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Problem 63 Hard Difficulty

ssm A 44-kg chandelier is suspended 1.5 m below a ceiling by three wires, each of which has the same tension and the same length of 2.0 m (see the drawing). Find the tension in each wire.

Answer

$1.92 \times 10^{2} \mathrm{N}$

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Video Transcript

in this problem, we're asked to determine the tension in three wires that are used to support a 44 kilograms chandelier. To begin, we have to understand that Newton's second law is gonna be the relationship that we use in order to determine the tension. Newton's second law states that the sun of all forces acting on the chandelier equal m times, eh? To determine the sum of the forces. It's best to start by drawing a visual representation of those forces that's called a free body diagram. So to do that all draw the four forces acting on the chandelier, starting with its weight, the weight of the chandelier ex downward towards the surface of the earth, and it has a strength of em times G. And if you plug in that the masses 44 kilograms, you'll find that the weight of this chandelier is 431 point to Newton's. Don't forget that on Earth, the acceleration due to gravity is 9.8 meters per square. Second, the other forces that are that you need to include in your free body diagram are the three forces do detention, each acting along one of the wires. Now, because all three wires are identical, they provide the same amount of tension so your arrows should act along the wires, but all be approximately the same length in your diagram, and you should label them all with the same letter, and I'll use capital T. Now notice that each wire acts in a different direction and at an angle with respect to the vertical. This means that each tension is gonna have two components, a horizontal and vertical component. Now I'm gonna look at the vertical components because I know that this chandelier is in equilibrium, meaning its acceleration is no longer changing. And it's zero that will allow me to relate the vertical components of the tension force to the objects. Wait. So returning to Newton's second law, um, I can express that the sum of the forces acting on the chandelier is equal to 343 vertical components. Attention. It's all right. It as three times t y. That's one vertical component attention for each wire minus because it's at acting in the opposite direction, the weight of the chandelier and that equals the mass of the chandelier multiplied by the acceleration. A couple of simplifications that we can make right away are on the right hand side of the equation. We know the mass of the chandelier is 44 we could also identify. Go now explicitly stated that the chandeliers acceleration is zero is at rest and remaining at rest. So the entire right hand side of this equation turns into zero. This was a louse is truly no the tension in the wire to the weight of the chandelier. Our equation becomes three times t y minus the weight of the chandelier, which I'll express is W equal zero. So we turn our attention now to determining this term. Attention in the vertical direction of any one of the wires on this page of provided a diagram of any one of the wires. Doesn't matter which one. It shows that the wire is supporting the chandelier at a height of 1.5 meters below the ceiling, and that the length of the chandeliers wire is two meters. This will set on angle between the wire, which provides the tension force and the vertical direction. Now the reason that I'm drawing this diagram is to highlight the relationship between the wires, tension and the vertical component of that tension, which I'm representing in green, a vertical arrow represents T y. And this blue arrow along the high pop news represents the total tension in the wire. The angle between the blue Arrow and the green arrow is the same as the angle between the wire and the vertical, Um, component. If we set up a triggered a metric relationship between the high partners and the vertical side of that triangle that I've drawn, we'll use the co sign function. We're co sign of the angle. Fada is equal to the adjacent side, 1.5 divided by the length of the high pot news, which is 2.0. Okay, we could express this equivalently in terms off the tension forces. So let me say that this equals do you, Jason, 10 of the adjacent tension, which is t y divided by the tension that acts along the high pot news or just capital team. So now we have a relationship between T y and the total tension which looks like this the total tension. I'm sorry me back up. Here's a different color. The tension in the vertical direction to you. Why is equal to 1.5 divided by two multiplied by the total attention in the wire? This is an important relationship for us to use in order to solve the problem. The equation that we arrived at on the first page. So we'll do our problem solving on the final stage. I'll rewrite the equation from page one three times. The vertical component of the tension equals the weight of the object. And instead of writing t y, I'll express it in the same way I'll make a substitution. I'll express it in the same way that I've, um, written here so three times t y or 1.5 divided by two times the total tension equals the weight of the chandelier. And from the first page, we know that the way to the Chanda Lier is 400 31 point to Newton's, and that equals I'll do a little bit of multiplication 4.5 divided by two times the total tension in the wire. And now it's just a matter of multiplying both sides of this equation by two and dividing by 4.5, and you can determine that the total tension in the wire is 100 and 92 millions