Find the tension T in each cable and the magnitude and directi...

Question

Find the tension T in each cable and the magnitude and direction of the force exerted on the strut by the pivot in each of the arrangements in Fig. E11.13. In each case let w be the weight of the suspended crate full of priceless art objects. The strut is uniform and also has weight w. Start each case with a free-body diagram of the strut.

Accepted Answer

Hey everyone in this problem, we have an ornamental structure consisting of a hinged rod and a supporting wire. Were asked to determine the tension t. In the supporting cable and the magnitude and direction of the force exerted on the rod by the hinge when you hang a decoration that has twice the weight as a supporting rod. Okay, We're told that a free body diagram is gonna be helpful here. Alright, We have our little diagram, we're just gonna kind of redraw it and include all of the forces. Alright, so we have the rod. Okay? And then we have the supporting cable. Now we know that this angle here is 15° and the angle down between the ground and the Rod is 60° And we have our weight hanging off the end of two W. Alright, so what other forces do we have here? Okay, well we have the rod. Okay, Which we know has weight W. Mhm. And so we're gonna have the weight of the rod pointing down here, magnitude W. Yeah, that's gonna be pointing from the center of mass, which is gonna be the center of that rod. Now, we also have the tension from this wire. Okay, so I'm gonna erase this wire just for a second, and you're just the force do detention T. Okay, and I get this angle is 15° and if we break this tension into its X. And Y components, Okay, the X component is going to be pointing to the right, so we have the tension in the X. Direction, and the Y component is going to be pointing down. So we have the tension in the Y direction. Ok, Alright, now, what other forces do we have? What we have the force exerted on the rod by the hinge. The hinges down here at the bottom we see it. And so we're gonna have the force exerted on the road by the hinge. Now, all of our why forces are pointing down. So we're going to assume that the force exerted on the rod is going to be pointing upwards to kind of counteract those forces pointing downwards. And we're gonna assume that the X component is pointing to the And so f is our force acting on the hinge. Now we want to find t the tension, we want to find the magnitude and direction of the force F. Which has components F X and F. Y. Okay, so these are the things in our diagram we want to find. All right, so let's take up into the right as our positive and we have our coordinate system here. No, this is an equilibrium problem. We know that we have no movement. Okay, this is remaining at rest, so we have no net external forces. So we can say that the sum of the forces is equal to zero. Okay, and that gives us two equations to work with because we have to consider the sum of the forces in the X direction equal to zero and the sum of the forces in the Y direction equal to zero. Alright. We also know that this is non rotating. Okay, we have this support wire so that this is stationary. It's not moving so it's non rotating which means we have no net external torque. Okay. And so we get that the sum of the torques is equal to zero. So we have these three equilibrium conditions or three equilibrium equations to satisfy. Mhm. So let's get started with the first. The sum of the forces in the X direction is equal to zero. Okay well which forces do we have in the extraction? Okay, in our positive extraction we have the tension and the X component of the tension T. X. And in the negative extraction we have this force um on the rod from the hinge fx. For the X component of that force It's going to be equal to zero. And so these two things are going to be equal the force, the X component of the force acting on the rod. By the hinge is going to be equal to the tension in the extraction. Alright, we can do the same in the y direction. So the sum of the forces in the y direction is equal to zero. In which y forces do we have while pointing upwards? In the positive y direction we have F. Y. The y component of the force of the hinge acting on the rod and then we have negative forces or forces acting in the negative Y direction with the force, the y component of the force of tension. Okay, We have the force negative two W. Relating to the object and then the fourth negative w. The weight of that rod and all of these added together. Give us zero. Okay, so we can simplify F, Y minus T. Y minus whoops. Not two but three W. is equal to zero. Alright. We're gonna label these equations one and two so we can come back to them if we need to. Okay. Alright so we've dealt with some of the forces equations. Let's move to the some of the torques. Okay, because in each of these equations that we've written we have more than one unknown in this equation. We don't know fx or T. X. And then this equation we don't know F. Y. Or T. Y. So we need another equation in order to try to find one of these values. So some of the torques is equal to zero. Now we need to choose a pivot for our torque. Okay? We want to choose a pivot to eliminate some of the unknown forces so we don't know the forces FX and fy and we don't know the force t So if we choose our pivot to be down here. Okay, we're gonna call this point A then fx and fy will not contribute to the torque at this point. So we won't have those in our equation. We're going to only have tension that we don't know which is going to allow us to solve for t. Okay, so let's go back and let's find the tension. Okay, we're gonna take the point a as our Alright, so now we have the torque relating to what while we have the torque at the center of mass. Then we have the torque kind of at the end of our rod, we have the torque at the center of the mass plus a torque at the end of the rod. This is actually going to be three torques. Okay, let's start with the center of mass, The center of mass. We have the weight acting okay, it's acting downwards. So you can imagine this is going to cause a counterclockwise rotation. Okay, so the torque is going to be positive. So we get the torque relating to the center of the mass is going to be our cm. Okay, that's gonna be the distance from our pivot to that center of mass. The force which is W times sine of the angle. Okay, well, what's the angle? Let's go up and working with all of these angles. The angle down here is 60°, which means that the angle here Okay, is going to be 30 degrees And similarly here we also have 30°. And so we're talking about our angle between that rod And that force we have 30°. Alright, now, torque at the end. Okay, well this is going to come from two places. Okay, We have torque relating to this object with force to W. And then we have the torque from the tension. When we're looking at the torque from the object with weight to w. Hey, this is pointing down just like the weight W at the center of mass, it's gonna cause counterclockwise rotation. So that's a positive torque we're gonna have plus. Okay, and it's gonna be our end. Okay? So the distance from the pivot to the end In terms of force which is two W. and the angle there was also 30°. And now the tension, the tension is pointing kind of down into the right, This is going to cause rotation clockwise. So this is gonna give us a negative torque. Okay? And we have an angle of 15° and so we have minus and again the distances from the end to the pivot In terms of force, which is attention t. sine of the angle 15° zero. Okay. And what you'll see now is that what we have unknown here is attention T. Okay. We want to solve for T. Let's take the length of the rod to be l. Then the distance to the center of mass will be L. Over two W. sign 30° in the distance to the end will be L. So we can replace all of these are values with their corresponding L. Value. No, because we have an L. In every turn we can divide by L. And we're gonna be left with T sign of 15° and we move this to the other side is equal to W Sign 30° over to Plus two w. sine of 30°. Okay. Alright, so T is going to be we have W Over two plus two W. Okay, that's gonna be 2.5. So 2.5 W. Sine of 30° divided by sine of 15° which gives attention. T. It's gonna be approximately 4. three times the weight W. Okay. And so that is the tension that we were looking for. T. Alright, so we found attention to you. Let's go back up to our problem and kind of remember what we're doing? We found tension T checkmark. Ok, now we need to find the magnitude and direction of the force exerted on the rod by the hinge. Okay, so we need to find the component fx the component F. Y. In order to find this force F. Okay and what we'll see is that those were related to the tension. Okay, now that we know the tension we can go back to equation one and two and solve. Okay, Alright, so let's go down give ourselves some more space And soft. So from equation one. Mhm We had that the force and the X component F. X was equal to the X component of the tension. T X. No, the X component of the tension. Let's go back and look at what angles we should be using. Okay, Alright. And what we see is that we have 30° here We have 15° in between here which means that the remaining angle Here's 45°. Okay. Alright, so when we write out the X component of tension we get that it is going to be equal to 4.83 W. Times coast of 45° Which gives us 3. W. Okay, and from equation two, let's do it up here, we have room and we get that F Y is going to be equal to the tension in the Y direction plus three W. Okay, tension in the Y direction K. Same thing 4.83 W. But in this case we have sine of 45 degrees. Okay. And we have a 45 degree angle. So cozy inside are actually gonna be equivalent plus three W. Which gives us a Y component of f The 6.415 W. Okay, so we have the y component here, The X component here and we just need to figure out the magnitude and direction. Alright, so let's try ourselves a triangle. We know that the Y component points upwards the X component points to the left and so the magnitude of the force is gonna be the hypotenuse here, this is tough. F X on this side and F Y on this side. Alright, this is gonna be our angle data prime. But what we want to know is the angle to the positive X axis. So our actual data is going to be here and you'll notice that data is going to be equal to 180 degrees minus data prime. Mhm. In our case data prime is going to be the inverse tangent of F Y over F X. So we get 180 degrees minus the inverse tangent of F Y. 6.415 W divided by F X 3.415 W. Which gives us an angle say to that is equal to 118.03°. So we found the direction of that force. Now we need to find the magnitude of that force. Okay, now we are going to go work up here, we are running out of space. This is um large problem. Okay, we're almost there. We just need to find the magnitude of this force. F. We have a right triangle. We know we can use pythagorean theorem. So f squared is going to be equal to F X squared plus F Y squared. F squared is equal to 3.415, W squared plus 6.415 W squared. And working this out, we take the absolute value. Okay, we take the positive route because we're looking for the magnitude we get 7.267 W. And so we found that the magnitude of the force is 7.267 W. Approximately at the direction of 118.3 degrees. Okay, let's go back up to our answer choices. And we see We found that the tension was approximately 4.8, 3 times the weight w. The magnitude of the force F was 7.27 W, with the direction approximately 118 degrees. Okay. And so we have answer. A thanks everyone for watching. I hope this video helped see you in the next one.

Key Concepts

Free-Body Diagram

Tension in Cables

Equilibrium of Forces

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