Centre of mass of two boaters in a small rowboat

[Andrei0408]

Homework Statement

Andy (80kg) and Ben (120kg) are in a rowboat (60kg) on a calm lake. Ben is near the bow of the boat and Andy is at the stern at 2m from Ben. How far does the boat move as they change the places? Neglect any horizontal force exerted by the water.

Relevant Equations

Centre of mass position vector = (m1*r1+m2*r2+...+mi*ri)/m1+m2+...+mi

If I calculate the cm pos vector using the two boys' masses, I get 2m, which is the distance between them, but I don't know how to find the boat's displacement when they switch places. How do I represent the fact that they switch places? In what form?

 

[Frigus]

If you take boat+2 persons as a system then no external force is involved when boys exchange their position and we know that Centre of mass remains at rest if no external force acts and it was initially at rest.
Now,
When Ben is near the bow of the boat and Andy is at the stern at 2m from Ben:-
Calculate the COM of the boat+Andy+Ben.
When Andy and ben exchange their position,
Calculate the COM of Andy and Ben then try to figure out where boat's COM should move so that COM of system stays at the same place.

 

[Andrei0408]

But when I calculate COM of boat+andy+ben , R = M * xboat + mB * xB + mA * xA / M + mA + mB, xB and xA are both going to be 2 because we take the position of the boys with regard to the water (I suppose)? And do I leave xboat as an unknown?

 

[Frigus]

Bow and stern refer to front and back of the boat.
Andy is 2m far from the ben and Is standing at stern,this tells us the length of the boat.
You can take any co-ordinate system but usually we take origin of coordinate system at one of the mass so that it's displacement from origin became 0 and it gets easier to solve the equation.

xB and xA are both going to be 2 because we take the position of the boys with regard to the water (I suppose)?

It's our choice to select the coordinate system.
I can't understand what coordinate system you are considering.
It would be easier if you select coordinate system in which y-axis is along the height of ben or Andy and x-axis along the boat.

And do I leave xboat as an unknown?

From the information given in question we are able to know the length of boat so we will be able to know the distance of boat's COM w.r.t to our coordinate system.

 

[Andrei0408]

Bow and stern refer to front and back of the boat.
If Andy is 2m far from the ben and Is standing at stern then this tells us the length of the boat.
You can take any co-ordinate system but usually we take origin of coordinate system at one of the masses so that it's displacement from origin became 0 and it gets easier to solve the equation.

It's our choice to select the coordinate system.
I can't understand what coordinate system you are considering.
It would be easier if you select coordinate system in which y-axis is along the height of ben or Andy and x-axis along the boat.

From the information given in question we are able to know the length of boat so we will be able to know the distance of boat's COM w.r.t to our coordinate system.

I believe I got it right, 0.31m, thank you once again!

 

[gneill]

I believe I got it right, 0.31m, thank you once again!

Can you show the details of the calculations you did to arrive at this result? I suspect that it is not correct. We can check your math.

 

[haruspex]

I believe I got it right, 0.31m, thank you once again!

It's correct.
You can think of their changing places as being a transfer of 40kg through 2m. To counter that, the whole system must move in the other direction by $\frac{2m\times 40kg}{120kg+80kg+60kg}=0.308m$.

 

[gneill]

I first looked at it as a shift of the center of mass (COM) with respect to the center of the boat. So, in the boat frame if the original COM was some distance d from the center in the first configuration, it must also be distance d from the center in the second (in the opposite direction, of course). A total shift of 2d then from one COM location to the other with respect to the center of the boat.

Turning that around, the boat has moved a distance 2d with respect to a fixed COM (COM is fixed in the lake frame of reference).

For the initial configuration I put the COM at distance:
$$d = \frac{-1m \times 80kg + 1m \times 120kg}{80kg + 120kg} = 0.2m$$
So a distance of 20 cm from the center of the boat. Double that to determine how far the boat must have moved to achieve the second configuration.

If my reasoning is faulty, I'd be grateful to know where it has gone astray.

 

[TSny]

For the initial configuration I put the COM at distance:
$$d = \frac{-1m \times 80kg + 1m \times 120kg}{80kg + 120kg} = 0.2m$$

The denominator should include the mass of the boat.

 

[gneill]

The denominator should include the mass of the boat.

Augh! Of course. Thanks.