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Q11.

Expert-verified
Found in: Page 69

### Physics Principles with Applications

Book edition 7th
Author(s) Douglas C. Giancoli
Pages 978 pages
ISBN 978-0321625922

# Determine the vector $\stackrel{\to }{\mathbf{A}}-\stackrel{\to }{\mathbf{C}}$ with given the vectors $\stackrel{\to }{\mathbf{A}}$ and $\stackrel{\to }{\mathbf{C}}$in Fig. 3–35.FIGURE 3-35

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

See the step by step solution

## Step 1. Significance of a vector

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.

## Step 2. Determination of components of vector A→

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

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

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

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

## Step 3. Determination of components of vector C→

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

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

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

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

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

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

## Step 4. Determination of the magnitude of A→-C→x

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

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

## Step 5. Determination of the magnitude of A→-C→y

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

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

## Step 6. Determination of the magnitude of the vector A→-C→

The magnitude of the vector can be calculated as

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

Substituting the values in the above equation,

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

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

## Step 7. Determination of the direction of the vectorA→-C→

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,

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

Thus, the direction of the vector $\stackrel{\to }{\mathbf{A}}-\stackrel{\to }{\mathbf{C}}$ is $53.054°$ from the positive x-axis in the anticlockwise direction.