Add mechanical engineering project.
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engg_mech/assignment/t_one.tex
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engg_mech/assignment/t_one.tex
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\documentclass{article}
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\usepackage{amsmath}
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\usepackage{amssymb}
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\usepackage{siunitx}
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\usepackage{graphicx}
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\usepackage{wrapfig}
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\graphicspath{{./images/}}
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\begin{document}
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\title{Engineering Mechanics}
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\author{Ahmad Saalim Lone, 2019BCSE017}
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\date{14 May, 2020}
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\maketitle
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\section*{Question 1}
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Triangle law of vector addition states that when two vectors are represented by
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two sides of a triangle in magnitude and direction taken in same order then
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third side of that triangle represents in magnitude and direction the resultant
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of the vectors.
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\begin{align*}
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P &= 48 N \\
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Q &= 60 N \\
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R^2 &= P^2 - Q^2 - 2PQ\cos{\ang{150}} \\
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R^2 &= 48^2 + 60^2 - 2\times 48 \times 60 \times (-0.866) \\
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R^2 &= 10892 \\
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R &= 104.36 \\
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\cos\theta &= \frac{{P^2 + R^2 - Q^2}}{2PR} \\
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\theta &= \ang{16.70} \\
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\text{R makes an angle of }(\ang{85} - \ang{16.70}) &= \ang{68.3}
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\end{align*}
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\section*{Question 2}
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\begin{align*}
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\text{For the $A = 80N$ force} \\
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A_x &= 80 \times \cos{\ang{40}} \\
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&= 61.28N \\
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A_y &= 80 \times \sin{\ang{40}} \\
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&= 51.42N \\
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\end{align*}
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\begin{align*}
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\text{For the $B = 120N$ force} \\
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B_x &= 120 \times \cos{\ang{70}} \\
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&= 41.04N \\
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B_y &= 120 \times \sin{\ang{70}} \\
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&= 112.76N \\
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\end{align*}
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\begin{align*}
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\text{For the $C = 150N$ force} \\
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C_x &= 150 \times \cos{\ang{165}} \\
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&= -122.87N \\
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C_y &= 150 \times \sin{\ang{165}} \\
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&= 86.036N \\
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\end{align*}
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\section*{Question 3}
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\begin{figure*}[h]
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\centering
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\includegraphics[width=0.5\textwidth]{t_one_1}
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\end{figure*}
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\begin{align*}
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\cos{\alpha} &= \frac{600}{650} \\
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&= 0.923 \\
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\sin{\alpha} &= \frac{250}{650} \\
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&= 0.384
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\end{align*}
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\begin{align*}
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T_1 + T_2 \sin \alpha + 360 \sin\ang{37} &= 480 N \\
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T_2 \cos \alpha &= 360 \cos \ang{37}
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\end{align*}
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From here,
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\begin{align*}
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T_1 = \text{The tension in cable } BC &= 143.724 N \\
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T_2 = \text{The tension in cable } AC &= 311.5 N
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\end{align*}
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\section*{Question 4}
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Draw Free body diagram of the lower pulley.
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\begin{figure*}[h]
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\centering
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\includegraphics[width=0.5\textwidth]{t_one_2}
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\end{figure*}
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\begin{align*}
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\cos{\theta} &= \frac{0.75}{2.514} = 0.3 \\
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\sin{\theta} &= \frac{2.4}{2.514} = 0.954 \\
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2P \cos{\theta} &= P \cos{\alpha} \\
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2P \sin{\theta} + P\sin{\alpha} &= mg \\
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\text{From this we get} \\
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P &= 738.825 N \\
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\alpha &= 53.13N
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\end{align*}
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\section*{Question 5}
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Moments about A
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\begin{align*}
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F_1 &= 250 \cos{\ang{30}} \times 2 = 433 Nm \\
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F_2 &= 300 \sin{\ang{60}} \times 5 = 1299.04 Nm \\
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F_3 &= 500 \times 1 Nm
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\end{align*}
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Moments about B
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\begin{align*}
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F_1 &= 0 Nm \\
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F_2 &= 300 \cos{\ang{60}} \times 4 = 600 Nm \\
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F_3 &= 0 Nm
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\end{align*}
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\section*{Question 6}
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\begin{center}
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Tension in $AD$ is $481N$ \\
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Tension in $AB$ is $T_1N$ \\
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Tension in $AC$ is $T_2N$ \\
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Length of $AD$ is $6.5m$ \\
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Length of $AB$ is $7m$ \\
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Length of $AC$ is $7.4m$ \\
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\end{center}
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On vertical components
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\begin{align*}
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P &= 481 \cos{\ang{30.51}} + T_1 \cos{\ang{35.87}} + T_2 \frac{5.6}{7.4} \\
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P &= 414.4 + 0.8 T_1 + 0.75 T_2
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\end{align*}
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On horizontal components
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\begin{align*}
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T_1 \sin{\ang{36.87}} N &= T_2 \frac{4.837}{7.4} \sin{\ang{29.74}}N \\
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0.6 \times T_1 &= 0.324 \times T_2 \\
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481 \sin{\ang{30.51}} N &= T_2 \frac{4.837}{7.4} \cos{\ang{29.74}} N \\
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242.2 &= 0.567 \times T_2 \\
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T_2 &= 430.6 \\
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T_1 &= 232.57 \\
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P &= 414.4 + 0.8 \times 242.57 + 0.75 \times 430.6 \\
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P &= 923.4 N
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\end{align*}
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\section*{Question 7}
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\begin{align*}
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\sum{F_A} &= 0 \\
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T_{AB} + T_{AC} + T_{AD} + P &= 0 \\
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& \text{where P has only one direction that is $\hat{\imath}$} \\
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\overrightarrow{AB} &= -(960mm)\hat{\imath} - (240mm)\hat{\jmath} + (380mm)\hat{k} \\
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AB &= 1060 mm \\
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\overrightarrow{AC} &= -(960mm)\hat{\imath} - (240mm)\hat{\jmath} - (320mm)\hat{k}\\
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AC &= 1040 mm \\
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\overrightarrow{AD} &= - (960mm)\hat{\imath} + (720mm)\hat{\jmath} - (220mm)\hat{k}\\
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AD &= 1220 mm \\
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\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) \\
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\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) \\
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\overrightarrow{T_{AD}} &= T_{AD} \cdot \hat{AD} = \frac{305}{1220} \times \overrightarrow{AD} = (-240 \hat{\imath} + 180 \hat{\jmath} - 55 \hat{k}) N
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\end{align*}
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We know
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\begin{align*}
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\sum F_A &= 0 \\
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-\frac{48}{53}T_{AB} - \frac{12}{13} T_{AC} - 240 + P &= 0 \;\text{(X-axis)} \\
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-\frac{12}{53}T_{AB} -\frac{3}{13} T_{AC} + 180 &= 0 \;\text{(Y-axis)} \\
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-\frac{19}{53}T_{AB} + \frac{4}{13} T_{AC} + 55 &= 0\;\text{(Z-axis)} \\
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\end{align*}
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By solving these equations, we get
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\begin{align*}
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T_{AB} &= 446.71 N \\
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T_{AC} &= 341.71 N \\
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P &= 960 N
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\end{align*}
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\section*{Question 8}
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Calculate the unit vector along each rope just like previous question. \\
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Coordinates of point $A, B, C, D$
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\begin{align*}
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A &= 4 \hat{k} m \\
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B &= -1.5 \hat{\imath} m - 2 \hat{\jmath} m \\
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C &= 2 \hat{\imath} m + 3 \hat{\jmath} m \\
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D &= 2.5 \hat{\jmath} m
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\end{align*}
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Position vectors of each rope is $reference = 0 \hat{\imath} + 0 \hat{\jmath} + 6 \hat{k}$
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\begin{align*}
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\overrightarrow{r_{AB}} &= -1.5 \hat{\imath} - 6 \hat{k} \\
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|r_{AB}| &= 6.5 \\
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\overrightarrow{r_{AC}} &= 2 \hat{\imath} - 3 \hat{\jmath} - 6 \hat{k} \\
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|r_{AC}| &= 7 \\
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\overrightarrow{r_{AD}} &= 2.5 \hat{\jmath} - 6 \hat{k} \\
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|r_{AD}| &= 6.5
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\end{align*}
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Now direction of forces are along unit vectors:
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\begin{align*}
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\hat{r_{AB}} &= -\frac{1.5}{6.5} \hat{\imath} - \frac{6}{6.5} \hat{k} \\
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\hat{r_{AC}} &= \frac{2}{7} \hat{\imath} - \frac{3}{7} \hat{\jmath} - \frac{6}{7} \hat{k} \\
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\hat{r_{AD}} &= \frac{2.5}{6.5} \hat{\jmath} - \frac{6}{7} \hat{k}
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\end{align*}
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\[
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\sum F_A = 0
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\]
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So,
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\begin{align*}
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-0.231 F_{AB} + 0.286 F_{AC} &= 0 \\
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-0.308 F_{AB} - 0.429 F_{AC} + 0.385 F_{AD} &= 0 \\
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-0.923 F_{AB} - 0.857 F_{AC} - 0.923 F_{AD} &= -800
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\end{align*}
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On solving these three equations, we get
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\begin{align*}
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F_{AB} &= 251.2N \\
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F_{AC} &= 202.9N \\
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F_{AD} &= 427.1 N
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\end{align*}
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\end{document}
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