Q28 E - University Physics with Modern Physics

Question

Let $\mathbf{\theta }$be the angle that the vector $\overrightarrow{\mathbf{A}}$ makes with the +x-axis, measured counterclockwise from that axis. Find angle $\mathbf{\theta }$ for a vector that has these components: (a) ${\mathbf{\theta A}}_{\mathbf{x}}\mathbf{=}\mathbf{-}\mathbf{2}\mathbf{.}\mathbf{00}\mathbf{ }\mathbf{m}\mathbf{,}{\mathbf{A}}_{\mathbf{y}}\mathbf{=}\mathbf{1}\mathbf{.}\mathbf{00}\mathbf{ }\mathbf{m}$; (b) ${\mathbf{A}}_{\mathbf{x}}\mathbf{=}\mathbf{2}\mathbf{.}\mathbf{00}\mathbf{ }\mathbf{m}\mathbf{,}{\mathbf{A}}_{\mathbf{y}}\mathbf{=}\mathbf{1}\mathbf{.}\mathbf{00}\mathbf{ }\mathbf{m}$; (c) ${\mathbf{A}}_{\mathbf{x}}\mathbf{=}\mathbf{-}\mathbf{2}\mathbf{.}\mathbf{00}\mathbf{ }\mathbf{m}\mathbf{,}{\mathbf{A}}_{\mathbf{y}}\mathbf{=}\mathbf{1}\mathbf{.}\mathbf{00}\mathbf{ }\mathbf{m}$; (d) ${\mathbf{A}}_{\mathbf{x}}\mathbf{=}\mathbf{-}\mathbf{2}\mathbf{.}\mathbf{00}\mathbf{\hspace{0.33em}}\mathbf{m}\mathbf{,}\mathbf{\hspace{0.33em}}{\mathbf{A}}_{\mathbf{y}}\mathbf{=}\mathbf{-}\mathbf{1}\mathbf{.}\mathbf{00}\mathbf{\hspace{0.33em}}\mathbf{m}$

Step-by-Step Solution

Answer

The angle made by the vectors can be summarised as, 

(a) $333.4°$ ,

(b) $26.6°$,

(c) $333.4°$, and

(d) $26.6°$ .

Step 1: Angle between two vectors

Here different coordinates of the vector are given. We are knowing that in coordinates x component of the vector will be first and then the y component. Now to find the angle of the vector with respect to the x-axis we just need to take the ratio of the y component to that of the x component and then take the tan inverse of that value.

 

This will give you the angle of the vector with the x-axis. 

 

It can be represented as 

 

 $\mathbf{\theta }\mathbf{=}{\mathbf{tan}}^{\mathbf{-}\mathbf{1}}\left(\frac{y}{x}\right)$…………………(1)

Step 2: (a) Calculation of the θ for A x = 2 . 00   m ,   A y = - 1 . 00   m

Substitute the given data in equation (1), and we get,

$$\begin{gathered} \theta ={\tan }^{-1}\left(\frac{A_y}{A_x}\right) \\ ={\tan }^{-1}\left(\frac{-1.00 m}{2.00 m}\right) \\ ={\tan }^{-1}\left(\frac{-1}{2}\right) \\ =-26.0° \\ =360°-26.0° \\ =333.4° \end{gathered}$$

Step 3: (b) Calculation of the θ for A x = - 2 . 00   m ,   A y = 1 . 00   m

Substitute the given data in equation (1), and we get,

$$\begin{gathered} \theta ={\tan }^{-1}\left(\frac{A_y}{A_x}\right) \\ ={\tan }^{-1}\left(\frac{-1.00 m}{2.00 m}\right) \\ ={\tan }^{-1}\left(\frac{-1}{2}\right) \\ =-26.0° \\ =360°-26.0° \\ =333.4° \end{gathered}$$ 

Step 4: (c) Calculation of the θ for A x = - 2 . 00   m ,   A y = 1 . 00   m

Substitute the given data in equation (1), and we get,

$$\begin{gathered} \theta ={\tan }^{-1}\left(\frac{A_y}{A_x}\right) \\ ={\tan }^{-1}\left(\frac{-1.00 m}{2.00 m}\right) \\ ={\tan }^{-1}\left(\frac{-1}{2}\right) \\ =-26.0° \\ =360°-26.0° \\ =333.4° \end{gathered}$$ 

Step 5: (d) Calculation of the θ for A x = - 2 . 00   m ,   A y = 1 . 00   m

Substitute the given data in equation (1), and we get,

$$\begin{gathered} \theta ={\tan }^{-1}\left(\frac{A_y}{A_x}\right) \\ ={\tan }^{-1}\left(\frac{-1.00 m}{-2.00 m}\right) \\ ={\tan }^{-1}\left(\frac{1}{2}\right) \\ =-26.0° \end{gathered}$$ 

 

 

Therefore the angle made by the vectors is (a)  $333.4°$, (b) $26.6°$ , (c) $333.4°$, and

(d) $26.6°$ .