Apps-On-Physics/C2/Inclined-Plane/English-timed
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| Time | Narration |
| 00:01 | Welcome to the Spoken Tutorial on Inclined Plane. |
| 00:05 | In this tutorial we will learn to, |
| 00:08 | Simulate the motion of a load on an inclined plane. |
| 00:12 | Resolve the vector components using basic trigonometry. |
| 00:17 | Calculate the vector components. |
| 00:20 | Here I am using,
Ubuntu Linux OS version 16.04 and Firefox Web Browser version 62.0.3 |
| 00:32 | To follow this tutorial, learner should be familiar with Apps on Physics. |
| 00:39 | For the pre-requisites tutorials please visit this site. |
| 00:44 | Let us define an inclined plane. |
| 00:47 | An inclined plane, is a flat supporting surface tilted at an angle. |
| 00:52 | It has one end higher than the other. |
| 00:56 | It is used for raising or lowering a load. |
| 01:00 | Use of an inclined plane provides greater mechanical advantage. |
| 01:05 | Examples of an inclined plane are ramps, slides, stairs, water slides and others. |
| 01:14 | Use the given link to download the Apps. |
| 01:18 | I have already downloaded Apps on Physics to my Downloads folder. |
| 01:23 | In this tutorial we will use,
Inclined Plane App. |
| 01:29 | After downloading, html5phen folder appears in the Downloads folder. |
| 01:35 | Double click on html5phen folder. |
| 01:39 | Now double-click on the phen folder. |
| 01:42 | In this folder, we see Apps in java script and htm format. |
| 01:48 | We will use the Apps with htm file format. |
| 01:52 | To open Inclined Plane press Ctrl, F keys simultaneously. |
| 01:58 | In the search bar type inclined plane. |
| 02:02 | Right click on inclinedplane_en.htm file. |
| 02:07 | Select the option Open With Firefox web Browser. |
| 02:12 | Inclined Plane App opens in the browser. |
| 02:16 | This is the interface of Inclined plane. |
| 02:20 | The green panel shows different parameters that we can change. |
| 02:25 | Reset button on the top of the green panel helps to edit values. |
| 02:30 | The yellow Start button is a toggle button for Start/Pause and Resume. |
| 02:37 | Slow motion check-box is used to observe the motion steadily. |
| 02:42 | Then we have Springscale and Force vectors radio buttons.
By default Springscale is selected. |
| 02:52 | Click on Start button. |
| 02:55 | Notice that a load is pulled by the springscale. |
| 02:59 | Click on the Pause button. |
| 03:02 | Now select Force vectors radio button. |
| 03:06 | Observe that there are five arrows pointing in different directions. |
| 03:11 | We can change the values of: Angle of inclination, Weight and Coefficient of friction in the white colour boxes. |
| 03:20 | Note that these values can be changed within certain limits. |
| 03:25 | Click on Reset button. |
| 03:28 | Here we can change the Angle of inclination from 0 degrees to 90 degrees. |
| 03:35 | Let’s change the Angle of inclination to 45 degrees and press Enter. |
| 03:41 | Now click on Start button. |
| 03:44 | When the load reaches the middle of the inclined plane click on Pause button. |
| 03:50 | Notice that the pink vector shows the force of gravity(mg). |
| 03:54 | It tries to pull the load towards the center of the Earth. |
| 03:59 | The blue and red vectors are the resolution vectors of gravity. |
| 04:04 | The red vector is perpendicular to the surface of the inclined plane. |
| 04:09 | The blue vector is parallel to the surface of the inclined plane. |
| 04:15 | If theta(θ) is 45 degrees, then this angle is 90 degrees.
As it is perpendicular to the surface of the earth. |
| 04:23 | From the triangle’s geometry this angle would be (90-θ). |
| 04:29 | To calculate the magnitude of the forces we need to know theta value. |
| 04:35 | Now these two lines are parallel and if we assume that this line is a transversal line.
Then angle (90-θ) is equal this angle through interior angle property. |
| 04:48 | Recall that red vector is perpendicular to the inclined plane. |
| 04:53 | Here the angle would be 90 degrees. |
| 04:57 | Let us assume this angle as x. |
| 05:01 | So 90 minus theta plus 90 plus x equals to 180
Therefore, x equals to theta |
| 05:12 | Using basic trigonometry we can resolve the parallel and perpendicular components.
Consider this right angle triangle. |
| 05:21 | Here the blue parallel component is equal to the black line. |
| 05:26 | Parallel force is Sin theta. |
| 05:29 | Sin theta equals to F(parallel) upon mg. |
| 05:34 | Let’s rearrange the equation.
F(parallel) equals to mg sin theta. |
| 05:41 | Similarly we can resolve the perpendicular component. |
| 05:46 | F(perpendicular) equals to mg cos theta. |
| 05:50 | Let us solve this numerical and verify the answers with the ones shown in the App. |
| 05:57 | Click on the Reset button to reset the App. |
| 06:01 | In the App change the values according to the numerical. |
| 06:06 | First let us convert 1.02 Kg into Newton and enter the value in the Weight box. |
| 06:14 | Next change the Angle of inclination to 30 degrees and press Enter. |
| 06:21 | Now click on Start button. |
| 06:24 | Again click on Pause button when the load reaches the center of the inclined plane. |
| 06:30 | Observe that the App has calculated the parameters. |
| 06:36 | Next we will calculate using the formulae. |
| 06:40 | Calculated value of the parallel component is 4.99 Newton and that of normal component is 8.65 Newton. |
| 06:51 | And the necessary force is equal to the parallel force but in the opposite direction. |
| 06:59 | Let us compare the answers with the ones shown in the App. |
| 07:04 | Observe that the calculated values are comparable to the measured values. |
| 07:10 | Let's observe the effect of friction. |
| 07:13 | Click on the Reset button. |
| 07:16 | In the Coefficient of friction box type 0.5 and press Enter. |
| 07:23 | Click on the Start button. |
| 07:26 | When the load reaches the middle of the inclined plane click on Pause button. |
| 07:31 | Notice that a black vector is added to the blue vector.
This vector represents Force of friction. |
| 07:40 | In the green panel Force of friction is measured as 4.3 Newton. |
| 07:46 | Notice that the Necessary force required to pull the load has changed to 9.3 Newton. |
| 07:53 | This is because the total necessary force is, sum of parallel and frictional forces. |
| 08:00 | As an assignment solve this numerical and compare your answer with the ones shown in the App. |
| 08:08 | Let us summarise |
| 08:10 | Using this App we have Simulated the motion of a load on an inclined plane. |
| 08:17 | Resolve the vector components using basic trigonometry. |
| 08:22 | Calculated the vector components. |
| 08:26 | These Apps are created by Walter-fendt and his team. |
| 08:31 | The video at the following link summarizes the Spoken Tutorial project.
Please download and watch it. |
| 08:39 | The Spoken Tutorial Projectteam, conducts workshops using spoken tutorials
and gives certificates. For more details, please write to us. |
| 08:48 | Please post your time queries on this forum |
| 08:53 | Spoken Tutorial Project is funded by MHRD, Government of India. |
| 08:59 | This is Himanshi Karwanje from IIT-Bombay.
Thank you for joining. |
