Add assignment three and four

elements_of_mech/assignment-work-and-heat/assign2.tex

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+\documentclass{article}
+\usepackage{amsmath}
+\usepackage{amssymb}
+\begin{document}
+\title{Assignment --- First Law of Thermodynamics}
+\author{Ahmad Saalim Lone, 2019BCSE017}
+\date{}
+\maketitle
+\section*{Question 1}
+First law of thermodynamics suggests that $\sum Q = \sum W$.
+\begin{align*}
+	Q_1 + Q_2 + Q_3 &= W_1 + W_2 \\
+	75 - 40 + Q_3 &= -15 + 44 \\
+	Q_3 &= -6 kJ
+\end{align*}
+i.e.\ from the system to the surroundings.
+\section*{Question 2}
+We know that,
+\begin{align*}
+	Q &= \Delta E + W \\
+	Q &= -50000 + \frac{-1000 \times 3600}{1000} kJ \\
+	Q &= -8600 kJ \\
+	Q &= -8.6 MJ
+\end{align*}
+\section*{Question 3}
+There is no heat transfer, therefore
+\begin{align*}
+	Q &=  \Delta E + W \\
+	W &= - \Delta E - \Delta V \\
+	W &= \int_{1}^{0} C_v \cdot dT \\
+	W &= -0.718 (T_2 - T_1) \\
+	W &= -50.26 \frac{kJ}{kg} \\
+	\text{Total Work} &= 2 \times (-50.26) = -100 kJ
+\end{align*}
+\section*{Question 4}
+\begin{align*}
+	1000 &= a + b \times 0.2 \\
+	200 &= a + b \times 1.2 \\
+	b &= -800 \\
+	a &= 1000 + 2 \times 800 = 1160 \\
+	\therefore P &= 1160 - 800 V \\
+	W &= \int_{V_1}^{V_2} P \cdot dV \\
+	  &= \int_{0.2}^{1.2} (1160 - 800V)dV \\
+	  &= 6000 kJ \\
+	V_1 &= 1.5 \times  1000 \times \frac{0.2}{1.5} - 85 \\
+		&= 215 \frac{kJ}{kg} \\
+	V_2 &= 1.5 \times 200 \times \frac{1.2}{1.5} - 85 \\
+		&= 155 \frac{kJ}{kg} \\
+	\Delta V &= V_2 - V_1 \\
+			 &= 40 \frac{kJ}{kg} \\
+	\Delta U &= m \Delta V  = 60 kJ \\
+	Q &= \Delta U + W \\
+	  &= 60 + 600 \\
+	  &= 660 kJ \\
+	U &= 1.5 PV - 85 \frac{kJ}{kg} \\
+	U &= 1.5 \left(\frac{1160 - 800 V}{1.5}\right)V - 85 \frac{kJ}{kg} \\
+	  &= 800 V^2 - 85 \frac{kJ}{kg} \\
+	\frac{\delta U}{\delta V} &= 1160 - 1600 V \\
+	\text{For maximum V,} \\
+	V_1 \rightarrow \frac{\delta U}{\delta V} &= 0 \\
+	V &= 0.725 \\
+	u_{\max} = 335.5 \frac{kJ}{kg} \\
+\end{align*}
+\begin{align*}
+	U_{\max} &= 1.5 \times u_{\max} = 503.25 kJ \\
+\end{align*}
+\section*{Question 5}
+\subsection*{Part a}
+\begin{align*}
+	Q &= \int_{273}^{373} C_p \cdot dT \\
+	t &= T - 273 K \\
+	\therefore t + 100 &= T - 173 \\
+	Q &= \int_{273}^{373} \left(2.093 + \frac{41.87}{T - 173}\right) \cdot dT \\
+	Q &= 238.32 J
+\end{align*}
+\subsection*{Part b}
+\begin{align*}
+	Q &= \Delta E + \int p\cdot dV \\
+	\Delta E &= Q - p(V_2 - V_1) \\
+	\Delta E &= 238.32 - 101.325 (0.0024 - 0.002) \times 1000 J \\
+	\Delta E &= 197.79J
+\end{align*}
+\end{document}

elements_of_mech/assignment-work-and-heat/assign3.tex

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+\documentclass{article}
+\usepackage{amsmath}
+\usepackage{amssymb}
+\begin{document}
+\title{Assignment --- Second Law of Thermodynamics}
+\author{Ahmad Saalim Lone, 2019BCSE017}
+\date{}
+\maketitle
+\section*{Question 1}
+We know that the net power output is the difference between the heat input and the heat rejected (cyclic device).
+\begin{align*}
+	W_{net,out} &= Q_H + Q_L \\
+	W_{out} &= 90 - 55 = 35 MW
+\end{align*}
+The thermal efficiency is the ratio of net work output and the heat output.
+\[
+	\eta_{thermal} = W_{out}\frac{W_{out}}{Q_H} =  \frac{35}{90} = 0.3889
+\]
+\section*{Question 2}
+\begin{align*}
+	\text{efficiency} = 1 - \frac{T_1}{T_2}  &= \frac{\text{work input}}{\text{work output}} \;\;\text{(must)}\\
+	1 - \frac{300}{1000} &\implies \frac{6}{1} \;\;\text{(claimed)} \\
+	0.7 &> 0.6 \\
+	\text{output} &> \text{claimed $\implies$ good}
+\end{align*}
+$\therefore$ we can agree to this claim.
+\section*{Question 3}
+\[
+	\eta = \frac{\text{output}}{\text{input}} = \frac{0.65}{0.65 + 0.4} = 0.619 = 61.9%
+\]
+now,
+\begin{align*}
+	\eta &= 1 - \frac{T_2}{T_1} \\
+	T_1 &= \frac{T_2}{1 - \eta} \\
+	T_1 &= 787.5 K
+\end{align*}
+$\therefore$, the temperature at which energy is absorbed is 787.5 K
+\section*{Question 4}
+\begin{align*}
+	\eta &= 1 - \frac{T_1}{T_2} \\
+\end{align*}
+where,
+\begin{align*}
+	\eta &= 0.6 \\
+	T_2 &= 800 K \\
+	T_1 &= T_{sink} \\
+	T_{sink} &= (1 - \eta)T_2 \\
+			  &= 324K
+\end{align*}
+Also,
+\begin{align*}
+	\eta &=  \frac{Q_1 - Q_{\text{rejected}}}{Q_1} \\
+	Q_{\text{rejected}} &= Q_1 (1 - \eta) \\
+							 &= 400(1 - 0.6) \\
+							 &= 160 kJ
+\end{align*}
+\section*{Question 5}
+Max COP will be achieved only in a Carnot refrigerator.
+\[
+	{(COP_{\max})}_{R} = \frac{Q_2}{Q_1 - Q_2} = \frac{T_2}{T_1 - T_2} = \frac{-20 + 273}{57} = 4.438
+\]
+Minimum Power $\to$ Max COP\@. We know that $P = \frac{Q}{\eta}$.
+\[
+	P = \frac{3 \times 57}{253} = 0.657 W
+\]
+\section*{Question 6}
+Max COP will be achieved only in a Carnot refrigerator.
+\begin{align*}
+	{(COP)}_{\text{Carnot refrigerator}} &= \frac{T_{low}}{T_{high} - T_{low}} = 1.37 \times 10^{-2} \\
+	{(COP)}_{\text{refrigerator}} &= \frac{Q_L}{W} = 1.37 \times 10^{-2} \\
+	Q_L &= 83.3 \\
+\end{align*}
+\section*{Question 7}
+For process $1-2$
+\begin{align*}
+	Q_{1-2} &= U_2 - U_1 + W_{1-2} \\
+	-732 &= U_2 - U_1 - 2.8 \times 3600 \\
+	U_2 - U_1 &= 9348kJ
+\end{align*}
+For process $2-1$
+\begin{align*}
+	Q_{2-1} &= U_1 - U_2 + W_{2-1} \\
+	-732 &= -9348 - 2.8 \times 3600 \\
+	Q_{2-1} &= -708 kJ
+\end{align*}
+Now,
+\[
+	Q_{2-1} = U_1 - U_2 + W_{2-1}
+\]
+For maximum work, $Q_{2-1} = 0$.
+\[
+	\therefore {(W_{2-1})}_{\max} = U_2 - U_1 = 9348kJ
+\]
+\section*{Question 8}
+\begin{align*}
+	{(COP)}_R &= \frac{268}{298 - 268} = 8.933 \\
+	{(COP)}_R &= \frac{5}{W} \\
+	W &= 0.56KW
+\end{align*}
+\section*{Question 9}
+\begin{align*}
+	\eta &= 1 - \frac{273+60}{273+671} = 0.64725 \\
+	\eta &= 0.3236 \\
+	\implies 1 - 0.3236 &= 0.6764 \\
+	\text{Ideal COP} &= \frac{305.2}{305.2 - 266.4} = 7.866 \\
+	\text{Actual COP} &= 3.923 = \frac{Q_3}{W} \\
+	if \;Q_3 &= 1kJ \\
+	\therefore W &= \frac{Q_3}{3.923} = 0.2549kJ \\
+	W &= Q_{\text{output}} - Q_{\text{input}} = \frac{1}{3.923} Q_1 - 0.6764Q_1 = 0.2549 \\
+	Q_{\text{out}} &= \frac{0.2549}{1 - 0.6764} = 0.7877 kJ \\
+\end{align*}
+\section*{Question 10}
+\begin{align*}
+	\frac{10}{W} &= \frac{293}{293-273} \\
+	or\;W &= 683 W
+\end{align*}
+\end{document}

engg_mech/assignment/t_four.tex

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+\documentclass{article}
+\usepackage{amsmath}
+\usepackage{amssymb}
+\usepackage{siunitx}
+\begin{document}
+\title{Engineering Mechanics}
+\author{Ahmad Saalim Lone, 2019BCSE017}
+\date{}
+\maketitle
+\section*{Question 1}
+Calculate Reactions:
+\begin{align*}
+	\sum M_J &= 0 \\
+	(12 kN)(4.8) + (12 kN)(2.4) - B(9.6) &= 0 \\
+	B &= 9 kN
+\end{align*}
+\begin{align*}
+	\sum F_y &= 0 \\
+	9 kN - 12 kN - 12 kN + J &= 0 \\
+	J &= 15kN
+\end{align*}
+Member CD:\@
+\begin{align*}
+	\sum F_y &= 0 \\
+	9 kN + F_{CD} &= 0 \\
+	F_{CD} &= 9 kN \to \text{compression}
+\end{align*}
+Member DF:\@
+\begin{align*}
+	\sum M_c &= 0 \\
+	F_{DF}1.8 m - 9kN \times 2.4m &= 0 \\
+	F_{DF} &= 12kN  \to \text{Tension}
+\end{align*}
+\section*{Question 2}
+Reactions:\@
+\[
+	A = N = 0
+\]
+DF member:\@
+\begin{align*}
+	\sum M_E &= 0 \\
+	(16 kN)(6m) - \frac{3}{5} F_{DF} (4m) &= 0 \\
+	F_{DF} &= 40 kN \to \text{Tension}
+\end{align*}
+
+EF member:\@
+\begin{align*}
+	\sum F&= 0 \\
+	16 kN \sin{\beta} - F_{EF} \cos{\beta} &= 0 \\
+	F_{EF} &= 16 \tan{\beta} \\
+		   &= 12 kN \to \text{Tension}
+\end{align*}
+
+EG member:\@
+\begin{align*}
+	\sum M_F &= 0 \\
+	16kN \times 9m + \frac{4}{5}F_{EG} \times 3m &= 0 \\
+	F_{EG} &= -60 kN \to \text{Compression}
+\end{align*}
+\section*{Question 3}
+Reactions
+\begin{align*}
+	\sum M_k &= 0 \\
+	36\times 2.4 - B \times 13.5 + 20 \times 9 + 20 \times 4.5 &= 0 \\
+	B &= 26.4kN \\
+\end{align*}
+\begin{align*}
+	\sum F_x &= 0 \\
+	K_x &= 36 \\
+\end{align*}
+\begin{align*}
+	\sum F_y &= 0 \\
+	26.4 - 20 -20 + K_y &= 0 \\
+	K_y &=  13.6 kN \uparrow \\
+\end{align*}
+\begin{align*}
+	\sum M_C &= 0 \\
+	36 \times 1.2 - 26.4 \times 2.25 - F_{AD} \times 1.2 &= 0 \\
+	F_{AD} &= 13.5 kN \to \text{compression} \\
+\end{align*}
+\begin{align*}
+	\sum M_A &= 0 \\
+	\left( \frac{8}{17}F_{CD}\right)(4.5) &= 0 \\
+	F_{CD} &= 0 \\
+\end{align*}
+\begin{align*}
+	\sum M_D &= 9 \\
+	\frac{15}{17} \times F_{CE} \times 2.4 - 26.4 \times 4.5 &= 0 \\
+	F_{CE} &= 56.1 kN \to \text{Tension}
+\end{align*}
+\section*{Question 4}
+
+Support reactions
+\begin{align*}
+	\sum M_I &= 0 \\
+	2\times 12 + 5\times 8 3\times 6 + 2\times 4 - A_y \times 16 &= 0 \\
+	A_y &= 5.625 kN
+\end{align*}
+\begin{align*}
+	\sum A_x &= 0 \\
+	A_x &= 0
+\end{align*}
+Method of joints: By inspection, members BN, NC, DO, OC, HJ, LE \& JG are zero force members \\
+Method of sections:
+\begin{align*}
+	\sum M_M &= 0 \\
+	4F_{CD} - 5.625 \times 4 &= 0 \\
+	F_{CD} &= 5.625 kN \to \text{Tension}
+\end{align*}
+\begin{align*}
+	\sum M_A &= 0 \\
+	4F_{CM} - 2\times 4 &= 0 \\
+	F_{CM} &= 2 kN \to \text{Tension}
+\end{align*}
+\end{document}

engg_mech/assignment/t_three.tex

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+\documentclass{article}
+\usepackage{amsmath}
+\usepackage{amssymb}
+\usepackage{siunitx}
+\begin{document}
+\title{Engineering Mechanics}
+\author{Ahmad Saalim Lone, 2019BCSE017}
+\date{}
+\maketitle
+\section*{Question 1}
+Applying the equations of the equilibrium to the FBD of the entire truss, we have
+\begin{align*}
+	\sum M_A &= 0 \\
+	N_C (2 + 2) - 4(2) - 3(1.5) &= 0 \\
+	N_C &= 3.125 kN
+\end{align*}
+\begin{align*}
+	\sum F_x &= 0 \\
+	3 - A_x &= 0 \\
+	A_x &= 3 kN
+\end{align*}
+\begin{align*}
+	\sum F_y &= 0 \\
+	A_y + 3.125 - 4 &= 0 \\
+	A_y &= 0.875 kN
+\end{align*}
+Method of joints \\
+Joint C:\@ Just assume it to be in equilibrium
+\begin{align*}
+	\sum F_y &= 0 \\
+	3.125 - F_{CD} \frac{3}{5} &= 0 \\
+	F_{CD} &= 5.21 kN  \to \text{Compression}
+\end{align*}
+\begin{align*}
+	\sum F_x &= 0 \\
+	5.208 \times \frac{4}{5} - F_{CB} &= 0 \\
+	F_{CB} &= 4.17kN  \to \text{Tension}
+\end{align*}
+Joint A:\@
+\begin{align*}
+	\sum F_y &= 0 \\
+	0.875 - F_{AD} \times \frac{3}{5} &= 0 \\
+	F_{AD} &= 1.46 kN  \to \text{Compression}
+\end{align*}
+
+\begin{align*}
+	\sum F_x &= 0 \\
+	F_{AB} - 3 - 1.458 \times \frac{4}{5} &= 0 \\
+	F_{AB} &= 4.167 kN  \to \text{Tension}
+\end{align*}
+Joint B
+\begin{align*}
+	\sum F_y &= 0 \\
+	F_{BD} &= 4 kN
+\end{align*}
+\begin{align*}
+	\sum F_x &= 0 \\
+	4.167 - 4.167 &= 0
+\end{align*}
+\section*{Question 2}
+Analyze equilibrium of joint D, C \& E.
+Joint D
+\begin{align*}
+	\sum F_x &= 0 \\
+	F_{DE} \times \frac{3}{5} - 600 &= 0 \\
+	F_{DE} &= 1 kN  \to \text{Compression}
+\end{align*}
+\begin{align*}
+	\sum F_y &= 0 \\
+	1000 \times \frac{4}{5} - F_{DC} &= 0 \\
+	F_{DC} &= 800 N  \to \text{Tension}
+\end{align*}
+Joint C
+\begin{align*}
+	\sum F_x &= 0 \\
+	F_{CE} &= 900 N  \to \text{Compression}
+\end{align*}
+\begin{align*}
+	\sum F_y &= 0 \\
+	F_{CB} &= 800 N  \to \text{Tension}
+\end{align*}
+Joint E
+\begin{align*}
+	\sum F_x &= 0 \\
+	F_{EB} &= 750 N  \to \text{Tension}
+\end{align*}
+\begin{align*}
+	\sum F_y &= 0 \\
+	F_{BA} &= 1.75 kN  \to \text{Compression}
+\end{align*}
+\section*{Question 3}
+Joint A
+\begin{align*}
+	\sum F_y &= 0 \\
+	F_{AL} &= 28.28 kN  \to \text{Compression}
+\end{align*}
+\begin{align*}
+	\sum F_x &= 0 \\
+	F_{AB} &= 20 kN  \to \text{Tension}
+\end{align*}
+Joint B
+\begin{align*}
+	\sum F_x &= 0 \\
+	F_{BC} &= 20 kN  \to \text{Tension}
+\end{align*}
+\begin{align*}
+	\sum F_y &= 0 \\
+	F_{BL} &= 0
+\end{align*}
+Joint L
+\begin{align*}
+	\sum F_x &= 0 \\
+	F_{LC} &= 0
+\end{align*}
+\begin{align*}
+	\sum F_y &= 0 \\
+	F_{LK} &= 28.28 kN \to \text{Compression}
+\end{align*}
+Joint C
+\begin{align*}
+	\sum F_x &= 0 \\
+	F_{CD} &= 20 kN  \to \text{Tension}
+\end{align*}
+\begin{align*}
+	\sum F_y &= 0 \\
+	F_{CK} &= 10 kN  \to \text{Tension}
+\end{align*}
+Joint K
+\begin{align*}
+	\sum F_x &= 0 \\
+	F_{KD} &= 7.454 kN
+\end{align*}
+\begin{align*}
+	\sum F_y &= 0 \\
+	F_{KJ} &= 23.57 kN  \to \text{Compression}
+\end{align*}
+Joint J
+\begin{align*}
+	\sum F_x &= 0 \\
+	F_{IJ} &= 23.57 kN
+\end{align*}
+\begin{align*}
+	\sum F_y &= 0 \\
+	F_{JD} &= 33.3 kN  \to \text{Tension}
+\end{align*}
+Now we know that there exists symmetry,
+\begin{gather*}
+	F_{AL} = F_{GH} = F{LK} = F{HI} = 28.3 kN \\
+	F_{AB} = F_{GF} = F_{BC} = F_{FE} = F_{CD} = F_{ED} = 20 kN \\
+	F_{BL} = F_{FH} = F_{LC} = F_{HE} = 0 \\
+	F_{CK} = F_{EI} = 10 kN \\
+	F_{KJ} = F_{IJ} = 23.6 kN \\
+	F_{KD} = F_{ID} = 7.45 kN
+\end{gather*}
+\section*{Question 4}
+To evaluate support reactions
+\begin{align*}
+	\sum M_E &= 0 \\
+	A_y &= \frac{4}{3} P
+\end{align*}
+\begin{align*}
+	\sum F_y &= 0 \\
+	E_y &= \frac{4}{3} P
+\end{align*}
+\begin{align*}
+	\sum F_x &= 0 \\
+	E_x &= P
+\end{align*}
+Methods of joints: By inspecting joint C, members CB \& CD are zero force members. Hence
+\[
+	F_{CB} = F_{CD} = 0
+\]
+Joint A
+\begin{align*}
+	\sum F_y &= 0 \\
+	F_{AB} &= 2.40 P  \to \text{Compression}
+\end{align*}
+\begin{align*}
+	\sum F_x &= 0 \\
+	F_{AF} &= 2P  \to \text{Tension}
+\end{align*}
+Joint B
+\begin{align*}
+	\sum F_x &= 0 \\
+	2.404P \times \frac{1.5}{\sqrt{3.25}} - P - F_{BF} \times \frac{0.5}{\sqrt{1.25}} - F_{BD} \times \frac{0.5}{\sqrt{1.25}} &= 0 \\
+	P - 0.447 F_{BF} - 0.447 F_{BD} &= 0
+\end{align*}
+\begin{align*}
+	\sum F_y &= 0 \\
+	2.404P \times \frac{1}{\sqrt{3.25}} - F_{BF} \times \frac{1}{\sqrt{1.25}} + F_{BD} \times \frac{1}{\sqrt{1.25}} &= 0 \\
+	1.33P - 0.8944 F_{BF} + 0.8944 F_{BD} &= 0
+\end{align*}
+Solving the above equations, we get
+\begin{align*}
+	F_{BD} &= 0.3727 P  \to \text{Compression}\\
+	F_{BF} &= 1.863 P  \to \text{Tension}
+\end{align*}
+Joint D
+\begin{align*}
+	\sum F_y &= 0 \\
+	F_{DE} &= 0.3727 P  \to \text{Compression}
+\end{align*}
+\section*{Question 5}
+FBD of Joint A
+\begin{gather*}
+	\frac{F_{AB}}{2.29} = \frac{F_{AC}}{2.29} = \frac{1.2}{1.2} kN \\
+	F_{AB} = 2.29 kN  \to \text{Tension} \\
+	F_{AC} = 2.29 kN  \to \text{Compression}
+\end{gather*}
+FBD of Joint F
+\begin{gather*}
+	\frac{F_{DF}}{2.29} = \frac{F_{EF}}{2.29} = \frac{1.2}{1.2} kN \\
+	F_{DF} = 2.29 kN  \to \text{Tension} \\
+	F_{EF} = 2.29 kN  \to \text{Compression}
+\end{gather*}
+FBD of Joint D
+\begin{gather*}
+	\frac{F_{BD}}{2.21} = \frac{F_{DE}}{0.6} = \frac{2.29}{2.29} kN \\
+	F_{DE} = 0.6 kN  \to \text{Compression}\\
+	F_{EF} = 2.21 kN \to \text{Tension}
+\end{gather*}
+FBD of Joint C
+\begin{align*}
+	\sum F_x &= 0 \\
+	F_{CE} &= 2.21 kN  \to \text{Compression}
+\end{align*}
+\begin{align*}
+	\sum F_y &= 0 \\
+	F_{CH} &= 1.2 kN  \to \text{Compression}
+\end{align*}
+FBD of Joint E
+\begin{align*}
+	\sum F_x &= 0 \\
+	F_{BH} &= 0
+\end{align*}
+\begin{align*}
+	\sum F_y &= 0 \\
+	F_{EJ} &= 1.2 kN  \to \text{Compression}
+\end{align*}
+\section*{Question 6}
+Zero Force Members
+Analyzing joint F:\@ Note that $DF$ and $EF$ are zero force members.
+\[
+	F_{DF} = F_{EF} = 0
+\]
+Analyzing joint D:\@ Note that $BD$ and $DE$ are zero force members.
+\[
+	F_{BD} = F_{DE} = 0
+\]
+FBD of joint A
+\begin{gather*}
+	\frac{F_{AB}}{2.29} = \frac{F_{AC}}{2.29} = \frac{1.2}{1.2} kN \\
+	F_{AB} = 2.29 kN  \to \text{Tension}\\
+	F_{AC} = 2.29 kN  \to \text{Compression}
+\end{gather*}
+FBD of joint B
+\begin{align*}
+	\sum F_x &= 0 \\
+	F_{BE} &= 2.7625 kN  \to \text{Tension}
+\end{align*}
+\begin{align*}
+	\sum F_y &= 0 \\
+	F_{BC} &= 2.25 kN  \to \text{Compression}
+\end{align*}
+FBD of joint C
+\begin{align*}
+	\sum F_x &= 0 \\
+	F_{CE} &= 2.21 kN \to \text{Compression}
+\end{align*}
+\begin{align*}
+	\sum F_y &= 0 \\
+	F_{CH} &= 2.86 kN  \to \text{Compression}
+\end{align*}
+FBD of joint E
+\begin{align*}
+	\sum F_x &= 0 \\
+	F_{EH} &= 0
+\end{align*}
+\begin{align*}
+	\sum F_y &= 0 \\
+	f_{EJ} &= 1.657 kN  \to \text{Tension}
+\end{align*}
+
+\end{document}

engg_mech/assignment/t_two.tex

@@ -0,0 +1,135 @@
+\documentclass{article}
+\usepackage{amsmath}
+\usepackage{amssymb}
+\usepackage{siunitx}
+\usepackage{graphicx}
+\usepackage{wrapfig}
+\graphicspath{{./images/}}
+\begin{document}
+\title{Engineering Mechanics}
+\author{Ahmad Saalim Lone, 2019BCSE017}
+\date{17 May, 2020}
+\maketitle
+\section*{Question 1}
+\subsection*{Question 1.a}
+We shall balance the torque about $C$. $\angle ABC = \theta$, Tension in cable $=T$.
+\begin{align*}
+	240 (0.4) + 240 (0.8) &= T\sin{\theta} \times 0.18 \\
+	T\sin{\theta} &= 1600 \\
+	T \frac{0.24}{0.3} &= 1600 \\
+	T &= 2000
+\end{align*}
+\subsection*{Question 1.b}
+On making FBD of bracket BCD.\@
+
+\begin{align*}
+	x-component &= N_x = T\sin{\theta} = 1600 \\
+	y-component &= N_y = T\cos{\theta} + 240 +240 = 1680
+\end{align*}
+
+\section*{Question 2}
+\subsection*{Question 2.a}
+We shall balance the torque about C
+\begin{align*}
+	P \times 7.5 &= T \times 5 \\
+	T &= 150 lb
+\end{align*}
+\subsection*{Question 2.b}
+\begin{align*}
+	N_x \text{(reaction at C along x-axis)} &= - (P + T\sin{\ang{37}}) = -190 lb \\
+	N_y \text{(reaction at C along y-axis)} &= - T \cos{\ang{37}} = -120 lb
+\end{align*}
+\section*{Question 3}
+\subsection*{Question 3.a}
+\[
+	\alpha = 0
+\]
+Balance torque about B
+\begin{align*}
+	N_a \times 20 &= 75 \times 10 \\
+	N_a &= 37.5 lb
+\end{align*}
+Balance torque about A
+\begin{align*}
+	N_b \times 20 &= 75 \times 10 \\
+	N_b &= 37.5 lb
+\end{align*}
+\subsection*{Question 3.b}
+\[
+	\alpha = \ang{90}
+\]
+Balance torque about A
+\begin{align*}
+	N_b \times 20 &= 75 \times 10 \\
+	N_b &= 37.5 lb
+\end{align*}
+Balance torque about B
+\begin{align*}
+	N_a \times 12 &= 75 \times 10 \\
+	N_a &= 62.5 lb
+\end{align*}
+\subsection*{Question 3.c}
+\[
+	\alpha = \ang{30}
+\]
+Balance torque about the mid point of horizontal rod
+\begin{align*}
+	N_a \times 10 &= {(N_b)}_y \times 10 \\
+	N_a &= {(N_b)}_y
+\end{align*}
+Balance torque about B
+\begin{align*}
+	N_a \times 20 &= 75 \times 10 \\
+	N_a &= 37.5 lb \\\\
+	N_a &= N_b \cos{\ang{30}} \\
+	N_b &= 43.30
+\end{align*}
+
+\section*{Question 4}
+\subsection*{Question 4.a}
+Balance torque about C
+\begin{align*}
+	120 \times 0.28 &= T \times  \frac{150}{250} \times 0.2 + T \times \frac{150}{390} \times 0.36 \\
+	33.6 &= T \times 0.26 \\
+	T &= 129.2 N \approx 130 N
+\end{align*}
+\subsection*{Question 4.b}
+\begin{align*}
+	{(N_c)}_x &= T \times \frac{200}{250} + T \times \frac{360}{390} \\
+	{(N_c)}_x &= 223 N \\
+	{(N_c)}_y &= 120 - \left(T \times  \frac{150}{250} + T \times  \frac{150}{390}\right) \\
+	{(N_c)}_y &= -7.21 N \\
+	N_c &= \sqrt{223^2 + 7.21^2} N \\
+	N_c &= 224 N
+\end{align*}
+
+\section*{Question 5}
+Balance torque about B
+\begin{align*}
+	{(N_a)}_y \times 8 &= 4000 \times 2 \\
+	{(N_a)}_y &= 1000 N
+\end{align*}
+FBD of Rod
+\begin{align*}
+	4000 &= {(N_A)}_y + {(N_B)}_y \\
+	4000 &= 1000 + {(N_B)}_y \\
+	{(N_B)}_y &= 3000 \\\\
+	{(N_B)}_y &= N_b \sin{\ang{60}} \\
+	N_B &= 3465 \\
+	{(N_A)}_x &= {(N_B)}_x \\
+	{(N_A)}_x &= N_B \sin{\ang{30}} \\
+	{(N_A)}_x &= 1732
+\end{align*}
+\section*{Question 6}
+First we have to calculate reaction at A
+\begin{align*}
+	\sum F_x &= 0 \\
+	4000 \cos{\ang{30}} &= A_x \\
+	A_x &= 3464 N
+\end{align*}
+\begin{align*}
+	\sum F_y &= 0 \\
+	6000 + 4000 \cos{\ang{30}} &= A_y \\
+	A_y &= 8000 N
+\end{align*}
+\end{document}

maths/assignment-matrices/assign4.tex

@@ -595,7 +595,7 @@ Doing transformations to form row echelon form, we get
 
 \section*{Question 15}
 \subsection*{Part i}
-Since $A + A^{-1} = 0$, $A$ must either be skew symmetric. If A is skew symmetric, we know that the rank of an odd order skew symmetric matrix must be even. $\therefore Rank \leq 2020$
+Since $A + A^T = 0$, $A$ must either be skew symmetric. If A is skew symmetric, we know that the rank of an odd order skew symmetric matrix must be even. $\therefore Rank \leq 2020$
 \subsection*{Part ii}
 Inverse does not exist as $A$ is singular matrix.
 \end{document}