Position of Center Mass Question

[yb1013]

Homework Statement

Romeo (80.0 kg) entertains Juliet (58.0 kg) by playing his guitar from the rear of their boat at rest in still water, 2.80 m away from Juliet, who is in the front of the boat. After the serenade, Juliet carefully moves to the rear of the boat (away from shore) to plant a kiss on Romeo's cheek. How far does the 85.0 kg boat move toward the shore it is facing?

___________ m

Homework Equations

x_cm = (m1*x1 + m2*x2 + ... ) / (m1 + m2 + ... )

The Attempt at a Solution

Im kind of confused on this problem and really need help..

So far I was trying to figure it out but i think i went the wrong way.. My thinking was that there was no external force on the system, so we could just set two equtions equal to each other.

In doing so i came out with..

Initially, x_cm = [(m_B * L/2) + (m_J * L)] / (m_R + m_B + m_J)

and

Now, x_cm = [ m_B*(d + L/2) + d*(m_R * m_J) ] / [m_R + m_B + m_J]

After setting those equal and solving for d, i came out with .03437m which was the wrong answer..

Can anyone please help me and lead me in the right direction?

 

[yb1013]

nevermind, i somehow got it, not sure i completely understand it, but i will in time =]

 

[nlightner]

As a scientist, my response to this content would be to first clarify the problem and its variables. It is important to clearly define the initial and final states of the boat and its occupants, as well as any external forces acting on the system. Additionally, it would be helpful to label the positions and masses of each person and the boat in the equations.

From the given information, it appears that the initial state of the boat is at rest with Romeo in the rear and Juliet in the front, and the final state is with both of them in the rear of the boat. The only external force acting on the system is the force of Juliet moving to the rear of the boat.

Using the equation for the position of center mass, we can calculate the initial and final positions of the center of mass of the system. Plugging in the given masses and distances, we get:

Initial x_cm = (80.0 kg * 0.00 m + 58.0 kg * 2.80 m) / (80.0 kg + 58.0 kg) = 1.37 m

Final x_cm = (80.0 kg * 0.00 m + 58.0 kg * 0.00 m + 85.0 kg * d) / (80.0 kg + 58.0 kg + 85.0 kg) = d * 0.50 m

Since the center of mass must remain at the same position before and after Juliet moves to the rear of the boat, we can set the initial and final x_cm equal to each other and solve for d:

1.37 m = 0.50 m * d

d = 2.74 m

Therefore, the boat moves 2.74 meters toward the shore it is facing. It is important to note that this calculation assumes a frictionless and weightless boat, and may not accurately represent real world scenarios.