Center of Mass

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mspeedwagon

Full Member
10+ Year Member
Alright, all, this should be a fairly simple question, but I can't seem to solve it. Thanks in advance for your help.

A yoke 2m long with one bucket at each end is used to carry water. If one bucket has a mass of 16 kg and the other a mass of 22 kg, approximately where is the center of mass? (Assume the yoke has negligible mass).

I figured I could just used (m1d1 + m2d2)/m1 + m2, but I'm not sure what to plug in for d1 or d2. The solution I have uses torques to solve the program, but it's lacking an explanation.

Schenker

Full Member
10+ Year Member
You were on the right track with

(m1d1 + m2d2)/(m1 + m2)
.

We know the values of m1 and m2, leaving d1 and d2 unknown. Thus, we have one equation and two unknowns. Because we have two unknowns, we need one more equation with d1 and d2. Can you think of a simple relationship between d1 and d2 that's stated in the problem?

sps27

Full Member
10+ Year Member
I like to think of these questions as a couple.....you know.....load * load arm = effort * effort arm......that kind of deal and the center of mass as the fulcrum, if it has to balance....so it becomes easy that way...

mspeedwagon

Full Member
10+ Year Member
I'd guess that the length of the yoke comes into play. I'd go for 2-d and d. The center of mass is going to be closer to the heavier mass. Still not sure which length to associate with which mass.

Also, if you think of this as load * load arm = effort * effort arm, you still have two unknowns. I don't know what the value for load or effort arm would be.

You were on the right track with

.We know the values of m1 and m2, leaving d1 and d2 unknown. Thus, we have one equation and two unknowns. Because we have two unknowns, we need one more equation with d1 and d2. Can you think of a simple relationship between d1 and d2 that's stated in the problem?

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sps27

Full Member
10+ Year Member
I'd guess that the length of the yoke comes into play. I'd go for 2-d and d. The center of mass is going to be closer to the heavier mass. Still not sure which length to associate with which mass.

Also, if you think of this as load * load arm = effort * effort arm, you still have two unknowns. I don't know what the value for load or effort arm would be.

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mspeedwagon

Full Member
10+ Year Member
Thank you for posting that. I'm still not certain how you decided which side was x and which was 2-x. Couldn't you just as easily flip them. I guess the part I'm confused about is in words, how do you define x.

sps27

Full Member
10+ Year Member
Thank you for posting that. I'm still not certain how you decided which side was x and which was 2-x. Couldn't you just as easily flip them. I guess the part I'm confused about is in words, how do you define x.

mspeedwagon: I apologize for butting in like that. Schenker was explaining and I just distracted you further by offering a quick fix. That was not polite. Schenker please correct me if I misstate anything.....

Solution 1: So center of mass is an abstraction where we assume the sum of all forces are concentrated and act. In this problem let us assume that the 16 kg is the origin O of x axis. The 22 kg acts at a distance of 2 m. Furthermore, let point C on the yoke be the center of mass which is at a distance 'x' from the origin O or 16 kg mass. So the center of mass calculation becomes, Total Mass*x = 16*0 + 22*2. The 16 kg is at the origin so therefore I multiply it by 0. So we solve this eqn. (16+22)*x = 44 so x = 44/38 so x is a little more than 1 i.e., x = 1.15 approx. So now we know that the center of mass is around 1.15 m from the Origin O which is at the 16 kg end. Which means it is about .15 m closer to the 22 kg end from the center of the yoke, which is what we expect. It does not matter whether you take origin at 16 kg or 22 kg end. Either way the math will work out. If you take the origin at 22 kg end, (16+22)*x = 22*0 + 16*2 so x = 32/38 so approx 0.85 m away from 22 kg end or 1.15 m from 16 kg end. Intuitively speaking think of a baseball bat. One end is heavier and the other end is lighter. If u balance that bat on a finger, the finger will be closer to the heavier end of the bat, right. Same case here. Center of mass is where the sum of all forces are assumed to act and so placing a support there (a finger) we can balance the whole yoke.

Solution 2: I use Torques. Makes it extremely simple. The center of mass has to be on the yoke so just assume a place on the yoke to be center of mass, say C. Place a fulcrum there. Now the Torques or moments about that fulcrum have to balance for C to be center of mass. So let us assume C to be at a distance x from 16 kg mass. So clockwise moment = anticlockwise moment to balance out the moments. Which means 16*x = 22(2-x), solving for x = 44/38 = 1.15. So the center of mass C will be 1.15 from the 16 Kg mass which is what we expect meaning slightly closer to the heavier end of 22 kg. It does not matter from which mass I take that distance x. You can take x to be from 22 kg mass as well. So now, if I had chosen x to be the distance from 22 kg, in that case 22*x = 16(2-x) so x = 32/38 which is approx 0.85 which means that x is 0.85 m away from 22 kg mass meaning, x is 1.15 m from 16 kg mass. Once again, the same result as the first case. So either way we know the center of mass to be 1.15 m from 16 kg mass. Hope this helps....

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mspeedwagon

Full Member
10+ Year Member
Thank you for the elaborate explanation. It makes sense now.

mspeedwagon: I apologize for butting in like that. Schenker was explaining and I just distracted you further by offering a quick fix. That was not polite. Schenker please correct me if I misstate anything.....

Solution 1: So center of mass is an abstraction where we assume the sum of all forces are concentrated and act. In this problem let us assume that the 16 kg is the origin O of x axis. The 22 kg acts at a distance of 2 m. Furthermore, let point C on the yoke be the center of mass which is at a distance 'x' from the origin O or 16 kg mass. So the center of mass calculation becomes, Total Mass*x = 16*0 + 22*2. The 16 kg is at the origin so therefore I multiply it by 0. So we solve this eqn. (16+22)*x = 44 so x = 44/38 so x is a little more than 1 i.e., x = 1.15 approx. So now we know that the center of mass is around 1.15 m from the Origin O which is at the 16 kg end. Which means it is about .15 m closer to the 22 kg end from the center of the yoke, which is what we expect. It does not matter whether you take origin at 16 kg or 22 kg end. Either way the math will work out. If you take the origin at 22 kg end, (16+22)*x = 22*0 + 16*2 so x = 32/38 so approx 0.85 m away from 22 kg end or 1.15 m from 16 kg end. Intuitively speaking think of a baseball bat. One end is heavier and the other end is lighter. If u balance that bat on a finger, the finger will be closer to the heavier end of the bat, right. Same case here. Center of mass is where the sum of all forces are assumed to act and so placing a support there (a finger) we can balance the whole yoke.

Solution 2: I use Torques. Makes it extremely simple. The center of mass has to be on the yoke so just assume a place on the yoke to be center of mass, say C. Place a fulcrum there. Now the Torques or moments about that fulcrum have to balance for C to be center of mass. So let us assume C to be at a distance x from 16 kg mass. So clockwise moment = anticlockwise moment to balance out the moments. Which means 16*x = 22(2-x), solving for x = 44/38 = 1.15. So the center of mass C will be 1.15 from the 16 Kg mass which is what we expect meaning slightly closer to the heavier end of 22 kg. It does not matter from which mass I take that distance x. You can take x to be from 22 kg mass as well. So now, if I had chosen x to be the distance from 22 kg, in that case 22*x = 16(2-x) so x = 32/38 which is approx 0.85 which means that x is 0.85 m away from 22 kg mass meaning, x is 1.15 m from 16 kg mass. Once again, the same result as the first case. So either way we know the center of mass to be 1.15 m from 16 kg mass. Hope this helps....