The Mystery of the "Man on a Boat" Problem with Easy Tips

[fartguy46]

Homework Statement

A man with mass m1 = 57 kg stands at the left end of a uniform boat with mass m2 = 179 kg and a length L = 3.1 m. Let the origin of our coordinate system be the man’s original location as shown in the drawing. Assume there is no friction or drag between the boat and water.

1) What is the location of the center of mass of the system? Answer: 1.18

2) If the man now walks to the right edge of the boat, what is the location of the center of mass of the system? Answer: 1.18

3) After walking to the right edge of the boat, how far has the man moved from his original location? (What is his new location?) Answer: not sure here

4) After the man walks to the right edge of the boat, what is the new location the center of the boat? Answer: not sure here

5) Now the man walks to the very center of the boat. At what location does the man end up?
Answer: not sure here

Relevant Equations

3) Delta X = L/(1+(m1/m2))? where m1 is the man's mass and m2 is the boat's mass
4) Delta X = L/(1+(m2/m1))?
5) I'm honestly not sure

Hey guys,

So with some help from classmates I have seen the way I am supposed to approach this problem starting from number 3, which is to use the above equation. But the problem is I have never seen this equation before in class! I don't know what the purpose of the equation is, where it was derived from, and how to utilize it for different scenarios. I'm trying to look for some practice problems but it's difficult given the fact I don't know what I'm supposed to be looking for in the first place.

Same thing with number 4. Then for number 5 I think I am supposed to utilize information obtained from 3 and 4, again not sure since I'm pretty lost on number 3 to begin with. Much help appreciated!

 

[RPinPA]

It's a conservation of momentum problem. When he starts walking with momentum mv, the boat receives an equal and opposite momentum. So it's moving backward in the water while he's moving forward. When he stops, it stops.

You weren't told how fast he's walking, but it turns out not to matter. Either pick some number to work with, like v = 1 m/s, or just leave his v as a variable. Then work out his motion and the boat's motion.

 

[haruspex]

starting from number 3, which is to use the above equation. But the problem is I have never seen this equation before in class!

It is never a good idea to employ equations for which you are uncertain of the circumstances in which they apply. Better to work from first principles.
What conservation law applies? What external horizontal forces act on the boat+man system? What does that tell you about the motion of their common mass centre?

 

[fartguy46]

How come velocity is at play here?

Also, so the conservation of momentum law should be applied here right? I can't think of any horizontal forces that would come into play here. Lastly, this would mean that the motion of the center of mass would be to the right?

 

[fartguy46]

It is never a good idea to employ equations for which you are uncertain of the circumstances in which they apply. Better to work from first principles.
What conservation law applies? What external horizontal forces act on the boat+man system? What does that tell you about the motion of their common mass centre?

Also, so the conservation of momentum law should be applied here right? I can't think of any horizontal forces that would come into play here. Lastly, this would mean that the motion of the center of mass would be to the right?

 

[RPinPA]

Also, so the conservation of momentum law should be applied here right? I can't think of any horizontal forces that would come into play here.

No external forces, but if he goes from standing still to moving, giving himself momentum, then he's going to be exerting a force on the boat.

Lastly, this would mean that the motion of the center of mass would be to the right?

No... and I just realized my suggestion above at #2 is unnecessarily complex. I mean, I've worked out similar problems using that approach, but there's a much easier approach.

Before he starts moving, the center of mass of boat + man is fixed. Not moving. There is no external force, so even when he starts walking, nothing will cause the center of mass to move. That's all you need to solve this problem. So it uses the answers to parts A and B.

 

[haruspex]

the conservation of momentum law should be applied here right?

Yes, and what is the momentum of the system initially, before the man moves?

can't think of any horizontal forces that would come into play here.

Quite so.

this would mean that the motion of the center of mass would be to the right?

Why, if there are no external forces?

 

[fartguy46]

No external forces, but if he goes from standing still to moving, giving himself momentum, then he's going to be exerting a force on the boat.
No... and I just realized my suggestion above at #2 is unnecessarily complex. I mean, I've worked out similar problems using that approach, but there's a much easier approach.

Before he starts moving, the center of mass of boat + man is fixed. Not moving. There is no external force, so even when he starts walking, nothing will cause the center of mass to move. That's all you need to solve this problem. So it uses the answers to parts A and B.

Alright so he does exert a force on the boat when he goes from standing to moving. I'm thinking that this force would have to do something with mass times velocity or mass times an acceleration of some sort correct? And what would such a force do? Would it end up moving the whole boat to the positive x direction?

 

[RPinPA]

Alright so he does exert a force on the boat when he goes from standing to moving. I'm thinking that this force would have to do something with mass times velocity or mass times an acceleration of some sort correct? And what would such a force do? Would it end up moving the whole boat to the positive x direction?

You have been given a number of hints which already answer this question, including in the answer you just quoted. Also you didn't have a diagram so I have no idea which way the positive x direction is.

 

[fartguy46]

Alright so the center of the mass cannot move since there are no external forces. I'm thinking maybe I need to create some type of equation with perhaps the momentum of the boat being equal to the momentum of the man? I'm trying to figure out how I would go about doing this since I need to make sure I include all possible information from both objects...

 

[RPinPA]

Alright so the center of the mass cannot move since there are no external forces. I'm thinking maybe I need to create some type of equation with perhaps the momentum of the boat being equal to the momentum of the man? I'm trying to figure out how I would go about doing this since I need to make sure I include all possible information from both objects...

That was my suggestion at #2. But you don't need to do that, as I said later.

The center of mass doesn't move. You should know where the boat is relative to the CM, and where the man is relative to the CM. So when you know where the CM is, you know where they are.

 

[fartguy46]

You have been given a number of hints which already answer this question, including in the answer you just quoted. Also you didn't have a diagram so I have no idea which way the positive x direction is.

Here's a photo:

Also my apologies I'm really trying to figure out based on previous replies how to utilize these hints. I'm really bad at physics for some reason compared to differential equations lol

 

[RPinPA]

And how do you calculate CM in the two situations? In other words, what is your calculation for parts 1 and 2?

 

[haruspex]

Alright so the center of the mass cannot move since there are no external forces.

Right.
How far is the man from that centre at the start? How far is he from it at the end?

 

[fartguy46]

And how do you calculate CM in the two situations? In other words, what is your calculation for parts 1 and 2?

So for part 1 I used the center of mass equation and did this: (57(0) + 179(1.55))/(179+57). Then for part 2 I assumed would be the same answer...and I just realized I don't know why...

 

[fartguy46]

And how do you calculate CM in the two situations? In other words, what is your calculation for parts 1 and 2?

I just found that if I do (0(179)+3.1(57))/(179+57) I would get 0.7487. Then I did 3.1-0.7487 and got 2.35. To summarize my thought process I realized that because the man is now on the other side, he would be at the point 3.1. Now the boat would be at the 0 point instead. I don't have any reasoning for this besides that since the man is on the other side the boat is now occupying the point 0.

 

[fartguy46]

Right.
How far is the man from that centre at the start? How far is he from it at the end?

So at the start he will be 1.18 units from the mass. Same if he moves to the end of the boat. I also replyed to someone else my current findings which I would appreciate if you would take a look at to see if I'm in the right thought process.

 

[haruspex]

So at the start he will be 1.18 units from the mass. Same if he moves to the end of the boat.

You mean, from the common mass centre.
And that common mass centre does not move, so how far has the man moved?

 

[jbriggs444]

Alright so he does exert a force on the boat when he goes from standing to moving. I'm thinking that this force would have to do something with mass times velocity or mass times an acceleration of some sort correct? And what would such a force do? Would it end up moving the whole boat to the positive x direction?

The nice thing about conservation laws is that they allow you to ignore the process and concentrate on the before and after states, already knowing that the conserved quantity is unchanged.

With no external forces, momentum is conserved.

With zero initial momentum, that means that total momentum is always zero.

But momentum is equal to total mass times velocity of the center of mass. Total mass is non-zero. So velocity of the center of mass must be zero. Wherever it starts, it stays there.

Knowing that the center of mass is stationary, all of the forces, velocities and accelerations become irrelevant. You can concentrate on the end states. And the formula for center of mass.

The center of mass is the weighted average position of a set of pieces. The weighting that is used is the masses of the pieces. [If the pieces are big, use the position of each piece's center of mass].

$$x_{cm}=\frac{x_1m_1 + x_2m_2 + ...}{m_{tot}}$$

My inclination is to write down some equations and solve algebraicly -- the brute force method. @haruspex appears to be aiming at an argument from symmetry. His way is likely to be simpler.

 

[RPinPA]

I just found that if I do (0(179)+3.1(57))/(179+57) I would get 0.7487. Then I did 3.1-0.7487 and got 2.35.

OK, that looks more like it. I'm not going to check your arithmetic, I'll take your word for it. What I was getting at was that is that if the CM of the boat alone is in the middle because of symmetry, then when you add the man to one end of the other, that shifts the CM of the system toward that end of the boat. So without checking the numbers, the answers above are plausible.

In the answers you gave in your original post, you said the CM was at 1.18 m in both cases. That makes no sense and also doesn't agree with what you just said.

To summarize my thought process I realized that because the man is now on the other side, he would be at the point 3.1.

That's fine. You're making a symmetry argument (I think), which is fine. The situation is the mirror image of the original situation, so the CM will be at the mirror image location. Instead of 0.75 m from the left, it's 0.75 m from the right (again just trusting your arithmetic without checking).

So at the start he will be 1.18 units from the mass.

Uh, what? He's at 0 m in the coordinate system. The CM is at 0.75 m in the same coordinate system by your calculation. How do you get a distance of 1.18 METERS between those two points?

 

[sojsail]

It sounded like you played around, without much understanding, with the data to come up with the 1.18 number. I would like to suggest that you rethink the solution, starting from #1. In post #19, @jbriggs444 gave a good formula for calculating c-o-m. Read his/her explanation again, I suggest you use that formula. Remember that initially the man and the left nose of the boat are at x=0. Do not be surprised if you again get 1.18 m for the com location.

For #2, review all of what post #19 said about com and conservation of momentum.

Then remember your new understanding of the formula when you go on to #3 and beyond. I will be busy tomorrow, but I will watch this problem when I can.

 

[berkeman]

Thread is closed until OP responds to the Mentors' PMs about fixing his username...