Calculating Mass in a Canoe Swap: A Physics Problem

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Homework Help Overview

The problem involves calculating the mass of Carmelita based on the movement of a canoe when she and Ricardo exchange seats. The scenario is set in a canoe with known masses for both individuals and the canoe itself, and it explores concepts related to the center of mass and internal forces during the exchange.

Discussion Character

  • Conceptual clarification, Assumption checking, Mathematical reasoning

Approaches and Questions Raised

  • Participants discuss using the center of mass formula and the implications of internal forces on the movement of the canoe. There are attempts to calculate Carmelita's mass using various equations and reasoning about the positions of the individuals relative to the center of mass.

Discussion Status

The discussion is active, with participants exploring different methods to approach the problem. Some have provided calculations and reasoning, while others question the assumptions made about the center of mass and the effects of the movements. There is no explicit consensus on the correct approach or solution yet.

Contextual Notes

Participants are working under the constraints of the problem as presented, with some expressing confusion over terms like center of mass (COM) and the implications of mass distribution in the canoe. The discussion includes varying interpretations of the movement of the center of mass and the distances involved.

  • #31
suspenc3 said:
...
*im pretty sure that the centre of mass of ricardo+carmelita+canoe does not change when they switch places
The "group" centre of mass does change.
Have you got a ruler handy. Balance it on your finger with two unequal weights (an eraser and a pencil ??) equidistant from the centre. Your finger will be at the COM of the ruler plus two weights. Now switch the positions of those weights. To keep it balanced, your finger will have to move position, so the COM of the "group" has changed.
The total movement of your finger will be twice the distance of it from the mid-point of the ruler.

Edit: could you say where you measured the COM from in each case ?
 
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  • #33
now that I think of it...Their answer makes no sence.

Why would The COM be in the center when the two masses are unequal?
 
  • #34
It's a question on conservation of momentum (and energy change). The OP question isn't.

Was there some bit in particular you meant :confused:
 
  • #35
Ah, you meant Q5.
 
  • #36
are they wrong?
 
  • #37
I would say, that solution is wrong!

They say that X is the position of the "group" COM, measured from the centre of the canoe.
Yet they mark it as being equidistant, 3.0m from M and m, even though thay also say it should be nearer to Cian.

Also, they give their eqns as,

X = M(-3.0m) + mc(0) + m(+3.0m)

But X is a distance, not a moment arm. They forgot to multiply X by the combined mass. MT = M + mc + m.

Even if you do that, if you substitute their answer of 68 kg back into those eqns, I got X = -0.202 from the 1st eqn and X = -0.209 from the 2nd eqn.

That last bit makes me believe that, yes, their answer is wrong.
 
  • #38
Correction:
They are not (totally) wrong!
The actual answer is m = 68.125 kg

Putting MT = 178.125 and using m = 68.125kg gives X = -0.2 (I only checked one eqn) I expect the other eqn wiil give the same result.
But it shows that the COM, X, is half the distance moved by the canoe. Which corresponds with my explanation of things :smile:
 

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