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\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}
