Center of Mass Rowboat Question

[sweetpete28]

Center of Mass Rowboat Question...Please Help!

Can someone please help with below question? I really don't know where to begin...

When empty, a 166 kg rowboat is symmetrical. A boy of mass 70.5 kg sits 2.1 m from the center of mass of the rowboat toward the front of the boat, and a girl of mass 68.8 kg sits 2.38 m from the center of mass of the rowboat toward the rear of the boat. Find x, distance from center mass of the emptyt rowboat to the center of mass of the rowboat-plus-kids system.


I know the equation for center of mass [xcm = m1x1 + m2x2 + ... / m1 + m2 + ...] but I don't know where to start since the L of the rowboat is not given.

Please help!

 

[cepheid]

Welcome to PF,

L of the boat doesn't really matter. The empty boat is symmetric, meaning that its mass is evenly distributed. That means that the centre of mass of the empty boat is right in the middle of the boat (at the geometric centre).

For convenience, you can measure positions from this point i.e. take x = 0 to be at the middle of the boat.

The other thing to note that the centre of mass is the point where you can consider all the mass to be located (i.e. the system is equivalent to a single mass at that point, in some sense). Therefore, the entire mass of the boat can be considered to be located at x = 0. That takes care of one of the three masses. The other two (the boy and the girl) are located at the stated positions relative to x = 0. So, you can solve for the position of centre of mass of this three-body system, which is what the problem is asking for.

 

[sweetpete28]

Thanks! That's what I thought, but I wasn't sure...

Ok. So: [(0)(166) + (2.1)(70.5) + (-2.38)(68.8)] / [166 + 70.5 + 68.8] = -.0514

So magnitude of distance b/t center of mass of empty boat and center of mass of rowboat + kids system = .0514...right?

 

[cepheid]

Thanks! That's what I thought, but I wasn't sure...

Ok. So: [(0)(166) + (2.1)(70.5) + (-2.38)(68.8)] / [166 + 70.5 + 68.8] = -.0514

So magnitude of distance b/t center of mass of empty boat and center of mass of rowboat + kids system = .0514...right?

I think that's the right way to do it.

 

[cepheid]

A check that gives me confidence in your answer: say instead of using the COM of the empty boat as x = 0, we instead decide to use the half-way point between the boy and the girl as x = 0. In this case, since the distance between them is 4.48 m, the half-way point is 2.24 m back from where the boy is. This means that the centre of the boat is 0.14 m ahead of the half-way point between the boy and the girl, like so:

B|------|----|-----------|G
       COM  x=0

Using this different coordinate system, the expression for the centre of mass is:

(2.24*(70.5) + (-2.24)*(68.8) + 166*(0.14)) / (70.5 + 68.8 + 166)

= 0.0885948248

So the centre of mass of the total system is 0.089 m ahead of the half-way point between the two people. This means its distance from the centre of the boat is:

0.14 - 0.0885948248 = 0.0514051752