Add assignment 2 engg_mech

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}