5th semester files

5th-Semester-Fall-2023/ME-For-ECEs/EquationSheet/ME-Equation-Sheet.aux

@@ -0,0 +1,13 @@
+\relax 
+\@writefile{toc}{\contentsline {section}{\numberline {I}Vectors}{1}{}\protected@file@percent }
+\@writefile{toc}{\contentsline {section}{\numberline {II}Statics}{1}{}\protected@file@percent }
+\@writefile{toc}{\contentsline {section}{\numberline {III}Dynamics}{1}{}\protected@file@percent }
+\@writefile{toc}{\contentsline {subsection}{\numberline {\mbox  {III-A}}Kinematic Equations}{1}{}\protected@file@percent }
+\@writefile{toc}{\contentsline {subsection}{\numberline {\mbox  {III-B}}Projectile Motion}{1}{}\protected@file@percent }
+\@writefile{toc}{\contentsline {section}{\numberline {IV}Heat Transfer}{1}{}\protected@file@percent }
+\@writefile{toc}{\contentsline {subsection}{\numberline {\mbox  {IV-A}}Conduction}{1}{}\protected@file@percent }
+\@writefile{toc}{\contentsline {subsection}{\numberline {\mbox  {IV-B}}Convection}{1}{}\protected@file@percent }
+\@writefile{toc}{\contentsline {subsection}{\numberline {\mbox  {IV-C}}Radiation}{1}{}\protected@file@percent }
+\@writefile{toc}{\contentsline {section}{\numberline {V}Fluids}{1}{}\protected@file@percent }
+\@writefile{toc}{\contentsline {section}{\numberline {VI}Gears}{2}{}\protected@file@percent }
+\gdef \@abspage@last{2}

5th-Semester-Fall-2023/ME-For-ECEs/EquationSheet/ME-Equation-Sheet.log

@@ -0,0 +1,285 @@
+This is pdfTeX, Version 3.141592653-2.6-1.40.24 (TeX Live 2022/CVE-2023-32700 patched) (preloaded format=pdflatex 2023.11.29)  6 DEC 2023 11:52
+entering extended mode
+ \write18 enabled.
+ %&-line parsing enabled.
+**/tmp/nvim.sharpe/dmzUoQ/0
+(/tmp/nvim.sharpe/dmzUoQ/0
+LaTeX2e <2022-06-01> patch level 5
+L3 programming layer <2022-12-17>
+(/usr/share/texlive/texmf-dist/tex/latex/ieeetran/IEEEtran.cls
+Document Class: IEEEtran 2015/08/26 V1.8b by Michael Shell
+-- See the "IEEEtran_HOWTO" manual for usage information.
+-- http://www.michaelshell.org/tex/ieeetran/
+\@IEEEtrantmpdimenA=\dimen140
+\@IEEEtrantmpdimenB=\dimen141
+\@IEEEtrantmpdimenC=\dimen142
+\@IEEEtrantmpcountA=\count185
+\@IEEEtrantmpcountB=\count186
+\@IEEEtrantmpcountC=\count187
+\@IEEEtrantmptoksA=\toks16
+LaTeX Font Info:    Trying to load font information for OT1+ptm on input line 5
+03.
+(/usr/share/texlive/texmf-dist/tex/latex/psnfss/ot1ptm.fd
+File: ot1ptm.fd 2001/06/04 font definitions for OT1/ptm.
+)
+-- Using 8.5in x 11in (letter) paper.
+-- Using PDF output.
+\@IEEEnormalsizeunitybaselineskip=\dimen143
+-- This is a 10 point document.
+\CLASSINFOnormalsizebaselineskip=\dimen144
+\CLASSINFOnormalsizeunitybaselineskip=\dimen145
+\IEEEnormaljot=\dimen146
+LaTeX Font Info:    Font shape `OT1/ptm/bx/n' in size <5> not available
+(Font)              Font shape `OT1/ptm/b/n' tried instead on input line 1090.
+LaTeX Font Info:    Font shape `OT1/ptm/bx/it' in size <5> not available
+(Font)              Font shape `OT1/ptm/b/it' tried instead on input line 1090.
+
+LaTeX Font Info:    Font shape `OT1/ptm/bx/n' in size <7> not available
+(Font)              Font shape `OT1/ptm/b/n' tried instead on input line 1090.
+LaTeX Font Info:    Font shape `OT1/ptm/bx/it' in size <7> not available
+(Font)              Font shape `OT1/ptm/b/it' tried instead on input line 1090.
+
+LaTeX Font Info:    Font shape `OT1/ptm/bx/n' in size <8> not available
+(Font)              Font shape `OT1/ptm/b/n' tried instead on input line 1090.
+LaTeX Font Info:    Font shape `OT1/ptm/bx/it' in size <8> not available
+(Font)              Font shape `OT1/ptm/b/it' tried instead on input line 1090.
+
+LaTeX Font Info:    Font shape `OT1/ptm/bx/n' in size <9> not available
+(Font)              Font shape `OT1/ptm/b/n' tried instead on input line 1090.
+LaTeX Font Info:    Font shape `OT1/ptm/bx/it' in size <9> not available
+(Font)              Font shape `OT1/ptm/b/it' tried instead on input line 1090.
+
+LaTeX Font Info:    Font shape `OT1/ptm/bx/n' in size <10> not available
+(Font)              Font shape `OT1/ptm/b/n' tried instead on input line 1090.
+LaTeX Font Info:    Font shape `OT1/ptm/bx/it' in size <10> not available
+(Font)              Font shape `OT1/ptm/b/it' tried instead on input line 1090.
+
+LaTeX Font Info:    Font shape `OT1/ptm/bx/n' in size <11> not available
+(Font)              Font shape `OT1/ptm/b/n' tried instead on input line 1090.
+LaTeX Font Info:    Font shape `OT1/ptm/bx/it' in size <11> not available
+(Font)              Font shape `OT1/ptm/b/it' tried instead on input line 1090.
+
+LaTeX Font Info:    Font shape `OT1/ptm/bx/n' in size <12> not available
+(Font)              Font shape `OT1/ptm/b/n' tried instead on input line 1090.
+LaTeX Font Info:    Font shape `OT1/ptm/bx/it' in size <12> not available
+(Font)              Font shape `OT1/ptm/b/it' tried instead on input line 1090.
+
+LaTeX Font Info:    Font shape `OT1/ptm/bx/n' in size <17> not available
+(Font)              Font shape `OT1/ptm/b/n' tried instead on input line 1090.
+LaTeX Font Info:    Font shape `OT1/ptm/bx/it' in size <17> not available
+(Font)              Font shape `OT1/ptm/b/it' tried instead on input line 1090.
+
+LaTeX Font Info:    Font shape `OT1/ptm/bx/n' in size <20> not available
+(Font)              Font shape `OT1/ptm/b/n' tried instead on input line 1090.
+LaTeX Font Info:    Font shape `OT1/ptm/bx/it' in size <20> not available
+(Font)              Font shape `OT1/ptm/b/it' tried instead on input line 1090.
+
+LaTeX Font Info:    Font shape `OT1/ptm/bx/n' in size <24> not available
+(Font)              Font shape `OT1/ptm/b/n' tried instead on input line 1090.
+LaTeX Font Info:    Font shape `OT1/ptm/bx/it' in size <24> not available
+(Font)              Font shape `OT1/ptm/b/it' tried instead on input line 1090.
+
+\IEEEquantizedlength=\dimen147
+\IEEEquantizedlengthdiff=\dimen148
+\IEEEquantizedtextheightdiff=\dimen149
+\IEEEilabelindentA=\dimen150
+\IEEEilabelindentB=\dimen151
+\IEEEilabelindent=\dimen152
+\IEEEelabelindent=\dimen153
+\IEEEdlabelindent=\dimen154
+\IEEElabelindent=\dimen155
+\IEEEiednormlabelsep=\dimen156
+\IEEEiedmathlabelsep=\dimen157
+\IEEEiedtopsep=\skip47
+\c@section=\count188
+\c@subsection=\count189
+\c@subsubsection=\count190
+\c@paragraph=\count191
+\c@IEEEsubequation=\count192
+\abovecaptionskip=\skip48
+\belowcaptionskip=\skip49
+\c@figure=\count193
+\c@table=\count194
+\@IEEEeqnnumcols=\count195
+\@IEEEeqncolcnt=\count196
+\@IEEEsubeqnnumrollback=\count197
+\@IEEEquantizeheightA=\dimen158
+\@IEEEquantizeheightB=\dimen159
+\@IEEEquantizeheightC=\dimen160
+\@IEEEquantizeprevdepth=\dimen161
+\@IEEEquantizemultiple=\count198
+\@IEEEquantizeboxA=\box51
+\@IEEEtmpitemindent=\dimen162
+\IEEEPARstartletwidth=\dimen163
+\c@IEEEbiography=\count199
+\@IEEEtranrubishbin=\box52
+) (/usr/share/texlive/texmf-dist/tex/latex/amsmath/amsmath.sty
+Package: amsmath 2022/04/08 v2.17n AMS math features
+\@mathmargin=\skip50
+
+For additional information on amsmath, use the `?' option.
+(/usr/share/texlive/texmf-dist/tex/latex/amsmath/amstext.sty
+Package: amstext 2021/08/26 v2.01 AMS text
+
+(/usr/share/texlive/texmf-dist/tex/latex/amsmath/amsgen.sty
+File: amsgen.sty 1999/11/30 v2.0 generic functions
+\@emptytoks=\toks17
+\ex@=\dimen164
+))
+(/usr/share/texlive/texmf-dist/tex/latex/amsmath/amsbsy.sty
+Package: amsbsy 1999/11/29 v1.2d Bold Symbols
+\pmbraise@=\dimen165
+)
+(/usr/share/texlive/texmf-dist/tex/latex/amsmath/amsopn.sty
+Package: amsopn 2022/04/08 v2.04 operator names
+)
+\inf@bad=\count266
+LaTeX Info: Redefining \frac on input line 234.
+\uproot@=\count267
+\leftroot@=\count268
+LaTeX Info: Redefining \overline on input line 399.
+LaTeX Info: Redefining \colon on input line 410.
+\classnum@=\count269
+\DOTSCASE@=\count270
+LaTeX Info: Redefining \ldots on input line 496.
+LaTeX Info: Redefining \dots on input line 499.
+LaTeX Info: Redefining \cdots on input line 620.
+\Mathstrutbox@=\box53
+\strutbox@=\box54
+LaTeX Info: Redefining \big on input line 722.
+LaTeX Info: Redefining \Big on input line 723.
+LaTeX Info: Redefining \bigg on input line 724.
+LaTeX Info: Redefining \Bigg on input line 725.
+\big@size=\dimen166
+LaTeX Font Info:    Redeclaring font encoding OML on input line 743.
+LaTeX Font Info:    Redeclaring font encoding OMS on input line 744.
+\macc@depth=\count271
+LaTeX Info: Redefining \bmod on input line 905.
+LaTeX Info: Redefining \pmod on input line 910.
+LaTeX Info: Redefining \smash on input line 940.
+LaTeX Info: Redefining \relbar on input line 970.
+LaTeX Info: Redefining \Relbar on input line 971.
+\c@MaxMatrixCols=\count272
+\dotsspace@=\muskip16
+\c@parentequation=\count273
+\dspbrk@lvl=\count274
+\tag@help=\toks18
+\row@=\count275
+\column@=\count276
+\maxfields@=\count277
+\andhelp@=\toks19
+\eqnshift@=\dimen167
+\alignsep@=\dimen168
+\tagshift@=\dimen169
+\tagwidth@=\dimen170
+\totwidth@=\dimen171
+\lineht@=\dimen172
+\@envbody=\toks20
+\multlinegap=\skip51
+\multlinetaggap=\skip52
+\mathdisplay@stack=\toks21
+LaTeX Info: Redefining \[ on input line 2953.
+LaTeX Info: Redefining \] on input line 2954.
+)
+(/usr/share/texlive/texmf-dist/tex/latex/l3backend/l3backend-pdftex.def
+File: l3backend-pdftex.def 2022-10-26 L3 backend support: PDF output (pdfTeX)
+\l__color_backend_stack_int=\count278
+\l__pdf_internal_box=\box55
+)
+(./ME-Equation-Sheet.aux)
+\openout1 = `ME-Equation-Sheet.aux'.
+
+LaTeX Font Info:    Checking defaults for OML/cmm/m/it on input line 5.
+LaTeX Font Info:    ... okay on input line 5.
+LaTeX Font Info:    Checking defaults for OMS/cmsy/m/n on input line 5.
+LaTeX Font Info:    ... okay on input line 5.
+LaTeX Font Info:    Checking defaults for OT1/cmr/m/n on input line 5.
+LaTeX Font Info:    ... okay on input line 5.
+LaTeX Font Info:    Checking defaults for T1/cmr/m/n on input line 5.
+LaTeX Font Info:    ... okay on input line 5.
+LaTeX Font Info:    Checking defaults for TS1/cmr/m/n on input line 5.
+LaTeX Font Info:    ... okay on input line 5.
+LaTeX Font Info:    Checking defaults for OMX/cmex/m/n on input line 5.
+LaTeX Font Info:    ... okay on input line 5.
+LaTeX Font Info:    Checking defaults for U/cmr/m/n on input line 5.
+LaTeX Font Info:    ... okay on input line 5.
+
+-- Lines per column: 58 (exact).
+
+LaTeX Warning: No \author given.
+
+
+LaTeX Warning: No \author given.
+
+
+LaTeX Warning: No \author given.
+
+
+! Package amsmath Error: Erroneous nesting of equation structures;
+(amsmath)                trying to recover with `aligned'.
+
+See the amsmath package documentation for explanation.
+Type  H <return>  for immediate help.
+ ...                                              
+                                                  
+l.16 \end{align}
+                $$
+Try typing  <return>  to proceed.
+If that doesn't work, type  X <return>  to quit.
+
+
+! Package amsmath Error: Erroneous nesting of equation structures;
+(amsmath)                trying to recover with `aligned'.
+
+See the amsmath package documentation for explanation.
+Type  H <return>  for immediate help.
+ ...                                              
+                                                  
+l.26 \end{align}
+                $$
+Try typing  <return>  to proceed.
+If that doesn't work, type  X <return>  to quit.
+
+
+Overfull \hbox (6.22469pt too wide) detected at line 26
+[][]
+ []
+
+
+Package amsmath Warning: Foreign command \over;
+(amsmath)                \frac or \genfrac should be used instead
+(amsmath)                 on input line 38.
+
+[1{/var/lib/texmf/fonts/map/pdftex/updmap/pdftex.map}
+
+
+] [2
+
+]
+(./ME-Equation-Sheet.aux) ) 
+Here is how much of TeX's memory you used:
+ 1978 strings out of 476182
+ 34856 string characters out of 5796580
+ 1869793 words of memory out of 6000000
+ 22338 multiletter control sequences out of 15000+600000
+ 539905 words of font info for 80 fonts, out of 8000000 for 9000
+ 1137 hyphenation exceptions out of 8191
+ 54i,14n,62p,236b,171s stack positions out of 10000i,1000n,20000p,200000b,200000s
+{/usr/share/texlive/texmf-dist/fonts/enc/dvips/base/
+8r.enc}</usr/share/texlive/texmf-dist/fonts/type1/public/amsfonts/cm/cmex10.pfb
+></usr/share/texlive/texmf-dist/fonts/type1/public/amsfonts/cm/cmmi10.pfb></usr
+/share/texlive/texmf-dist/fonts/type1/public/amsfonts/cm/cmmi5.pfb></usr/share/
+texlive/texmf-dist/fonts/type1/public/amsfonts/cm/cmmi7.pfb></usr/share/texlive
+/texmf-dist/fonts/type1/public/amsfonts/cm/cmr10.pfb></usr/share/texlive/texmf-
+dist/fonts/type1/public/amsfonts/cm/cmr5.pfb></usr/share/texlive/texmf-dist/fon
+ts/type1/public/amsfonts/cm/cmr7.pfb></usr/share/texlive/texmf-dist/fonts/type1
+/public/amsfonts/cm/cmsy10.pfb></usr/share/texlive/texmf-dist/fonts/type1/publi
+c/amsfonts/cm/cmsy7.pfb></usr/share/texlive/texmf-dist/fonts/type1/urw/times/ut
+mr8a.pfb></usr/share/texlive/texmf-dist/fonts/type1/urw/times/utmri8a.pfb>
+Output written on ME-Equation-Sheet.pdf (2 pages, 120288 bytes).
+PDF statistics:
+ 67 PDF objects out of 1000 (max. 8388607)
+ 40 compressed objects within 1 object stream
+ 0 named destinations out of 1000 (max. 500000)
+ 1 words of extra memory for PDF output out of 10000 (max. 10000000)
+

5th-Semester-Fall-2023/ME-For-ECEs/EquationSheet/ME-Equation-Sheet.tex

@@ -0,0 +1,146 @@
+\documentclass{IEEEtran}
+\usepackage{amsmath}
+
+\title{ME for ECEs Equation Sheet}
+\begin{document}
+\maketitle
+
+\section{Vectors}
+$$\vec{v}_R = \sum v_x \hat{x} + \sum v_y \hat{y} + \sum v_z \hat{z}$$
+
+$$\|\vec{v}\| = \sqrt{v_x^2 + v_y^2 + v_z^2}$$
+
+$$\begin{align}
+    \vec{u} \cdot \vec{v} &= \|\vec{u}\| \|\vec{v}\| \cos(\theta) \\
+    &= (u_x v_x) + (u_y v_y) + (u_z v_z)
+\end{align}$$
+
+$$\begin{align}
+    \vec{u} \times \vec{v} &= 
+    \begin{vmatrix}
+        \hat{x} & \hat{y} & \hat{z} \\
+        u_x & u_y & u_z \\
+        v_x & v_y & v_z
+    \end{vmatrix} \\
+    &= (u_y v_z - u_z v_y)\hat{x} - (u_x v_z - u_z v_x)\hat{y} + (u_x v_y - u_y v_x)\hat{z}
+\end{align}$$
+
+\section{Statics}
+$$\vec{F} = m \vec{a}$$
+$$\vec{M} = \vec{r} \times \vec{F}$$
+$$\vec{r} = \vec{s}_f - \vec{s}_0$$
+$$\sum F = 0$$
+$$\sum M = 0$$
+
+\section{Dynamics}
+
+\subsection{Kinematic Equations}
+$$\vec{s}_f - \vec{s}_0 = \vec{v}_0 t + {1\over2} \vec{a} t^2$$
+$$\vec{v}_f^2 = \vec{v}_0^2 + 2 \vec{a} \cdot (\vec{s}_f - \vec{s}_0)$$
+$$\vec{v}_f = \vec{v}_0 + \vec{a}t$$
+$$\vec{s}_f = \vec{s}_0 + \vec{v}t$$
+
+\subsection{Projectile Motion}
+$$\vec{g} = -9.81\hat{y}\left[{\text{m} \over \text{s}^2}\right] = -32.2\hat{y} \left[{ \text{ft} \over \text{s}^2}\right]$$
+Without air resistance:
+$$v_{x_0} = v_{x_f}$$
+
+
+\section{Heat Transfer}
+
+\subsection{Conduction}
+$$q_x'' = k {dT \over dx}$$
+$${dT \over dx} = {T_2 - T_1 \over L}$$
+$$q_x = q_x'' A$$
+Where:
+\begin{itemize}
+    \item[$q_x''$] is heat flux
+    \item[$q_x$] is heat rate
+    \item[$T$] is temperature
+    \item[$L$] is length
+    \item[$A$] is the contact area
+\end{itemize}
+
+\subsection{Convection}
+$$q'' = hA(T_s - T_\infty)$$
+Where:
+\begin{itemize}
+    \item[$h$] is the heat transfer coefficient
+    \item[$A$] is the contact area between the surface and the fluid
+    \item[$T_s$] is the surface temperature
+    \item[$T_\infty$] is the fluid temperature very far away from the surface
+\end{itemize}
+
+\subsection{Radiation}
+$$q_\text{ideal}'' = \sigma T_s^4$$
+$$q_\text{real}'' = \varepsilon \sigma T_s^4$$
+Where:
+\begin{itemize}
+    \item[$T_s$] is the absolute temperature
+    \item[$\sigma$] is the Stefan-Boltzmann constant
+    \item[$\varepsilon$] is the emissivity
+\end{itemize}
+
+\section{Fluids}
+\noindent
+Pascal's Law:
+$${F_1 \over A_1} = {F_2 \over A_2}$$
+Density:
+$$\rho = {m \over v}$$
+Specific Weight:
+$$\gamma = {mg \over v} = \rho g$$
+Pressure in a fluid:
+$$P_2 = P_1 + \rho g z$$
+Bernoulli Equation:
+$$p_1 + {1\over2}\rho v_1^2 + \rho g h_1 = p_2 + {1\over2}\rho v_2^2 + \rho g h_2$$
+Fluid Velocity Between Plates:
+$${U \over b} = {u \over y}$$
+Where:
+\begin{itemize}
+    \item[$U$] is the velocity of the moving plate
+    \item[$b$] is the distance between the plates
+    \item[$u$] is the velocity of the fluid at between the plates at some distance above the stationary plate
+    \item[$y$] is the distance above the stationary plate.
+\end{itemize}
+Continuity Equation
+$$A_1 v_1 \Delta t = A_2 v_2 \Delta t$$
+$$\dot{m} = \rho_1 A_1 v_1 = \rho_2 A_2 v_2$$
+
+\section{Gears}
+$$N = P d$$
+$$N = {d \over m}$$
+$$c = {d_1 + d_2 \over 2} = {N_1 + N_2 \over 2 P} = {(N_1 + N_2) m \over 2}$$
+Where:
+\begin{itemize}
+    \item[$N$] is the number of teeth
+    \item[$d$] is the pitch diameter
+    \item[$c$] is the center distance
+    \item[$P$] is the diametral pitch (Customary)
+    \item[$m$] is the module (SI)
+\end{itemize}
+\noindent
+Gear Ratio:
+$$R = {T_2 \over T_1} = {N_2 \over N_1} = {d_2 \over d_1} = {\omega_1 \over \omega_2}$$
+$$\omega = {\pi \over 30}\text{RPM}$$
+
+\noindent
+Big Gear to Small Gear:
+\begin{itemize}
+    \item Speed increases
+    \item Torque decreases
+\end{itemize}
+
+\noindent
+Small Gear to Big Gear:
+\begin{itemize}
+    \item Speed decreases
+    \item Torque increases
+\end{itemize}
+
+\noindent
+Power:
+$$P = \omega T$$
+If no power losses:
+$$P_\text{in} = P_\text{out}$$
+$$\omega_1 T_1 = \omega_2 T_2$$
+\end{document}

5th-Semester-Fall-2023/ME-For-ECEs/Homework/Homework-01.md

@@ -0,0 +1,57 @@
+# Aidan Sharpe - Homework 1
+
+1. $\vec{F}$ has magnitude $800$N and makes an angle of $35^\circ$ with the y-axis in the second quadrant.
+    
+    $\vec{F}_x = \lVert \vec{F} \rVert \sin(-35^\circ) = 800 \times -0.5736 = -458.861$
+
+    $\vec{F}_y = \lVert \vec{F} \rVert \cos(-35^\circ) = 800 \times 0.8192 = 655.322$
+
+    $\vec{F} = -458.861\hat{i} + 655.322\hat{j}$
+
+1. The force, $\vec{F}$, has magnitude $6.6$kN and slope is $-\frac{5}{12}$.
+    
+    $\theta = \arctan \left(-\frac{5}{12} \right) = -0.3948 = -22.620^\circ$
+
+    $\vec{F}_x = 6600\cos(-22.620^\circ) = 6600 \times 0.9231 = 6092.308$
+
+    $\vec{F}_y = 6600\sin(-22.620^\circ) = 6600 \times -0.3846 = -2538.462$
+
+    $\vec{F} = 6092.308\hat{i} - 2538.462\hat{j}$
+
+1. $F_1 = 500$N and $F_2 = 350$-N. $F_1$ is in the direction of the x-axis, and $F_2$ makes an angle $60^\circ$ with the x-axis.
+
+    $\vec{R} = \vec{F}_1 + \vec{F}_2$
+
+    $\vec{F}_1 = 500\hat{i} + 0\hat{j}$
+    
+    $\vec{F}_2 = 350\cos(60^\circ)\hat{i} + 350\sin(60^\circ)\hat{j} = 175\hat{i} + 303.109\hat{j}$
+
+    $\vec{R} = 675\hat{i} + 303.109\hat{j}$
+
+    $\theta_R = \arctan(\frac{303.109}{675}) = 0.4221 = 24.182^\circ$
+
+1. $F_y = 70$lbs. The slope of $\vec{F}$ is $\frac{12}{5}$.
+    
+    $\frac{70}{F_x} = \frac{12}{5}$
+
+    $70 \times 5 = 12 F_x$
+
+    $F_x = \frac{70 \times 5}{12} = 29.166$
+
+    $\vec{F} = F_x\hat{i} + F_y\hat{j}$
+
+    $\lVert \vec{F} \rVert = \sqrt{F_x^2 + F_y^2} = \sqrt{29.166^2 + 70^2} = 75.833$N
+
+1. $\vec{F}_1$ has magnitude $2$kN and makes an angle $30^\circ$ with the x-axis. $\vec{F}_2$ has magnitude $3$kN and has slope $-\frac{4}{3}$.
+
+    $\vec{F}_1 = 2000\cos(30^\circ)\hat{i} + 2000\sin(30^\circ)\hat{j} = 1732.051\hat{i} + 1000\hat{j}$
+
+    $\theta_{F_2} = \arctan(-\frac{4}{3}) = -0.9273 = -53.130^\circ$
+
+    $\vec{F}_2 = 3000\cos(\theta_{F_2})\hat{i} + 3000\sin(\theta_{F_2})\hat{j} = 1800\hat{i} - 2400\hat{j}$
+
+    $\vec{R} = \vec{F}_1 + \vec{F}_2 = (1732.051 + 1800)\hat{i} + (1000 - 2400)\hat{j} = 3532.051\hat{i} - 1400\hat{j}$
+
+    $\theta = \arctan(-\frac{1400}{3532.051}) = -0.3774 = -21.622^\circ$
+
+    $\lVert \vec{R} \rVert = \sqrt{3532.051^2 + (-1400)^2} = 3799.393$

5th-Semester-Fall-2023/ME-For-ECEs/Homework/Homework-02.md

@@ -0,0 +1,109 @@
+# Aidan Sharpe - Homework 2
+
+## 1.
+a) Point $O$ exists at the origin, and point $B$ exists at $(16, 0)$. A $5$kN force, $\vec{F}$, makes an angle of $\frac{5\pi}{6}$ with the $\hat{x}$ direction at $(12, -15)$.
+
+### a) Find the moment, $M_O$, about point $O$.
+
+$$M_O = \vec{r} \times \vec{F}$$
+$$\vec{r} = (12 - 0)\hat{x} + (-15 - 0)\hat{y} = 12\hat{x} - 15\hat{y}$$
+$$\vec{F} = 5000\cos\left(\frac{5\pi}{6}\right)\hat{x} + 5000\sin\left(\frac{5\pi}{6}\right)\hat{y} = -4330.13\hat{x} + 2500\hat{y}$$
+$$\vec{r} \times \vec{F} = 
+\begin{vmatrix} 
+\hat{x} & \hat{y} & \hat{z} \\
+12 & -15 & 0 \\
+-4330.13 & 2500 & 0 \\
+\end{vmatrix} = -34951.91\hat{z}$$
+$$M_O = -34951.91\hat{z}\text{[N m]}$$
+
+### b) Find the moment, $M_B$, about point $B$
+
+$$M_B = \vec{r} \times \vec{F}$$
+$$\vec{r} = (12 - 16)\hat{x} + (-15 - 0)\hat{y} = -4\hat{x} - 15\hat{y}$$
+$$\vec{F} = 5000\cos\left(\frac{5\pi}{6}\right)\hat{x} + 5000\sin\left(\frac{5\pi}{6}\right)\hat{y} = -4330.13\hat{x} + 2500\hat{y}$$
+$$\vec{r} \times \vec{F} = 
+\begin{vmatrix} 
+\hat{x} & \hat{y} & \hat{z} \\
+-4 & -15 & 0 \\
+-4330.13 & 2500 & 0 \\
+\end{vmatrix} = -74951.91\hat{z}$$
+$$M_B = -74951.91\hat{z}\text{[N m]}$$
+
+## 2
+A $250$N force is applied at an angle $\frac{5\pi}{12}$ with respect to the $\hat{x}$ direction at a distance $30$mm above and $200$mm to the right of the center of a bolt. Find the moment, $M_B$, about the center of the bolt.
+
+$$M_B = \vec{r} \times \vec{F}$$
+Convert distance to meters:
+$$\vec{r} = 0.2\hat{x} + 0.03\hat{y}$$
+$$\vec{F} = 250\cos\left(\frac{5\pi}{12}\right)\hat{x} + 250\sin\left(\frac{5\pi}{12}\right)\hat{y} = 64.70\hat{x} + 241.48\hat{y}$$
+$$\vec{r} \times \vec{F} =
+\begin{vmatrix} 
+\hat{x} & \hat{y} & \hat{z} \\
+0.2 & 0.03 & 0 \\
+64.70 & 241.48 & 0 \\
+\end{vmatrix} = 46.36\hat{z}$$
+$$M_B = 46.36\hat{z}\text{[N m]}$$
+
+## 3
+Consider a bar made up of two segments, $\overline{AB}$ and $\overline{BC}$, each with length, $1.6$m. Segment $\overline{AB}$ makes an angle $\frac{\pi}{2}$ with the $\hat{x}$ direction, and segment $\overline{BC}$ makes an angle $\frac{3\pi}{4}$ with the $\hat{x}$ direction. A $30$N force, $\vec{P}$, is applied perpendicular to $\overline{BC}$. Determine the moment, $M_B$, about point $B$, and $M_A$ about point $A$.
+
+### a) Find the moment, $M_B$, about point $B$
+$$M_B = \vec{r} \times \vec{F}$$
+$$\vec{r} = 1.6\cos\left(\frac{3\pi}{4}\right)\hat{x} + 1.6\sin\left(\frac{3\pi}{4}\right)\hat{y} =-1.13\hat{x} + 1.13\hat{y}$$
+$$\vec{F} = 30\cos\left(\frac{\pi}{4}\right)\hat{x} + 30\sin\left(\frac{\pi}{4}\right)\hat{y} = 21.21\hat{x} + 21.21\hat{y}$$
+$$\vec{r} \times \vec{F} =
+\begin{vmatrix} 
+\hat{x} & \hat{y} & \hat{z} \\
+-1.13 & 1.13 & 0 \\
+21.21 & 21.21 & 0 \\
+\end{vmatrix} = -47.93\hat{z}$$
+$$M_B = -47.93\hat{z}\text{[N m]}$$
+
+### b) Find the moment, $M_A$, about point $A$
+$$M_A = \vec{r} \times \vec{F}$$
+$$\vec{r} = 1.6\cos\left(\frac{3\pi}{4}\right)\hat{x} + \left[1.6 + 1.6\sin\left(\frac{3\pi}{4}\right)\right]\hat{y} = -1.13\hat{x} + 2.73\hat{y}$$
+$$\vec{F} = 30\cos\left(\frac{\pi}{4}\right)\hat{x} + 30\sin\left(\frac{\pi}{4}\right)\hat{y} = 21.21\hat{x} + 21.21\hat{y}$$
+$$\vec{r} \times \vec{F} =
+\begin{vmatrix} 
+\hat{x} & \hat{y} & \hat{z} \\
+-1.13 & 2.73 & 0 \\
+21.21 & 21.21 & 0 \\
+\end{vmatrix} = -81.87\hat{z}$$
+$$M_A = -81.87\hat{z}\text{[N m]}$$
+
+## 4
+Consider an arm holding a ball and the moments acting on the elbow. The forearm makes an angle, $-35^\circ$ with the $\hat{x}$ direction. The ball weighs 8lbs, and its center of gravity is a distance of $13$ inches away in the $\hat{x}$ direction. The forearm weighs $5$lbs, and its center of gravity is $6$ inches away along the forearm. A third, balancing force of tension acts $2$ inches down the forearm. Find the force of tension, $\vec{T}$, such that the moment about the elbow, $M_O$, is $0$.
+
+$$M_O = (\vec{r}_T \times \vec{T}) + (\vec{r}_G \times \vec{G}) + (\vec{r}_A \times \vec{A}) = 0$$
+Calculate radial vectors:
+$$\vec{r}_G = 6\cos(-35^\circ)\hat{x} + 6\sin(-35^\circ)\hat{y} = 4.91\hat{x} - 3.44\hat{y}$$
+$$\vec{r}_A = 13\hat{x} + 13\tan(-35^\circ)\hat{y} = 13\hat{x} - 9.10\hat{y}$$
+$$\vec{r}_T = 2\hat{x} + 2\tan(-35^\circ)\hat{y} = 2\hat{x} - 1.4\hat{y}$$
+
+Assign force vectors:
+$$\vec{G} = -5\hat{y}$$
+$$\vec{A} = -8\hat{y}$$
+
+Evaluate known cross products:
+$$\vec{r}_G \times \vec{G} = 
+\begin{vmatrix}
+\hat{x} & \hat{y} & \hat{z} \\
+4.91 & -3.44 & 0 \\
+0 & -5 & 0
+\end{vmatrix} = -24.55\hat{z}$$
+$$\vec{r}_A \times \vec{A} =
+\begin{vmatrix}
+\hat{x} & \hat{y} & \hat{z} \\
+13 & -9.10 & 0 \\
+0 & -8 & 0
+\end{vmatrix} = -104\hat{z}$$
+
+Combine all knowns:
+$$0 = (\vec{r}_T \times \vec{T}) - 24.55\hat{z} - 104\hat{z}$$
+$$\therefore \vec{r}_T \times \vec{T} = 128.55\hat{z}$$
+
+Since $\vec{T}$ only acts in the $\hat{y}$ direction:
+$$128.55 = r_{T_x} T_y - r_{T_y} T_x = (2)T_x-(-1.4)(0)$$
+$$\therefore 128.55 = (2)T_y$$
+$$\therefore T_y = 64.28$$
+$$\vec{T} = 64.28\hat{y} \text{[lb in]}$$

5th-Semester-Fall-2023/ME-For-ECEs/Homework/Homework-03.md

@@ -0,0 +1,107 @@
+# Homework 3 - Aidan Sharpe
+
+## 1
+A 50[kg] homogeneous smooth sphere rests on a $30^\circ$ incline and against a vertical wall.
+
+Since the sphere is in static equilibrium:
+$$\sum \vec{F} = 0$$
+$$\therefore \sum \vec{F}_x = \sum \vec{F}_y = 0$$
+$$\sum \vec{F}_y = \vec{F}_g + \vec{F}_{A_y}$$
+$$\sum \vec{F}_x = \vec{F}_B + \vec{F}_{A_x}$$
+Since $\vec{F}_A$ is normal to the surface at $30^\circ$ to the $-\hat{x}$ direction:
+$$\vec{F}_{A_x} = F_A \cos(60^\circ)\hat{x}$$
+$$\vec{F}_{A_y} = F_A \sin(60^\circ)\hat{y}$$
+
+Find the force due to gravity:
+$$\vec{F}_g = (50)(-9.8)\hat{y} = -490\hat{y}$$
+
+Solve for $\|\vec{F}_A\|$:
+$$-490 + F_A \sin(60^\circ) = 0$$
+$$\therefore \|\vec{F}_A\| = {490 \over \sin(60^\circ)} = 565.8 \text{[N]}$$
+$$\vec{F}_B + 565.8\cos(60^\circ) = 0$$
+$$\therefore \vec{F}_B = -282.9\hat{x} \text{[N]}$$
+
+Since $F_{A_x} = -F_B$ and $F_{A_y} = -F_g$:
+$$\vec{F}_A = 282.9\hat{x} + 490\hat{y} \text{[N]}$$
+
+## 2
+A uniform 150[kg], 15[m] long pole is supported by vertical walls spaced 12[m] apart at points $A$ and $B$. A vertical tension force is applied 5[m] from point $A$ (10[m] from point $B$). Find the reactions at $A$ and $B$.
+
+#### Assumptions:
+The moment, $M_g$, acts at the center of mass. The pole pivots around the point where the cable is attached. The forces at points $A$ and $B$ act strictly in the horizontal direction. There is no friction between the pole and the walls.
+
+#### Find $F_A$ and $F_B$:
+$$\sum \vec{M} = 0 = \vec{M}_g + \vec{M}_A + \vec{M}_B$$
+Find the angle that the pole makes:
+$$\theta = \arccos\left({12 \over 15}\right) = 0.6435$$
+
+Find the about the cable due to gravity, $M_g$:
+$$\vec{M}_g = \vec{r}_g \times \vec{F}_g$$
+
+Since $\vec{r}_g$ is 2.5[m] along the bar from the point of tension:
+$$\vec{r}_g = 2.5 \cos(\theta)\hat{x} + 2.5 \sin(\theta)\hat{y} = 2\hat{x} + 1.5\hat{y}$$
+$$\vec{F}_g = mg = (150)(-9.8)\hat{y} = -1470\hat{y}$$
+
+Plug in and evaluate $\vec{M}_g$:
+$$\vec{M}_g =
+\begin{vmatrix}
+\hat{x} & \hat{y} & \hat{z} \\
+2 & 1.5 & 0 \\ 
+0 & -1470 & 0 
+\end{vmatrix} = -2940\hat{z}$$
+
+Find the moment about the cable attachment due to the support from point $A$, $\vec{M}_A$:
+$$\vec{M}_A = \vec{r}_A \times \vec{F}_A$$
+
+Since $\vec{r}_A$ is -5[m] along the pole from the point of tension:
+$$\vec{r}_A = -5\cos(\theta)\hat{x} - 5\sin(\theta)\hat{y} = -4\hat{x} - 3\hat{y}$$
+
+Since $\vec{F}_A$ is unknown, but known to only act in the $\hat{x}$ direction:
+$$\vec{M}_A =
+\begin{vmatrix}
+\hat{x} & \hat{y} & \hat{z} \\
+-4 & -3 & 0 \\
+F_A & 0 & 0
+\end{vmatrix} = 3\vec{F}_A \hat{z}$$
+
+Find the moment about the cable attachment due to the suport from point $B$, $\vec{M}_B$:
+$$\vec{M}_B = \vec{r}_B \times \vec{F}_B$$
+
+Since $\vec{r}_B$ is 10[m] along the pole from the point of tension:
+$$\vec{r}_B = 10\cos(\theta)\hat{x} + 10\sin(\theta)\hat{y} = 8\hat{x} + 6\hat{y}$$
+
+Since $\vec{F}_B$ is unknown, but known to only act in the $\hat{x}$ direction:
+$$\vec{M}_B =
+\begin{vmatrix}
+\hat{x} & \hat{y} & \hat{z} \\
+8 & 6 & 0 \\
+F_B & 0 & 0
+\end{vmatrix} = -6F_B \hat{z}$$
+
+To find the values $F_A$ and $F_B$:
+$$\sum \vec{M} = 0 = \vec{M}_A + \vec{M}_B + \vec{M}_g$$
+$$\therefore -2940 + 3F_A - 6F_B = 0$$
+$$\sum \vec{F}_x = 0 =\vec{F}_A + \vec{F}_B$$
+$$\therefore\vec{F}_A = -\vec{F}_B$$
+$$\therefore -2940 = 9\vec{F}_B$$
+$$\therefore \vec{F}_B = -326.6\hat{x}$$
+$$\therefore \vec{F}_A = 326.6\hat{x}$$
+
+## 4
+If the car accelerates at 2.75[$\text{m}/\text{s}^2$] for 3[m] and then maintains speed for 4[m], find the time it takes to travel the entire distance.
+
+$$v^2 = 2a\Delta p = 2(2.75)(3) = 16.5 \text{[m/s]}$$
+$$v = at$$
+$$t_\text{decline} = {16.5 \over 2.75} = 6\text{[s]}$$
+$$\Delta p = vt$$
+$$t_\text{coasting} =  {4 \over 16.5} = 0.242\text{[s]}$$
+$$t = t_\text{decline} + t_\text{coasting} = 6.242\text{[s]}$$
+
+## 5
+A ball is thrown upwards with an initial velocity of 30[m/s] at the edge of a 60[m] high cliff. Find the maximum height above the ground, $h$, and the total time, $t$, before the ball hits the ground.
+$$0^2 = (30)^2 + 2(-9.8)\Delta h$$
+$${900 \over (2)(9.8)} = \Delta h = 45.918 \text{[m]}$$
+$$h = 60 + \Delta h = 105.918 \text{[m]}$$
+$$\Delta h_\text{final} = 30 + {1 \over 2}(-9.8)t^2$$
+$$-60 = 30 - 4.9t^2$$
+$$\therefore t = +\sqrt{{-90 \over -4.9}} = 4.2857 \text{[s]}$$

5th-Semester-Fall-2023/ME-For-ECEs/Homework/Homework-04.md

@@ -0,0 +1,37 @@
+# Homework 4 - Aidan Sharpe
+
+## 1
+The wall of a house is 7[m] wide, 6[m] high, and 0.3[m] thick with $k=0.6$[W/($m \cdot K$)]. The surface temperature on the inside of the wall is 16$^\circ$C and the temperature on the outside is 6$^\circ$. Find the heat flux through the wall and the total heat loss through it.
+
+Heat flux is given by
+$$q''_x = -k{dT \over dx}$$
+$\Delta T = 10$, $k = 0.6$, $\Delta x$ = 0.3
+$$q''_x= -0.6 {10 \over 0.3} = -20\text{[W/m$^2$]}$$
+
+Total heat loss is the product of heat flux and the area:
+$$-20 (6 \times 7) = -840\text{[W]}$$
+
+## 2
+A 20[mm] diameter copper pipe is used to carry heated water. The external surface of the pipe is subjected to a convective heat transfer coefficient of $h = 6$[W/(m$^2 \cdot$ K)]. Find the heat loss by convection per meter length of the pipe when the external surface temperature is 80$^\circ$C and the surroundings are at 20$^\circ$C. Assuming black body radiation, what is the heat loss by radiation?
+
+$$\dot{Q} = hA(T-T_f)$$
+$h=6$, $A=2\pi 0.02 l$, $T = 80$, $T_f = 20$
+$$\dot{Q} = 6 (2 \pi 0.02 x) (80 - 20) = 45.2389x$$
+Heat loss due to convection is $45.2389$[W/m].
+
+$$q'' = \sigma T_s^4 = 5.67 \times 10^{-8} (80 + 273.15) = 2 \times 10^{-5} \text{[W/m$^2$]}$$
+
+The total heat loss is the product of surface area and thermal flux:
+$$2\times 10^{-5} (2\pi 0.02x) = 2.516 x \text{[W]}$$
+
+## 3
+A plate 0.3[m] long and 0.1[m] wide, with a thickness of 12[mm] is made from stainless steel ($k = 16$[W/(m $\cdot$ K)]), the top surface is exposed to a 20$^\circ$C airstream. In an experiment, the plate is heated by an electrical heater (also 0.3[m] by 0.1[m]) positioned on the underside of the plate and the temperature of the plate adjacent to the heater is maintained at 100$^\circ$C. A voltmeter and ammeter are connected to the heater and these read 200[V] and 0.25[A] respectively. Assuming that the plate is perfectly insulated on all sides except the top surface, what s the convective heat transfer coefficient?
+
+Heat transfer from heater to plate:
+$$P = I V = 200 \times 0.25 = 50\text{W}]$$
+$$q'' = {50 \over 0.3 \times 0.1} = 1666.667 \text{[W/m$^2$]}$$
+
+Heat transfer from the plate to the air:
+$$q'' = 1666.667 = {d\dot{Q} \over dA}$$
+$$\dot{Q} = h A(T_\text{plate} - T_\text{air})$$
+$$h = {q'' \over T_\text{plate} - T_\text{air}} = {1666.667 \over 100 - 20} = 20.833\text{[W/(m$^2 \cdot$K)]}$$

5th-Semester-Fall-2023/ME-For-ECEs/Homework/Homework-05.md

@@ -0,0 +1,22 @@
+# Homework 5 - Aidan Sharpe
+
+## 1
+If the specific weight, $\gamma$, of a substance is 8.2[kN/m$^3$], what is its mass density, $\rho$, in kg/m$^3$?
+
+On Earth, 9810[kg] weighs 1[kN], so an object that weighs 8.2[kN] will have a mass of 80442[kg]. Therefore, $\rho = 80442$[kg/m$^3$].
+
+## 2
+A fluid flowing between two parallel plates has a viscosity, $\mu = 0.62$[Ns/m$^2$], and density, $\rho = 1250$[kg/m^3]. Calculate the intensity of shear stress, $\tau$, in pascals at $y = 3$[cm], assuming a straight-line viscosity distribution, given that the top plate has a velocity of 100[cm/s] and the fluid is 6[cm] thick.
+
+Velocity at $y = 3$[cm]:
+$$\tau = \mu{du \over dy} = 0.62\left(0.03 \times {1 \over 0.06}\right) = 0.31\text{[Pa]}$$
+
+## 3
+A cube with side length 10[cm] is placed into two different liquids. In the first liquid, the top of the cube is $h_1 = 1$[cm] above the surface. In the second liquid, the top of the cube is $h_2$ above the surface. If the densities of the liquids are known to be $\rho_1 = 1000$[kg/m$^3$], and $\rho_2 = 1300$[kg/m$^3$] respectively, find $h_2$.
+
+Displaced mass for liquid 1, same as mass of cube:
+$$0.1 \times 0.1 \times (0.1 - 0.01) \times 1000 = 0.9\text{[kg]}$$
+
+Displaced mass for liquid 2:
+$$0.1 \times 0.1 \times (0.1 - h_2) \times 1300 = 0.9$$
+$$\therefore h_2 = 0.031\text{[m]} = 3.1\text{[cm]}$$

5th-Semester-Fall-2023/ME-For-ECEs/Homework/Homework-06.md

@@ -0,0 +1,46 @@
+# Homework 6 - Aidan Sharpe
+
+## 1
+A hose lying on the ground has water coming out of it at a speed of 5.4 meters/sec. You lift the nozzle of the hose to a height of 1.3 meters above the ground. At what speed does the water now come out of the hose? (Note: Atmospheric pressure acts on the fluid at both points). List any assumptions.
+
+$$v_1 = 5.4 \text{[m/s]}$$
+$$h_1 = 0 \text{[m]}$$
+$$h_2 = 1.3 \text{[m]}$$
+$${1\over2}\rho v_1^2 + \rho g h_1 = {1\over2} \rho v_2^2 + \rho g h_2$$
+$${1\over2}(5.4^2) + 9.81 (0) = {1\over 2} v_2^2 + 9.81 (1.3)$$
+Solve for $v_2$ algebraically:
+$$v_2 = 1.911 \text{[m/s]}$$
+
+## 2
+A dam is holding back the water in a lake. There is a small hole 1.4 meters below the surface of the lake. At what speed does water exit the hole? List any assumptions. (Note: if the hole is small, what is a safe assumption for the particles at the top of the lake)
+$$P = \rho g h$$
+$$\rho = 1000 \left[{\text{kg} \over \text{m}^3}\right]$$
+$$h = 1.4 \text{[m]}$$
+$$P = 9.81(1000)(1.4) = 13720 \text{[Pa]}$$
+$$P_1 + {1\over2}\rho v_1^2 + \rho g h_1 = P_2 + {1\over2}\rho v_2^2 + \rho g h_2$$
+$${1\over2}\rho v_2^2 = P_1$$
+$$v_2 = 5.23 \text{[m/s]}$$
+
+## 3
+A basketball is floating in a bathtub full of water. The basketball has a mass of 0.5 kg and a diameter of 22 cm.
+
+### a)
+What is the buoyancy force? Draw a Free Body Diagram and apply equations.
+$$F_b = mg = 4.9 \text{[N]}$$
+
+### b)
+What is the volume of water displaced by the ball? (consider it a sphere and use Archimedes principle)
+
+$$F_b = \rho v g$$
+$$v = {4.9 \over \rho g} = 5 \times 10^{-4} \text{[m}^3]$$
+
+### c)
+What is the average density of the basketball?
+$$d = {0.5 \over {4\over3}\pi r^3} = {0.5 \over {4\over3}\pi {0.11}^3} = 89.68 \left[{\text{kg} \over \text{m}^3}\right]$$
+
+## 4
+You need to extend a 2 cm diameter pipe, but you have only a 1.10 cm diameter pipe on hand. You make a fitting to connect these pipes end to end. If the water is flowing at 2.50 cm/sec in the wide pipe, how fast will it be flowing through the narrow one?
+
+$$v_1 A_1 = v_2 A_2$$
+$$0.025 (\pi 0.02^2) = v_2 (\pi 0.0055^2)$$
+$$v_2 = 0.083 \text{[m/s]} = 8.3 \text{[cm/s]}$$

5th-Semester-Fall-2023/ME-For-ECEs/Notes/Fluids.md

@@ -0,0 +1,45 @@
+# Fluid Mechanics
+
+##### Density
+$$\rho = {m \over V}$$
+
+##### Specific weight
+$$\gamma = {m g \over V} = \rho V$$
+
+##### Viscosity
+- Resistance to flow
+- Ability to resist shear force
+
+### Example
+Consider a fluid between two plates a distance, $b$, apart. One is moving with velocity, $\vec{u}$, and the other is fixed in place. The moving plate applies a force, $\vec{p}$ to the fluid. The fluid has continuously changing velocity between the two plates.
+
+The velocity gradient between the two plates is given by:
+$${du \over dy}$$
+
+The shearing stress, $T$, is given by:
+$$\vec{T} = {\vec{p} \over A} = \mu {du \over dy}$$
+
+Where $\mu$ is the absolute viscosity.
+
+### Pressure in a Fluid
+$$p_2 = p_1 + \rho g z$$
+
+##### Pascal's Law
+$${F_1 \over A_1} = {F_2 \over A_2}$$
+
+##### Archemedes Principle
+The upward force of bouyancy, $F_B$, is given by the weight of the displaced fluid.
+
+### Continuity Equation
+$$A_1 \vec{v}_1 \Delta t = A_2 \vec{v}_2 \Delta t$$
+
+Conservation of mass principle applied to a steady flow. The mass flow rate $\dot{m}$ is given by:
+$$\dot{m} = \rho_1 A_1 \vec{v}_1 = \rho_2 A_2 \vec{v}_2$$
+
+Flowrate:
+$$\vec{Q} = A \vec{v}$$
+
+### Bernoulli's Equation
+$$p_1 + {1\over2}\rho v_1^2 + \rho g h_1 = p_2 + {1\over2}\rho v_2^2 + \rho g h_2$$
+
+An increase in speed of a fluid occurs with a decrease of static pressure or decrease in potential energy.

5th-Semester-Fall-2023/ME-For-ECEs/Notes/Gears.md

@@ -0,0 +1,15 @@
+## Circular pitch
+$P_c$ is the arc distance between two adjacent teeth. The pitch diameter is used for calculating the distance between the centers of gears.
+
+Center distance:
+$$c = {d_1 + d_2 \over 2} = {(N_1 + N_2)m \over 2}$$
+
+Diametral pitch:
+$$P = {N \over d}$$
+Where:
+- $N$ is the number of teeth
+
+Module:
+$$m = {d \over N}$$
+
+

5th-Semester-Fall-2023/ME-For-ECEs/Notes/HeatTransfer.md

@@ -0,0 +1,73 @@
+# Heat Transfer
+
+Heat energy in transit due to a temperature difference
+
+1. Conduction - A stationary medium that has a temperature gradient
+1. Convection - heat transfer between a surface and a moving fluid when they are at different temperatures
+1. Thermal radiation - energy emission in the form of electromagnetic waves
+    - radiation between two surfaces at different temperatures
+    - all surfaces emit thermal radiation
+
+### Conduction
+At the atomic and molecular level, energy is transferred from more energetic particles to less energetic ones. This energy transfer always happens in the direction of decreasing temperature. The net transfer of energy by random molecular motion is called diffusion of energy.
+
+In solids, conduction may be due to atomic activity due to vibrations. Materials that are particularly good at heat transfer are considered conductors. These conductors can conduct heat tranfer with lattice wave vibrations and by the translational motion of free elections. Non-conductors can only conduct heat with lattice wave vibrations.
+
+Since the molecules in liquids are closer together, interactions are stronger and more frequent. 
+
+Fourier's Law for Heat Conduction
+$$q_x'' = -k{dT \over dx}$$
+Where:
+
+$q_x''$ is the heat flux, the rate of heat transfer per unit area
+
+$k$ is the thermal conductivity
+
+### Convection
+Two mechanisms:
+1. Diffusion - Random molecular motion
+1. Bulk transfer - Energy transferred by the macroscopic motion of the fluid
+
+Convection between a fluid in motion and a bounding surface. At the surface, the velocity of the particles is zero. This is called the hydrodynamic velocity boundry.
+
+If the temperature of the surface is greater than the temperature of the outer flow, convection heat transfer occurs from the surface to the outer flow.
+
+The thermal boundary layer is the region of a fluid through which the temperature varies from the surface temperature to the outer flow temperature. At the surface, the fluid velocity is zero, so heat transfer can only occur by diffusion.
+
+### Radiation
+Thermal energy radiation is emitted by matter that is at a finite temperature.
+- Can occur from liquids and gases
+- The emission may be attributed to changes in electron configurations of the molecules
+- Electromagnetic waves transport the energy
+- Doesn't require the persence of another material.
+    - Most efficient in a vacuum
+
+### Stefan-Boltzmann Law
+Gives the maximum flux at which radiation may be emitted
+$$q'' = \sigma T_s^4$$
+Where:
+
+$T_s$ is the absolute temperature
+
+$\sigma$ is the Stefan-Boltzmann constant
+
+For a real surface:
+$$q'' = \varepsilon \sigma T_s^4$$
+Where $\varepsilon$ is the emissivity $(0 \le \varepsilon \le 1)$.
+
+If radiation is incident upon a surface, a portion will be absorbed. The rate of absorbtion per unit surface area:
+$$q''_\text{abs} = \alpha q''_\text{inc}$$
+Where $(0 \le \alpha \le 1)$.
+
+Radiation emission decreases the thermal energy of the matter, and radiation absorbtion increases the thermal energy of the matter.
+
+*Special case where we have the net exchange between a small surface and a much larger surface completely surounding it. Assume $\alpha = \varepsilon$.
+$$q'' = {q\over A} = \varepsilon \sigma (T_s^4 - T_\text{sur}^4)$$
+
+The net rate of radiation per unit area.
+
+### Example
+$$q_\text{rad} = h_r A(T_s - T_\text{sur})$$
+$$q_\text{conv} = hA(T_s - T_\infty)$$
+$$q_\text{total} = q_\text{conv} + q_\text{rad}$$
+$$q_\text{total} = hA(T_s - T_\infty) + \varepsilon A \sigma(T_s - T_\text{sur})$$

5th-Semester-Fall-2023/ME-For-ECEs/WriteUp/me-write-up.aux

@@ -0,0 +1,5 @@
+\relax 
+\bibstyle{apacite}
+\@writefile{toc}{\contentsline {section}{Inventing the Zipper}{1}{}\protected@file@percent }
+\@writefile{toc}{\contentsline {section}{How Zippers Zip}{3}{}\protected@file@percent }
+\gdef \@abspage@last{4}

5th-Semester-Fall-2023/ME-For-ECEs/WriteUp/me-write-up.log

@@ -0,0 +1,187 @@
+This is pdfTeX, Version 3.141592653-2.6-1.40.24 (TeX Live 2022/CVE-2023-32700 patched) (preloaded format=pdflatex 2023.11.29)  29 NOV 2023 21:35
+entering extended mode
+ \write18 enabled.
+ %&-line parsing enabled.
+**/tmp/nvim.sharpe/oF9ZGB/0
+(/tmp/nvim.sharpe/oF9ZGB/0
+LaTeX2e <2022-06-01> patch level 5
+L3 programming layer <2022-12-17>
+(/usr/share/texlive/texmf-dist/tex/latex/apa/apa.cls
+Document Class: apa 2008/12/09 American Psychological Association format v1.3.4
+
+
+Class apa Warning: Using BibTeX with apacite for citations and references.
+
+
+Class apa Warning: Using default mode (doc).
+
+(/usr/share/texlive/texmf-dist/tex/latex/base/article.cls
+Document Class: article 2021/10/04 v1.4n Standard LaTeX document class
+(/usr/share/texlive/texmf-dist/tex/latex/base/size11.clo
+File: size11.clo 2021/10/04 v1.4n Standard LaTeX file (size option)
+)
+\c@part=\count185
+\c@section=\count186
+\c@subsection=\count187
+\c@subsubsection=\count188
+\c@paragraph=\count189
+\c@subparagraph=\count190
+\c@figure=\count191
+\c@table=\count192
+\abovecaptionskip=\skip47
+\belowcaptionskip=\skip48
+\bibindent=\dimen140
+)
+(/usr/share/texlive/texmf-dist/tex/latex/apacite/apacite.sty
+Package: apacite 2013/07/21 v6.03 APA citation
+\c@BibCnt=\count193
+\bibleftmargin=\skip49
+\bibindent=\skip50
+\bibparsep=\skip51
+\bibitemsep=\skip52
+\biblabelsep=\skip53
+)
+\b@level@one@skip=\skip54
+\e@level@one@skip=\skip55
+\b@level@two@skip=\skip56
+\e@level@two@skip=\skip57
+\b@level@three@skip=\skip58
+\e@level@three@skip=\skip59
+\b@level@four@skip=\skip60
+\e@level@four@skip=\skip61
+\b@level@five@skip=\skip62
+\e@level@five@skip=\skip63
+\c@APAenum=\count194
+\@text@par@indent=\skip64
+\gr@box=\box51
+\gr@boxwidth=\skip65
+\gr@boxheight=\skip66
+
+(/usr/share/texlive/texmf-dist/tex/latex/tools/bm.sty
+Package: bm 2022/01/05 v1.2f Bold Symbol Support (DPC/FMi)
+\symboldoperators=\mathgroup4
+\symboldletters=\mathgroup5
+\symboldsymbols=\mathgroup6
+Package bm Info: No bold for \OMX/cmex/m/n, using \pmb.
+LaTeX Font Info:    Redeclaring math alphabet \mathbf on input line 149.
+)
+\c@appendix=\count195
+)
+(/usr/share/texlive/texmf-dist/tex/latex/csquotes/csquotes.sty
+Package: csquotes 2022-09-14 v5.2n context-sensitive quotations (JAW)
+
+(/usr/share/texlive/texmf-dist/tex/latex/etoolbox/etoolbox.sty
+Package: etoolbox 2020/10/05 v2.5k e-TeX tools for LaTeX (JAW)
+\etb@tempcnta=\count196
+)
+(/usr/share/texlive/texmf-dist/tex/latex/graphics/keyval.sty
+Package: keyval 2022/05/29 v1.15 key=value parser (DPC)
+\KV@toks@=\toks16
+)
+\csq@reset=\count197
+\csq@gtype=\count198
+\csq@glevel=\count199
+\csq@qlevel=\count266
+\csq@maxlvl=\count267
+\csq@tshold=\count268
+\csq@ltx@everypar=\toks17
+
+(/usr/share/texlive/texmf-dist/tex/latex/csquotes/csquotes.def
+File: csquotes.def 2022-09-14 v5.2n csquotes generic definitions (JAW)
+)
+Package csquotes Info: Trying to load configuration file 'csquotes.cfg'...
+Package csquotes Info: ... configuration file loaded successfully.
+
+(/usr/share/texlive/texmf-dist/tex/latex/csquotes/csquotes.cfg
+File: csquotes.cfg 
+))
+Package csquotes Info: Checking for multilingual support...
+Package csquotes Info: ... none found.
+
+(/usr/share/texlive/texmf-dist/tex/latex/l3backend/l3backend-pdftex.def
+File: l3backend-pdftex.def 2022-10-26 L3 backend support: PDF output (pdfTeX)
+\l__color_backend_stack_int=\count269
+\l__pdf_internal_box=\box52
+)
+(./me-write-up.aux)
+\openout1 = `me-write-up.aux'.
+
+LaTeX Font Info:    Checking defaults for OML/cmm/m/it on input line 10.
+LaTeX Font Info:    ... okay on input line 10.
+LaTeX Font Info:    Checking defaults for OMS/cmsy/m/n on input line 10.
+LaTeX Font Info:    ... okay on input line 10.
+LaTeX Font Info:    Checking defaults for OT1/cmr/m/n on input line 10.
+LaTeX Font Info:    ... okay on input line 10.
+LaTeX Font Info:    Checking defaults for T1/cmr/m/n on input line 10.
+LaTeX Font Info:    ... okay on input line 10.
+LaTeX Font Info:    Checking defaults for TS1/cmr/m/n on input line 10.
+LaTeX Font Info:    ... okay on input line 10.
+LaTeX Font Info:    Checking defaults for OMX/cmex/m/n on input line 10.
+LaTeX Font Info:    ... okay on input line 10.
+LaTeX Font Info:    Checking defaults for U/cmr/m/n on input line 10.
+LaTeX Font Info:    ... okay on input line 10.
+ (/usr/share/texlive/texmf-dist/tex/latex/apa/english.apa
+File: english.apa 2006/01/01
+)
+\c@maskedRefs=\count270
+
+(/usr/share/texlive/texmf-dist/tex/latex/apacite/english.apc
+File: english.apc 2013/07/21 v6.03 apacite language file
+LaTeX Info: Redefining \BPBI on input line 129.
+LaTeX Info: Redefining \BHBI on input line 130.
+)
+
+Class apa Warning: Abstract not defined.
+
+Using title for running head
+LaTeX Font Info:    External font `cmex10' loaded for size
+(Font)              <10.95> on input line 18.
+LaTeX Font Info:    External font `cmex10' loaded for size
+(Font)              <8> on input line 18.
+LaTeX Font Info:    External font `cmex10' loaded for size
+(Font)              <6> on input line 18.
+! Undefined control sequence.
+l.18 ...ritain in the second half of the 18$^\text
+                                                  {th}$ Century, textile pro...
+The control sequence at the end of the top line
+of your error message was never \def'ed. If you have
+misspelled it (e.g., `\hobx'), type `I' and the correct
+spelling (e.g., `I\hbox'). Otherwise just continue,
+and I'll forget about whatever was undefined.
+
+! Undefined control sequence.
+l.18 ... finished products. On September 20$^\text
+                                                  {th}$, 1755, Charles Fredr...
+The control sequence at the end of the top line
+of your error message was never \def'ed. If you have
+misspelled it (e.g., `\hobx'), type `I' and the correct
+spelling (e.g., `I\hbox'). Otherwise just continue,
+and I'll forget about whatever was undefined.
+
+[1
+
+{/var/lib/texmf/fonts/map/pdftex/updmap/pdftex.map}] [2]
+Underfull \vbox (badness 4913) has occurred while \output is active []
+
+ [3]
+[4] (./me-write-up.aux) ) 
+Here is how much of TeX's memory you used:
+ 2330 strings out of 476182
+ 35254 string characters out of 5796580
+ 1854793 words of memory out of 6000000
+ 22666 multiletter control sequences out of 15000+600000
+ 519044 words of font info for 54 fonts, out of 8000000 for 9000
+ 1137 hyphenation exceptions out of 8191
+ 46i,5n,52p,875b,149s stack positions out of 10000i,1000n,20000p,200000b,200000s
+</usr/share/texlive/texmf-dist/fonts/type1/public/amsf
+onts/cm/cmmi8.pfb></usr/share/texlive/texmf-dist/fonts/type1/public/amsfonts/cm
+/cmr10.pfb></usr/share/texlive/texmf-dist/fonts/type1/public/amsfonts/cm/cmr12.
+pfb></usr/share/texlive/texmf-dist/fonts/type1/public/amsfonts/cm/cmr17.pfb></u
+sr/share/texlive/texmf-dist/fonts/type1/public/amsfonts/cm/cmsl10.pfb>
+Output written on me-write-up.pdf (4 pages, 70358 bytes).
+PDF statistics:
+ 42 PDF objects out of 1000 (max. 8388607)
+ 25 compressed objects within 1 object stream
+ 0 named destinations out of 1000 (max. 500000)
+ 1 words of extra memory for PDF output out of 10000 (max. 10000000)
+

5th-Semester-Fall-2023/ME-For-ECEs/WriteUp/me-write-up.tex

@@ -0,0 +1,41 @@
+\documentclass{apa}
+\linespread{2.0}
+
+\usepackage{csquotes}
+
+\author{Aidan Sharpe}
+\affiliation{ME for ECEs}
+\title{Unzipping Zippers}
+
+\begin{document}
+\maketitle
+
+It's late November again, and most days are cold enough to really start bundling up. College students across the country take advantage of their Thanksgiving weekend to make the trek home and pick up their winter coats. As millions of cups of hot chocolate are prepared around the world, people are having another shared experience: stuck zippers. 
+
+While a stuck zipper is likely in the top ten most infuriating experiences everyone has, most of the time the humble zipper works flawlessly. Over the course of the past century and a half, the zipper was invented a couple times, but has not seen much in terms of revision. It is an overwhelmingly popular mechanism with a fairly intricate design and several varieties. Let's explore the zipper's history and unzip how it works.
+
+\section{Inventing the Zipper}
+When the industrial revolution spread across Great Britain in the second half of the 18$^\text{th}$ Century, textile products were, for the first time in human history, able to be produced consistently at scale. Suddenly, there was a large gap between how fast fabric could be produced and shaped and how fast it could be sewed into finished products. On September 20$^\text{th}$, 1755, Charles Fredrick Weisenthal was granted a patent for a needle pointed on both ends so it did not have to be turned around when sewing. This device inspired the first sewing machines invented in 1790 by Thomas Saint.
+
+These machines had a critical problem, however. Rather than weaving back and forth through the fabric, a loops of thread were joined on the underside of the fabric. Called a \enquote{chain stitch}, once the thread broke at any point, every loop made before it would pull out in rapid succession. This made machine stitched fabrics prone to coming apart, limiting the potential of early sewing machines. 
+
+In 1844, English inventor, John Fisher made the chain stitch obsolete by inventing a machine that used a second thread to lock the loops made by the first thread in place. This stitch was given the rather uncreative name, the \enquote{lock stitch}. Several inventors would go on to refine this design. One man from Massachusetts, Elias Howe, invented a variation on the lock stitch sewing machine in which the fabric was held vertically rather than flat on a table.
+
+Just a few years prior, in 1851, Elias Howe had received a patent for \enquote{Improvement in Fastenings for Garments}. This design was the precursor to the zipper, although he never made much of an attempt to market it. The design consisted of sliding clips tied together by a string. The clips were able to slide along the edge of the fabric, and could be bunched up to open the seam or spread apart to close the seam.
+
+Over four decades later, Whitcomb L. Judson invented a mechanism that functioned similar to modern zippers, the only difference being how the two sides connect. The same zipping motion is used, but used to quickly connect hooks to loops rather than using a zipper geometry. This design would evolve over the following decades into the modern zipper that gets its name from the iconic sound it makes.
+
+\section{How Zippers Zip}
+Zippers are devices that exhibit strength in compression, but can be pulled apart at an angle and rejoined at some point in the future. Importantly, zippers are quick to both unzip and rezip. There are three parts that are critical to the function of the zipper: the two sets of teeth, and the slider. Most zippers also include stops at the ends to keep the slider attached to the mechanism, but these components are not critical to how the zipper actually works. They come in several forms, but some of the more common types are metal molded zippers, plastic molded zippers, and coil zippers. 
+
+All three of these designs use identical slider geometry, usually molded from metal or plastic. The role of the slider is to join and separate the two sets of teeth. The slider has two channels at the top and a single, slightly wider channel at the bottom. As the slider travels up the zipper, the two sets of teeth are drawn into the two channels at the top, joined in the middle, and then exit the slider through the single channel at the bottom. Unzipping works in a similar way, but the geometry of the slider has one more trick. When transitioning from the lower channel to the upper two channels, the paths separate gradually, leaving a small spike of material in the middle at the top. This spike serves three purposes: dividing the two channels, pulling apart the zipper, and connecting the front and back of the slider. 
+
+Plastic and metal molded zippers have the same geometry, but are made from different materials. The cast teeth are crimped onto the fabric, and have a special shape such that they mesh together easily at an angle but rigid when the two strips are parallel. These zippers are fairly common on coats, jackets, and jeans.
+
+Coil zippers, on the other hand, are made from a continuous coil of material, usually nylon. The coil wraps around the edge of the zipper and is shaped on the inside to form small bumps on the top and bottom. These bumps interface with the inside of the coil above it, locking the zipper in place.
+These zippers are fairly common on book bags, pencil cases, and suitcases. They are also easily covered by another piece of fabric to create so called \enquote{invisible zippers}.
+
+That begs the question: what causes a stuck zipper? Zippers can get stuck for a whole host of reasons, but perhaps the most common are fabric getting caught under the slider causing it to be trapped between teeth and the teeth being damaged in the first place. While zipper repair does exist, it can be difficult, and usually a damaged zipper leads to the item being discarded.
+
+Zippers are quite universal invention that most people overlook. Although they have only existed in their modern form for about a century, they have undoubtedly become a ubiquitous fastening mechanism for clothing around the world. It is safe to say that the zipper is here to stay.
+\end{document}