Is this center of mass question missing information?

[toesockshoe]

Homework Statement

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.

Homework Equations

xcmi=xcmf[/B]

The Attempt at a Solution

Let the mass of the man be $m_1$.
There are no external forces so, $x_{icm} = x_{fcm}$.[/B]
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 doesn't 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?

 

[SammyS]

Homework Statement

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.

Homework Equations

xcmi=xcmf[/B]

The Attempt at a Solution

Let the mass of the man be $m_1$.
There are no external forces so, $x_{icm} = x_{fcm}$.[/B]
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 doesn't 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:

$x_{icm} = \frac{m_1L+(M_b+M_1)R}{M_1+M_b}$

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

set the 2 equal, and you have:
$$\frac{m_1L+(M_b+M_1)R}{M_1+M_b} = \frac{m(L-d) + (M_1+M_b)(R+x)}{M_1+M_b}$$

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

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 M_b .

You have 4 different masses in your equations.

 

[toesockshoe]

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 M_b .

You have 4 different masses in your equations.

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

 

[SammyS]

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

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 ?

 

[toesockshoe]

That part is now fixed.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 ?

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

 

[SammyS]

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

For what ?

 

[toesockshoe]

For what ?

nevermind. is d+x=y?

 

[SammyS]

nevermind. is d+x=y?

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.

 

[toesockshoe]

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.

ok... so is the new CoM final:

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

ugh i can't 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.

 

[SammyS]

ok... so is the new CoM final:

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

that is still 2 unknowns though.

You're missing a final ' } ' .

That looks good.

(Why change variable names now?)

 

[toesockshoe]

You're missing a final ' } ' .

That looks good.

(Why change variable names now?)

im sorry i don't understand... where did i change the variable names?

 

[SammyS]

im sorry i don't understand... where did i change the variable names?

M_boat was M_b .

M_dog was M₁ .

Right ?

 

[toesockshoe]

M_boat was M_b .

M_dog was M₁ .

Right ?

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.

 

[SammyS]

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.

Not really.

Treat d as a known quantity.

 

[toesockshoe]

Not really.

Treat d as a known quantity.

but R and y are unknown.

 

[haruspex]

but R and y are unknown.

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.

 

[toesockshoe]

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.

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.

is that relationship:

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

 

[SammyS]

is that relationship:

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

Not quite right.

 

[toesockshoe]

Not quite right.

ok what is the relationship?

 

[SammyS]

ok what is the relationship?

What is the equation that you're solving?

 

[toesockshoe]

What is the equation that you're solving?

I'm trying to find a relation between x y and the two masses

 

[SammyS]

What is the equation that you're solving?

Well, I thought you were equating the two forms you have for x_cm:

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

then solving for y .

 

[theodoros.mihos]

For simpler equations you can set $x_{icm}=x_{fcm} = 0$

 

[insightful]

Another simplification is to start the dog at the center of mass of the boat.

 

[haruspex]

is that relationship:

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

Did the dog move through a distance y?

 

[toesockshoe]

Did the dog move through a distance y?

no, mdog*x=mboat*y right?

 

[haruspex]

no, mdog*x=mboat*y right?

Yes.