How to find the center of mass and unknown weight

[JakeD23]

Homework Statement

A canoe of mass 38 kg lies at rest in still water. A man and a woman are at opposite ends of the canoe 4.0 m apart and symmetrically located with respect to the canoe’s centre (which is also its centre of mass). The mass of the man is 65 kg and the woman’s mass is smaller.

The two people then change places and the man observes that the canoe shifts a distance 0.20 m relative to the water. What is the woman’s mass? [57 kg]

Homework Equations

mr=Ʃm_ir_i

The Attempt at a Solution

I'm guessing the solution is found through a quadratic. I tried substituting in the know data for before and after the boat and making the position of the center of mass negative for the second equation but nothing seemed to help. I've done the problem 4 times and am no closer to solving it no matter how I visual the problem. I just can't seem to find the right answer

 

[ehild]

Show your work in detail, please.

ehild

 

[Gavran]

m_man⋅x_man+m_woman⋅x_woman+m_canoe⋅x_canoe=(m_man+m_woman+m_canoe)⋅x_cm
m_man<m_woman
m_man⋅(x_man+4m+0,20m)+m_woman⋅(x_woman-4m+0,20m)+m_canoe⋅(x_canoe+0,20m)=(m_man+m_woman+m_canoe)⋅x_cm
m_man>m_woman
m_man⋅(x_man+4m-0,20m)+m_woman⋅(x_woman-4m-0,20m)+m_canoe⋅(x_canoe-0,20m)=(m_man+m_woman+m_canoe)⋅x_cm
m_man⋅x_man+m_woman⋅x_woman+m_canoe⋅x_canoe=m_man⋅(x_man+3,8m)+m_woman⋅(x_woman-4,2m)+m_canoe⋅(x_canoe-0,20m)