Hey, everyone. So in the previous few videos, we've dealt with rigid bodies, not just tiny point masses, but rather extended objects. We assumed that they had what's called uniform mass distribution, which means that the mass is evenly spread out, and we can assume that the center of mass and the weight force acts in the middle. However, in some cases, you won't have that; you'll have what's called non-uniform mass distributions, like the problem we're going to work out here. So, I want to go ahead and show you how that works, and we're going to do this example together. Alright? So let's check this out here.
Unless your problems otherwise state, you can usually assume that a rigid body has uniform mass distribution. Remember, this uniformity just means that the mass is evenly spread out. For example, if I had a textbook, which I'm just going to imagine as a little rectangle, then if this mass is evenly distributed, one side of the textbook wouldn't be heavier than the other. This is good because we can usually just assume that the center of mass is in the middle, and that's where the weight force acts. We assume the weight force acts on its center of mass. Now, if you don't have uniform mass distribution, you cannot assume the location of its center of mass, and that's usually what these problems involve.
There are generally two broad types of problems that you're going to see. In some, you'll be given the center of mass, the location of it, and you'll be asked to calculate something else, like a force or a torque. In other problems, you might be given some information and asked to calculate where the center of mass is, and that's exactly what our problem will deal with. Alright? So let's just go ahead and get started with this.
In this problem, we have an 80 kilogram man who is 2 meters tall. So, I've got m=80 kg and l=2 m. But, we're told explicitly in this problem that human bodies generally do not have uniform mass distribution. This kind of makes sense if you think about it; our mass is a bit unevenly distributed, skewed or heavier towards our topside due to our arms and torso, and then we have our legs. As human beings, we do not usually have uniform mass distribution. We want to calculate how far from the man's head the center of mass is.
We've got two scales underneath this board, and remember scales measure the normal force. We're actually told what those normal forces are. The left one measures 320, so I'm going to call this normal left equals 320, and this one here measures 480. This right normal force is a little bigger, exerting more force, and this tells us that our mass is skewed more towards our head. There's also another force that's acting, which is the weight force of the man. The board has no mass, but the man does; he's 80 kilograms. We're going to draw that weight force, and if this were a uniform body, we would draw it in the center, but it's not uniform. If you didn't know that, you could sort of guess that because the normal forces are skewed toward the right, more of the weight force and more of the center of mass is skewed to that side.
Here's what we have: our mg force, and what we want to figure out is the distance between this center of mass and where the mg acts, and that's my x distance. So this is really what we want to find here, what x is. Usually, in these types of problems where you're given some kind of unknown distance, you want to write all of the other distances in terms of that variable. So if the distance here that you're looking for is x, then the other distance that I have to use will be this one here, which is really just the entire length minus the x. So, the whole length is 2, and this piece is x, this will be 2 minus x. These are our distances here. We want to calculate what this x is. This is a total equilibrium problem.
We know that the sum of forces and torques is going to be 0. Let's go ahead and get started with the easier one, sum of forces is 0. We've got two forces that point up, so you've got n_left and n_right, and then one that points down, which is mg, and that should equal 0. If you plug in these numbers quickly, 320 and 480, and then we're just going to use 10 for g. This is going to be 80 times 10. All you're going to get here is you're going to get 800 minus 800 equals 0. That's cute, but it doesn't really tell us anything. All it just says is that the forces cancel. This equation tells us nothing about the location of the center of mass; all it's just telling us is that the guy isn't lifting off into space or plummeting through the ground.
So, let's just go ahead and move on to the sum of all torques. The sum of all torques is equal to 0. Remember, when you are measuring torques or calculating torques, you have to pick a reference point, and you usually want to pick one of your reference points where one of the forces is acting. The reason for that is you want one of the torques to cancel out because its distance to the axis of rotation will be 0. That's a trick you can use to make one of your terms always cancel out. It doesn't matter which point you choose; you'll still get the right answer as long as you keep track of all the signs. But I think the easiest one to pick will be the first one here. So, we've got our object, and we've got a force that acts right here, that's n_left, and because this is the axis of rotation, this force will not produce a torque because there's no distance, their torque from n_left is equal to 0. What about the other forces? We've got an mg that's going to act somewhere over here, and this mg will produce a clockwise rotation. If you're pivoting it around at this point, like my hand over here, and you have a force this way, it's going to want to turn you clockwise. Remember, a clockwise rotation will be a negative torque, so this is going to be the torque of mg and that's going to be a negative. The last one here is going to be this torque over here from the normal from the normal right force, and that's going to produce a counterclockwise torque, and that's going to be a positive torque. So, we've got negative torque mg plus and we've got positive torque from n_r, and this equals 0. What I'm going to do is I'm just going to move this to the other side so that it becomes positive.
Remember, torques are f r sin theta. So, we've got normal force times some distance times the sin of theta equals the force here is mg, so we've got mg times r times the sin of theta. We just have to figure out what are the sin and r's that we plug in. If you look at these forces, remember we have to figure out where's the r vector. The r vector points from the axis of rotation out to that force. So this is going to be r_mg like this, and then the other one is going to be this one, and this is going to be r_nr. Notice how both of these r vectors make 90-degree angles with their forces, and so therefore the sine of the angle is just going to be 90 for both of them. Remember, the sine of 90 is just 1, so we can kind of just cancel that out, right? So we really just need to figure out what these lengths are, the r_mg, and the r for the normal force.
If you look at the first diagram that we drew, you'll notice that this distance right here from the feet to where the mg acts is going to be that distance, 2 minus x. So this is going to be 2 minus x here, that's what we plug into this equation. This is going to be 2 minus x, and then, for the mg, I'm sorry, and then for the right force, this is actually just going to be equal to 2 because it's the entire length of the man. So now, I'm just going to start plugging in some numbers. This is going to be 480 times 2, remember the sine goes away, equals, and this is going to be 80 times 10 for g, and this is going to be 2 minus x and then this is also going to be 1. This equation simplifies to 960 equals 1600 minus 800x.
Now all we have to do is solve for x. All the physics is done; we just have to do some algebra and cleaning up. We'll move the 800x to the right, and when you subtract 960 from 1600, you'll get 640. So your x distance is going to be 640 divided by 800, and if you cancel out a 0, you'll notice that these things are divisible by eights, this is going to be 0.8 meters. That is your answer. Going back to our diagram, what this tells us is that the center of mass is actually 0.8 meters away from the head, and that makes sense because we said that this distance here is going to be closer to the head than it is to the feet. If the human body is 2 meters long, then the middle point would be 1 meter, but the center of mass is closer towards the head, which means that this distance over here is actually going to be 1.2 meters.
So that should make some sense here. Alright, guys, that's it for this one. Let me know if you have any questions.