November 2023 – Physicspupil

November 27, 2023November 27, 2023

202311270959 Exercise 1.4.33

Simplify:

$(AB)^{-1}(AC^{-1})(D^{-1}C^{-1})^{-1}D^{-1}$ .

_{Extracted from H. Anton & C. Rorres. (2010). Elementary Linear Algebra Application Version (10e)}

Answer. $B^{-1}$

Roughwork.

Let

$\begin{aligned} A_{m\times n}&:m\textrm{-by-}n \\ B_{n\times o}&:n\textrm{-by-}o \\ (C^{\mathrm{T}})_{n\times p}}&:n\textrm{-by-}p \\ (D^{\mathrm{T}})_{q\times n}&:q\textrm{-by-}n \\ C_{p\times n}&:p\textrm{-by-}n \\ D_{n\times q}&:n\textrm{-by-}q \\ (AB)_{m\times o}&:m\textrm{-by-}o \\ ((AB)^{\mathrm{T}})_{o\times m}&:o\textrm{-by-}m \\ ((AC)^{\mathrm{T}})_{m\times p}&:m\textrm{-by-}p \\ (D^{\mathrm{T}}C^{\mathrm{T}})_{q\times p}&:q\textrm{-by-}p \\ ((D^{\mathrm{T}}C^{\mathrm{T}})^{\mathrm{T}})_{p\times q}&:p\textrm{-by-}q \\ \end{aligned}$

such that after simplification the original array of matrices (here transpose in place of inverse) should give an $o$ -by- $n$ matrix. For merely satisfying this requirement one candidate can be $B_{n\times o}$ ‘s transpose, i.e., $(B^{\mathrm{T}})_{o\times n}$ , but not much evidence than coincidence here.

Observe, that the simplified expression, in general, should $\textrm{\scriptsize{NOT}}$ necessarily be a span (/linear combination) of matrices in the list above, for none any pair of matrices have necessarily the same dimension for which matrix addition is possible, apart from square matrices.

Hence, assume $AB$ , $C$ , and $D$ to be three square matrices which are invertible.

Note, that the simplified expression should be an arrangement of matrix multiplication amongst some of $8$ matrices $A$ and $A^{-1}$ , $B$ and $B^{-1}$ , $C$ and $C^{-1}$ , and $D$ and $D^{-1}$ .

Consider, by simplification the matrix product should have strictly less than $r=7$ factors, i.e., at most $6$ ; and in adding an extra identity matrix $I$ to the preceding (*such as to avoid terms $AA^{\mathrm{-1}}=A^{\mathrm{-1}}A=I$ to go nearby) we have $n=9$ matrices at hand; the possible outcomes is a permutation

$\text{}^9P_{6}=9^6=531441$

with replacement, minus

$8\times 6=48$ .

The chance of getting by is

$\displaystyle{\frac{1}{531441-48}=\frac{1}{531393} \approx 0.000188\%$ .

This problem is not to be attempted.

November 9, 2023November 12, 2023

202311091202 Exercise 23.2.8

A block of weight $30\,\mathrm{N}$ rests on a smooth plane inclined at an angle of $20^\circ$ to the horizontal. Find the least horizontal force required to keep it in equilibrium and the reaction of the plane.

_{Extracted from A. Godman & J. F. Talbert. (1973). Additional Mathematics Pure and Applied in SI Units.}

Roughwork.

As always, begin with free-body diagrams. Here we combine the two in one force diagram below.

The equation of motion for block $m$ is

$\begin{aligned} m\mathbf{a}_{\parallel} & =\mathbf{F}_{\textrm{net},m} \\ & = \mathrm{}_M\mathbf{N}_m + \mathrm{}_E\mathbf{W}_m + \mathbf{f}_m \\ \end{aligned}$

along the plane; and for wedge $M$

$\begin{aligned} M\mathbf{a}_{\perp} & =\mathbf{F}_{\textrm{net},M} \\ & = \mathrm{}_E\mathbf{N}_M + \mathrm{}_m\mathbf{N}_M + \mathrm{}_E\mathbf{W}_M + \mathbf{f}_M \\ \end{aligned}$

along the earth $E$ . Intuitively, when block $m$ slides leftward down the slope, wedge $M$ will slide rightward on the earth.

This problem is not to be attempted.

November 8, 2023December 10, 2023

202311081445 Pastime Exercise 006

The figure below is a representation of the $\mathbf{E}$ -field by electric field lines in a family $R$ of infinitely many quadratic functions

$y_t(x_{t'})=a_tx_{t'}^2+b_tx_{t'}+c_t$

Roughwork.

A tangent to any quadratic curve at some point in the locus gives the $\mathrm{\pm ve}$ direction of the electric force experienced by a (positive) test charge placed there. I.e.,

$\displaystyle{T(x_{t'},y_{t})=\frac{\mathrm{d}}{\mathrm{d}x}\big( y_t(x_{t'})\big) = 2a_tx_{t'}+b_{t}}$ ,

the slope of tangent.

WLOG we work with the first quadrant. First, begin with the repulsive force $\mathrm{}_Q\mathbf{F}_{q}$ acting on the test charge $+q$ due to point charge $+Q$ .

$\begin{aligned} \mathrm{}_QF_{q,x}^2+\mathrm{}_QF_{q,y}^2 & = \mathrm{}_QF_{q}^2 \\ \mathrm{}_QF_{q,y} & = \bigg(\frac{y_t}{x_{t'}+d/2}\bigg)\mathrm{}_QF_{q,x} \\ \cdots\cdots & \cdots\cdots\\ \mathrm{}_QF_{q,x} & = \frac{\mathrm{}_QF_{q}}{\sqrt{1+\Big(\frac{y_t}{x_{t'}+d/2}\Big)^2}} \\ \mathrm{}_QF_{q,y} & = \bigg(\frac{y_t}{x_{t'}+d/2}\bigg) \Bigg(\frac{\mathrm{}_QF_{q}}{\sqrt{1+\Big(\frac{y_t}{x_{t'}+d/2}\Big)^2}}\Bigg) \\ \cdots\cdots & \cdots\cdots\\ \mathrm{}_Q\mathbf{F}_{q} & = \mathrm{}_QF_{q,x}\,\hat{\mathbf{i}}+ \mathrm{}_QF_{q,y}\,\hat{\mathbf{j}} \\ \end{aligned}$

Next, continue with the attractive force $\mathrm{}_{-Q}\mathbf{F}_{q}$ acting on the test charge $+q$ due to point charge $-Q$ .

$\begin{aligned} \mathrm{}_{-Q}F_{q,x}^2+\mathrm{}_{-Q}F_{q,y}^2 & = \mathrm{}_{-Q}F_{q}^2 \\ \mathrm{}_{-Q}F_{q,y} & = \bigg(\frac{y_t}{d/2-x_{t'}}\bigg)\mathrm{}_{-Q}F_{q,x} \\ \cdots\cdots & \cdots\cdots\\ \mathrm{}_{-Q}F_{q,x} & = \frac{\mathrm{}_{-Q}F_{q}}{\sqrt{1+\Big(\frac{y_t}{d/2-x_{t'}}\Big)^2}} \\ \mathrm{}_{-Q}F_{q,y} & = \bigg(\frac{y_t}{d/2-x_{t'}}\bigg) \Bigg(\frac{\mathrm{}_{-Q}F_{q}}{\sqrt{1+\Big(\frac{y_t}{d/2-x_{t'}}\Big)^2}}\Bigg) \\ \cdots\cdots & \cdots\cdots\\ \mathrm{}_{-Q}\mathbf{F}_{q} & = \mathrm{}_{-Q}F_{q,x}\,\hat{\mathbf{i}}+ \mathrm{}_{-Q}F_{q,y}\,\hat{\mathbf{j}} \\ \end{aligned}$

Adding $\mathrm{}_{Q}\mathbf{F}_{q}$ and $\mathrm{}_{-Q}\mathbf{F}_{q}$ will give the resultant electric force $\mathbf{F}_{E}(x_{t'},y_t)$ .

Note that

$\begin{aligned} \mathrm{}_QF_q & = k\frac{Q}{\mathrm{}_Qr_{q}^2} \\ \mathrm{}_{-Q}F_q & = k\frac{-Q}{\mathrm{}_{-Q}r_{q}^2} \\ \end{aligned}$

where

$\begin{aligned} \mathrm{}_Qr_q & = \sqrt{(d/2+x_{t'})^2+y_t^2} \\ \mathrm{}_{-Q}r_q & = \sqrt{(d/2-x_{t'})^2+y_t^2} \\ & \\ & \\ \end{aligned}$

are the distances of a test charge at $q(x_{t'},y_t)$ from point charges $+Q$ at $A(-d/2,0)$ and $-Q$ at $B(d/2,0)$ .

The smallest angle between $\mathbf{F}_{E}(x_{t'},y_{t})$ and the level should be equal to the slope of tangent $2a_tx_{t'}+b_{t}$ at point $q(x_{t'},y_t)$ .

As of the vertex of each parabola, the $x$ -coordinate is

$\displaystyle{0=x_{t'}=-\frac{b_{t}}{2a_{t}}}$

s.t. $b_{t}=0$ , and the $y$ -coordinate $c_{t}=y_t(0)$ .

I guess, under correction, the separation distance $d$ is none any parameter of the loci.

Visualisation is $\textrm{\scriptsize{NOT}}$ mathematical $\textrm{\scriptsize{BUT}}$ conceptual.

November 7, 2023November 30, 2023

202311070929 Exercise 21.1.2

Find (by drawing or calculation) the resultant of the following forces, in magnitude and direction.

(a)

(b)

(c)

_{Extracted from A. Godman & J. F. Talbert. (1973). Additonal Mathematics Pure and Applied in SI units.}

(a)

So that the two forces are labelled as vectors $\mathbf{OA}$ and $\mathbf{OB}$ , let their initial points for both be $O$ and the terminal points of each $A$ and $B$ .

In Cartesian coordinates,

$\begin{aligned} \mathbf{OA} & = |\mathbf{OA}|\,\hat{\mathbf{i}} \\ & = OA\,\hat{\mathbf{i}} \\ & = 12\,\mathrm{N}\,\hat{\mathbf{i}} \\ \mathbf{OB} & = |\mathbf{OB}|\,\hat{\mathbf{j}} \\ & = OB\,\hat{\mathbf{j}} \\ & = 10\,\mathrm{N}\,\hat{\mathbf{j}} \\ \end{aligned}$

With, on purpose, no diagrams being provided, let also the resultant force $\mathbf{OC}$ be directed from the same initial point $O$ to some terminal point we let be $C$ . Then,

$\begin{aligned} \mathbf{OC} & = \mathbf{OA} + \mathbf{OB} \\ & = 12\,\mathrm{N}\,\hat{\mathbf{i}} + 10\,\mathrm{N}\,\hat{\mathbf{j}} \\ OC & = |\mathbf{OC}| \\ & = \sqrt{12^2+10^2} \\ & = 15.6\,\mathrm{N}\qquad\textrm{(3 s.f.)} \\ \end{aligned}$

Trying in polar coordinates,

$\begin{aligned} \mathbf{OA} & = |\mathbf{OA}|\,\hat{\mathbf{r}} \\ & = 12\,\mathrm{N}\,\hat{\mathbf{r}} \\ \mathbf{OB} & = |\mathbf{OB}|\,\hat{\mathbf{r}} +\frac{\pi}{2}\,\hat{\boldsymbol{\theta}}\\ & = \bigg( 10\,\hat{\mathbf{r}} + \frac{\pi}{2}\,\hat{\boldsymbol{\theta}}\bigg) \,\mathrm{N} \\ \end{aligned}$

Recall the formulae for addition of vectors

$\begin{aligned} r_3 & = \sqrt{r_1^2+2r_1r_2\cos(\theta_2-\theta_1)+r_2^2} \\ \theta_3 & = \theta_1+\arctan \bigg(\frac{r_2\sin (\theta_2-\theta_1)}{r_1+r_2\cos (\theta_2-\theta_1)}\bigg) \\ \end{aligned}$

beside the rectangular form I am stupid enough to do the polar. Quit.

(b)

By tail-to-tip method, draw

$\begin{aligned} OC & = \sqrt{OB^2+BC^2} \\ & = \sqrt{OB^2+OA^2} \\ & = \sqrt{40^2+30^2} \\ & = 50\,\mathrm{N} \\ \tan\theta & = \frac{30}{40} \\ \theta & = \tan^{-1} \bigg(\frac{30}{40}\bigg) \\ & = 36.9^\circ\qquad\textrm{(3 s.f.)} \\ \end{aligned}$

Or, by parallelogram method, draw

which makes no matter how you visualise, only are the calculations the matters.

(c)

Draw a labelled diagram below:

$\begin{aligned} \mathbf{OA} & = 4\cos 30^\circ \,\hat{\mathbf{i}} + 4\sin 30^\circ \,\hat{\mathbf{j}} \\ & = 4\bigg(\frac{\sqrt{3}}{2}\bigg)\,\hat{\mathbf{i}} + 4\bigg(\frac{1}{2}\bigg) \,\hat{\mathbf{j}} \\ & = \big(2\sqrt{3}\,\hat{\mathbf{i}} +2\,\hat{\mathbf{j}}\big)\,\mathrm{N} \\ \mathbf{OB} & = \big( 6\,\hat{\mathbf{i}}\big)\,\mathrm{N} \\ \end{aligned}$

Let $\mathbf{OC}$ be the resultant vector,

$\begin{aligned} \mathbf{OC} & = \mathbf{OA} + \mathbf{OB} \\ & = \big( 2\sqrt{3}+6\big) \,\hat{\mathbf{i}} + 2\,\hat{\mathbf{j}} \\ \end{aligned}$

Write $OA$ , $OB$ , and $OC$ the magnitudes of respective vectors $\mathbf{OA}$ , $\mathbf{OB}$ , and $\mathbf{OC}$ ,

$\begin{aligned} OA & = 4 \\ OB & = 6 \\ OC & = |\mathbf{OC}| \\ & = \sqrt{(2\sqrt{3}+6)^2+2^2} \\ & = \sqrt{52+24\sqrt{3}} \\ & = 2\sqrt{13+6\sqrt{3}} \\ \end{aligned}$

With a diagram in mind,

apply cosine law,

$\begin{aligned} \cos \angle COB & = \frac{OC^2+OB^2-OA^2}{2(OC)(OB)} \\ & = \frac{(52+24\sqrt{3})+(36)-(16)}{2(52+24\sqrt{3})(6)} \\ \angle COB & = \cdots \\ \end{aligned}$

This problem is not to be attempted.

November 6, 2023December 19, 2023

202311061327 Exercise 36.11

Two plane mirrors are parallel to each other and spaced $20\,\mathrm{cm}$ apart. A luminous point is placed between them and $5\,\mathrm{cm}$ from one mirror. Determine the distance from each mirror of the three nearest images in each.

_{Extracted from Daniel Schaum. (1961). Schaum’s Ouline of Theory and Problems of College Physics. (6e)}

Roughwork.

Ans. $5,\, 35,\, 45\,\mathrm{cm}$ ; $15,\, 25,\, 55\,\mathrm{cm}$

This problem is not to be attempted.