Chapter 3, Problem 5P

V is a vector 24.8 units in magnitude and points at an angle of 23.4° above the negative axis, (a) Sketch this vector, (b) Calculate V_xand V_y. (c) Use V_xand V_y.andVy to obtain (again) the magnitude and direction of [Note: Part (c) is a good way to check if you've resolved your vector correctly.]

Solution

1. Sketch of the vector $\left|\overrightarrow{V}\right|=24.8\text{units}$ and points at an angle of $\theta \text{=23}{\text{.4}}^o$ above the negative x -axis
2. $V_x$ and $V_y$
3. The magnitude $\left|\overrightarrow{V}\right|$ and direction $\theta$ using $V_x$ and $V_y$ .

Solution:

1. The sketch of the vector is shown below:

  

2. The components $V_x$ and $V_y$ are -22.8 units and 9.85 units respectively.
3. The magnitude $\left|\overrightarrow{V}\right|$ is 24.8 units and the direction $\theta \text{=23}{\text{.4}}^o$ above the negative x-axis.

Explanation:

The magnitude of the vector $\left|\overrightarrow{V}\right|=24.8\text{units}$ and points at an angle of $\theta \text{=23}{\text{.4}}^o$ above the negative x-axis

In this case, the direction θ is given above negative x-axis, this angle is counted from the negative-axis as follows:

  

Formula used:

The x-component $V_x$ of vector $\overrightarrow{V}$ is defined as:

  $V_x=\left|\overrightarrow{V}\right|\cos \theta$

The y-component of $V_y$ vector $\overrightarrow{V}$ is defined as:

  $V_y=\left|\overrightarrow{V}\right|\sin \theta$

Where:

• $\left|\overrightarrow{V}\right|$ is the magnitude of the vector $\overrightarrow{V}$
• $\theta$ is the direction of the vector  $\overrightarrow{V}$ measured respect to the positive x -axis counter-clockwise.

The magnitude of vector can be obtained by:

  $\left|\overrightarrow{V}\right|=\sqrt{\left(V_x{}^2\right)+{\left(V_y\right)}^2}$

And, the direction θ is defined as:

  $\theta ={\tan }^{-1}\left(\frac{V_y}{V_x}\right)$

Calculation:

Substitute the values

  $\begin{array}{l}{V}_x=24.8\cos \left({180}^o-{23.4}^o\right)\\ V_x=-22.8\text{units}\\ V_y=24.8\sin \left({180}^o-{23.4}^o\right)\\ V_y=9.85\text{units}\end{array}$

The magnitude of vector can be obtained by:

  $\begin{array}{l}\left|\overrightarrow{V}\right|=\sqrt{\left(V_x{}^2\right)+{\left(V_y\right)}^2}\\ \left|\overrightarrow{V}\right|=\sqrt{{\left(-22.8\right)}^2+{\left(9.85\right)}^2}\\ \left|\overrightarrow{V}\right|=24.8\end{array}$

And, the direction θ is defined as:

  $\theta ={\tan }^{-1}\left(\frac{V_y}{V_x}\right)$

  $\theta ={\tan }^{-1}\left(\frac{9.85}{-22.8}\right)=-{23.4}^o$