How far does the canoe move during this process?

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The discussion focuses on calculating the movement of a canoe when a woman walks from one end to the other, emphasizing the concept of the center of mass. The participants clarify that, since there are no external forces acting on the system, the center of mass remains unchanged, meaning the canoe must move to maintain this balance. The initial and final positions of the center of mass are calculated using the formula for the center of mass, leading to the conclusion that the canoe moves in response to the woman's movement. The final calculation suggests that the canoe moves approximately 1.29 meters during the process. This confirms the understanding of the relationship between the positions of the woman and the canoe in maintaining the center of mass.
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[SOLVED] Center of Mass

Homework Statement



A 45.0-kg woman stands up in a 60.0-kg canoe of length 5.00 m. She walks from a point 1.00 m from one end to a point 1.00 m from the other end.

yf_Figure_8_44.jpg


If you ignore resistance to motion of the canoe in the water, how far does the canoe move during this process?


2. The attempt at a solution

So she is moving to the other side of the canoe.

What do they mean by "If you ignore resistance to motion of the canoe in the water, how far does the canoe move during this process?"

That the water does not move the canoe?

Ma = 45kg
Mb = 60kg

There is no applied external force, so velocity is constant..(is this right)

How do I proceed...

(MaX + MbX) / (Ma+Mb) for finding center of mass, so

45X + 60X / 105 ...
 
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Right, you've answered one of your own questions. There's no applied external force, so we can forget about any outside forces on the system. You know the equation for the center of mass, so you want to find the initial center of mass of the system, the system being the lady and the canoe. Next you want to find the center of mass of the system after the lady has walked one meter from the end of the canoe. The difference in the two cm's is the distance the conoe moves, right? Don't forget that we assume the canoe is of uniform mass, in other words, we take its center of mass, not to be confused with the center of mass of the system, to be right in the middle of the canoe. Does this help?
 
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Better to write (MaXa+MbXb)/(Ma+Mb) as center of mass, where a is woman and b is canoe. When the woman (and canoe) move to new positions Y, the new center of mass is (MaYa+MbYb)/(Ma+Mb). But since there is no force from the water, the center of mass doesn't move. Why? So those two centers of mass are equal. Subtract the two and get zero. What does this tell you about the relation between (Xa-Ya) and (Xb-Yb)?
 
After simplifying I get that

MaXa+MbXb = MaYa+MbYb

this should mean that there was no movement in the center of mass.

I see, the rest was plug in the equation above. And solve for Yb, and since it's the center of mass, Ya (Position of person) must be added 1.5 for her to reach the other side of the canoe. :)
 
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Hey is the answer to this problem 1.29 m? Sorry for the huge bump, I didn't want to post something that was already posted. I just need someone to confirm this please?
 
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