2.3 Algebra of vectors (Page 2/8)

University physics volume 1 Page 2 / 8

{\begin{matrix} R_{x} = A_{x} + B_{x}, \\ R_{y} = A_{y} + B_{y}, \\ R_{z} = A_{z} + B_{z} . \end{matrix}

Analytical methods can be used to find components of a resultant of many vectors. For example, if we are to sum up $N$ vectors ${\vec{F}}_{1}, {\vec{F}}_{2}, {\vec{F}}_{3}, \dots, {\vec{F}}_{N}$ , where each vector is ${\vec{F}}_{k} = F_{k x} \hat{i} + F_{k y} \hat{j} + F_{k z} \hat{k}$ , the resultant vector ${\vec{F}}_{R}$ is

\begin{matrix} {\vec{F}}_{R} = {\vec{F}}_{1} + {\vec{F}}_{2} + {\vec{F}}_{3} + … + {\vec{F}}_{N} & = \sum_{k = 1}^{N} {\vec{F}}_{k} = \sum_{k = 1}^{N} (F_{k x} \hat{i} + F_{k y} \hat{j} + F_{k z} \hat{k}) \\ = (\sum_{k = 1}^{N} F_{k x}) \hat{i} + (\sum_{k = 1}^{N} F_{k y}) \hat{j} + (\sum_{k = 1}^{N} F_{k z}) \hat{k} . \end{matrix}

Therefore, scalar components of the resultant vector are

{\begin{matrix} F_{R x} = \sum_{k = 1}^{N} F_{k x} = F_{1 x} + F_{2 x} + … + F_{N x} \\ F_{R y} = \sum_{k = 1}^{N} F_{k y} = F_{1 y} + F_{2 y} + … + F_{N y} \\ F_{R z} = \sum_{k = 1}^{N} F_{k z} = F_{1 z} + F_{2 z} + … + F_{N z} . \end{matrix}

Having found the scalar components, we can write the resultant in vector component form:

{\vec{F}}_{R} = F_{R x} \hat{i} + F_{R y} \hat{j} + F_{R z} \hat{k} .

Analytical methods for finding the resultant and, in general, for solving vector equations are very important in physics because many physical quantities are vectors. For example, we use this method in kinematics to find resultant displacement vectors and resultant velocity vectors, in mechanics to find resultant force vectors and the resultants of many derived vector quantities, and in electricity and magnetism to find resultant electric or magnetic vector fields.

Analytical computation of a resultant

Three displacement vectors

\vec{A}

\vec{B}

, and

\vec{C}

in a plane ( [link] ) are specified by their magnitudes A = 10.0, B = 7.0, and C = 8.0, respectively, and by their respective direction angles with the horizontal direction

α = 35 °,

β = −110 °

, and

γ = 30 °

. The physical units of the magnitudes are centimeters. Resolve the vectors to their scalar components and find the following vector sums: (a)

\vec{R} = \vec{A} + \vec{B} + \vec{C}

, (b)

\vec{D} = \vec{A} - \vec{B}

, and (c)

\vec{S} = \vec{A} - 3 \vec{B} + \vec{C}

Strategy

First, we use [link] to find the scalar components of each vector and then we express each vector in its vector component form given by [link] . Then, we use analytical methods of vector algebra to find the resultants.

Solution

We resolve the given vectors to their scalar components:

\begin{array}{l} {\begin{array}{l} A_{x} = A cos α = (10.0 cm) cos 35 ° = 8.19 cm \\ A_{y} = A sin α = (10.0 cm) sin 35 ° = 5.73 cm \end{array} \\ {\begin{array}{l} B_{x} = B cos β = (7.0 cm) cos (−110 °) = −2.39 cm \\ B_{y} = B sin β = (7.0 cm) sin (−110 °) = −6.58 cm \end{array} \\ {\begin{array}{l} C_{x} = C cos γ = (8.0 cm) cos 30 ° = 6.93 cm \\ C_{y} = C sin γ = (8.0 cm) sin 30 ° = 4.00 cm \end{array} \end{array} .

For (a) we may substitute directly into [link] to find the scalar components of the resultant:

{\begin{array}{l} R_{x} = A_{x} + B_{x} + C_{x} = 8.19 cm - 2.39 cm + 6.93 cm = 12.73 cm \\ R_{y} = A_{y} + B_{y} + C_{y} = 5.73 cm - 6.58 cm + 4.00 cm = 3.15 cm \end{array} .

Therefore, the resultant vector is $\vec{R} = R_{x} \hat{i} + R_{y} \hat{j} = (12.7 \hat{i} + 3.1 \hat{j}) cm$ .

For (b), we may want to write the vector difference as

\vec{D} = \vec{A} - \vec{B} = (A_{x} \hat{i} + A_{y} \hat{j}) - (B_{x} \hat{i} + B_{y} \hat{j}) = (A_{x} - B_{x}) \hat{i} + (A_{y} - B_{y}) \hat{j} .

Then, the scalar components of the vector difference are

{\begin{array}{l} D_{x} = A_{x} - B_{x} = 8.19 cm - (−2.39 cm) = 10.58 cm \\ D_{y} = A_{y} - B_{y} = 5.73 cm - (−6.58 cm) = 12.31 cm \end{array} .

Hence, the difference vector is $\vec{D} = D_{x} \hat{i} + D_{y} \hat{j} = (10.6 \hat{i} + 12.3 \hat{j}) cm$ .

For (c), we can write vector $\vec{S}$ in the following explicit form:

\begin{matrix} \vec{S} & = \vec{A} - 3 \vec{B} + \vec{C} = (A_{x} \hat{i} + A_{y} \hat{j}) - 3 (B_{x} \hat{i} + B_{y} \hat{j}) + (C_{x} \hat{i} + C_{y} \hat{j}) \\ = (A_{x} - 3 B_{x} + C_{x}) \hat{i} + (A_{y} - 3 B_{y} + C_{y}) \hat{j} . \end{matrix}

Then, the scalar components of $\vec{S}$ are

{\begin{array}{l} S_{x} = A_{x} - 3 B_{x} + C_{x} = 8.19 cm - 3 (−2.39 cm) + 6.93 cm = 22.29 cm \\ S_{y} = A_{y} - 3 B_{y} + C_{y} = 5.73 cm - 3 (−6.58 cm) + 4.00 cm = 29.47 cm \end{array} .

The vector is $\vec{S} = S_{x} \hat{i} + S_{y} \hat{j} = (22.3 \hat{i} + 29.5 \hat{j}) cm$ .

Significance

Having found the vector components, we can illustrate the vectors by graphing or we can compute magnitudes and direction angles, as shown in [link] . Results for the magnitudes in (b) and (c) can be compared with results for the same problems obtained with the graphical method, shown in [link] and [link] . Notice that the analytical method produces exact results and its accuracy is not limited by the resolution of a ruler or a protractor, as it was with the graphical method used in [link] for finding this same resultant.

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Source: OpenStax, University physics volume 1. OpenStax CNX. Sep 19, 2016 Download for free at http://cnx.org/content/col12031/1.5

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