Clutch Prep is now a part of Pearson
Ch 15: Rotational EquilibriumWorksheetSee all chapters

# Center of Mass & Simple Balance

See all sections
Sections
Equilibrium in 2D - Ladder Problems
Equilibrium with Multiple Supports
Torque & Equilibrium
Review: Center of Mass
Beam / Shelf Against a Wall
Center of Mass & Simple Balance
Equilibrium with Multiple Objects
More 2D Equilibrium Problems

Concept #1: Center of Mass & Simple Balance

Transcript

Hey guys! In this video I'm going to talk about the relationship between an object's center of mass and whether the object will balance itself on a surface, whether the object would stay balanced or tilt at the edge of the surface. Let's check it out. First of all, remember that an object's weight mg always acts on the objectÕs center of gravity. It's called center of gravity because that's where gravity acts. For most of you, most of the time center of gravity means the same thing as center of mass. If your professor has made a big deal about the difference between the two, then you need to know the difference between them. I'm not going to talk about it in this video. For a vast majority of you guys and for a vast majority of physics problems, all you need to know is that the two things are really the same. I'm going to call this center of gravity or center of mass. In fact, some of you will never really see a problem where they are different. Remember also that if an object has what's called uniform mass distribution, this means that mass is evenly distributed in an object. For example, if you have a bar, this means that you have the same amount of mass in every piece of the bar as opposed, so this is a uniform mass distribution, as opposed to if you have a bar that has way more mass here than in other parts. This is not a uniform mass distribution. Guess what? A vast majority of physics problems will be like this. It will be uniform mass distribution. That's good news. If you have uniform mass distribution, the objectÕs center of mass will be on its geometric center. What geometric center means is it's going to be in the middle. It's just going to be dead in the center right there and what that means is that mg will act here. mg always acts on the center of gravity and the center of gravity is almost always in the middle. It is in the middle if you have uniform mass distribution. If you have an object sticking out of a surface like this, it will tilt if its center of mass is located beyond the supportÕs edge. That's two situations. Here I got the same bar on two desks, but this one is located here. The center of mass is within the table. In this case it's right down the middle and then here, it is beyond the table. What that means is that here, the object will not tilt. You can try this at home. But the acceleration will be zero and this is at equilibrium. It won't tilt. Here the object will tilt. There will be an acceleration. That is not zero and this is not equilibrium. If you want an object not to tilt, you want this situation here and this is static equilibrium. Some questions will ask what's the farthest you can place this object so that it doesn't tilt and we're going to solve these problems using center of mass equation which I'll show you here, which is actually much simpler. These are not torque problems, though they show up in the middle of a bunch of torque equilibrium questions. The equation here is that, let's say if you have two objects, m1 is here and then m2 is here and you want to find the center of mass between them. The x position of the center of mass will be given by the sum of mx divided by the sum of m. What this means for two objects just to be very clear it's something like m1x1 + m2x2 divided by m1 + m2. If you had three objects, you keep going, m1x1, m2x2, m3x3. The masses in x is the x position of that object.

Let's check out this example here. Here we have a 20-kilogram plank that is 10 meters long, so mass of plank 20, length of plank 10. It's supported by two small blocks right here, one, two. One is at its left edge, so this is considered to be all the way at the left even though it's a little wide here, you can just think of it being right here at the very left. The other one is 3 meters from its right edge, so the right edge of the plank is here. This is 3 meters away. The entire thing is 10 meters, so if this is 3, this distance has to be 7. A 60-kilogram person walks on the plank so this guy right here IÕm going to call it big M = 60. I want to know what is the farthest the person can get to the right of the rightmost support before the plank tips. I want to know how far he can go to the right of this. I want to know what is this distance here. The idea is this is not really an equilibrium question weÕre gonna solve with torque. Instead, it's an equilibrium question we're going to solve with center of mass equation. The idea is as this person changes position, the center of mass of this system will change. The system here is made up of plank + person. You can imagine if the guy is somewhere over here, the center of mass of the two will be somewhere like here. If this thing was really long and the guy was here, you would imagine that the center of mass between the two would be somewhere here which means it would definitely tip because it's past the rightmost support point. It's past the edge. What you want to find the rightmost he can go, the farthest he can go, is you want to know what position does he have to have so that the center of mass of the system of the combination of the two will end up here. This is the farthest that the center of mass can be before this thing tips. Basically we want to set the systemÕs center of mass to be at this point right which is 7 meters from the left. The idea is if the center of mass can be as far as 7, what must x be? This distance here we're going to call this x. What must x be to achieve that? That's what we're going to do.

What we're going to do to solve this is weÕre going to expand the Xcm equation. I have two objects so it's going to be m1x1 + m2x2 / m1 + m2 and this equals 7. The tricky part here is going to be not the masses but the xÕs, the distances. The first mass is 20. It's the mass of the plank. The x of the plank is where the plank is. The plank is an extended body, so where the plank is really the plankÕs center of mass which because the plank has uniform mass distribution, it doesn't say this in the question but we can assume it. Because it has uniform mass distribution, I'm going to assume this happens in the middle, little mg. The guy has big Mg over here. This happens at a distance of 5 meters right down the middle. I'm going to put a 5 here. What about the guy? The guyÕs position is over here which is 7 + x. I hope you see this is x and this whole thing here is 7. This entire distance from the left is 7 + x, so that's what we're going to do here. m2 is the guy, 60 (7 + x) divided by the two masses which are 20 and 60 this equals to 7. This is a setup. If you got here, youÕre 99% done. We just got to get x out here by using algebra. IÕm gonna multiply these two, 100. IÕm going to distribute the 60. 60*7 is 420 + 60x. This is 80. If I multiply 7*80, I get 560. That' going to be 560. I forgot that this is 60x of course. I'm going to send these two guys to the other side so I'm going to get 60x = 560 - these two which is 520 and the answer here is or the result here is 40. I have x equals 40/60 and 40/60 is 4/6 or 2/3 which is 0.67 meters. This means that x is 0.67 meters. It's how much farther he can go beyond that point. That's not much. Even though this bar is 10 meters long and it's supported here, the guy can only walk a little bit more and that's because he's much heavier than the bar. This should make some sense if you can somehow picture a 10-meter long or a 30-foot long bar. You can only walk a few steps beyond its 7-meter point or 70% length of the bar before the bar starts tipping if you are much heavier than the bar. That's it. That's how you would find this and I hope it makes sense. Let me know if you have any questions and let's keep going.

Concept #2: Non-Uniform Mass Distributions (Find Center of Mass)

Practice: A 70 kg, 1.90 m man doing push-ups holds himself in place making 20° with the floor, as shown. His feet and arms are, respectively, 1.15 m below and 0.4 m above from his center of mass. You may model him as a thin, long board, and assume his arms and feet are perpendicular to the floor. How much force does the floor apply to each of his hands? (Use g=10 m/s2.)

BONUS: How much force does the floor apply to each of his feet?