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Ceda EI b4987269ea Add EVS project. 2020-05-15 07:10:38 +05:30
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\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{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}

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\begin{center}
\section{Environmental Studies Assignment 1}
\subsection{Ahmad Saalim Lone, 2019BCSE017}
\end{center}
## Question 1
### a) Explain the forest resources in nature and also discuss the over-exploitation methods of forest resources. How the different forests activities help to the human needs?
### Solution
Forest are an important renewable resource. There are many types of forests such as:
- Equatorial Moist Evergreen or Rainforest
- Mediterranean Forests
- Tropical deciduous forest
- Temperate board-leaved deciduous and mixed forest
- Warm temperature broad-leaved deciduous forest
- Coniferous Forest
It is estimated that about 10% of world area is covered by forests. Forests provide food, medicine and other products to the tribal people and play a vital role in the life and economy. Forests contribute substantially to the national economy with increasing population and thus increasing demand of fuel, wood, rubber, paper and expansion of area under urban development and industries has head to over-exploitation of forest covers.
### b) Classify the energy resources in nature. How renewable energy sources play an important role to save the environment?
### Solution
On the basis of availability and renewability of natural resources. They are classified in two categories:
- Renewable resources
- Non-renewable resources
Some notes on renewable and non renewable resources are:
- Renewable resources can replenish themselves naturally. E.g. Soil, Water, etc.
- Non-renewable resources cannot be replenished once they get exhausted in a definite amount of time.
- Renewable resources play an important role to save environment. This renewable energy resources help in decreasing the negative impact on the environment. They are clean sources of energy.
## Question 2
### a) Discuss the different water resources on the living planet. Explain the sustainable water management approach.
### Solution
Water comprises 71% of our total area. Somewhere around 97% of that water is saline which leaves only 3% fresh water. Some fresh water sources are:
- Surface Water
- Standing Surface Water
- Flowing Surface Water
- Underground Water
Sustainable water management means the ability to meet the water needs of the present without compromising the ability of future generations to do the same. Water substantially also means effective and holistic management of water resources.
### b) What are the problems associated with floods, drought and dams? How they affect the environment/humans? Highlight the national and international water conflicts.
### Solution
Effects of some of the natural disasters such as
- **Floods**: Water submerges the low lying areas, cultivation of land gets affected, colossal damage, monetary loss.
- **Drought**: Almost no rain for a very long time causes draughts. No irrigation, plant and animal life suffers and thus humans get affected.
Unequal amount of rain distribution causes disputes among various lands such as
- National Disputes: sharing of Kaveri water between Karnataka and Tamil Nadu, sharing of Krishna water between Karnataka and Andhra Pradesh.
- International Disputes: Indus River between India and Pakistan. Colorado River between Mexico and United States of America.
## Question 3
### a) What are the food resources? How you can minimize world food problems? Discuss the types of nutrition.
### Solution
Food is what we eat and need to live. Food can be of two types to humans:
1. **Primary**: Primary comprises of vegetables and plants. E.g. Vegetables, herbs, fruits, etc.
2. **Secondary**: Secondary comprises of other animals such as chicken, goat, sheep, etc.
In many developing countries, population is increasing rapidly and the government is unable to meet the daily food requirements, which results in diseases and deficiencies such as malnutrition. We can minimize world food issues by supporting domestic food production, stabilize and guarantee fair prices to farmers, maintain supply chains and entertain new and useful public policies.
### Discuss the types of agriculture in India. Which factors affect the modern agriculture in India with reference to fertilizers-pesticides and water logging?
### Solution
Different farming practices in India include:
- **Subsistence Farming**: In this type of farming nearly all the crops or livestock raised are used to maintain the farmers' own use. It is also referred to as small scale production.
- **Shifting Cultivation**: Shifting Cultivation means migratory shifting agriculture. Under this system, a plot is cultivated for few years and when the crop yield declines because of soil exhaustion and is deserted, we shift to another plot for the same and back and forth over the years.
- **Plantation Agriculture**: In this type of agriculture cash crops are cultivated. A single crop like rubber, sugar cane, coffee, tea is grown. These are used majorly for export.
Modern agriculture has its pros and cons. Using fertilizers produce a lot of yield but at the same time cause soil pollution, irrigation is important for plants but the excess of it can cause soil erosion. So, balance must be prevailed in order to maintain soil quality and prevent health hazards. Long term thinking and approach is needed for better overall development.
## Question 4
### a) What are the important mineral resources in nature? What types of mining activities damages the environment?
### Solution
Mineral resources are defined as an occurrence of natural, solid, inorganic or fossilized organic material in/on the earth's crust in such quantity and quality that it is feasible and reasonable to extract it out.
Mineral resources are divided into metallic and non-metallic resources. Mineral resources are the most important natural resources and determine a country's industrial and economic growth by supplying raw material to economy's primary, secondary and tertiary sectors.
Some reasons that affect modern agriculture system affect the mineral resources such as over exploitation, waste, not recycling of materials. This not only affects us but will surely affect the coming generations.
### b) Write short notes on the following:
#### i. Chipko Movement
#### Solution
Chipko Movement (or Chipko Andolan) was a forest conservation movement in India. It began in 1970s in Uttarakhand, then a part of Uttar Pradesh and went on to become a rallying point for many future environmental movements. Chipko Andolan is a movement that practised methods of Satyagraha where both male and female activists from Uttarakhand played vital roles including Gaura Devi, Suraksha Devi, Sudesha Devi, Bachni Devi and Chandi Prasad Bhatt and others. These people after hearing that trees are being cut down in forests reached there and hugged the trees and proved their point to the whole world.
#### ii. Sardar Sarovar Dam
#### Solution
It is a Gravity Dam on the Narmada River near Navagam, Gujarat in India. Four states -- Gujarat, Madhya Pradesh, Maharashtra and Rajasthan receive water and electricity from the dam. Foundation stone was laid out by Prime Minister Jawaharlal Nehru on 5^{th} April 1961. Narmada Bacho Andolan was initiated by environmentalists, tribal people and human right activists against the construction of the dam on the Narmada river. It had a controversial issue due to problems in the form of displacement of local people, loss of livelihood, flood deforestation, etc. Thus, by passing through many controversies, the dam was built.