Shadowing Practice: Equilibrium of Rigid Bodies (2D - Coplanar Forces) | Mechanics Statics | (Solved examples) - Learn English Speaking with Video
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Let’s talk about equilibrium of rigid bodies. Imagine we have a flower pot hung up like this.
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There is a weight to this flower pot which is straight down. So what’s keeping it still, in other words, in equilibrium? It’s the rope that’s attached to it, which is creating an opposite force, in other words, a tension in the rope that’s holding it in place.
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Now imagine we have an object like this, and there are a bunch of forces and moments effecting it.
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If we sum all the forces and moments, meaning we find the resultant force and the couple moment, and they are both equal to zero, then that object is in equilibrium. In simple terms, the object is not moving. So this equation means that the sum of all the forces acting on the object is equal to zero. We can break this equation down to forces in the x direction and y direction. So all the forces in the x direction added together must equal zero and the same for y axis forces. This equation means the sum of all the moments, including ones being applied, and ones being created by the forces, is equal to zero. Today, we are going to look at just two-dimensional problems, so just objects in the x and y plane. When we look at an object, we need to look at something called support reactions. These reactions at the supports are the forces and couple moments that keep an object from moving. They create opposite forces and moments so that an object doesn’t translate or rotate. So in our flower pot example, the rope is a support reaction. The rope creates a tension which keeps the flower pot from falling. Let’s explore more support reactions. Here we have a bar. there are 2 support reactions. One is a roller and the other is a pin. Let’s focus on just the roller first. Assume that our object can move, and we apply a force from the left, notice how our bar moves. That’s because a roller can only prevent vertical movement. If we apply a force from top, the roller prevents the bar from moving down by creating an opposite force.
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So if we’re given an object with a roller as a support, that roller can only exert a vertical force which prevents the object from translating in the vertical direction. Now, let’s look at the pin side. With a pin, the bar can rotate, but it can’t move vertically or horizontally. So that means at a pin, it would exert 2 forces, one in the x direction and another in the y direction.
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In all of supports we just looked at, rotation can occur, however, imagine we fix our beam to the ground or to a wall. Now it can’t move in the horizontal direction or the vertical direction, and it also cannot rotate. In this case, we’d have 2 reactions, one in the x axis and another in the y axis, and we would also have a couple moment because rotation cannot occur. In other words, an opposite couple moment must be developed so that the bar doesn’t rotate due to the moments created by forces. Your textbooks will contain a list of these supports and what type of reactions these supports develop, so please take a look at those. Lastly, to solve the problems you face, you must know how to break forces into components and how to figure out moments created by forces.
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If you don’t remember, or need a refresh, please check the description.
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Now let’s get to some examples and see how we can actually solve for unknown reactions.
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Let’s take a look at this problem where we need to find the reactions at pin A and the tension in the rope. The first step is to draw a free body diagram. Since we have a pin at A, we know there will be 2 reactions, Ax and Ay. Next, we look at the forces in our diagram.
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We have a 26kN force, and we draw it broken into it’s x and y components.
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Then we have a 40 kN force straight down. Lastly, we have the tension in the rope, which can also be broken into it’s x and y components. It is time for us to write our equations of equilibrium.
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First, we will consider the x-axis forces. What we are saying is that for this object to stay in equilibrium, the sum of all the forces in the x direction must equal to zero. We will assume right to be positive. So we have Ax, then we have the 10kN force, but notice how our 10kN force is negative, that’s because we assumed right to be positive and our force is facing left. Lastly, we have the x component of the tension in the rope. All of this added together is equal to zero. Next, an equation for y axis forces. We will assume up to be positive. It’s the same as before, but now we are just looking at the vertical forces. We now have 2 equations with 3 unknowns, which means we need one more equation, which is going to be a moment equation. This is usually the case with most questions, you will need 3 equations, 1 for x axis forces, 1 for y axis forces, and then one moment equation. We want to eliminate as many unknowns as possible, so we are going to write our moment equation about point A, which gets rid of Ax and Ay. Remember, where ever we calculate the moment about, we ignore the forces applied at that point since their line of action will go through that point. So for example, if we picked point B, then we ignore the forces at point B. Now we will pick clockwise to be positive and write our moment equation about point A. The 10kN force, which is our x component won’t create a moment since it’s line of action goes through point A, the same goes for the x component of the tension in the cable. So only the y components create moments. With this, we now have 3 equations with 3 unknowns, so you can solve them anyway you like.
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Solving gives us the reactions at pin A and the tension in the cable. Notice that we got a negative value for AX, all it means is that its opposite to the direction we assumed.
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Let’s take a look at this problem where we need to find the reactions at the roller A and pin B. Our first step is to draw a free body diagram. Since it’s a roller at A, it will only have a vertical reaction. At B, we will have both the x and y reactions.
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The distributed load has to be expressed as a resultant force, which can be found by multiplying the load by the length, giving us 12 kN, and it will be placed at the center, so 2 m away from point B. Now we can write our equations of equilibrium.
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First, the x-axis forces. Note that the reaction at A is at an angle of 30 degrees, so we can break it into x and y components. Next, the y-axis forces.
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We see that we have 2 equations with 3 unknowns, so we need one more equation, which will be a moment equation about point B. We pick point B because there are 2 unknowns there, and we always want to get rid of as many unknowns as possible. We will assume clockwise to be positive. Focusing on the force at A, we have 2 components, both with create a moment. For the x component, we have a perpendicular distance of 3sin 30 degrees. For the y component, we have 4m plus 3 cos 30 degrees. The only other force is the 12 kN force, which is 2 m away from point B.
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We now have 3 equations with 3 unknowns. Solving gives us the reactions at A and B.
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In this question, we need to figure out the reactions at A and at B. Let’s draw a free body diagram. Since it’s a smooth collar at A, we will have a force perpendicular to the rod, and a moment created about A. At point B, which is a smooth contacting surface, we will have a reaction force that acts perpendicular to the surface. Since the forces are perpendicular, all that happens is that the given ratio triangles just flip. There is also a 100 N force that’s straight down, and we have a moment of 20 nm applied at the bend. Now we are going to write our equations of equilibrium. Let’s start with an equation for x-axis forces.
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Note that we are using the x components of the forces.
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Next, an equation for y-axis forces. From these two equations, we can solve for forces FA and FB. Now we still need to figure out the moment created about the smooth collar. For that, we can write a moment equation about point A and we will assume clockwise to be positive. So at A, the reactions at A are ignored since their line of action goes through point A, however, we still need to consider the moment about point A.
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That moment must be included in our equation. Next, we have the 100N force, and the perpendicular distance is 0.3 m. After that, we have the 20 Nm applied at the bend, it’s a clockwise moment, so it’ll be positive. Finally, we have the x and y components of our force at B.
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For the x component, the perpendicular distance is 0.2 m and for the y component, it’s 0.6 m.
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The x component creates a clockwise moment and the y component creates a counter clockwise moment.
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Solving gives us the moment at A, but notice that it’s a negative value.
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All it means is that the moment is opposite to our assumption, so it’s actually counter clockwise.
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Let’s take a look at one last example. In this example, we have a weight attached to a rod and on one end, it’s pinned, and at the other end, we have a spring. We need to figure out the angle created when the system is in equilibrium. First, we have to draw a free body diagram.
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At pin A, we have 2 reactions. We also have the moment applied at A. Note that this moment is not created about pin A by other forces, it is something that was manually applied. Now we look at the weight of the cylinder, that’s mass times the acceleration due to gravity.
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Here, I think it requires a bit of intuition, but the angle that’s created at this point, is the same angle that’s created here because if we place our coordinate system like this and move it, you can see that it’d be the same. Lastly, we have the force of the spring that’s straight upwards.
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Now we have to put on our thinking caps, so initially, the spring has an unstretched length of 1 m. In the second state, when everything moved, the spring stretched a bit. How much did it stretch? Well, let’s draw a right-angle triangle like this. Notice how the length of stretch is the opposite side. So we can use sine to figure it out. Now that’s the length the spring stretched, so to find the force, we use hooks law. The stiffness is 600 N/m. Let’s simplify.
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Now also notice that when we draw this force, the angle that it creates is once again the same.
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Now we can figure out this angle, in fact, we actually only need a single moment equation about point A. That would get rid of the 2 unknown reactions, and only leave the angle to be found since we know all the other forces. Let’s go over this equation. First, we have the 600 Nm moment applied at A. Then we look at the weight of the cylinder. Only the y component of the weight creates a moment since the x component’s line of action goes through point A. The perpendicular distance is 1.5 m. The same is true for the force of the spring. Only the y component does any work and the perpendicular distance is 3 m. Note that we assumed clockwise moments to be positive so the spring force creates a counter clockwise moment. Now we can solve for theta. We are looking for values between 0 degrees and 90 degrees. If you graph it, you will see that there are 2 such values, and those are our answers. Always remember that if an object is in equilibrium, the sum of the forces must equal zero. With that, I hope this video helped and if it did, please consider sharing it with your friends and class mates, they too might find it helpful. Thanks for watching and best of luck with your studies!
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Vocabulario y notas de pronunciación para esta lección
Esta lección de conversación de nivel C1 se basa en el vídeo “Equilibrium of Rigid Bodies”. Las palabras que más se repiten: force, moment, equation, reaction, component. Este vídeo tiene 32 frases y 2137 palabras para practicar shadowing. La parte hablada dura 11:28. El hablante habla rápido, unas 186 palabras por minuto, así que espera sonidos enlazados y reducidos. Solo el 79 % de las palabras está entre las 3.000 más comunes del inglés, por lo que el vocabulario es exigente.
Vocabulario clave de este vídeo
Las 15 palabras más avanzadas del vídeo, con su pronunciación y significado:
| Palabra | Pronunciación | Significado |
|---|---|---|
| equation sustantivo | /ɪˈkweɪ.ʒən/ | ecuación |
| axis sustantivo | /ˈæksɪs/ | eje |
| clockwise adverbio | /ˈklɒkwaɪz/ | en el sentido de las agujas del reloj, en sentido horario |
| equilibrium sustantivo | /ɛkwɪˈlɪbɹɪəm/ | equilibrio |
| perpendicular adjetivo | /ˌpɜː.pənˈdɪk.jə.lə(ɹ)/ | perpendicular |
| roller sustantivo | /ˈɹoʊlɚ/ | rodillo |
| rope sustantivo | /ɹoʊp/ | soga, maroma |
| diagram sustantivo | /ˈdaɪ.ə.ɡɹæm/ | diagrama |
| rotate verbo | /ˈɹoʊ.teɪt/ | girar |
| exert verbo | /ɪɡˈzɝt/ | ejercer |
| attach verbo | /əˈtæt͡ʃ/ | adjuntar, atachar |
| triangle sustantivo | /ˈtɹaɪˌæŋɡəl/ | triángulo |
| translate verbo | /tɹænzˈleɪt/ | traducir, trasladar |
| cylinder sustantivo | /ˈsɪləndɚ/ | cilindro |
| collar sustantivo | /ˈkɑlɚ/ | cuello |
Phrasal verbs que vas a escuchar
| Palabra | Significado |
|---|---|
| figure out verbo | averiguar, captar |
| go through verbo | atravesar |
| go over verbo | sentar, pasar |
Gramática en este vídeo
Las estructuras que más usa el hablante, con las palabras exactas del vídeo:
| Estructura | En el vídeo |
|---|---|
| Voz pasiva be + participio pasado — importa lo que ocurre, no quién lo hace | being applied · must be developed · be broken |
| Oraciones de relativo who / which + oración — información extra sobre una persona o cosa | pot which is · rope, which can · unknowns, which means |
Pronunciación a tener en cuenta
El hablante usa 10 contracciones y formas reducidas, como can't, doesn't, don't. Dilas en su forma corta, tal como las oyes.
- Los sonidos de “sh” y “zh”: equation /ɪˈkweɪ.ʒən/, rotation /ɹoʊˈteɪʃən/, intuition /ˌɪn.tjuːˈɪʃ.ən/, refresh /ɹɪˈfɹɛʃ/, acceleration /əkˌsɛl.əˈɹeɪ.ʃən/
- Palabras largas — cuida el acento: equilibrium /ɛkwɪˈlɪbɹɪəm/, perpendicular /ˌpɜː.pənˈdɪk.jə.lə(ɹ)/, vertically /ˈvɝ.tɪ.kəli/, intuition /ˌɪn.tjuːˈɪʃ.ən/, manually /ˈmænj(u)əliː/
Sonidos difíciles para hispanohablantes:
- /z/ — sonora, no /s/: clockwise /ˈklɒkwaɪz/, exert /ɪɡˈzɝt/, resultant /ɹɪˈzʌltənt/, translate /tɹænzˈleɪt/, upwards /ˈʌpwɚdz/
Cómo practicar con este vídeo
- Escucha el vídeo entero una vez sin hablar y anota las palabras que no conoces.
- Empieza a velocidad 0,75×, haz shadowing frase por frase y vuelve a la velocidad normal cuando te resulte fácil.
- Grábate y compara con el original, prestando atención a palabras como equation, axis, clockwise.
¿Qué es la Técnica de Shadowing?
Shadowing es una técnica de aprendizaje de idiomas respaldada por la ciencia, desarrollada originalmente para la formación de intérpretes profesionales y popularizada por el políglota Dr. Alexander Arguelles. El método es simple pero poderoso: escuchas audio en inglés nativo y lo repites en voz alta de inmediato, como una sombra que sigue al hablante con solo 1-2 segundos de retraso. A diferencia de la escucha pasiva o los ejercicios de gramática, el shadowing obliga a tu cerebro y músculos de la boca a procesar y reproducir simultáneamente patrones de habla reales. Las investigaciones muestran que mejora significativamente la precisión de la pronunciación, la entonación, el ritmo, el habla conectada, la comprensión auditiva y la fluidez al hablar, convirtiéndola en una de las metodologías más efectivas para la preparación del IELTS Speaking y la comunicación en inglés en el mundo real.