Skip to content

Commit 894d930

Browse files
WxMichelleYangWxMichelleYang
committed
the example questions used as examples in prompt and test
1 parent e253e86 commit 894d930

File tree

14 files changed

+216
-0
lines changed

14 files changed

+216
-0
lines changed

‎wizard/example_questions/q1.tex

+7
Original file line numberDiff line numberDiff line change
@@ -0,0 +1,7 @@
1+
A builder pulls with a force of \SI{300}{\N} on a rope attached to a building as shown in figure~\ref{A1:fig:Q1a}. What are the horizontal and vertical components of the force exerted by the rope at the point A?
2+
\begin{figure}
3+
\centering
4+
\includegraphics[width=0.6\columnwidth]{problem1_1a.png}
5+
\caption{A builder applying \SI{300}{\N} to a building using a rope.}
6+
\label{A1:fig:Q1a}
7+
\end{figure}

‎wizard/example_questions/q2.tex

+7
Original file line numberDiff line numberDiff line change
@@ -0,0 +1,7 @@
1+
Four forces act on a bolt as shown in figure~\ref{A1:fig:Q2}. Determine the resultant of the forces on the bolt.
2+
\begin{figure}
3+
\centering
4+
\includegraphics[width=0.65\columnwidth]{problem1_2.png}
5+
\caption{A bolt with four forces acting on it.}
6+
\label{A1:fig:Q2}
7+
\end{figure}

‎wizard/example_questions/q3.tex

+11
Original file line numberDiff line numberDiff line change
@@ -0,0 +1,11 @@
1+
A car is being towed with two ropes as shown in figure~\ref{A1:fig:Q3}. If the resultant of the two forces is a \SI{30}{\N} force parallel to the long axis of the car, find:
2+
\begin{enumerate}
3+
\item the tension in each of the ropes, if $\alpha = \ang{30}$.
4+
\item the value of $\alpha$ such that the tension in rope, $T_2$ is minimal.
5+
\end{enumerate}
6+
\begin{marginfigure}[-20mm]
7+
\centering
8+
\includegraphics[width=\columnwidth]{problem1_3.png}
9+
\caption{A car with two tow ropes attached.}
10+
\label{A1:fig:Q3}
11+
\end{marginfigure}

‎wizard/example_questions/q4.tex

+7
Original file line numberDiff line numberDiff line change
@@ -0,0 +1,7 @@
1+
A \SI{30}{\N} force acts on the end of a \SI{3}{\m} lever as shown in figure~\ref{A1:fig:Q4a}. Determine the moment of the force about the point O.
2+
\begin{figure}
3+
\centering
4+
\includegraphics[width=0.8\columnwidth]{problem1_4a.png}
5+
\caption{A lever with a single force acting on it.}
6+
\label{A1:fig:Q4a}
7+
\end{figure}

‎wizard/example_questions/q5.tex

+13
Original file line numberDiff line numberDiff line change
@@ -0,0 +1,13 @@
1+
Find the real and imaginary parts of:
2+
3+
$
4+
\begin{array}[h!]{lll}
5+
{\rm (a)}\hskip5pt 8+3\,i\hskip24pt&
6+
{\rm (b)}\hskip5pt 4-15\,i\hskip24pt&
7+
{\rm (c)}\hskip5pt \cos\theta-i\,\sin\theta\\
8+
\noalign{\vskip12pt}
9+
{\rm (d)}\hskip5pt i^2&
10+
{\rm (e)}\hskip5pt i\,(2-5\,i)&
11+
{\rm (f)}\hskip5pt (1+2\,i)(2-3\,i)
12+
\end{array}
13+
$

‎wizard/example_questions/q6.tex

+13
Original file line numberDiff line numberDiff line change
@@ -0,0 +1,13 @@
1+
Write each of the following expressions as a complex number in the form $x+i\,y$:
2+
3+
$
4+
\begin{array}[h!]{lll}
5+
{\rm (a)}\hskip5pt (5-i)(2+3\,i)\hskip24pt&
6+
{\rm (b)}\hskip5pt (3-4\,i)(3+4\,i)\hskip24pt&
7+
{\rm (c)}\hskip5pt (1+2\,i)^2\\
8+
\noalign{\vskip12pt}
9+
{\rm (d)}\hskip5pt \displaystyle{10\over4-2\,i}&
10+
{\rm (e)}\hskip5pt \displaystyle{3-i\over4+3\,i}&
11+
{\rm (f)}\hskip5pt \displaystyle{1\over i}
12+
\end{array}
13+
$

‎wizard/example_questions/q7.tex

+10
Original file line numberDiff line numberDiff line change
@@ -0,0 +1,10 @@
1+
Define $z=(5+7\,i)(5+b\,i)$.
2+
3+
$
4+
\begin{array}[h!]{l}
5+
{\rm (a)}\hskip5pt {\rm If}\ b\ {\rm and}\ z\ {\rm are\ both\ real,\ find}\ b.\\
6+
\noalign{\vskip12pt}
7+
{\rm (b)}\hskip5pt {\rm If}\ {\rm Im}(b)={4\over5},\ {\rm and}\ z\ {\rm is\ pure\ imaginary,\ find}\
8+
{\rm Re}(b).
9+
\end{array}
10+
$

‎wizard/example_questions/s1.tex

+23
Original file line numberDiff line numberDiff line change
@@ -0,0 +1,23 @@
1+
The angle $\alpha$ between the rope and the horizontal line can be calculated using the relation between $\tan\alpha$ and the lengths of the sides of the right angled triangle as shown in figure~\ref{A1:fig:Q1b},
2+
\begin{align*}
3+
\tan\alpha &=\frac{\SI{6}{\m}}{\SI{8}{\m}}\\
4+
&=0.75 \, ,
5+
\end{align*}
6+
from which $\alpha=\ang{36.87}$.
7+
\begin{marginfigure}
8+
\centering
9+
\includegraphics[width=\columnwidth]{problem1_1b.png}
10+
\caption{The right angle triangle formed by the rope.}
11+
\label{A1:fig:Q1b}
12+
\end{marginfigure}
13+
14+
The horizontal component, $F_H$, is the magnitude of the force projected onto the horizontal axis,
15+
\begin{align*}
16+
F_H &=\SI{300}{\N} \cos\alpha \\
17+
&= \SI{240}{\N} \,,
18+
\end{align*} and the vertical component, $F_V$, is the magnitude of the force projected onto the vertical axis,
19+
\begin{align*}
20+
F_V &=\SI{300}{\N} \sin\alpha \\
21+
&= \SI{180}{\N}\, .
22+
\end{align*}
23+
\clearpage

‎wizard/example_questions/s2.tex

+19
Original file line numberDiff line numberDiff line change
@@ -0,0 +1,19 @@
1+
To obtain the resultant force, the projection of the resultant on the vertical and horizontal axes is determined as the sum of the projections of the four forces onto the respective axes,
2+
\begin{align*}
3+
R_H &= F_1 \cos\ang{30} - F_2\sin\ang{20} + F_4 \cos\ang{15}\\
4+
&= \SI{199.1}{\N} \,,\\
5+
R_V &= F_1 \sin\ang{30} +F_2 \cos\ang{20}-F_3 -F_4\sin\ang{15}\\
6+
&= \SI{14.3}{\N} \, .
7+
\end{align*}
8+
The angle with the horizontal then follows from
9+
\begin{align*}
10+
\frac{R_V}{R_H} &= \frac{R\sin\alpha}{R\cos\alpha}\\
11+
&= \tan\alpha \\
12+
&=0.0718\, .
13+
\end{align*}
14+
Hence, $\alpha=\ang{4.1}$. The magnitude of the resultant follows from,
15+
\begin{align*}
16+
|R| &=\sqrt{R_H^2+R_V^2}\\
17+
&= \SI{199.6}{\N}
18+
\end{align*}
19+
\clearpage

‎wizard/example_questions/s3.tex

+34
Original file line numberDiff line numberDiff line change
@@ -0,0 +1,34 @@
1+
In order for the resultant force to be a force with magnitude \SI{30}{\N} oriented along the horizontal, the following conditions must be true,
2+
\begin{align}
3+
T_2 \cos\alpha + T_1 \cos\ang{20} &= \SI{30}{\N} \\
4+
T_2 \sin\alpha - T_1 \sin\ang{20} &= 0 \,.
5+
\end{align}
6+
Hence,
7+
\begin{align}
8+
T_2 &= T_1\frac{\sin\ang{20}}{\sin\alpha} \label{A1:eq:Q3:T1}\\
9+
\left( T_1 \frac{\sin\ang{20}}{\sin\alpha} \right) \cos\alpha &+ T_1\cos\ang{20} = \SI{30}{\N} \nonumber \\
10+
T_1 &= \frac{\SI{30}{\N}}{\frac{\sin\ang{20}}{\tan\alpha} + \cos\ang{20}} \,. \label{A1:eq:Q3:T2}
11+
\end{align}
12+
When $\alpha=\ang{30}$, this yields
13+
\begin{align*}
14+
T_1&=\SI{19.58}{\N}\\
15+
T_2&=\SI{13.39}{\N}
16+
\end{align*}
17+
18+
Substituting the expression for $T_1$, equation~\ref{A1:eq:Q3:T1}, into the expression for $T_2$, equation~\ref{A1:eq:Q3:T2} allows us to determine $T_2$ as a function of $\alpha$,
19+
\begin{align}
20+
T_2 &=\frac{\SI{30}{\N}}{\frac{\sin\ang{20}}{\tan\alpha} + \cos\ang{20}} \frac{\sin\ang{20}}{\sin\alpha} \nonumber \\
21+
&=\frac{\SI{30}{\N}} {\sin\ang{20}\frac{\cos\alpha}{\sin\alpha} + \cos\ang{20}} \frac{\sin\ang{20}}{\sin\alpha} \nonumber \\
22+
&=\frac{\SI{30}{\N}} {\sin\ang{20}\frac{\cos\alpha}{\sin\alpha}\frac{\sin\alpha}{\sin\ang{20}} + \cos\ang{20}\frac{\sin\alpha}{\sin\ang{20}}} \nonumber \\
23+
&=\frac{\SI{30}{\N}} {\cancel{\sin\ang{20}} \frac{\cos\alpha} {\cancel{\sin\alpha}} \frac{\cancel{\sin\alpha}}{\cancel{\sin\ang{20}}} + \cos\ang{20}\frac{\sin\alpha}{\sin\ang{20}}} \nonumber \\
24+
&=\frac{\SI{30}{\N}}{\cos\alpha + \frac{\sin\alpha}{\tan\ang{20}}} \,.
25+
\end{align}
26+
Examining the form of this, we can see that $T_2$ will be minimal at $\alpha_{min}$ when $\left(\cos\alpha + \frac{\sin\alpha}{\tan\ang{20}}\right)$ is maximal. Searching for maximal stationary points in the usual manner,
27+
\begin{align}
28+
\frac{\mathrm{d}}{\mathrm{d}\alpha} \left(\cos\alpha + \frac{\sin\alpha}{\tan\ang{20}}\right) &= 0 \nonumber\\
29+
&= -\sin\alpha_{min} +\frac{\cos\alpha_{min}}{\tan\ang{20}} \nonumber \\
30+
\tan\alpha_{min} &= \frac{1}{\tan\ang{20}} \label{A1:eq:Q3:alpha}\\
31+
\Rightarrow\,\alpha_{min} &= \ang{70} \nonumber
32+
\end{align}
33+
Hence, $T_2$ is minimal when $T_2$ is perpendicular to $T_1$.\sidenote[][]{This is true for all angles of $T_2$, see if you can convince yourself of this by examining equation~\ref{A1:eq:Q3:alpha} analytically.}
34+
\clearpage

‎wizard/example_questions/s4.tex

+14
Original file line numberDiff line numberDiff line change
@@ -0,0 +1,14 @@
1+
We can make many problems easier by choosing a suitable set of axes. Looking at the special directions of this system, we can choose to orient the x-axis parallel to the line connecting point $O$ and the point where the force acts and the y-axis perpendicular to this, as shown in figure~\ref{A1:fig:Q4b}. The moment of the force is then given by
2+
\begin{align*}
3+
M &= F_y\, d \\
4+
&= \SI{30}{\N} \sin\ang{20} \SI{3}{\m}\\
5+
&= \SI{30.8}{\N} \,.
6+
\end{align*}
7+
\begin{figure}
8+
\centering
9+
\includegraphics[width=0.8\columnwidth]{problem1_4b.png}
10+
\caption{The x-axis oriented parallel to the line between $O$ and the force's point of action.}
11+
\label{A1:fig:Q4b}
12+
\end{figure}
13+
We can think of this in two ways. From one viewpoint the moment is the component of the force parallel to the y-axis, multiplied by the distance from the pivot. From the other viewpoint, the moment is the whole force multiple by the perpendicular distance from the pivot. Note that these are equivalent, it just depends where you choose to group the $\sin\ang{20}$ term. This is because only the component of the force that is non-parallel to the line joining the point of action and the pivot will induce rotation.
14+
\clearpage

‎wizard/example_questions/s5.tex

+20
Original file line numberDiff line numberDiff line change
@@ -0,0 +1,20 @@
1+
For a complex number $a+i\,b$, in which $a$ and $b$ are real, the real and imaginary parts
2+
are given by ${\rm Re}(a+i\,b)=a$ and ${\rm Im}(a+i\,b)=b$, respectively. Thus,
3+
4+
$
5+
\begin{array}[h!]{l}
6+
{\rm (a)}\hskip5pt {\rm Re}(8+3\,i)=8\, ,\qquad{\rm Im}(8+3\,i)=3\, .\\
7+
\noalign{\vskip12pt}
8+
{\rm (b)}\hskip5pt {\rm Re}(4-15\,i)=4\, ,\qquad {\rm Im}(4-15\,i)=-15\, .\\
9+
\noalign{\vskip12pt}
10+
{\rm (c)}\hskip5pt {\rm Re}(\cos\theta-i\,\sin\theta)=\cos\theta\, ,\qquad {\rm
11+
Im}(\cos\theta-i\,\sin\theta)=-\sin\theta\, .\\
12+
\noalign{\vskip12pt}
13+
{\rm (d)}\hskip5pt i^2=-1.\quad {\rm Re}(i^2)=-1\, ,\qquad {\rm Im}(i^2)=0\, .\\
14+
\noalign{\vskip12pt}
15+
{\rm (e)}\hskip5pt i\,(2-5\,i)=5+2\,i.\qquad {\rm Re}(5+2\,i)=5\, ,\qquad {\rm Im}(5+2\,i)=2\, .\\
16+
\noalign{\vskip12pt}
17+
{\rm (f)}\hskip5pt (1+2\,i)(2-3\,i)=2-3\,i+4\,i+6=8+i\, .\qquad {\rm Re}(8+i)=8\, ,\qquad {\rm
18+
Im}(8+i)=1\, .
19+
\end{array}
20+
$

‎wizard/example_questions/s6.tex

+20
Original file line numberDiff line numberDiff line change
@@ -0,0 +1,20 @@
1+
Applying the rules for the multiplication and division of
2+
complex numbers yields:
3+
4+
$
5+
\begin{array}[h!]{lll}
6+
{\rm (a)}\hskip5pt (5-i)(2+3\,i)=10+15\,i-2\,i+3=13+13\,i\, .\\
7+
\noalign{\vskip12pt}
8+
{\rm (b)}\hskip5pt (3-4\,i)(3+4\,i)=9+12\,i-12\,i+16=25\, .\\
9+
\noalign{\vskip12pt}
10+
{\rm (c)}\hskip5pt (1+2\,i)^2=(1+2\,i)(1+2\,i)=1+2\,i+2\,i-4=-3+4\,i\, .\\
11+
\noalign{\vskip12pt}
12+
{\rm (d)}\hskip5pt \displaystyle{{10\over4-2\,i}={10\over4-2\,i}\times{4+2\,i\over4+2\,i}=
13+
{40+20\,i\over16+8\,i-8\,i+4}={40+20\,i\over20}=2+i\, .}\\
14+
\noalign{\vskip12pt}
15+
{\rm (e)}\hskip5pt \displaystyle{{3-i\over4+3\,i}={3-i\over4+3\,i}\times{4-3\,i\over4-3\,i}=
16+
{12-9\,i-4\,i-3\over16-12\,i+12\,i+9}={9-13\,i\over25}={9\over25}-i\,{13\over25}\, .}\\
17+
\noalign{\vskip12pt}
18+
{\rm (f)}\hskip5pt \displaystyle{{1\over i}={1\over i}\times{-i\over-i}=-i\, .}
19+
\end{array}
20+
$

‎wizard/example_questions/s7.tex

+18
Original file line numberDiff line numberDiff line change
@@ -0,0 +1,18 @@
1+
We have that $z=(5+7\,i)(5+b\,i)=25+5b\,i+35\,i-7b$.
2+
3+
$
4+
\begin{array}[h!]{l}
5+
{\rm (a)}\hskip5pt {\rm If}\ b\ {\rm and}\ z\ {\rm are\ both\ real},\ {\rm then\ the \ imaginary\
6+
parts\ of\ both\ quantities\ vanish.}\ {\rm Thus,}\\
7+
\noalign{\vskip12pt}
8+
\hskip20pt {\rm Im}(z)=35+5b=0, {\rm so}\ b=-7.\\
9+
\noalign{\vskip12pt}
10+
{\rm (b)}\hskip5pt {\rm If}\ {\rm Im}(b)={4\over5},\ {\rm and}\ z\ {\rm is\ pure\ imaginary,}\
11+
{\rm then\ the\ real\ part\ of}\ z\ {\rm vanishes\!:}\\
12+
\noalign{\vskip12pt}
13+
\hskip20pt{\rm Re}(z)=25+5\bigl[i\,{\rm Im}(b)\bigr]i-7\,{\rm Re}(b)=25-4-7\,{\rm Re}(b)=21-7\,{\rm
14+
Re}(b)=0,\\
15+
\noalign{\vskip12pt}
16+
\hskip20pt{\rm so}\ {\rm Re}(b)=3.
17+
\end{array}
18+
$

0 commit comments

Comments
 (0)