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Q11.

Expert-verifiedFound in: Page 69

Book edition
7th

Author(s)
Douglas C. Giancoli

Pages
978 pages

ISBN
978-0321625922

**Determine the vector $\overrightarrow{\mathbf{A}}-\overrightarrow{\mathbf{C}}$**** with given the vectors $\overrightarrow{\mathbf{A}}$**** and $\overrightarrow{\mathbf{C}}$**

**in Fig. 3–35.**

**FIGURE 3****-35**

The magnitude of the vector $\overrightarrow{\mathbf{A}}-\overrightarrow{\mathbf{C}}$** **is $64.636$, and the direction of this vector is$53.054\xb0$ from the positive *x*-axis in the anticlockwise direction.

**A vector can be used to define the magnitude along with the direction of a physical quantity.**

**The resultant of two or more vectors can be calculated with the help of the addition or the subtraction of the horizontal and vertical components of the individual vectors.**

** **

The component of vector A in the *x*-direction can be calculated with the help of figure 3-35 given in the question as

$\begin{array}{c}{A}_{x}=44\mathrm{cos}28\xb0\\ =38.85\end{array}$

The component of vector A in the *y*-direction can be expressed as

$\begin{array}{c}{A}_{y}=44\mathrm{sin}28\xb0\\ =20.657\end{array}$

The inclination of the vector C from the *x*-axis in figure 3-35 can be given as

$\begin{array}{c}\varphi =90\xb0+90\xb0+90\xb0\\ =270\xb0\end{array}$

Here, $\varphi $ is the angle of inclination of vector C with the positive *x*-axis in the clockwise direction.

The component of vector C in the *x*-direction can be calculated with the help of figure 3-35 given in the question as

$\begin{array}{c}{C}_{x}=31\mathrm{cos}\varphi \\ =31\mathrm{cos}270\xb0\\ =0\end{array}$

The component of vector C in the –*y-*direction can be expressed as

$\begin{array}{c}{C}_{y}=31\mathrm{sin}\varphi \\ =31\mathrm{sin}270\xb0\\ =-31\end{array}$

The magnitude of the component of vector $\left(\overrightarrow{\mathbf{A}}-\overrightarrow{\mathbf{C}}\right)$ in the *x*-direction can be expressed as

$\begin{array}{c}\left|{\left(\overrightarrow{\mathbf{A}}-\overrightarrow{\mathbf{C}}\right)}_{x}\right|={A}_{x}-{C}_{x}\\ =38.85-0\\ =38.85\end{array}$

The component of vector $\left(\overrightarrow{\mathbf{A}}-\overrightarrow{\mathbf{C}}\right)$ in the *y*-direction can be expressed as

$\begin{array}{c}\left|{\left(\overrightarrow{\mathbf{A}}-\overrightarrow{\mathbf{C}}\right)}_{y}\right|={A}_{y}-{C}_{y}\\ =20.657-\left(-31\right)\\ =51.657\end{array}$

** **

The magnitude of the vector can be calculated as

$\left|\overrightarrow{\mathbf{A}}-\overrightarrow{\mathbf{C}}\right|=\sqrt{{\left(\left|{\left(\overrightarrow{\mathbf{A}}-\overrightarrow{\mathbf{C}}\right)}_{x}\right|\right)}^{2}+{\left(\left|{\left(\overrightarrow{\mathbf{A}}-\overrightarrow{\mathbf{C}}\right)}_{y}\right|\right)}^{2}}$.

Substituting the values in the above equation,

$\begin{array}{c}\left|\overrightarrow{\mathbf{A}}-\overrightarrow{\mathbf{C}}\right|=\sqrt{{\left(38.85\right)}^{2}+{\left(51.657\right)}^{2}}\\ =64.636\end{array}$.

Thus, the magnitude of the vector $\overrightarrow{\mathbf{A}}-\overrightarrow{\mathbf{C}}$ is $64.636$.

** **

The direction of the vector can be expressed as

$\begin{array}{c}\mathrm{tan}\theta =\frac{{A}_{y}-{C}_{y}}{{A}_{x}-{C}_{x}}\\ \theta ={\mathrm{tan}}^{-1}\left\{\frac{{A}_{y}-{C}_{y}}{{A}_{x}-{C}_{x}}\right\}\end{array}$

Substituting the values in the above expression,

$\begin{array}{c}\theta ={\mathrm{tan}}^{-1}\left(\frac{51.657}{38.85}\right)\\ =53.054\xb0\end{array}$

Thus, the direction of the vector $\overrightarrow{\mathbf{A}}-\overrightarrow{\mathbf{C}}$** **is $53.054\xb0$ from the positive *x*-axis in the anticlockwise direction.

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