Chapter 3, Problem 9P

Three vectors are shown in Fig. 3-35 Q. Their magnitudes are given in arbitrary units. Determine the sum of the three vectors. Give the resultant in terms of (a) components, (b) magnitude and angle with the + x axis.

Figure 3—35
Problems 9, 10, 11, 12, and 13. Vector magnitudes are given in arbitrary units.

Solution

To Determine:

1. The resultant $\overrightarrow{R}$ of the sum of three vectors $\overrightarrow{A}+\overrightarrow{B}+\overrightarrow{C}$ in terms of its components.
2. The resultant $\overrightarrow{R}$ of the sum of three vectors $\overrightarrow{A}+\overrightarrow{B}+\overrightarrow{C}$ in terms of its magnitude and angle (direction) with the + x-axis.

Solution:

1. The x-component $R_x$ of the resultant $\overrightarrow{R}$ is 24.0 arbitrary units and the y-component $R_y$ of the resultant $\overrightarrow{R}$ is 11.6 arbitrary units.
2. The magnitude $\left|\overrightarrow{R}\right|$ of the resultant vector $\overrightarrow{R}$ \theta {\text{'}}_A$$ is 76.7 arbitrary units with a direction θ_R counterclockwise of 47.6° respect to +x-axis.

Explanation:

Section 1:

Given: The magnitude $\left|\overrightarrow{A}\right|$ is 44.0 with a direction of $28.0˚$ counterclockwise from the positive x -axis. For $\left|\overrightarrow{B}\right|$ the magnitude is 26.5with a direction $\theta {\text{'}}_B$ of $56.0˚$ clockwise from the negative x -axis and the magnitude of $\left|\overrightarrow{C}\right|$ is 31.0with a direction of $\theta {\text{'}}_C$ of $270˚$ on the negative y-axis.

Formula used:

The resultant $\overrightarrow{R}$ in terms of its components can be defined as:

The x-component of the resultant $R_x$ can be obtained using:

  $R_x=A_x+B_x+C_x=\left|\overrightarrow{A}\right|\cos ({\theta }_A)+\left|\overrightarrow{B}\right|\cos ({\theta }_B)+\left|\overrightarrow{C}\right|\cos ({\theta }_C)(1)$

The y-component of the resultant $R_y$ can be obtained using:

  $R_y=A_y+B_y+C_y=\left|\overrightarrow{A}\right|\sin ({\theta }_A)+\left|\overrightarrow{B}\right|\sin ({\theta }_B)+\left|\overrightarrow{C}\right|\sin ({\theta }_C)(2)$

Where:

• ${\theta }_A$ , ${\theta }_B$ and ${\theta }_C$ are the directions for vectors $\overrightarrow{A},\overrightarrow{B}\text{and}\overrightarrow{C}$ respectively measured counterclockwise from the + x -axis.
• $A_y$ , $B_y$ and $C_y$ are the y-components for vectors $\overrightarrow{A},\overrightarrow{B}\text{and}\overrightarrow{C}$ respectively.
• $A_x$ , $B_x$ and $C_x$ are the x-components for vectors $\overrightarrow{A},\overrightarrow{B}\text{and}\overrightarrow{C}$ respectively.
• $\left|\overrightarrow{A}\right|$ , $\left|\overrightarrow{B}\right|$ and $\left|\overrightarrow{C}\right|$ are the magnitudes for vectors $\overrightarrow{A},\overrightarrow{B}\text{and}\overrightarrow{C}$ respectively.

The resultant $\left|\overrightarrow{R}\right|$ of the sum of three vectors $\overrightarrow{A}+\overrightarrow{B}+\overrightarrow{C}$ in terms of its magnitude $\left|\overrightarrow{R}\right|$ can be obtained using:

  $\left|\overrightarrow{R}\right|=\sqrt{R_x{}^2+R_y{}^2}$

Similarly, the direction (angle) θ_Ris defined as:

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

Where:

• $R_x$ is the x -component of the resultant vector $\overrightarrow{R}$ .
• $R_y$ is the y -component of the resultant vector $\overrightarrow{R}$ .

Calculation:

Replacing data given into equation (1):

  $R_x=\left(44.0\right)\cos ({28}^o)+\left(26.5\right)\cos \left({180}^o-{56}^o\right)+\left(31.0\right)\cos ({270}^o)$

  $R_x=24.0$

Similarly, inserting data given into equation (2):

  $R_y=44.0\sin ({28.0}^{\circ })+\left(26.5\right)\sin ({180}^{\circ }-{56.0}^{\circ })+\left(31.0\right)\sin ({270}^{\circ })$

  $R_y=11.6$

Therefore, the x -component $R_x$ of the resultant $\overrightarrow{R}$ is 24.0 arbitrary units and the y-component $R_y$ of the resultant $\overrightarrow{R}$ is 11.6 arbitrary units.

Section 2: Substitute the values in formula for finding the magnitude of the resultant:

  $\left|\overrightarrow{R}\right|=\sqrt{{24.0}^2+{11.6}^2}=26.7$

Similarly, inserting results (4) and (6) into equation (8):

  ${\theta }_R={\tan }^{-1}\left(\frac{11.6}{24.0}\right)={25.8}^{\circ }$

Conclusion:

The x -component $R_x$ of the resultant $\overrightarrow{R}$ is 24.0 arbitrary units and the y-component $R_y$ of the resultant $\overrightarrow{R}$ is 11.6 arbitrary units. The magnitude $\left|\overrightarrow{R}\right|$ of the resultant vector $\overrightarrow{R}$ is 26.7 arbitrary units with a direction ${\theta }_R$ counterclockwise of $25.8˚$ respect to + x -axis.