A 350-N, uniform, 1.50-m bar is suspended horizontally by two ver...

Question

A 350-N, uniform, 1.50-m bar is suspended horizontally by two vertical cables at each end. Cable A can support a maximum tension of 500.0 N without breaking, and cable B can support up to 400.0 N. You want to place a small weight on this bar. (a) What is the heaviest weight you can put on without breaking either cable, and (b) where should you put this weight?

Accepted Answer

Hey everyone in this problem, we have two cables connect vertically each extremity of a uniform horizontal wooden board of 15 kg mass to the ceiling of the living room. The length l of the board is one m, The left and right cables can withstand maximum tensions of 200 newtons and newtons respectively. We want to place a flower pot on the board and were asked what is the maximum weight you can place without breaking either cable? And where should you place it? Alright, so let's draw a little diagram first. So we have this horizontal wooden board, okay, And it's gonna be held up by these cables, okay? On either end. So we get the tension from the cable on the left side, when we get the tension from the cable on the right side. Alright, we know the distance of this board Between these cables is one m and now we want to kind of figure out what other forces we have to draw here. Okay, well, the board has a mass, which means that we have this force downwards due to the weight of the board. Okay, We'll call it W. B. And if we put a flower pot on here, put a flower pot on here, then we're also gonna have the weight of the flowers, and I've drawn the flowerpot here, that's not necessarily where it will go. Um We've drawn it here, just to indicate that we do have that weight from the flowers. Okay? Alright, so we have this problem and this is an equilibrium problem. Have these cables holding this board, we don't want it to be moving. Okay, there's no motion. And so we know that the sum of the forces in both the X. Direction and in the Y direction will be equal to zero. Okay. And this is also non rotating. So we know that the sum of the torques is equal to zero. Okay, So we have these equilibrium conditions that must be satisfied. Okay, what we'll see is that we don't have any forces acting in the X. Direction, Only in the Y direction K. We'll take our coordinate system to be the typical coordinate system right into the leftist, positive. Okay. And so we're actually not going to be using the first equation because we don't have forces in the extraction. Alright, so let's start with the first part of this problem. Okay, what is the maximum weight we can place? Well, let's start with the sum of the forces. Okay, Because we have the weight of the flower. Okay, so the sum of the forces. And in this case it's the Y direction. Because that's the only direction we have forces zero. What forces do we have? Well, acting in the positive Y direction. We have the tension from the left end cable. Okay, We have the tension from the right end cable and then acting in the negative Y direction. We have the weight from the board W. B. And we have the weight from the flowers, W. F. And what we're trying to find is the maximum weight of the flowers. Okay, so we're trying to find W. F. Alright, so let's isolate WF. WF is going to be equal to the tension from the left. The tension from the right minus the weight of the board. The tension from the left side. Okay, it has a maximum of 200 newtons. And on the right, it has a maximum of 300 Nunes. And we were told in the problem and then we have the weight of the board. Okay, well the weight is going to be M G. Okay, the mass of the board is 15 kg and G is gravitational acceleration to be multiplied by 9.81 m/s, kilogram meters per second. This is a unit of newton as well. And so we end up with 352. newtons. Okay, And we see the answer choices are rounded to one decimal place. So we're just gonna approximate 352.9 newtons. Okay, And because we've used the maximum tension in the left and the right cables, this is going to give us the maximum weight that we can place. Okay, so wFr maximum weight is 352.9 newtons. And so we can see that we're an answer choice. Either A or B. So let's move on to part two and figure out Which one it'll be Now in Part two. We are asked to determine where we should place the flowerpot. Now, if we think of this problem is the left cable can withstand less maximum tension than the right cable. The right cable can withstand more. So it makes sense that the flower pot should be closer to the right. Okay, so if we were looking at A. And B, we would kind of assume that B. Might be the right answer because it's further from the left cable which holds less tension. But let's figure this out mathematically figure out exactly where it should be. Now we've used the sum of the forces in the y direction is equal to zero, but we haven't used our other equal room condition of the some of the torques equal to zero. So let's go ahead and use that equation and try to figure out what that distance should be. Okay, so the some of the torques is equal to zero and we have to choose a pivot. Okay, now the answer choices are giving us um distances from the left cable. So let's choose the left cable as our pivot. Okay, you can choose the pivot to be anywhere. You'll get the same answer. You'll just have to kind of work with the numbers a little bit more. If we take from the left end then the distances from our pivot is going to be the distance from the left end, which is what the answers are given. Okay, so we're gonna take the left end as our pivot. Alright, so what torques do we have? Okay, well, if we're working from the left and the torque from this tension in the left cable is going to be zero. Okay, we don't have to worry about that. It's not going to contribute, Hey, we have the torque from these weights W B and W. F. Okay, you can see that they're kind of pushing down. You can imagine that they would cause clockwise rotation of our board. And so those are gonna be negative torques and then we have the torque from the right end. This is pushing up. This would cause counterclockwise rotation, so that's going to be a positive torque. And so we end up with negative the torque from the board and negative the torque from the flowers, plus the torque from the right cable equal to zero. Okay, Alright. And remember that this distance is one m. Okay. And so the Weight of the board is going to be at 0. m, kids gonna be halfway from that left end to the right end because it's acting at the center of mass and so the tension or the torque due to the board is gonna be negative 0.5 m. Okay. The distance to the pivot times the force which is the way to the board times sine of the angle. Okay, and this is acting at 90 degrees Katie angle between that force and the board is 90 degrees. Alright then we give the tension or the torque. Sorry, due to the flowers. Okay, In this distance we don't know, we're gonna call it X. And that's going to be the distance from the left end which we've chosen as a pivot to where this weight of the flower force is acting. Okay, So that is a value that we're trying to find. Okay, we're trying to find this value X. Then it's going to be times of force which is going to be the weight of the flowers, times the angle and again The angle between the the weight of the flowers and that board is 90°. Okay, they're perpendicular sign of 90. And finally the torque related to the right end cable. This is acting the full distance away from the left end. This is one m from our pivot. So we have one m times of force, which is attention from the right side, Times the sine of the angle. And again, this is acting perpendicular early to the board. We get side of 90° and all of this is equal to zero. Alright, so let's move Actually let's divide by sine of 90 degrees. Okay, sign of 90 degrees is gonna be one anyways, but we can divide that term. Okay, And we're gonna move this term with the X. That we're looking for over to the other side. So we're gonna get X times W. F. Is equal to one m times attention on the right side -0.5 m times the weight of the board. All right. Now, in order to isolate X. We'll need to divide by the weight of the flowers. Okay? So X is going to be equal to one m times attention from the right minus 0.5 m times the weight of the board. All divided by the weight of the flowers. And if we plug in the values we know pay the tension on the right side. The maximum is 300 newtons. The weight of the board. The mass was 15 kg. Okay. And just like we calculated it here, the weight is going to be the mass times the gravitational acceleration. So we get 15 kg times 9.81 m per second squared. And I think we made a mistake up here with our units. We did. I'm going to correct this. So when we did our gravitational acceleration, we had meters per second. This should be meters per second squared. Ok, kilogram meters per second squared. That is what gives us a newton. Okay, Alright, continuing down here. Okay, so we have the weight of the board 15 kg times 9.81 m per second squared, divided by the weight of the flowers. In the maximum weight of the flowers we just found was 352.9 newtons. So this is going to give us one m times 300 newtons. This is 300 newton meters minus 0. m times 15 kg times 9.81 m per second squared. While kilogram meter per second squared is a newton. Okay. And then we have times meter. So we get 73.575 Newton m Divided by 352.9 Newtons. The unit of Newton will cancel when we divide and we're left with the unit of m and we get 0.64 m. And so the distance our flowers should be from the left end. Okay, because we've chosen that as our pivot is 0.64 m. Alright? So if we go back up to our answer traces, we found That the maximum weight we can place on our board is 352.9 newtons and it should be placed 0. m from the left cable. So we have answer choice b. Thanks everyone for watching. I hope this video helped see you in the next one.

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