Torque Due to Weight - Video Tutorials & Practice Problems

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Torque Due to Weight

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Hey, guys. So in this video, we're gonna talk about an objects weight force MG producing a torque on itself. In other words, on objects own weight will produce a torque on that object. Let's check it out. So imagine you have a bar like this that is free, that that can rotate about this point. So if it does spin anyway, it would be around here kind of like your arm going like this. Right now, there's a weight force that pulls in the middle like this mg because if you have mass, you have wait. So there's something pulling down on you, which means you can think of it in terms off. The weight of your arm is pulling down like this or pushing down like this. And because you're fixed here, you're it's trying to cause it to spin like this. Okay, it's trying to cause this. So knowing eso being aware of the torque and knowing how to calculate the torque due to weight is really important because it's gonna come up in a lot of questions. All right, so I want to remind you, however few things real quick, so an objects weight, mg always act on the objects center of gravity center of gravity. Now, for most of you, the terms center of gravity and center of mass will be the same thing. So you can also think of this a center of mass. Okay, most professors will actually not even draw a distinction between the two. Now, the way this works really briefly is that they are different Onley if your gravitational field is not the same everywhere, so there's like more gravity here than here. For most of you, you're not gonna have to worry about that. And you can think that center of gravity really just means the same thing. The center of mass. If those two things were different, then you have to look at your center of gravity, not your center of mass. But again, if you're Professor has made a distinction about that, you used to that by now. Um, if not, don't worry about it. If this sounds foreign, it's probably because he hasn't talked about this and you don't have to worry about it. Okay, so but the official answer, the correct answer is center of gravity. It's just that they're the same. Most of the time. Um, if on object has what's called uniforms mass distribution, uniformed mass distribution means that the mass of an object is evenly distributed throughout. Okay, you can think of this is even mass distribution very well distributed. It's the difference between a bar that has roughly the same mass. Everywhere were exactly the same amount of mass everywhere versus a bar that's much heavier. Let's stay on this piece right here. Right. Um, most of the time you're gonna have uniformed mass distribution, and in this case, your center of gravity, Your center of gravity or center of mass, if they are the same is going to be on the Objects Geometric center. What the heck is geometric center? It's just fancy for its in the middle, right? So if you have, let's say this bar is 4 m long. Length equals 4 m. It means that the M G is going to pull at a distance of 2 m from the edges right down the middle. Okay, long story short mg pulls in the middle. That's how it works for a vast majority of physics problems. Okay, so let's look at this part here. We have a bar a cylindrical rod that has a mass M equals 20 and the length equals four. And it has one of its ends fixed to an axis right here that is mounted on the floor. The rod makes 37 above the horizontal right there. And suppose you have massive 80. So that's your own personal masculine, and you're gonna stand on the other end of the rod. So you're up here with a mass? Let's call it big M of 80 kg. We wanna know that Net torque. OK, so torque nets, Um, that's what we wanna know that is produced on the rod about its axes due to the to wait forces. Okay, So much torque you produced on the Rod about it's axis. That's always how this works. This is just sort of extra language. Um, torques are always produced about an axis of rotation, um, due to the weight forces. So what we're doing is we're doing the torque of do too little m g plus the torque due to big M G. And it says you may assume the Rod has uniformed mass distribution. What? That means that the weight force on the torque on the Rod will happen right in the middle. If this question didn't say that, you are going to assume that anyway. Cool. So unless the question says that the object doesn't have uniformed mass distribution, you assume that it has uniformed mass distribution. It's kind of like friction back in the day. If it didn't say anything about friction, you just assume no friction, Okay? It's fixed in place, so it does not move or rotate. So this object, the rod isn't gonna move sideways or up and down. And actually, this particular case, it doesn't even spin now, just to be clear, it doesn't mean that there is no torques on it. It just means that for whatever reason, something stuck here. It's made to not spin. Okay, you don't do anything with that. You still calculate the network on it. All right, So we can calculate these two guys individually torque of little MG torque of big M G. And then we're gonna put them together. I'm gonna leave a little space for the signs. And remember, torque torque is f R sine of theta. In this case, the force gets replaced by M. G because that is the force. So the torque of little MGI is mg are sign of things. Now there are two forces in the they are different. Right? Um this is they are for little m, and this is the or little mg. And this is the are for this is big mg the our vector for big M G and the angles air different as well. This is the angle for little mg, the angle for big m G. Okay, so what we gotta do is plug in these numbers and the signs. Okay, so remember the steps draw our vector, figure out the data on, then plug it into the equation. So little mg acts right in the middle because this is uniformed mass distribution. Um, the our vector is a narrow from the axis of rotation right here to the point where the force happened. So this is our four mg, okay. And the other m g is gonna have a are that goes all the way to the end of the bar because you stand on the edge right here. This is our big m G. Okay, so our little mgs half of the bar because M G acts in the middle. So the length of the bars four. So you're our mgs to your are big m G is four. Okay, so we got the we got the arse figured outs. Um, what about the angles? What about the angles? So do we use this 37 here? And the answer is no. You do not use that. 37. Be careful. That is the wrong answer. Let me draw it and show you why. So for this here we have our from the access to M g. And then this is your actual force, which is m g. This is the angle that I need. That's the angle right here between these two, right? Not this other angle. Here. Now notice that this forms a triangle. If this is a 37 it's a right triangle. This is a 90. So this has to be 90 minus 37. So this is 53. Very, very careful. We're gonna use 53 instead. Okay. The same thing happens for the other mg. The only difference that the arrows longer but that doesn't matter are big. M g m g is this way. This 37th still shows up right here. But that's still not the angle we want. We want 53. Okay. Be very careful. The angle that you're being given is not the one that you're supposed to use. That happens quite a bit now. Finally, before we plug all this stuff in here, we've got the angle. We got ours. We confined the efs just mass times gravity. Now, let's look at the sign. Okay, So we're gonna do is pretend that this rod is your arm, right? And you're being pushed down this way by the first force, which causes you or tries to cause a rotation this way, which is like this, which is counter clockwise. It's positive. It follows the unit circle. And that's the one in the middle. The one at the end does the same thing. It tries to cause you to spend this way. They're both giving you a positive torque because they try toe, make you spin in the positive rotational direction. Okay, so I'm going to say here that this torque is positive and that this torque is positive, so they get positives in both. Okay, So the first mass is a little mg has a mass of 20 I'm gonna use 10 for gravity are for the little guy is too sign of 53. That's the same angle for both. When you put all of this together, you get 3 13 3 Newton meter. Here you have positive B M instead of 20 is 80. Grab him Rounding into attend the are the distance instead of a two. It's a four, and the angle is 53 as well. When you multiply all of this, you get 25 04 They're both positive. So the network is just both of these guys added, plus 3 +13 plus positive +2504 gives us a positive +2817 Newton meter. Cool. So that's how this works the way Do Thio due to the weight force. I'm sorry. The torque due to the weight force. Really important how this works. A lot of stuff is gonna come up later. So that's it for this one. Let me let you have any questions and let's keep going

2

Problem

Problem

Two kids play on a seesaw that has mass 20 kg, length 3 m, and its fulcrum at its mid-point. The seesaw is originally horizontal, when the kids sit at the edge of opposite ends (m,_{LEFT} = 25 kg, m,_{RIGHT} = 30 kg). Calculate the Net Torque from the 3 weights acting on the seesaw, immediately after the kids sit (simultaneously) on their respective places.

A

75 N•m clockwise

B

75 N•m counterclockwise

C

150 N•m clockwise

D

150 N•m counterclockwise

3

Problem

Problem

A guy standing straight up stretches out his arm horizontally while holding a 60 lb (27.2 kg) barbell. His arm is 64 cm long and weighs 45 N. Calculate the Net Torque that the barbell and the weight of his arm produce about his shoulder. You may assume that his arm has uniform mass distribution.

A

−156 N•m

B

−171 N•m

C

−185 N•m

D

−312 N•m

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