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Write force OA in component form.
$${F}_{x}=10\mathrm{sin}{30}^{0}=10X\frac{1}{2}=5\phantom{\rule{1em}{0ex}}N$$ $${F}_{y}=10\mathrm{cos}{30}^{0}=10X\frac{\sqrt{3}}{2}=5\sqrt{3}\phantom{\rule{1em}{0ex}}N$$ $$\mathbf{F}=-{F}_{x}\mathbf{i}-{F}_{y}\mathbf{j}=-5\mathbf{i}-5\sqrt{3}\mathbf{j}\phantom{\rule{1em}{0ex}}N$$
Write force OA in component form.
$${F}_{x}=10\mathrm{sin}{30}^{0}=10X\frac{1}{2}=5\phantom{\rule{1em}{0ex}}N$$ $${F}_{y}=10\mathrm{cos}{30}^{0}=10X\frac{\sqrt{3}}{2}=5\sqrt{3}\phantom{\rule{1em}{0ex}}N$$ $$\mathbf{F}={F}_{x}\mathbf{i}-{F}_{y}\mathbf{j}=5\mathbf{i}-5\sqrt{3}\mathbf{j}\phantom{\rule{1em}{0ex}}N$$
Graphical method is meticulous and tedious as it involves drawing of vectors on a scale and measurement of angles. More importantly, it does not allow algebraic operations that otherwise would give a simple solution. We can, however, extend algebraic techniques to vectors, provided vectors are represented on a rectangular coordinate system . The representation of a vector on coordinate system uses the concept of unit vector and component.
Now, the stage is set to design a frame work, which allows vector addition with algebraic methods. The frame work for vector addition draws on two important concepts. The first concept is that a vector can be equivalently expressed in terms of three component vectors :
$$\begin{array}{l}\mathbf{a}={a}_{x}\mathbf{i}+{a}_{y}\mathbf{j}+{a}_{z}\mathbf{k}\\ \mathbf{b}={b}_{x}\mathbf{i}+{b}_{y}\mathbf{j}+{b}_{z}\mathbf{k}\end{array}$$
The component vector form has important significance. It ensures that component vectors to be added are restricted to three known directions only. This paradigm eliminates the possibility of unknown direction. The second concept is that vectors along a direction can be treated algebraically. If two vectors are along the same line, then resultant is given as :
When θ = 0°, cosθ = cos 0° = 1 and,
$$\begin{array}{l}\Rightarrow R=\surd ({P}^{2}+2PQ+{Q}^{2})=\surd \left\{{(P+Q)}^{2}\right\}=P+Q\end{array}$$
When θ = 180°, cosθ = cos 180° = -1 and,
$$\begin{array}{l}\Rightarrow R=\surd ({P}^{2}-2PQ+{Q}^{2})=\surd \left\{{(P-Q)}^{2}\right\}=P-Q\end{array}$$
Thus, we see that the magnitude of the resultant is equal to algebraic sum of the magnitudes of the two vectors.
Using these two concepts, the addition of vectors is affected as outlined here :
1: Represent the vectors ( a and b ) to be added in terms of components :
$$\begin{array}{l}\mathbf{a}={a}_{x}\mathbf{i}+{a}_{y}\mathbf{j}+{a}_{z}\mathbf{k}\\ \mathbf{b}={b}_{x}\mathbf{i}+{b}_{y}\mathbf{j}+{b}_{z}\mathbf{k}\end{array}$$
2: Group components in a given direction :
$$\begin{array}{l}\mathbf{a}+\mathbf{b}={a}_{\mathrm{x}}\mathbf{i}+{a}_{\mathrm{y}}\mathbf{j}+{a}_{\mathrm{z}}\mathbf{k}+{b}_{\mathrm{x}}\mathbf{i}+{b}_{\mathrm{y}}\mathbf{j}+{b}_{\mathrm{z}}\mathbf{k}\\ \Rightarrow \mathbf{a}+\mathbf{b}=({a}_{\mathrm{x}}+{b}_{\mathrm{x}})\mathbf{i}+({a}_{\mathrm{y}}+{b}_{\mathrm{y}})\mathbf{j}+({a}_{\mathrm{z}}+{b}_{\mathrm{z}})\mathbf{k}\end{array}$$
3: Find the magnitude and direction of the sum, using analytical method
Find the angle that vector √3 i - j makes with y-axis.
From graphical representation, the angle that vector makes with y-axis has following trigonometric function :
$$\begin{array}{c}\mathrm{tan}\theta =\frac{{a}_{x}}{{a}_{y}}\end{array}$$
Now, we apply the formulae to find the angle, say θ, with y-axis,
$$\begin{array}{c}\mathrm{tan}\theta =-\frac{\surd 3}{1}=-\surd 3=\mathrm{tan}120\xb0\end{array}$$
$$\begin{array}{c}\Rightarrow \theta =120\xb0\end{array}$$
In case, we are only interested to know the magnitude of angle between vector and y-axis, then we can neglect the negative sign,
$$\begin{array}{c}\mathrm{tan}\mathrm{\theta \u2019}=\surd 3=\mathrm{tan}60\xb0\end{array}$$
$$\begin{array}{c}\Rightarrow \mathrm{\theta \u2019}=60\xb0\end{array}$$
If a vector makes angles α,β and γ with x,y and z axes of a rectangular coordinate system, then prove that :
$$\begin{array}{c}{\mathrm{cos}}^{2}\alpha +{\mathrm{cos}}^{2}\beta +{\mathrm{cos}}^{2}\gamma =1\end{array}$$
The vector can be expressed in terms of its component as :
$$\begin{array}{c}\mathbf{a}={a}_{x}\mathbf{i}+{a}_{y}\mathbf{j}+{a}_{z}\mathbf{k}\end{array}$$
where ${a}_{x}=a\mathrm{cos}\alpha ;\phantom{\rule{4pt}{0ex}}{a}_{y}=a\mathrm{cos}\beta ;\phantom{\rule{4pt}{0ex}}{a}_{z}=a\mathrm{cos}\gamma $ .
The magnitude of the vector is given by :
$$\begin{array}{l}a=\surd ({{a}_{x}}^{2}+{{a}_{y}}^{2}+{{a}_{z}}^{2})\end{array}$$
Putting expressions of components in the equation,
$$\begin{array}{l}\Rightarrow a=\surd ({a}^{2}{\mathrm{cos}}^{2}\alpha +{a}^{2}{\mathrm{cos}}^{2}\beta +{a}^{2}{\mathrm{cos}}^{2}\gamma )\end{array}$$
$$\begin{array}{l}\Rightarrow \surd ({\mathrm{cos}}^{2}\alpha +{\mathrm{cos}}^{2}\beta +{\mathrm{cos}}^{2}\gamma )=1\end{array}$$
Squaring both sides,
$$\begin{array}{l}\Rightarrow {\mathrm{cos}}^{2}\alpha +{\mathrm{cos}}^{2}\beta +{\mathrm{cos}}^{2}\gamma =1\end{array}$$
Find the components of weight of a block along the incline and perpendicular to the incline.
The component of the weight along the incline is :
$$\begin{array}{c}{W}_{x}=\mathrm{mg}\phantom{\rule{2pt}{0ex}}\mathrm{sin}\theta =100\phantom{\rule{2pt}{0ex}}x\phantom{\rule{2pt}{0ex}}\mathrm{sin}30\xb0=100\phantom{\rule{2pt}{0ex}}x\phantom{\rule{2pt}{0ex}}\frac{1}{2}=50\phantom{\rule{2pt}{0ex}}N\end{array}$$
and the component of weight perpendicular to incline is :
$$\begin{array}{c}{W}_{y}=\mathrm{mg}\phantom{\rule{2pt}{0ex}}\mathrm{cos}\theta =100\phantom{\rule{2pt}{0ex}}x\phantom{\rule{2pt}{0ex}}\mathrm{cos}30\xb0=100\phantom{\rule{2pt}{0ex}}x\phantom{\rule{2pt}{0ex}}\frac{\surd 3}{2}=50\surd 3\phantom{\rule{2pt}{0ex}}N\end{array}$$
The sum of magnitudes of two forces acting at a point is 16 N. If the resultant of the two forces is 8 N and it is normal to the smaller of the two forces, then find the forces.
We depict the situation as shown in the figure. The resultant force R is shown normal to small force ${F}_{1}$ . In order that the sum of the forces is equal to R, the component of larger force along the x-direction should be equal to smaller force :
$$\begin{array}{c}{F}_{2}\mathrm{sin}\theta ={F}_{1}\end{array}$$
Also, the component of the larger force along y-direction should be equal to the magnitude of resultant,
$$\begin{array}{c}{F}_{2}\mathrm{cos}\theta =R=8\end{array}$$
Squaring and adding two equations, we have :
$$\begin{array}{c}{F}_{2}^{2}={F}_{1}^{2}+64\end{array}$$
$$\begin{array}{c}\Rightarrow {F}_{2}^{2}-{F}_{1}^{2}=64\end{array}$$
However, according to the question,
$$\begin{array}{c}\Rightarrow {F}_{1}+{F}_{2}=16\end{array}$$
Substituting, we have :
$$\begin{array}{c}\Rightarrow {F}_{2}-{F}_{1}=\frac{64}{16}=4\end{array}$$
Solving,
$$\begin{array}{c}\Rightarrow {F}_{1}=6\phantom{\rule{2pt}{0ex}}N\\ \Rightarrow {F}_{2}=10\phantom{\rule{2pt}{0ex}}N\end{array}$$
More illustrations on the subject are available in the module titled Resolution of forces
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