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