# Homework Help: Is this center of mass question missing information?

1. May 27, 2015

### toesockshoe

1. The problem statement, all variables and given/known data
A dog of mass M is standing on a raft so that he is a distance L from the shore. He walks a distance d on the boat toward the shore and then stops. The boats has a mass of Mb. Assume no fricton between the boat and the water. How far is the dog from shore when he stops moving?

I think that the distance between the edge raft and the shore should be given. If i am wrong please tell me where I am going wrong.

2. Relevant equations

xcmi=xcmf

3. The attempt at a solution

Let the mass of the man be $m_1$.
There are no external forces so, $x_{icm} = x_{fcm}$.

Note: we can not assume the person walks to the edge of the boat.
Because we are only interested in the change in the positions, it doesnt matter where the origin is.... cause we only care about the change, so I put the origin at the shore.

Here is where I got lost: I let the distance between the shore and the right edge of the boat (assuming the shore is to the right of the boat.... so the closer edge) is R.
also, I denoted the horizontal displacement of the boat as x.

so:

$x_{icm} = \frac{M_1L+(M_b+M_1)R}{M_1+M_b}$

$x_{fcm} = \frac{M_1(L-d) + (M_1+M_b)(R+x)}{M_1+M_b}$

set the 2 equal, and you have:
$$\frac{M_1L+(M_b+M_1)R}{M_1+M_b} = \frac{M_1(L-d) + (M_1+M_b)(R+x)}{M_1+M_b}$$

here is the problem: I have 2 unknowns and one equation. Where did I go wrong?

Last edited: May 27, 2015
2. May 27, 2015

### SammyS

Staff Emeritus
First of all, there is enough information given to solve the problem.

What are all of those masses?

There are only two objects with mass involved here. The dog (or person) has mass, M . The boat (or raft) has mass Mb .

You have 4 different masses in your equations.

3. May 27, 2015

### toesockshoe

sorry, i messed up on the latex. i believe i changed it correctly now. there are only 2 masses.

4. May 27, 2015

### SammyS

Staff Emeritus
That part is now fixed.

There are still a number of difficulties with you equations.

To make this easier to fix-up, let L be the initial position of the dog (as you have it). However, to truly get the center of mass of the combination, R should represent the initial position of the CoM of the boat (without the dog) .

Then your equation for the position of the initial position of the CoM of the pair is almost correct. It should be
$\displaystyle \ (x_{cm})_i = \frac{M_1 L+M_b R}{M_1+M_b} \$​

The dog moves a distance of d relative to the boat. Note: If you then say the position of the dog is L-d, that means you are saying that's your final answer is L-d , but that's incorrect.

Think of it this way. If the dog moves a distance of y towards the shoreline and the boat moves a distance of x away from the shoreline, the how are x and y related to d ?

5. May 27, 2015

### toesockshoe

x*the mass of the boat+dog=-y*the mass of the dog right?

6. May 28, 2015

### SammyS

Staff Emeritus
For what ?

7. May 28, 2015

### toesockshoe

nevermind. is d+x=y?

8. May 28, 2015

### SammyS

Staff Emeritus
If x and y are both positive, the d = x + y.

The dog will be at L - y from shore. The boat will be R + x from shore.

Since you want y in your answer, replace x with d - y, so the boat's CoM will be R + (d - y) from shore.

Now find an expression for the final CoM for the combination of dog & boat.

9. May 28, 2015

### toesockshoe

ok... so is the new CoM final:

$\frac{(L-y)M_{dog}+(R+d-y)M_{boat}}{M_{dog}+M_{boat}}$ ?

ugh i cant get latex to work: here:

((L-x)Md + (R+d-y)Mb)/(Md + Mb)...... Mb stands for the mass of hte boat and Md stands for hte mass of the dog.

that is still 2 unknowns though.

10. May 28, 2015

### SammyS

Staff Emeritus
You're missing a final ' } ' .

That looks good.

(Why change variable names now?)

11. May 28, 2015

### toesockshoe

im sorry i dont understand.... where did i change the variable names?

12. May 28, 2015

### SammyS

Staff Emeritus
Mboat was Mb .

Mdog was M1 .

Right ?

13. May 28, 2015

### toesockshoe

oh sorry, yes yes. ok so how can i continue? if i set it equal to the initial center of mass equation, then I have 2 unknowns and 1 equation.

14. May 28, 2015

### SammyS

Staff Emeritus
Not really.

Treat d as a known quantity.

15. May 28, 2015

### toesockshoe

but R and y are unknown.

16. May 28, 2015

### haruspex

You know y in terms of x and d.
When you equate initial and final mass centres, maybe, just maybe, R will vanish?

I feel working with explicit mass centres has made this unnecessarily complex, though. If the dog is distance x closer to the shore and the boat is distance y further away, there's a simple relationship between x, y, and the two masses.

17. May 28, 2015

### toesockshoe

is that relationship:

mdog * x = m(boat+dog)*y?

18. May 28, 2015

### SammyS

Staff Emeritus
Not quite right.

19. May 28, 2015

### toesockshoe

ok what is the relationship?

20. May 28, 2015

### SammyS

Staff Emeritus
What is the equation that you're solving?

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