second-sem/engg_mech/assignment/t_one.tex

\documentclass{article}
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\begin{document}
\title{Engineering Mechanics}
\author{Ahmad Saalim Lone, 2019BCSE017}
\date{14 May, 2020}
\maketitle
\section*{Question 1}

Triangle law of vector addition states that when two vectors are represented by
two sides of a triangle in magnitude and direction taken in same order then
third side of that triangle represents in magnitude and direction the resultant
of the vectors.

\begin{align*}
	P &= 48 N \\
	Q &= 60 N \\
	R^2 &= P^2 - Q^2 - 2PQ\cos{\ang{150}} \\
	R^2 &= 48^2 + 60^2 - 2\times 48 \times 60 \times (-0.866) \\
	R^2 &= 10892 \\
	R &= 104.36 \\
	\cos\theta &= \frac{{P^2 + R^2 - Q^2}}{2PR} \\
	\theta &= \ang{16.70} \\
	\text{R makes an angle of }(\ang{85} - \ang{16.70}) &= \ang{68.3}
\end{align*}

\section*{Question 2}

\begin{align*}
	\text{For the $A = 80N$ force} \\
	A_x &= 80 \times \cos{\ang{40}} \\
		&= 61.28N \\
	A_y &= 80 \times \sin{\ang{40}} \\
		&= 51.42N \\
\end{align*}

\begin{align*}
	\text{For the $B = 120N$ force} \\
	B_x &= 120 \times \cos{\ang{70}} \\
		&= 41.04N \\
	B_y &= 120 \times \sin{\ang{70}} \\
		&= 112.76N \\
\end{align*}

\begin{align*}
	\text{For the $C = 150N$ force} \\
	C_x &= 150 \times \cos{\ang{165}} \\
		&= -122.87N \\
	C_y &= 150 \times \sin{\ang{165}} \\
		&= 86.036N \\
\end{align*}

\section*{Question 3}

\begin{figure*}[h]
	\centering
	\includegraphics[width=0.5\textwidth]{t_one_1}
\end{figure*}

\begin{align*}
	\cos{\alpha} &= \frac{600}{650} \\
				 &= 0.923 \\
	\sin{\alpha} &= \frac{250}{650} \\
				 &= 0.384
\end{align*}
\begin{align*}
	T_1 + T_2 \sin \alpha + 360 \sin\ang{37} &= 480 N \\
	T_2 \cos \alpha &= 360 \cos \ang{37}
\end{align*}

From here,
\begin{align*}
	T_1 = \text{The tension in cable } BC  &= 143.724 N \\
	T_2 = \text{The tension in cable } AC  &= 311.5 N
\end{align*}

\section*{Question 4}

Draw Free body diagram of the lower pulley.
\begin{figure*}[h]
	\centering
	\includegraphics[width=0.5\textwidth]{t_one_2}
\end{figure*}

\begin{align*}
	\cos{\theta} &= \frac{0.75}{2.514} = 0.3 \\
	\sin{\theta} &= \frac{2.4}{2.514} = 0.954 \\
	2P \cos{\theta} &= P \cos{\alpha} \\
	2P \sin{\theta} + P\sin{\alpha} &= mg \\
	\text{From this we get} \\
	P &= 738.825 N \\
	\alpha &= 53.13N
\end{align*}

\section*{Question 5}

Moments about A
\begin{align*}
	F_1 &= 250 \cos{\ang{30}} \times 2 = 433 Nm \\
	F_2 &= 300 \sin{\ang{60}} \times 5 = 1299.04 Nm \\
	F_3 &= 500 \times 1 Nm
\end{align*}

Moments about B
\begin{align*}
	F_1 &= 0 Nm \\
	F_2 &= 300 \cos{\ang{60}} \times 4 = 600 Nm \\
	F_3 &= 0 Nm
\end{align*}

\section*{Question 6}

\begin{center}
Tension in $AD$ is $481N$ \\
Tension in $AB$ is $T_1N$ \\
Tension in $AC$ is $T_2N$ \\
Length of $AD$ is $6.5m$ \\
Length of $AB$ is $7m$ \\
Length of $AC$ is $7.4m$ \\
\end{center}

On vertical components

\begin{align*}
	P &= 481 \cos{\ang{30.51}} + T_1 \cos{\ang{35.87}} + T_2 \frac{5.6}{7.4} \\
	P &= 414.4 + 0.8 T_1 + 0.75 T_2
\end{align*}

On horizontal components

\begin{align*}
	T_1 \sin{\ang{36.87}} N &= T_2 \frac{4.837}{7.4} \sin{\ang{29.74}}N \\
	0.6 \times T_1 &= 0.324 \times T_2 \\
	481 \sin{\ang{30.51}} N &= T_2 \frac{4.837}{7.4} \cos{\ang{29.74}} N \\
	242.2 &= 0.567 \times T_2 \\
	T_2 &= 430.6 \\
	T_1 &= 232.57 \\
	P &= 414.4 + 0.8 \times 242.57 + 0.75 \times 430.6 \\
	P &= 923.4 N
\end{align*}

\section*{Question 7}
\begin{align*}
	\sum{F_A} &= 0 \\
	T_{AB} + T_{AC} + T_{AD} + P &= 0 \\
								 & \text{where P has only one direction that is $\hat{\imath}$} \\
	\overrightarrow{AB} &= -(960mm)\hat{\imath} - (240mm)\hat{\jmath} + (380mm)\hat{k} \\
	AB &= 1060 mm \\
	\overrightarrow{AC} &= -(960mm)\hat{\imath} - (240mm)\hat{\jmath} - (320mm)\hat{k}\\
	AC &= 1040 mm \\
	\overrightarrow{AD} &= - (960mm)\hat{\imath} + (720mm)\hat{\jmath} - (220mm)\hat{k}\\
	AD &= 1220 mm \\
	\overrightarrow{T_{AB}} &= T_{AB} \cdot \hat{AB} = T_{AB} \left(-\frac{48}{53} \hat{\imath} - \frac{12}{53} \hat{\jmath} + \frac{19}{53} \hat{k} \right) \\
	\overrightarrow{T_{AC}} &= T_{AC} \cdot \hat{AC} = T_{AC} \left(-\frac{12}{13} \hat{\imath} - \frac{3}{13} \hat{\jmath} - \frac{4}{13} \hat{k}\right) \\
	\overrightarrow{T_{AD}} &= T_{AD} \cdot \hat{AD} = \frac{305}{1220} \times \overrightarrow{AD} = (-240 \hat{\imath} + 180 \hat{\jmath} - 55 \hat{k}) N
\end{align*}

We know
\begin{align*}
	\sum F_A &= 0 \\
	-\frac{48}{53}T_{AB} - \frac{12}{13} T_{AC} - 240 + P &= 0 \;\text{(X-axis)} \\
	-\frac{12}{53}T_{AB} -\frac{3}{13} T_{AC} + 180 &= 0  \;\text{(Y-axis)} \\
	-\frac{19}{53}T_{AB} + \frac{4}{13} T_{AC} + 55 &= 0\;\text{(Z-axis)} \\
\end{align*}

By solving these equations, we get

\begin{align*}
	T_{AB} &= 446.71 N \\
	T_{AC} &= 341.71 N \\
	P &= 960 N
\end{align*}

\section*{Question 8}

Calculate the unit vector along each rope just like previous question. \\
Coordinates of point $A, B, C, D$
\begin{align*}
	A &= 4 \hat{k} m \\
	B &= -1.5 \hat{\imath} m - 2 \hat{\jmath} m \\
	C &= 2 \hat{\imath} m + 3 \hat{\jmath} m \\
	D &= 2.5 \hat{\jmath} m
\end{align*}

Position vectors of each rope is $reference = 0 \hat{\imath} + 0 \hat{\jmath} + 6 \hat{k}$
\begin{align*}
	\overrightarrow{r_{AB}} &= -1.5 \hat{\imath} - 6 \hat{k} \\
	|r_{AB}| &= 6.5 \\
	\overrightarrow{r_{AC}} &= 2 \hat{\imath} - 3 \hat{\jmath} - 6 \hat{k} \\
	|r_{AC}| &= 7 \\
	\overrightarrow{r_{AD}} &= 2.5 \hat{\jmath} - 6 \hat{k} \\
	|r_{AD}| &= 6.5
\end{align*}

Now direction of forces are along unit vectors:

\begin{align*}
	\hat{r_{AB}} &= -\frac{1.5}{6.5} \hat{\imath} - \frac{6}{6.5} \hat{k} \\
	\hat{r_{AC}} &= \frac{2}{7} \hat{\imath} - \frac{3}{7} \hat{\jmath} - \frac{6}{7} \hat{k} \\
	\hat{r_{AD}} &= \frac{2.5}{6.5} \hat{\jmath} - \frac{6}{7} \hat{k}
\end{align*}

\[
	\sum F_A = 0
\]
So,

\begin{align*}
	-0.231 F_{AB} + 0.286 F_{AC} &= 0 \\
	-0.308 F_{AB} - 0.429 F_{AC} + 0.385 F_{AD} &= 0 \\
	-0.923 F_{AB} - 0.857 F_{AC} - 0.923 F_{AD} &= -800
\end{align*}

On solving these three equations, we get

\begin{align*}
	F_{AB} &= 251.2N \\
	F_{AC} &= 202.9N \\
	F_{AD} &= 427.1 N
\end{align*}

\end{document}