Canoe and Man Motion: Finding the Relationship Between Distance Traveled

In summary, the man's displacement on the canoe is related to the canoe's displacement on the water, with the man's displacement being equal in magnitude but opposite in direction to the canoe's displacement. This is because, in the lake frame, the center of mass of the man and canoe do not move despite their individual movements.
  • #1
alex.pasek
12
0

Homework Statement


A 55 kg man is standing on a 65 kg and 4.0 m length canoe that floats without friction on water.
The man walks from a 0.75m from one end of the canoe to another point 0.75 from the other end of the canoe.
What distance does the canoe move?

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Homework Equations


Center of mass equation.

The Attempt at a Solution


A tried looking for the ΔXCM, the change on the center of mass, but I do not know how to compute the canoe's CM.

Mod note: Fixed misspelling of "canoe."
 
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  • #2
alex.pasek said:
I do not know how to compute the canoe's CM.
The center of mass for the canoe does not matter, only its displacement does.
 
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  • #3
So; I should compute CMo = CMf?
CMo=(1/M)⋅∑xn ⋅ mn
CMf=(1/M)⋅∑xn ⋅ mn

And calculate ΔCM=Cf - Co

With this I get ΔCM= -0.33 but I do not know how to proceed forward.
 
  • #4
alex.pasek said:
So; I should compute CMo = CMf?
CMo=(1/M)⋅∑xn ⋅ mn
CMf=(1/M)⋅∑xn ⋅ mn

And calculate ΔCM=Cf - Co

With this I get ΔCM= -0.33 but I do not know how to proceed forward.

Well, first of all, you should define all of your quantities, I cannot read your mind to know what Cf and Co are and what sets them apart from CMo and CMf.

Second, the difference in the center of mass is a length and must be given in terms of a length unit.
 
  • #5
Sorry :frown: Here goes the complete reasoning;

CMo=(1/M)⋅∑xn ⋅ mn= (1/120 kg)⋅[(55⋅0.75 kg⋅m)+(65⋅2 kg⋅m)] = 1.43 m
CMf=(1/M)⋅∑xn ⋅ mn= (1/120 kg)⋅[(55⋅ 3.25 kg⋅m)+(65⋅2 kg⋅m)] = 2.57 m

And calculate ΔCM=CMf-CMo= (2.57-1.43)m = 1.14 m

Would be this the displacement of the boat?
 
  • #6
alex.pasek said:
Sorry :frown: Here goes the complete reasoning;

CMo=(1/M)⋅∑xn ⋅ mn= (1/120 kg)⋅[(55⋅0.75 kg⋅m)+(65⋅2 kg⋅m)] = 1.43 m
CMf=(1/M)⋅∑xn ⋅ mn= (1/120 kg)⋅[(55⋅ 3.25 kg⋅m)+(65⋅2 kg⋅m)] = 2.57 m

And calculate ΔCM=CMf-CMo= (2.57-1.43)m = 1.14 m

Would be this the displacement of the boat?

What you have computed is the change in the center of mass in the boat frame. How does this relate to how far the boat moves? What assumptions go into that relation?
 
  • #7
So, If I applied the reference frame, I guess I should undo it. Then by substracting it to each center of mass I should be given the result; so:
dcanoo= (xocanoo-ΔCM)= (2 - 1.14)m = 0.86 m
Would it be correct?
 
  • #8
alex.pasek said:
So, If I applied the reference frame, I guess I should undo it. Then by substracting it to each center of mass I should be given the result; so:
dcanoo= (xocanoo-ΔCM)= (2 - 1.14)m = 0.86 m
Would it be correct?

No. What is true about the center of mass of both man and boat (assuming that there is no transfer of momentum to the water)?
 
  • #9
Orodruin said:
No. What is true about the center of mass of both man and boat (assuming that there is no transfer of momentum to the water)?
I think the momentum is conserved, no externa forces are acting on the system.
 
  • #10
So, assuming that there is no momentum to begin with, what happens with the center of mass?
 
  • #11
Orodruin said:
So, assuming that there is no momentum to begin with, what happens with the center of mass?
The conoe does not move? :oldeek:
 
  • #12
No, the center of mass does not move. You have computed that the center of mass moves by 1.14 m (1.15 m with correct rounding) it a frame that is attached to the canoe. How far must the canoe move for this to happen at the same time as the CoM does not move in the lake frame?
 
  • #13
What if I tried a different approach, as CMo=CMf=1.43 m And calling the displacement "x";
1.43 ⋅ 120 m⋅kg = 55 ⋅ 3.25 m⋅kg + 65 ⋅ x kg ; solving for x i get x ≈-0.11 m
And this would be the displacement of the canoe, I think.
 
  • #14
alex.pasek said:
What if I tried a different approach, as CMo=CMf=1.43 m And calling the displacement "x";
1.43 ⋅ 120 m⋅kg = 55 ⋅ 3.25 m⋅kg + 65 ⋅ x kg ; solving for x i get x ≈-0.11 m
And this would be the displacement of the canoe, I think.
No, this is not correct. Why switch methods when you are basically done?
 
  • #15
I do not understand what is actually happening within the lake's reference frame. I do not know where is the origin for that frame :(
 
  • #16
alex.pasek said:
I do not understand what is actually happening within the lake's reference frame. I do not know where is the origin for that frame :(
You do not need an origin (you can pick one wherever you like).

It may be beneficial to start looking in the lake frame, where the center of mass does not move. What does this tell you about the displacements of the man and boat, respectively?
 
  • #17
So, if in the lake's frame the CM of the man and the boat does not move even if they do. This may suggest the displacement of the man is somehow proportional to the boat's?
I mean, if the man walks to the right, then the boar moves to the left?
 
  • #18
alex.pasek said:
So, if in the lake's frame the CM of the man and the boat does not move even if they do. This may suggest the displacement of the man is somehow proportional to the boat's?
I mean, if the man walks to the right, then the boar moves to the left?
Correct, but how much?
 
  • #19
The distance the man walks minus the change in the center of mass (in the boat's frame)?
 
  • #20
alex.pasek said:
The distance the man walks minus the change in the center of mass (in the boat's frame)?
I suggest you only look at the lake frame. What is the relation between how far the canoe moves and how far the man moves in that frame?
 

What is the center of mass of a canoe?

The center of mass of a canoe is the point where the weight of the canoe is evenly distributed and balanced. It is also known as the center of gravity.

Why is it important to know the center of mass of a canoe?

Knowing the center of mass of a canoe is important for stability and maneuverability. It helps to determine how the canoe will behave in different water conditions and how much weight it can carry without tipping over.

How is the center of mass of a canoe calculated?

The center of mass of a canoe is calculated by finding the midpoint between the bow and stern of the canoe, and then finding the midpoint between the widest points of the canoe. The center of mass is located at the intersection of these two midpoints.

Does the center of mass of a canoe change?

The center of mass of a canoe can change depending on the weight distribution of the canoe. It will shift towards the heavier side if there is more weight on one side of the canoe, and towards the lighter side if there is more weight on the other side.

How can the center of mass of a canoe be adjusted?

The center of mass of a canoe can be adjusted by redistributing the weight of the passengers and cargo. Moving weight towards the center of the canoe can help to lower the center of mass and increase stability.

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