Calculating Mass in a Canoe Swap: A Physics Problem

[suspenc3]

are they wrong?

 

[Fermat]

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) + m_c(0) + m(+3.0m)

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

 

[Fermat]

Correction:
They are not (totally) wrong!
The actual answer is m = 68.125 kg

Putting M_T = 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