="1655375812403" $\mathbf{+}\mathit{x}\mathbf{}\mathit{a}\mathit{x}\mathit{i}\mathit{s}$is,

$$\begin{gathered} \theta =180+51 \\ ={231}^{°} \end{gathered}$$

If the is measured in clockwise direction the angle from $\mathbf{+}\mathit{x}\mathbf{}\mathit{a}\mathit{x}\mathit{i}\mathit{s}$is,

${\mathbf{360}}^{\mathbf{°}}\mathbf{-}{\mathbf{231}}^{\mathbf{°}}\mathbf{=}{\mathbf{129}}^{\mathbf{°}}$.

Therefore, the required angle is ${\mathbf{129}}^{\mathbf{°}}$when measured in clockwise direction.