Find the tension in each cord in Fig. E5.7 if the weight of the suspended object is w.

Question

Diagram showing forces on a suspended object with angles 20° and 35° for cords A and B.

Accepted Answer

Hey, everyone. Welcome back in this problem. We are asked to find the tension in chords A B and C for an object with weight W suspended by cords as shown below. So we have this diagram, we have chord A that makes a 20 degree angle with the ceiling or the top of This object. We have Chord B which makes a 35° angle. And then A and B join with chord C that is hanging vertically attached to the weight. W. All right. So we want to find tension in all three of these courts, A B and C. How can we do that? We have to consider two different points. We can draw free body diagrams but one is not going to be enough. Okay. What do I mean? Well, for Court W oh Sorry, wait W cord C. Okay. We can draw a free body diagram of this object. And if we draw a free body diagram of the object, we know that we have the weight W acting downwards. And the only other thing acting on this weight or the object directly is this chord C. And so we get this tension in chord C, we're gonna take up to be positive, we have an equilibrium situation. So we know that the sum of the forces is going to be zero. So let's look in the Y direction, since our forces are only in the Y direction, we get that the sum of the forces in the Y direction is zero. In the positive Y direction, we have the tension force and chord C and in the negative Y direction, we have the weight of the object W and effing downwards. And so we get T C minus W is equal to zero, which tells us that the tension force in court C is equal to the weight. W. Alright, so we've done, we've got one of them done. Okay. We found the tension in one chord. Now let's move up to the joint. Okay. So we start with the object and the things that are immediately acting on the object. Now we're going to move up to the joint right here between all three of these chords. Now, if we draw a free body diagram for that joint, what is it going to look like? We have the joint here. We have the tension in chord B. We have the tension in court A and then we have the tension in court, see, acting straight downwards. Now we're going to take up into the right as positive again. But what that means is that we need to break these A and B tensions into X and Y components. So the tension in A is gonna have some Y component, we can call it T A Y acting upwards. Okay. The tension in B is going to have some Y component as well acting upwards. Okay. The tension in B is going to have an X component pointing to the right. And you can imagine if I draw the roof or the ceiling that this is hanging from up here very lightly. This angle here makes the 35 degree angle okay. So using the Z rule, this angle here is also going to be 35 degrees. Okay. The angle between um tension B and its X component and doing the same for the A tension. Okay. We have some x component of that tension and the angle is going to be 20°. Okay. So this is gonna be our free body diagram for that joint. Yeah, that we've drawn on our diagram where A B and C meet. Now, in this case, we're gonna use the same ideas and we're gonna use that sum of forces is equal to zero. However, in this case, we have to consider two directions, we have forces acting in both the X direction and the Y direction. So let's start with the forces in the X direction. The sum of the forces in the X direction is going to be equal to zero at equilibrium acting in the positive X direction. We have the X component of the tension from chord B and in the negative X direction, we have the tension and the, the X component of the tension from Corday. So we get TB X minus T A X is equal to zero. And this is gonna tell us that these two tensions are going to be equal. A T V X is equal to T A X. Alright. What about the Y direction? OK. We're kind of at a standstill here. We can't do anything more. We know that the X components are equal but we can't do anything else. So in the Y direction, we have some of the force is equal to zero as well. In the positive Y direction, we have the tension, the Y component of the tension from court A, the Y component of the tension from court B and in the negative Y direction, we have the tension in court C and all of this is equal to zero. Now, we know that the tension in chord C is equal to W and we're gonna move that to the right hand side. And so we get that the tension in the Y direction of A, the tension in the Y direction of B is equal to W. All right. Now, let's go ahead and try to write these out in terms of tension, a intention be okay. We're trying to find T A and T B. So let's go ahead and write these components. Okay. And both are X direction equation and our Y direction equation in terms of TA and TB. So that we can relate these two equations right now, T B X and T A X are completely separate from T A Y and T V Y. But we know that they're related through TA and TB. So T B X is going to be equal to the tension T B Times cosign of 35°. And similarly, TA X is going to be T A times cosign of 20 degrees. If we wanted to isolate one of these terms, let's go ahead and choose T B. We can write T B is equal to T A, Kassian of 20° divided by Kassian 35°. Now, instead of having a relationship between the x components of the tension, we have a relationship between the tension a and attention be themselves. All right. So let's go back to this equation on the right hand side and do the same thing when we're looking in the Y direction instead of having cosign, we're gonna have sign. So we're gonna have T A Sign of 20° plus TV. Sign of 35° is equal to the weight. W now let's call This guy star. Okay, that's the equation TB is equal to T.A. cosine of 20° divided by cosine of 35°. Okay, let's go ahead and substitute that into our equation on the right hand side, we have T A sign of 20 degrees plus TV. Sign of 35 degrees is equal to W Subsequuting that we're gonna get ta sign of plus T A coast of Times sine of divided by cosign, 35 degrees is equal to W okay, we've just substituted what we found using the X direction into the Y direction. And now our only unknown is th all right, let's give ourselves some more space to work. If we factor out the T A, we have that T A times sine of 20 degrees plus Kassian of degrees times sine of 35° divided by oops Kassian of 35° is equal to W Okay. And if you work out sign of 20 plus coast of 20, sorry. Yeah. Sign of 20 plus coast 20 sign 35 over coast 35. You're gonna get that. This is just equal to one. Okay. Work that out on your calculator. And so the tension in A is just going to be equal to the weight W as well. Alright, so we found T C, we found T A. Now let's go to T be looking back at the start equation. We have the T V is equal to T A coast of 20 divided by coast of 35. We know that ta is W so we can set that in. Now, we have the T B is equal to W times co sign of 12 degrees divided by Kassian of 35 degrees. And if we work this out on our calculator, we get that T B is equal to approximately 1.147 times the weight of that object. And so now we have our attention and a, our attention and be and our attention and see if we go back to our answer choices. We found that the tension in A was equal to the weight of the object. W We found that the attention and be was approximately 1. times the weight of that object. And we found that the tension in C was also equal to the weight of the object. W and so we have answer choice. D thanks everyone for watching. I hope this video helped see you in the next one.

Find the tension in each cord in Fig. E5.7 if the weight of the suspended object is w.

Verified Solution

Key Concepts

Tension in Cords

Equilibrium of Forces

Resolving Forces