Skip to main content

General Physics I:

Sunil Karna

Section 2.2 Examples

Example 2.2.1.

Write the vector equations for the given diagrams.
Solution.
  1. Do your self:
  2. Do your self:
  3. From figure,
    we can write resultant of \(\vec{a}\) and \(\vec{b}\) as \(\vec{a}+\vec{b}\) given by dotted lines. Hence, \((\vec{a}+\vec{b})+(\vec{c}+\vec{d})=\vec{e}\)
  4. From figure,
    we can write
    \begin{equation*} \vec{r}_{1}=\vec{a}+\vec{b}, \end{equation*}
    \begin{equation*} \vec{r}_{2}=\vec{r}_{1}+\vec{c}=(\vec{a}+\vec{b})+\vec{c}, \end{equation*}
    and
    \begin{equation*} \vec{e}=\vec{r}_{2}+\vec{d} \end{equation*}
    \begin{equation*} \therefore \vec{e}=\vec{a}+\vec{b}+\vec{c}+\vec{d} \end{equation*}

Example 2.2.2.

Find the magnitude and direction of
  1. \(\displaystyle \vec{a}+\vec{b}\)
  2. \(\displaystyle \vec{a}-\vec{b}\)
if \(\vec{a}=3\hat{i}+4\hat{j} +2\hat{k}\) and \(\vec{b}=4\hat{i}-\hat{j} -2\hat{k}.\)
Solution.
  1. \(\vec{a}+\vec{b} =7\hat{i}+3\hat{j}\)
    \begin{equation*} | \vec{a}+\vec{b} |= \sqrt{(\vec{a}+\vec{b})\cdot (\vec{a}+\vec{b})} = \sqrt{49+9}=\sqrt{58} \end{equation*}
  2. \(\vec{a}-\vec{b} =-\hat{i}+5\hat{j}+4\hat{k}\)
    \begin{equation*} | \vec{a}-\vec{b} |= \sqrt{(\vec{a}-\vec{b})\cdot (\vec{a}-\vec{b})} = \sqrt{1+25+16}=\sqrt{42} \end{equation*}

Example 2.2.3.

The magnitude of two vectors \(\vec{a}\) and \(\vec{b}\) are 20 and 40, respectively. Vector \(\vec{a}\) is pointed along \(20^{o}\) north of east and vector \(\vec{b}\) is pointed along \(230^{o}\) south of west. Find the magnitude and direction of
  1. \(\displaystyle \vec{a}+\vec{b}\)
  2. \(\displaystyle \vec{a}-\vec{b}\)
Solution.
This problem can be solved by two different ways, one way is using head and tail rules as shown in figure,
here
\begin{equation*} \vec{c}=\vec{a}+\vec{b} \end{equation*}
its magnitude can be found by using relation
 1 
see appendix
\begin{equation*} c =\sqrt{a^{2}+b^{2}+2ab\cos\theta} \end{equation*}
\begin{equation*} =\sqrt{20^{2}+40^{2}+2\times 20\times 40\times \cos 150^{o}} = 24.78 \end{equation*}
its direction can be found by using relation
\begin{equation*} \phi=\tan^{-1}\left(\frac{b\sin\theta}{a+b\cos\theta}\right) = -53^{o}-20^{o}=-73^{o}\quad \text{(S of W)} \end{equation*}
We add \(20^{o}\) in \(\phi\) to find the direction of resultant vector from x-axis.
Another way is to resolved these two vectors into component vectors and then find resultant of the corresponding components along x and y axes. From figure,
we have
\begin{equation*} a_{x}=a\cos\theta_{1} = 20\cos 20^{o}=18.79;\quad a_{y}=a\sin\theta_{1}=20\sin 20^{o} = 6.84 \end{equation*}
\begin{equation*} b_{x}=b\cos\theta_{2}= 40\cos 50^{o}=25.71;\quad b_{y}=b\sin\theta_{2}=40\sin 50^{o}=30.64 \end{equation*}
\begin{equation*} \therefore\quad R_{x}=a_{x}-b_{x} = -6.92 \quad \text{and}\quad R_{y}=a_{y}-b_{y} =-23.8 \end{equation*}
Hence \(R = 24.78\text{,}\) and \(\phi = \tan^{-1}\left(\frac{R_{y}}{R_{x}}\right)= 73.78^{o}\) (S of W).

Example 2.2.4.

Find the magnitude of a vector \(\vec{c}\) in the vector equation \(\vec{a}+2\vec{b}-3\vec{c}=-4\hat{j}\text{,}\) where \(\vec{a} =\hat{i}-2\hat{k}\) and \(\vec{b} =-\hat{j}+\hat{k}\text{.}\)
Solution.
\begin{equation*} \vec{a}+2\vec{b}-3\vec{c}=-4\hat{j} \end{equation*}
\begin{equation*} \text{or,}\quad (\hat{i}-2\hat{k}) +2(-\hat{j}+\hat{k})-3\vec{c}=-4\hat{j} \end{equation*}
\begin{equation*} \text{or,}\quad -3\vec{c}=-4\hat{j}-\hat{i}+2\hat{k} +2\hat{j}-2\hat{k}=-\hat{i} -2\hat{j} \end{equation*}
\begin{equation*} \text{or,}\quad \vec{c} = \frac{1}{3}(\hat{i}+2\hat{j}) \end{equation*}
\begin{equation*} \therefore \quad |\vec{c}| = c = \sqrt{\frac{1}{9}(1+4)} = \sqrt{\frac{5}{9}} =\frac{1}{3}\sqrt{5} \end{equation*}

Example 2.2.5.

Show that the distance between points in a plane do not change by rotating a coordinate system.
Solution.
Let p and q be two points in space. Now from figure,
we have -
\begin{equation*} x'=OL'=r\cos\phi \end{equation*}
\begin{equation*} \text{and} \quad y'= PL'=r\sin\phi\text{.} \end{equation*}
\begin{equation*} \text{Also} \quad x =OL=r\cos (\theta+\phi) \end{equation*}
\begin{equation*} \text{and} \quad y= PL=r\sin (\theta+\phi) \end{equation*}
\begin{equation*} \text{Now,} \quad x=r(\cos\theta\cos\phi-\sin\theta\sin\phi)=r\cos\phi\cos\theta-r\sin\phi\sin\theta \end{equation*}
\begin{equation*} \therefore\quad x = x'\cos\theta -y'\sin\theta \end{equation*}
\begin{equation*} \text{Again,} \quad y=r(\sin\theta\cos\phi-\cos\theta\sin\phi)=r\cos\phi\sin\theta+r\sin\phi\cos\theta \end{equation*}
\begin{equation*} \therefore\quad y = x'\sin\theta +y'\cos\theta \end{equation*}
On solving these equations we get -
\begin{equation*} x' = x\cos\theta + y\sin\theta \end{equation*}
\begin{equation*} \text{and}\quad y' = -x\sin\theta + y\cos\theta \end{equation*}
The distance between points \(p(x_{1},y_{1})\) and \(q(x_{2},y_{2})\) in xy coordinate system is given by
\begin{equation*} d=pq=\sqrt{(x_{2}-x_{1})^{2}+(y_{2}-y_{1})^{2}} \end{equation*}
\begin{equation*} =\sqrt{[(x'_{2}-x'_{1})\sin\theta + (y'_{2}-y'_{1})\cos\theta}]^{2} =\sqrt{(x'_{2}-x'_{1})^{2}+ (y'_{2}-y'_{1})^{2}} \end{equation*}
which is the distance between points \(p(x'_{1},y'_{1})\) and \(q(x'_{2},y'_{2})\) in x’ y’ coordinate system, i.e., the distance between two points is invariant upon the rotation of a coordinate system.

Example 2.2.6.

Given the vectors in figure,
Do the following:
  1. From the scale on the diagram, estimate the magnitude of each vector, its x-component, and its y-component.
  2. From the numbers on the diagram, calculate each vector and its magnitude.
  3. Construct the vector sum of these vectors.
  4. From the above components calculate the resultant of the four vectors.
  5. Calculate the magnitude of the resultant vector.
  6. What is the angle the resultant makes with the x-axis?
Solution.
  1. \begin{equation*} A = 10\,m; \quad A_{x} =A\sin37^{o}= 6.018 \approx 6\,m; \end{equation*}
    \begin{equation*} A_{y}=A\cos37^{o}=7.986 \approx 8\,m \end{equation*}
    \begin{equation*} B = 10m;\quad B_{x}= B\cos45^{o}=7.07 \approx -7\,m; \end{equation*}
    \begin{equation*} B_{y} = B\sin45^{o}= 7.07 \approx 7\,m \end{equation*}
    \begin{equation*} C = 10m;\qquad C_{x} = -8m; \qquad C_{y} = -6\,m \end{equation*}
    \begin{equation*} D = 10m;\qquad D_{x} = 9m; \qquad D_{y} = -5\,m \end{equation*}
    \begin{equation*} \left[\theta_{c}\approx 36.87^{o}; \quad \theta_{d}\approx 30^{o} \quad\text{from x-axis}\right] \end{equation*}
  2. \begin{equation*} \vec{A} = 6\,m\hat{i} +8\,m\hat{j}; \quad A = 10\,m \end{equation*}
    \begin{equation*} \vec{B} = -7\,m\hat{i} +7\,m\hat{j}; \quad B =10\,m \end{equation*}
    \begin{equation*} \vec{C} = -8\,m\hat{i} -6\,m\hat{j};\quad C = 10 \end{equation*}
    \begin{equation*} \vec{D} = 8.66\,m\hat{i} -5\,m\hat{j}; \quad D =10\,m \end{equation*}
    Note: Distance between points, \(d=\sqrt{x^{2}+y^{2}}\text{.}\)
  3. \begin{equation*} \vec{R} = \vec{A}+\vec{B}+\vec{C}+\vec{D} \end{equation*}
    \begin{equation*} =(6-7-8+8.66)\hat{i}+(8+7-6-5)\hat{j} =-0.34\hat{i}+4\hat{j} \end{equation*}
    The resultant vector is plotted in figure,
  4. \begin{equation*} R = -0.34m\hat{i} +4m\hat{j} \end{equation*}
  5. \begin{equation*} R = \sqrt{-0.34^{2}+4^{2}}=4.01\,m \end{equation*}
  6. \begin{equation*} \theta = \tan^{-1}\left(\frac{4}{-0.34}\right) =(180-85)=95^{o} \end{equation*}

Example 2.2.7.

Two points \(A\) and \(B\) in the xy plane have the respective Cartesian and polar coordinates \((3 \,m, - 4\, m)\) and \((15 \,m, 30^{o})\text{.}\)
  1. What are the polar coordinates of the first point?
  2. What are the rectangular coordinates of the second point?
  3. What is the distance between the two points?
Solution.
  1. \((5m,-53.13^{o})\)
    \begin{equation*} A=\sqrt{x^{2}+y^{2}} ; \quad \theta = tan^{-1}\left(\frac{-4}{3}\right) \end{equation*}
  2. \((13\,m,7.5\,m)\)
    \begin{equation*} x=B\cos\theta;\quad y=B\sin\theta \end{equation*}
  3. \begin{equation*} AB=\sqrt{(x_{2}-x_{1})^{2}+(y_{2}-y_{1})^{2}}=15.24\,m. \end{equation*}

Example 2.2.8.

Prove that \(|\vec{A}+\vec{B} | = |\vec{A}-\vec{B}| \text{,}\) only when \(\vec{A}\) and \(\vec{B} \) are perpendicular to each other.
Solution.
Consider
\begin{equation*} \vec{r} = r_{x}\hat{i} + r_{y}\hat{j} + r_{z}\hat{k} \end{equation*}
then
\begin{equation*} |\vec{r}| = \sqrt{\vec{r}\cdot\vec{r}}= \sqrt{r_{x}^{2} + r_{y}^{2}+r_{z}^{2}}. \end{equation*}
Now,
\begin{equation*} \vec{A}+\vec{B} = (A_{x}\hat{i}+ A_{y}\hat{j} +A_{z}\hat{k} ) + (B_{x}\hat{i}+ B_{y}\hat{j} +B_{z}\hat{k}) \end{equation*}
\begin{equation*} = (A_{x}+B_{x})\hat{i} + (A_{y}+B_{y})\hat{j} + (A_{z}+B_{z})\hat{k} \end{equation*}
and
\begin{equation*} \vec{A}\cdot\vec{B} = |\vec{A}||\vec{B}|\cos\theta = A_{x}B_{x} + A_{y}B_{y} + A_{z}B_{z} \end{equation*}
if \(\theta = 0\text{.}\) but,
\begin{equation*} \vec{A}\cdot\vec{B} = |\vec{A}||\vec{B}|\cos\theta = 0 \end{equation*}
if \(\theta = 90^{o}\text{.}\) Therefore,
\begin{equation*} |\vec{A}+\vec{B}| =\sqrt{(\vec{A}+\vec{B})\cdot (\vec{A}+\vec{B})} \end{equation*}
\begin{equation*} = \sqrt{(A_{x}^2 + B_{x}^2+ 2 A_{x}B_{x}\cos\theta) + (A_{y}^2 + B_{y}^2+ 2 A_{y}B_{y}\cos\theta) + (A_{z}^2 + B_{z}^2+ 2 A_{z}B_{z}\cos\theta)} \end{equation*}
\begin{equation} \therefore \quad |\vec{A}+\vec{B}| = \sqrt{(A_{x}^2 + A_{y}^2 + A_{z}^2) + (B_{x}^2 + B_{y}^2 + B_{z}^2)} \tag{2.2.1} \end{equation}
Similarly,
\begin{equation*} |\vec{A}-\vec{B}| =\sqrt{(\vec{A}-\vec{B})\cdot (\vec{A}-\vec{B})} \end{equation*}
\begin{equation*} = \sqrt{(A_{x}^2 + B_{x}^2- 2 A_{x}B_{x}\cos\theta) + (A_{y}^2 + B_{y}^2- 2 A_{y}B_{y}\cos\theta) + (A_{z}^2 + B_{z}^2-2 A_{z}B_{z}\cos\theta)} \end{equation*}
\begin{equation} \therefore \quad |\vec{A}-\vec{B}| = \sqrt{(A_{x}^2 + A_{y}^2 + A_{z}^2) + (B_{x}^2 + B_{y}^2 + B_{z}^2) } \tag{2.2.2} \end{equation}
Hence from (2.2.1)–(2.2.2), it is clear that
\begin{equation*} |\vec{A}+\vec{B}| = |\vec{A}-\vec{B}| \end{equation*}
If and only if \(\vec{A}\) and \(\vec{B}\) are perpendicular.

Example 2.2.9.

Find the condition at which \(|\vec{A}-\vec{B}| = |\vec{A}| - |\vec{B}| \text{.}\)
Solution.
\begin{equation*} |\vec{A}-\vec{B}| =\sqrt{(\vec{A}-\vec{B})\cdot (\vec{A}-\vec{B})} \end{equation*}
\begin{equation*} = \sqrt{(A_{x}^2 + B_{x}^2- 2 A_{x}B_{x}\cos\theta) + (A_{y}^2 + B_{y}^2- 2 A_{y}B_{y}\cos\theta) + (A_{z}^2 + B_{z}^2-2 A_{z}B_{z}\cos\theta)} \end{equation*}
\begin{equation*} = \sqrt{(A_{x}^2 + A_{y}^2 + A_{z}^2 )+ (B_{x}^2 + B_{y}^2 + B_{z}^2)- 2\cos\theta(A_{x}B_{x}+A_{y}B_{y}+A_{z}B_{z})} \end{equation*}
If \(\theta = 0^{o}\text{,}\) then \(\cos\theta = 1 \text{.}\) Therefore,
\begin{equation*} |\vec{A}-\vec{B}| = \sqrt{(A_{x}^2 + A_{y}^2 + A_{z}^2 )+ (B_{x}^2 + B_{y}^2 + B_{z}^2)- 2(A_{x}B_{x}+A_{y}B_{y}+A_{z}B_{z})} \end{equation*}
or,
\begin{equation*} |\vec{A}-\vec{B}| = \sqrt{|\vec{A}^{2}|+ |\vec{B}^{2}|- 2|\vec{A}| |\vec{B}|} = \sqrt{(|\vec{A}| - |\vec{B}|)^{2}}= |\vec{A}| - |\vec{B}| \end{equation*}
PrevTopNext