\n",
" How to use this notebook?

\n", " This notebook is made of text cells and code cells. The code cells have to be**executed** to see the result of the program.

To execute a cell, simply select it and click on the \"play\" button (►) in the tool bar just above the notebook, or type

It is important to execute the code cells in their order of appearance in the notebook.\n", "

"
]
},
{
"cell_type": "markdown",
"metadata": {
"toc-hr-collapsed": true
},
"source": [
"\n", " This notebook is made of text cells and code cells. The code cells have to be

To execute a cell, simply select it and click on the \"play\" button (►) in the tool bar just above the notebook, or type

`shift + enter`

. It is important to execute the code cells in their order of appearance in the notebook.\n", "

\n", "\n", "# Learning goals\n", "\n", "After using this notebook, you should be able to:\n", "* Analyze how the position of the cable influences the tension force in the jeans problem\n", "* Describe how the counterweight influences the position of the cable in the jeans problem\n", "* Use Python to make mathematical calculations and write simple functions\n", "\n" ] }, { "cell_type": "markdown", "metadata": {}, "source": [ "

\n", "\n", "---" ] }, { "cell_type": "markdown", "metadata": {}, "source": [ "# The problem\n", "\n", "The question we are trying to answer in this notebook is the following: \n", "**Estimate which counterweight allows to suspend wet jeans (3kg) on the cable in the position illustrated below.**\n", "\n", "\n", "

Figure 1: The suspended jeans situation

\n", "\n", "\n", "The activities below allow you to find out the answer to this question by exploring how the counterweight affects the position of the jeans suspended on the cable. " ] }, { "cell_type": "markdown", "metadata": { "toc-hr-collapsed": false }, "source": [ "\n", "\n", "---" ] }, { "cell_type": "markdown", "metadata": {}, "source": [ "# Activity 1: Experimenting with the virtual lab\n", "\n", "The \"Suspended object\" virtual lab will allow you to **experiment with different counterweights** to see how it affects the position of the jeans suspended on the cable. \n", - "We will follow a 3-step process highlighting the important steps in experimenting, whether in the lab or with a virtual lab like today.\n", + "We will follow a 3-step process highlighting the important steps in experimenting, whether in a physical lab or with a virtual lab like today.\n", "\n", "## 1. Plan\n", - "Before starting, it is useful to think about which value(s) for the counterweight you would be interested to try out (based on your answer to the clicker question). \n", - "In any experiment, the planning step is essential and involves identifying testing scenarios as well as the results you expect from them.\n", + "In any experiment, the planning step is essential and involves identifying testing scenarios as well as the results you expect from them. \n", + "It is therefore useful to start by thinking about which value(s) for the counterweight you would be interested to experiment. \n", "\n", "

\n",
"\n",
"In the cell below, **list the different values** you would like to test for the counterweight. \n",
- "This cell is for just note taking, you can refer to these values when you will carry out the experiment in the next step.\n",
+ "This cell is for just note taking, you can refer to these values when you carry out the experiment in the next step.\n",
" \n",
"

"
]
},
{
"cell_type": "raw",
"metadata": {},
"source": [
"- value 1:\n",
"- value 2:\n",
"- value 3:"
]
},
{
"cell_type": "markdown",
"metadata": {},
"source": [
"## 2. Experiment\n",
"Now is time to experiment! \n",
"A key to this exciting step of the process is to be as systematic as possible with your investigations and to carefully observe the results.\n",
"\n",
"\n",
"\n",
"**Execute the code cell below** (click on it then click on the \"play\" button in the tool bar above) to launch the virtual lab. \n",
"Try out the different values you have listed above and **observe**: \n",
"* the position of the cable \n",
"* the angle that the cable makes with the horizon \n",
"* the **forces** that apply on the object and **how they evolve** as you change the counterweight\n",
" \n",
"

"
]
},
{
"cell_type": "code",
"execution_count": null,
"metadata": {},
"outputs": [],
"source": [
"# Let's import the custom library with the virtual lab\n",
"from lib.suspendedobject import *\n",
"\n",
"# The virtual lab sets up a situation with an object of mass 3kg, suspended on a cable between poles 1.5m high and 5m apart\n",
"SuspendedObjectLab(mass_object = 3, height = 1.5, distance = 5).launch();"
]
},
{
"cell_type": "markdown",
"metadata": {},
"source": [
"## 3. Reflect\n",
"\n",
"The goal of the reflection step is for you to look back and identify your key take-aways from the experiment. \n",
"In other words this is when you learn!\n",
"\n",
"So let's come back to our original question: \n",
- "\"Which counterweight do you think would allow to suspend the jeans (3kg) on the cable in the position illustrated below?\" \n",
+ "**Which counterweight do you think would allow to suspend the jeans (3kg) on the cable in the position illustrated below?** \n",
"More importantly, **can you explain why**?\n",
"\n",
"\n",
"\n",
"\n",
"\n",
"\n",
"Execute the code cell below and **post a message in the chat** with:\n",
"* the value you found for the mass of the counterweight \n",
"* a brief explanation of **why** the mass of the counterweight differs from the mass of the jeans. \n",
"\n",
"**Compare your answer with the others** and vote for the best explanation using the \"thumb up\" icon.\n",
"\n",
"*This chat is completely anonymous.* \n",
"*You might need to accept the cookies to see the text box where to post your message.*\n",
"\n",
"

\n"
]
},
{
"cell_type": "code",
"execution_count": null,
"metadata": {},
"outputs": [],
"source": [
"from IPython.display import IFrame\n",
"IFrame('https://speakup.epfl.ch/room/12989', 400, 500)"
]
},
{
"cell_type": "markdown",
"metadata": {
"toc-hr-collapsed": false
},
"source": [
"\n", "\n", - "Congrats, you have finished Activity 1! \n", - "Optionnally, you can continue with Activity 2 and programming activities below if you want.\n", + "Congratulations, you have completed Activity 1! \n", + "You can continue with the programming exercises in Activity 2 below if you want.\n", "\n", "

\n", "\n", "---" ] }, { "cell_type": "markdown", "metadata": {}, "source": [ "# Activity 2: Looking more closely at the tension in the cable\n", "\n", "In the lecture, we have seen that, in this situation, the value of the tension in the cable (represented by the plain red arrows in the virtual lab above) is defined by:\n", "\n", "$\n", "\\begin{align}\n", "\\lvert\\vec{T}\\rvert = \\frac{\\frac{1}{2}.m.g}{sin(\\alpha)}\n", "\\end{align}\n", "$, where $m$ is the mass of the suspended object, $\\alpha$ the angle that the cable makes with the horizon and $g$ the gravity of earth.\n", "\n", "In the following we are going to use Python to **compute the tension in the cable** for **different values of the angle $\\alpha$**. \n", "For this, we need to import some useful Python libraries. " ] }, { "cell_type": "markdown", "metadata": { "toc-hr-collapsed": false }, "source": [ "

\n",
" Activity

\n", "\n", "**Execute the code cell below** so that the necessary libraries get imported.\n", " \n", "

"
]
},
{
"cell_type": "code",
"execution_count": null,
"metadata": {},
"outputs": [],
"source": [
"# We need Numpy for some mathematical functions (e.g. sinus)\n",
"import numpy as np\n",
"\n",
"# We will do some plotting with Bokeh\n",
"from bokeh.plotting import figure\n",
"\n",
- "# We will use a custom library with visualizations and tools developed for this exercise\n",
- "from lib.suspendedobject import *\n",
- "\n",
"# And display a message once all libraries are imported\n",
"print(\"Libraries imported.\")"
]
},
{
"cell_type": "markdown",
"metadata": {
"tags": []
},
"source": [
"## Defining the constants of our physics problem\n",
"\n",
"In the above equation we have two constants:\n",
"- $g$ the gravity of earth, which is 9.81m.s$^{-2}$\n",
"- $m$ the mass of the suspended object, which is 3 kg in the case of the jeans\n",
"\n",
"\n", "\n", "**Execute the code cell below** so that the necessary libraries get imported.\n", " \n", "

\n",
" Activity

\n", "\n", "**Execute the code cell below** so that these two constants get defined in Python.\n", " \n", "

"
]
},
{
"cell_type": "code",
"execution_count": null,
"metadata": {},
"outputs": [],
"source": [
"# Let's define the value of gravity\n",
"g = 9.81\n",
"\n",
"# And the mass of the jeans\n",
"m = 3\n",
"\n",
"# Display the value of the constants to check they are well defined\n",
"print(\"gravity:\", g, \", jeans_mass:\", m)"
]
},
{
"cell_type": "markdown",
"metadata": {},
"source": [
"## Creating a function to compute the tension in the cable\n",
"\n",
"Let's define a Python function that represents the equation of the tension. \n",
"Its input parameters are:\n",
"* `g`: the gravity of earth\n",
"* `m`: the mass of the jeans\n",
"* `alpha`:the angle that the cable makes with the horizon\n",
"\n",
"It returns the value of the mass of the counterweight as computed with the equation $\n",
"\\begin{align}\n",
"\\frac{\\frac{1}{2}.m.g}{sin(\\alpha)}\n",
"\\end{align}\n",
"$\n",
"\n",
"\n", "\n", "**Execute the code cell below** so that these two constants get defined in Python.\n", " \n", "

\n",
" Activity

\n", "\n", "In the code cell below, **complete the code of the function ``tension_norm``** by implementing the equation above. \n", "Here is some syntax you will need:\n", "* multiplication: ``*``\n", "* division: ``/``\n", "* sinus function $sin(x)$: ``np.sin(x)``\n", "\n", "You can also use parentheses to indicate the order of operations.\n", " \n", "Then **execute the code cell** so that this function gets defined in Python.\n", "

\n"
]
},
{
"cell_type": "code",
"execution_count": null,
"metadata": {},
"outputs": [],
"source": [
"# Then let's define the function\n",
"def tension_norm(g, m, alpha):\n",
" tension = 1 # REPLACE \"1\" BY YOUR EQUATION HERE\n",
" return tension\n",
"\n",
"# And display a message once it is defined\n",
"print(\"Function defined.\")"
]
},
{
"cell_type": "markdown",
"metadata": {},
"source": [
"\n", "\n", "In the code cell below, **complete the code of the function ``tension_norm``** by implementing the equation above. \n", "Here is some syntax you will need:\n", "* multiplication: ``*``\n", "* division: ``/``\n", "* sinus function $sin(x)$: ``np.sin(x)``\n", "\n", "You can also use parentheses to indicate the order of operations.\n", " \n", "Then **execute the code cell** so that this function gets defined in Python.\n", "

\n",
"\n",
"**Solution** - you can check your answer with the solution by clicking on the \"...\" below.\n",
" \n",
"

"
]
},
{
"cell_type": "markdown",
"metadata": {
"jupyter": {
"source_hidden": true
},
"tags": []
},
"source": [
"\n",
" Solution

\n", "\n", "Your function should look like the following:\n", "\n", "

"
]
},
{
"cell_type": "markdown",
"metadata": {
"tags": []
},
"source": [
"## Computing the tension in the cable using our function\n",
"\n",
"Now let's use this Python function to compute the norm of the tension in the cable for a given angle of $\\alpha$.\n",
"\n",
"Because our function takes the sinus of our angle $\\alpha$, we first need to convert it from degrees into radians. \n",
"This is where the custom library we have imported gets useful as it provides us with a function `degrees_to_radians` which we can use to convert our angle.\n",
"\n",
"The code cell below defines a value for $\\alpha$ in degrees, converts it to radians, then computes the tension on the cable using our previously defined function with `g` = 9.81 m.s$^{-2}$ and `m` = 3 kg and prints the result."
]
},
{
"cell_type": "code",
"execution_count": null,
"metadata": {},
"outputs": [],
"source": [
- "# First choose an angle alph, e.g. 2 degrees\n",
- "alpha = 2\n",
+ "# First choose an angle alph, e.g. 1.5 degrees\n",
+ "alpha = 1.5\n",
"\n",
"# Then convert it to radians\n",
"alpha_radians = degrees_to_radians(alpha)\n",
"\n",
"# Then compute the tension using our equation with g = 9.81 m.s-2 and m = 3 kg\n",
"T = tension_norm(g, m, alpha_radians)\n",
"\n",
"# And print the result\n",
"print(\"Norm of the tension in the cable: T =\", T, \"N\")"
]
},
{
"cell_type": "markdown",
"metadata": {},
"source": [
"\n", "\n", "Your function should look like the following:\n", "\n", "

```
def tension_norm(g, m, alpha):\n",
" tension = (1/2 * m * g) / np.sin(alpha)\n",
" return tension\n",
"
```

\n",
"\n",
"\n",
" Activity

\n", "\n", "**Execute the code cell** to see the value of the corresponding tension force $\\lvert\\vec{T}\\rvert$, expressed in Newtons. \n", " \n", "You can compare the value you obtain to the force exerted by earth gravity on the jeans, which weight 3 kg: $\\lvert\\vec{F}\\rvert = $ 29 $N$ \n", "Is the **tension in the cable bigger or smaller than the weight** of the jeans on the cable? Why? \n", "You can experiment with different values for $\\alpha$.\n", " \n", "Note down your answer in the cell below, then you can check the solution by clicking on the \"...\" below.\n", "\n", "

"
]
},
{
"cell_type": "raw",
"metadata": {},
"source": [
"Just note down your answer here (this cell is for note taking)."
]
},
{
"cell_type": "markdown",
"metadata": {},
"source": [
"\n", "\n", "**Execute the code cell** to see the value of the corresponding tension force $\\lvert\\vec{T}\\rvert$, expressed in Newtons. \n", " \n", "You can compare the value you obtain to the force exerted by earth gravity on the jeans, which weight 3 kg: $\\lvert\\vec{F}\\rvert = $ 29 $N$ \n", "Is the **tension in the cable bigger or smaller than the weight** of the jeans on the cable? Why? \n", "You can experiment with different values for $\\alpha$.\n", " \n", "Note down your answer in the cell below, then you can check the solution by clicking on the \"...\" below.\n", "\n", "

\n",
"\n",
"**Solution** - you can check your answer with the solution by clicking on the \"...\" below.\n",
" \n",
"

"
]
},
{
"cell_type": "markdown",
"metadata": {
"jupyter": {
"source_hidden": true
},
"tags": []
},
"source": [
"\n",
" Solution

\n", "\n", "If you have chosen an angle $\\alpha$ smaller than 30$^\\circ$ (which is quite likely), then you will notice that the tension $\\lvert\\vec{T}\\rvert$ is bigger than the weight of the jeans.\n", " \n", "This is because such small angles make the cable more horizontal whereas the weight of the jeans is purely vertical, therefore a higher tension will be necessary to create the vertical resulting force that compensates for the weight of the jeans.\n", " \n", "\n", "\n", "Mathematically speaking, remember that the tension is defined by the following equation: \n", "$\n", "\\begin{align}\n", "\\lvert\\vec{T}\\rvert = \\frac{\\frac{1}{2}.m.g}{sin(\\alpha)}\n", "\\end{align}\n", "$\n", "\n", "Because the weight of the jeans is $\\lvert\\vec{F}\\rvert = m.g$, the tension can also be written: \n", "$\n", "\\begin{align}\n", "\\lvert\\vec{T}\\rvert = \\frac{\\frac{1}{2}}{sin(\\alpha)}.\\lvert\\vec{F}\\rvert\n", "\\end{align}\n", "$\n", "\n", "As a consequence:\n", "* For angles between 0$^\\circ$ and 30$^\\circ$, $sin(\\alpha)$ will be between 0 and $\\frac{1}{2}$ and therefore we will have $\\lvert\\vec{T}\\rvert > \\lvert\\vec{F}\\rvert$\n", "* For angles between 30$^\\circ$ and 90$^\\circ$ (although such high values do not really make sense in the real situation), $sin(\\alpha)$ will be between $\\frac{1}{2}$ and 1 and therefore we will have $\\lvert\\vec{T}\\rvert < \\lvert\\vec{F}\\rvert$\n", "\n", "

"
]
},
{
"cell_type": "markdown",
"metadata": {
"tags": []
},
"source": [
"## Looking at how the tension in the cable evolves with the angle $\\alpha$\n",
"\n",
"Now we want to look at **how the tension evolves** when the angle $\\alpha$ changes. \n",
"Thanks to our previously defined Python function, we are going to compute the tension in the cable for 100 different values of $\\alpha$, from 20$^\\circ$ to 0$^\\circ$, and plot the result on a graph. \n",
"\n",
"\n", "\n", "If you have chosen an angle $\\alpha$ smaller than 30$^\\circ$ (which is quite likely), then you will notice that the tension $\\lvert\\vec{T}\\rvert$ is bigger than the weight of the jeans.\n", " \n", "This is because such small angles make the cable more horizontal whereas the weight of the jeans is purely vertical, therefore a higher tension will be necessary to create the vertical resulting force that compensates for the weight of the jeans.\n", " \n", "\n", "\n", "Mathematically speaking, remember that the tension is defined by the following equation: \n", "$\n", "\\begin{align}\n", "\\lvert\\vec{T}\\rvert = \\frac{\\frac{1}{2}.m.g}{sin(\\alpha)}\n", "\\end{align}\n", "$\n", "\n", "Because the weight of the jeans is $\\lvert\\vec{F}\\rvert = m.g$, the tension can also be written: \n", "$\n", "\\begin{align}\n", "\\lvert\\vec{T}\\rvert = \\frac{\\frac{1}{2}}{sin(\\alpha)}.\\lvert\\vec{F}\\rvert\n", "\\end{align}\n", "$\n", "\n", "As a consequence:\n", "* For angles between 0$^\\circ$ and 30$^\\circ$, $sin(\\alpha)$ will be between 0 and $\\frac{1}{2}$ and therefore we will have $\\lvert\\vec{T}\\rvert > \\lvert\\vec{F}\\rvert$\n", "* For angles between 30$^\\circ$ and 90$^\\circ$ (although such high values do not really make sense in the real situation), $sin(\\alpha)$ will be between $\\frac{1}{2}$ and 1 and therefore we will have $\\lvert\\vec{T}\\rvert < \\lvert\\vec{F}\\rvert$\n", "\n", "

\n",
" Activity

\n", "\n", "**Execute the code cell below** to see the resulting graph.\n", "

\n"
]
},
{
"cell_type": "code",
"execution_count": null,
"metadata": {},
"outputs": [],
"source": [
"# Let's define the boundaries of the graph\n",
"a_start = 20 # start the angle at 20°\n",
"a_stop = 0 # stop at O°\n",
"\n",
"# Now we generate 100 possible values for the angle alpha between a_start and a_stop (excluded)\n",
"a_deg = np.linspace(start=a_start, stop=a_stop, num=100, endpoint = False)\n",
"print(\"The 100 different values for the angle alpha:\\n\", a_deg, \"\\n\")\n",
"\n",
"# Since our previously defined function works with angles expressed in radians, we first convert our list of angles from degrees to radians\n",
"a_rad = degrees_to_radians(a_deg)\n",
"\n",
"# Then we compute the value of the tension for all the 100 possible angles alpha using our previously defined function\n",
"t = tension_norm(g, m, a_rad)\n",
"print(\"The corresponding values for the tension in the cable:\\n\", t)\n",
"\n",
"# Finally we generate the plot and customize its appearance\n",
"fig = figure(title='Tension in the cable', x_axis_label = 'Angle ⍺ (°)', y_axis_label = 'Tension T (N)', x_range=(a_deg[0],0), width=500, height=400, toolbar_location=None)\n",
"fig.line(a_deg, t, color=\"red\", line_width=2)\n",
"show(fig)"
]
},
{
"cell_type": "markdown",
"metadata": {},
"source": [
"\n", "\n", "**Execute the code cell below** to see the resulting graph.\n", "

\n",
" Activity

\n", "\n", "What happens to the tension when the angle $\\alpha$ gets closer to 0$^\\circ$? \n", "Could we ever have $\\alpha$ = 0$^\\circ$, i.e. **could we ever pull the cable taught completely horizontally**?\n", " \n", "Note down your answer in the cell below, then you can check the solution by clicking on the \"...\" below.\n", "

\n"
]
},
{
"cell_type": "raw",
"metadata": {},
"source": [
"Just note down your answer here (this cell is for note taking)."
]
},
{
"cell_type": "markdown",
"metadata": {},
"source": [
"\n", "\n", "What happens to the tension when the angle $\\alpha$ gets closer to 0$^\\circ$? \n", "Could we ever have $\\alpha$ = 0$^\\circ$, i.e. **could we ever pull the cable taught completely horizontally**?\n", " \n", "Note down your answer in the cell below, then you can check the solution by clicking on the \"...\" below.\n", "

\n",
"\n",
"**Solution** - you can check your answer with the solution by clicking on the \"...\" below.\n",
" \n",
"

"
]
},
{
"cell_type": "markdown",
"metadata": {
"jupyter": {
"source_hidden": true
},
"tags": []
},
"source": [
"\n",
" Solution

\n", "\n", "The curve on the graph indicates that the tension in the cable is getting infinitely big as the angle $\\alpha$ gets closer to 0$^\\circ$. \n", "This means that we will never be able to pull the cable taught horizontally as this would mean an infinite tension (the cable would break before that).\n", "\n", " \n", "From a mathematical point of view, remember that the tension is defined by the following equation: \n", "$\n", "\\begin{align}\n", "\\lvert\\vec{T}\\rvert = \\frac{\\frac{1}{2}.m.g}{sin(\\alpha)}\n", "\\end{align}\n", "$\n", "\n", "When $\\alpha$ tends toward 0, $sin(\\alpha)$ also tends toward 0, therefore the limit of the expression defining the tension in the cable is:\n", "\n", "$\n", "\\begin{align}\n", "\\lim_{\\alpha\\to0}\\; \\lvert\\vec{T}\\rvert = \\lim_{\\alpha\\to0}\\;\\frac{\\frac{1}{2}.m.g}{sin(\\alpha)} = +\\infty\n", "\\end{align}\n", "$\n", "\n", "You can find out the detailed answer to this question below, in the [solution section of the notebook](#Solution:-Is-it-possible-to-pull-the-cable-taut-completely-horizontally?).\n", "\n", "

"
]
},
{
"cell_type": "markdown",
"metadata": {},
"source": [
"\n", "\n", "The curve on the graph indicates that the tension in the cable is getting infinitely big as the angle $\\alpha$ gets closer to 0$^\\circ$. \n", "This means that we will never be able to pull the cable taught horizontally as this would mean an infinite tension (the cable would break before that).\n", "\n", " \n", "From a mathematical point of view, remember that the tension is defined by the following equation: \n", "$\n", "\\begin{align}\n", "\\lvert\\vec{T}\\rvert = \\frac{\\frac{1}{2}.m.g}{sin(\\alpha)}\n", "\\end{align}\n", "$\n", "\n", "When $\\alpha$ tends toward 0, $sin(\\alpha)$ also tends toward 0, therefore the limit of the expression defining the tension in the cable is:\n", "\n", "$\n", "\\begin{align}\n", "\\lim_{\\alpha\\to0}\\; \\lvert\\vec{T}\\rvert = \\lim_{\\alpha\\to0}\\;\\frac{\\frac{1}{2}.m.g}{sin(\\alpha)} = +\\infty\n", "\\end{align}\n", "$\n", "\n", "You can find out the detailed answer to this question below, in the [solution section of the notebook](#Solution:-Is-it-possible-to-pull-the-cable-taut-completely-horizontally?).\n", "\n", "

\n", "\n", "---" ] }, { "cell_type": "markdown", "metadata": {}, "source": [ "# What have you learned so far?\n", "\n", "

\n",
" Activity

\n", "\n", "Write **2 things you have learned** about the **tension force** in a cable that is used to suspend an object:\n", " \n", "

"
]
},
{
"cell_type": "raw",
"metadata": {},
"source": [
"- \n",
"- "
]
},
{
"cell_type": "markdown",
"metadata": {},
"source": [
"\n", "\n", "Write **2 things you have learned** about the **tension force** in a cable that is used to suspend an object:\n", " \n", "

\n",
"\n",
"Can you identify **other real-life situations** in which cables are used to suspend objects or in which cables are taut between poles? \n",
"Write 2 ideas:\n",
"\n",
"

"
]
},
{
"cell_type": "raw",
"metadata": {},
"source": [
"- \n",
"- "
]
},
{
"cell_type": "markdown",
"metadata": {},
"source": [
"\n", "\n", "Congratulations, you have completed Activity 2! \n", "Below are some optional exercises if you want to continue.\n" ] }, { "cell_type": "markdown", "metadata": {}, "source": [ "

\n", "\n", "---" ] }, { "cell_type": "markdown", "metadata": {}, "source": [ "# Additional exercises (optional)" ] }, { "cell_type": "markdown", "metadata": {}, "source": [ "## How does our Python function behave with $\\alpha$ = 0?\n", "\n", "What happens with our piece of Python code when we say that the cable it taut completely horizontal, i.e. the angle $\\alpha$ = 0$^\\circ$, which is actually impossible in real life? \n", "\n", "

\n",
" Activity

\n", "\n", "In the cell below, call your ``tension_norm`` function with a value of $0$ for `alpha` and execute the cell. \n", "What happens? \n", "

"
]
},
{
"cell_type": "code",
"execution_count": null,
"metadata": {},
"outputs": [],
"source": [
"# Compute the tension with g = 9.81 m.s-2, m = 3 kg and alpha = 0°\n",
"T = 1 # REPLACE 1 WITH THE CALL TO YOUR FUNCTION tension_norm HERE\n",
"\n",
"# And print the result\n",
"print(\"Norm of the tension in the cable: T =\", T, \"N\")"
]
},
{
"cell_type": "markdown",
"metadata": {},
"source": [
"\n", "\n", "In the cell below, call your ``tension_norm`` function with a value of $0$ for `alpha` and execute the cell. \n", "What happens? \n", "

\n",
" Solution

\n", "\n", "Here is how to call your function:\n", "

"
]
},
{
"cell_type": "markdown",
"metadata": {},
"source": [
"## What are the consequences?\n",
"\n",
"It is important to know that high tension in a cable is very dangerous and can lead to serious accidents, especially when the cable is not well adapted for its intended use (i.e. not strong enough) or when the cable is progressively deteriorating with use. \n",
"\n",
"\n", "\n", "Here is how to call your function:\n", "

`T = tension_norm(g, m, 0)`

\n",
" \n",
"When you execute this code, two things happen:\n",
"* You get a warning message which says `RuntimeWarning: divide by zero encountered in double_scalars`\n",
"* The function nonetheless returns a value which is `inf`\n",
"\n",
"When the angle $\\alpha$ = 0 then $sin(\\alpha)$ = 0 and therefore we divide `(.5 * jeans_mass * gravity)` by 0 so on one hand, mathematically speaking, we know that the result should be $+\\infty$. On the other hand, we also know that usually division by 0 is not well supported by computers.\n",
"\n",
"Actually, division by 0 is not supported in standard Python. \n",
"You can create a code cell (click on the `+` icon in the toolbar above) and try to execute the following computation to see what happens: ``(0.5 * 3 * 9.81)/0``\n",
" \n",
"Now, because in the calculation we use the function `np.sin()` from the Numpy library, our data is automatically converted to Numpy types, which support division by zero and returns the \"real\" result which is $+\\infty$. \n",
"By convention, Numpy also generates a warning message but this can be deactivated when not necessary. \n",
"If you are curious, you can take a look at [the errors generated by Numpy for floating-point calculations](https://docs.scipy.org/doc/numpy/reference/generated/numpy.seterr.html#numpy.seterr).\n",
"\n",
"\n",
" Activity

\n", "\n", "Execute the cell below to see the video and watch the first minute (sound is not necessary). \n", "What happens at time 0:53? Why?\n", "

"
]
},
{
"cell_type": "code",
"execution_count": null,
"metadata": {},
"outputs": [],
"source": [
"from IPython.display import YouTubeVideo\n",
"YouTubeVideo('KIbd5zBek5s', 600, 337, start=45, mute=1)"
]
},
{
"cell_type": "markdown",
"metadata": {},
"source": [
"\n", "\n", "Execute the cell below to see the video and watch the first minute (sound is not necessary). \n", "What happens at time 0:53? Why?\n", "

\n",
" Solution

\n", "\n", "At time 0:53, the cable desintegrates completely. This is probably because it has grown weaker and weaker after repeated use until it cannot withstand the high tension anymore.\n", " \n", "

"
]
},
{
"cell_type": "markdown",
"metadata": {},
"source": [
"## Improving the `tension_norm` function\n",
"\n",
"As you have noticed, each time you want to call your `tension_norm` function with an angle expressed in degrees, you have first to convert your angle from degrees to radians. \n",
"A better design would be to do the conversion of the angle inside the function so that you can pass it angles in degrees directly.\n",
"\n",
"\n", "\n", "At time 0:53, the cable desintegrates completely. This is probably because it has grown weaker and weaker after repeated use until it cannot withstand the high tension anymore.\n", " \n", "

\n",
" Activity

\n", "\n", "Define a new function ``tension_norm_degrees`` which converts `alpha` from degrees to radians using the ``degrees_to_radians`` function, then calls the ``tension_norm`` function to compute the norm of the tension and returns the result. \n", " \n", "Then **execute the code cell** so that this function gets defined and then tested with an angle $\\alpha$ of 20$^\\circ$. \n", "If your function works correctly, the result of the execution should be ``Tension norm: T = 43.023781748399834 N``.

\n", "You can further check your answer with the solution by clicking on the \"...\" below. \n", "\n", "

"
]
},
{
"cell_type": "code",
"execution_count": null,
"metadata": {},
"outputs": [],
"source": [
"# Define your function\n",
"def tension_norm_degrees(g, m, alpha):\n",
" # WRITE YOUR CODE HERE\n",
" return\n",
"\n",
"# Let's test it: the execution of these lines should give T = 43.023781748399834 N\n",
"alpha = 20\n",
"T = tension_norm_degrees(g, m, alpha) \n",
"print(\"Tension norm: T =\", T, \"N\")"
]
},
{
"cell_type": "markdown",
"metadata": {},
"source": [
"\n", "\n", "Define a new function ``tension_norm_degrees`` which converts `alpha` from degrees to radians using the ``degrees_to_radians`` function, then calls the ``tension_norm`` function to compute the norm of the tension and returns the result. \n", " \n", "Then **execute the code cell** so that this function gets defined and then tested with an angle $\\alpha$ of 20$^\\circ$. \n", "If your function works correctly, the result of the execution should be ``Tension norm: T = 43.023781748399834 N``.

\n", "You can further check your answer with the solution by clicking on the \"...\" below. \n", "\n", "

\n",
" Solution

\n", "\n", "Your function could look like the following:\n", "\n", "

Note: Another way to test such a function would be to use [Python assertions](https://zetcode.com/python/assert/). \n", "

However, comparing float values can get tricky. These can be made simpler by the use of more elaborate libraries such as ``pytest``, which support [approximations](https://randycoulman.com/blog/2018/06/19/comparing-floats-in-tests/).\n", "

"
]
},
{
"cell_type": "markdown",
"metadata": {},
"source": [
"\n", "\n", "Your function could look like the following:\n", "\n", "

```
def tension_norm_degrees(g, m, alpha):\n",
" apha_rad = degrees_to_radians(alpha)\n",
" tension = tension_norm(g, m, apha_rad)\n",
" return tension\n",
"
```

\n",
"\n",
"Of course the three lines in this function can be combined into just one: \n",
"\n",
"```
def tension_norm_degrees(g, m, alpha):\n",
" return tension_norm(g, m, degrees_to_radians(alpha))\n",
"
```

\n",
" \n",
"Note: Another way to test such a function would be to use [Python assertions](https://zetcode.com/python/assert/). \n", "

However, comparing float values can get tricky. These can be made simpler by the use of more elaborate libraries such as ``pytest``, which support [approximations](https://randycoulman.com/blog/2018/06/19/comparing-floats-in-tests/).\n", "

\n", "\n", "---" ] }, { "cell_type": "markdown", "metadata": { "toc-hr-collapsed": true, "toc-nb-collapsed": true }, "source": [ "# Solution: Is it possible to pull the cable taut completely horizontally?\n", "\n", "In this section we demonstrate the physical and mathematical reasoning to answer this question.\n", "\n", "### Goal\n", "\n", "For the cable to be taut completely horizontally means that the angle $\\alpha$ that the cable makes with the horizon is 0$^\\circ$. \n", "In other words, the question could be reframed as: which mass $m_{cw}$ should we put as a counterweight to get $\\alpha$ = 0$^\\circ$?\n", "If we can find a 'reasonable' value for the counterweight then it will mean that it is possible to pull the cable taut completely horizontally.\n", "\n", "To answer this question, we therefore look for an expression relating the **mass of the counterweight $m_{cw}$** and the **angle $\\alpha$** that the cable makes with the horizon.\n", "\n", "\n", "### Method\n", "\n", "Given that the system is in static equilibrium, the sum of external forces exerted on the system will be zero, so using Newton's second law should be easy. The force that the counterweight exerts on the system will involve the mass of the counterweight so we should be able to rewrite Newton's second law to get an expression of the form $m_{cw} = ...$.\n", "\n", "### Hypotheses and simplifications\n", "\n", "We make the following assumptions and simplifications:\n", "* the jeans are considered as positioned exactly mid-way between the poles so the tension is equal on both sides of the cable\n", "* we represent the jeans by the point at which they are suspended\n", "* the cable is considered as rigid (not bended), with a negligible mass\n", "* the pulley is considered as perfect, without mass nor friction\n", "* we consider the static equilibrium obtained after changing the weight, once the system is stabilized\n", "\n", "### Resolution\n", "\n", "First, let's draw a diagram and represent the different forces involved.\n", "\n", "\n", "The *forces applied on the jeans* are:\n", "* the weight: $\\vec F_j = m_j \\vec g$ \n", "* the force exerted by the cable on each side of the jeans: assuming the jeans are suspended at the exact center of the cable, then the tension applied on each of the two sides is equally distributed with two tensions of equal magnitude $T$ on each side but with opposite directions on the $x$ axis, which combine into a vertical resulting tension $\\vec T_r$\n", "\n", "From Newton's second law in a static equilibrium we can write: $\\sum \\vec F_j = \\vec 0$ \n", "With the forces on the jeans we get: $\\vec F_j + \\vec T_r = 0$ \n", "\n", "If we project on $x$ and $y$ axes, we get: \n", "$\\left\\{\\begin{matrix} F_{jx} + T_{rx} = 0 \\\\ F_{jy} + T_{ry} = 0\\end{matrix}\\right. $\n", "\n", "Since $\\vec T_r$ is the result of two tensions of same magnitude $T$ on both sides of the jeans but of opposite directions on the $x$ axis, we can decompose $T_{rx}$ and $T_{ry}$ into the $x$ and $y$ components of $T$: \n", "$\\left\\{\\begin{matrix} T_{rx} = T_{x} - T_{x} \\\\ T_{ry} = 2.T_{y} \\end{matrix}\\right. $\n", "\n", "\n", "Now we can use this in our previous equations, which gives: \n", "$\\left\\{\\begin{matrix} F_{jx} + T_{x} - T_{x} = 0 \\\\ F_{jy} + 2.T_{y} = 0\\end{matrix}\\right. $\n", "\n", "\n", "Since the two $T_{x}$ cancel out we cannot know their value yet and so we focus on the second equation: \n", "$\\begin{align} F_{jy} + 2.T_y = 0 \\end{align}$\n", "\n", "The component of the weight on the y axis is $F_{jy} = - m_j.g$, which gives us: \n", "$\\begin{align} - m_j.g + 2.T_y = 0 \\end{align}$\n", "\n", "Using the angle $\\alpha$ we can get the tension $T_y$ expressed as a function of T since $sin(\\alpha) = \\frac{T_y}{T}$, therefore $T_y = T.sin(\\alpha)$\n", "\n", "By replacing $T_y$ by this expression in the above equation we get: \n", "$\\begin{align} - m_j.g + 2.T.sin(\\alpha) = 0 \\end{align}$\n", "\n", "From there we can get $T$, and this is equation number $(1)$: \n", "\n", "$\n", "\\begin{align}\n", "T = \\frac{m_j.g}{2.sin(\\alpha)}\n", "\\end{align}\n", "$\n", "\n", " \n", "\n", "We now want to make the mass of the counterweight appear in this expression. \n", "So we will now look at the forces applied on the *counterweight*.\n", "\n", " \n", "\n", "The forces applied on the *counterweight* are:\n", "* the weight: $\\vec F_{cw} = m_{cw} \\vec g$ \n", "* the force exerted by the cable: a simple pulley simply changes the direction of the tension so the tension applied on the counterweight is therefore of the same magnitude $T$ as before but with a different direction - we will note it $\\vec T$\n", "\n", "From Newton's second law in a static equilibrium we can write: $\\sum \\vec F_{cw} = \\vec 0$ \n", "With the forces involved in our problem : $\\vec F_{cw} + \\vec T = \\vec 0$ \n", "\n", "All forces being vertical, there is no need to project on $x$ so we get: $- F_{cw} + T = 0$ \n", "We replace the weight by its detailed expression: $-m_{cw}.g + T = 0$ \n", "Now we can express $T$ as a function of the other parameters, which is equation number $(2)$: $T = m_{cw}.g$ \n", "\n", " \n", "\n", "Let's now summarize what we have so far with equations $(1)$ and $(2)$: \n", "\n", "$\n", "\\begin{align}\n", "\\left\\{\\begin{matrix}T = \\frac{m_j.g}{2.sin(\\alpha)} \\\\ T = m_{cw}.g\\end{matrix}\\right. \n", "\\end{align}\n", "$\n", "\n", "These two equations combined give us:\n", "\n", "$\n", "\\begin{align}\n", "\\frac{m_j.g}{2.sin(\\alpha)} = m_{cw}.g\n", "\\end{align}\n", "$\n", "\n", "This allow us to find the mass of the counterweight as a function of the *mass of the jeans* and of the *angle that the cable makes with the horizon*: \n", "\n", "

\n",
"\n",
"$$m_{cw} = \\frac{m_j}{2.sin(\\alpha)}$$\n",
"\n",
"

"
]
},
{
"cell_type": "markdown",
"metadata": {},
"source": [
"### Conclusion\n",
"\n",
"When the cable approaches the horizon, $\\alpha$ is really small i.e. really close to zero. \n",
"This means that $sin(\\alpha)$ is also close to zero, which means in turn that $m_{cw}$ is very big.\n",
"Therefore, **the more we want the cable to be close to the horizon, the bigger $m_{cw}$ we will need!**\n",
"\n",
"Mathematically speaking, the limit of the expression defining the mass of the counterweight when alpha tends to 0 is:\n",
"\n",
"$\n",
"\\begin{align}\n",
"\\lim_{\\alpha\\to0}m_{cw} = \\lim_{\\alpha\\to0}\\frac{m_j}{2.sin(\\alpha)} = +\\infty\n",
"\\end{align}\n",
"$\n",
"\n",
"So basically, we would need to put an **infinitely heavy counterweight** to make the angle null and that is why it is impossible to get the cable taut so that it is absolutely straight (the cable would break before that!)..."
]
},
{
"cell_type": "markdown",
"metadata": {},
"source": [
"### Numerical application\n",
"\n",
"We have already defined above the constant `m` which represents the mass of our jeans (3 kg).\n",
"\n",
"Let's define a Python function that represents the above equation. \n",
"The function takes two input parameters, `m` which represents the mass of our jeans and `alpha`, the angle that the cable makes with the horizon. \n",
"It returns the mass of the counterweight as computed with the equation above."
]
},
{
"cell_type": "code",
"execution_count": null,
"metadata": {},
"outputs": [],
"source": [
"# Define the function\n",
"def counterweight_mass(m, alpha):\n",
" return m / (2 * np.sin(alpha))\n",
"\n",
"# And display a message once it is defined\n",
"print(\"Function defined.\")"
]
},
{
"cell_type": "markdown",
"metadata": {},
"source": [
"For an angle $\\alpha$ of $1.5^\\circ$, we need to put a counterweight of:"
]
},
{
"cell_type": "code",
"execution_count": null,
"metadata": {},
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"source": [
"# Choose a value for angle alpha\n",
"alpha = 1.5\n",
"\n",
"# Then convert it to radians\n",
"alpha_radians = degrees_to_radians(alpha)\n",
"\n",
"# Compute the mass of the counterweight using our function\n",
"m_cw = counterweight_mass(m, alpha_radians)\n",
"\n",
"# And print the result\n",
"print(\"Mass of the counterweight: m_cw =\", m_cw, \"kg\")"
]
}
],
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