Finding the Center of Mass of a Man-Boat System

[InertialRef]

Homework Statement

A man with mass m1 = 69.0 kg stands at the left end of a uniform boat with mass m2 = 163.0 kg and a length L = 2.8 m. Let the origin of our coordinate system be the man’s original location as shown in the drawing. Assume there is no friction or drag between the boat and water.

After the man walks to the right edge of the boat, what is the new location the center of the boat?

I have calculated center of mass to be at 0.98 m.

Homework Equations

Xcm = [(m1r1)+(m2r2)+...+(mNrN)]/[m1+m2+...+mN]

The Attempt at a Solution

What I did was plug in the numbers given in the equation to get this:

[(69*2.8)+(163*x)]/(69+163) = 0.98

Center of mass does not change, so that is what I set the equation to. x = new distance of the center of the boat. Once I found what x was equal to, I subtracted it from 1.4 as follows:

1.4 - x = new position

But this reasoning gives me the wrong answer. Could someone please help me with where I have gone wrong?

 

[tiny-tim]

Hi InertialRef!

[(69*2.8)+(163*x)]/(69+163) = 0.98

No, that assumes he walks 2.8 m to the right, he doesn't (because the boat moves him back a bit).

Instead, find the old and new positions on the boat of the centre of mass.