Boat-man problem, velocity of center of mass

In summary, the conversation is discussing a problem where a man of mass 84.4 kg is moving from the back to the top of a boat of mass 425 kg, which is moving without friction on ice with a velocity of 4.16 m/s. The man travels for 18.2 m with a velocity of 2.08 m/s. The goal is to find the distance traveled by the boat while the man moves. The attempt at a solution involves using the equation v_{cm} = (v_{1}m_{1} + v_{2}m_{2})/(m_{1}+m_{2}) to find the velocity of the boat to be 0.41 m/s. However
  • #1
Dansuer
81
1

Homework Statement



A man of mass 84.4 Kg is in the back of a boat of mass 425 kg which is moving without friction on ice with velocity 4.16 m/s. the man moves from the back to the top of the boat with velocity 2.08 m/s traveling for 18.2m. How much distance does the boat travel while the man moves?

Homework Equations



v[tex]_{cm}[/tex] = (v[tex]_{1}[/tex]m[tex]_{1}[/tex] + v[tex]_{2}[/tex]m[tex]_{2}[/tex])/(m[tex]_{1}[/tex]+m[tex]_{2}[/tex])

The Attempt at a Solution



The way i try to solve this was that i put V[tex]_{cm}[/tex] = 0 because i choose a coordinate system with velocity V[tex]_{cm}[/tex]. The equation on top gives the velocity of the boat to be 0.41 m/s. Subtract this to the velocity of the boat relative to the ice gives v=3.7 m/s
Now i calculated the distance travelled: 18.2 : 2.08 = x : 3.7 which is 32.37

but it's not the same results as the book gives as a solution. What did i do wrong?
 
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  • #2
why are you taking the frame at velocity of CM. take frame at v=0
then you don't need to find the relative speeds, just use them as given
 
  • #3
You're right i could've done that. But the answer to the problem does not change. And it's not the same as the book. Is the book wrong or i am ?
 
  • #4
I need to sleep now ... its 3AM here ...

But you can post your answer and i'll tell you your mistake as soon as i wake up.
will that work?
 
  • #5
of course, no hurry:smile:

You can find my answer in the point 3 of my first post

thanks
 

What is the Boat-man problem?

The Boat-man problem is a physics problem that involves a boat with two people on it, one at the front and one at the back. The person at the front is pushing the boat forward with a constant force, while the person at the back is pulling the boat backwards with a constant force. The question is, what is the velocity of the center of mass of the boat?

Why is the Boat-man problem important?

The Boat-man problem is important because it demonstrates the concept of center of mass and how it behaves in a system with multiple forces acting on it. It also has real-world applications, as it can be used to understand the motion of objects like boats or cars.

How do you calculate the velocity of the center of mass in the Boat-man problem?

The velocity of the center of mass in the Boat-man problem can be calculated using the principle of conservation of momentum. This states that the total momentum of a system remains constant, so the sum of the momenta of the boat and the two people must equal the final velocity of the center of mass.

What factors affect the velocity of the center of mass in the Boat-man problem?

The velocity of the center of mass in the Boat-man problem is affected by several factors, including the individual masses of the boat and the two people, the forces they are exerting, and the distance between the two forces. Changing any of these variables will result in a different velocity for the center of mass.

Can the velocity of the center of mass in the Boat-man problem ever be zero?

Yes, it is possible for the velocity of the center of mass in the Boat-man problem to be zero. This would occur if the two people on the boat are exerting equal and opposite forces, cancelling each other out and resulting in no net force on the boat. In this case, the boat would remain stationary and the velocity of the center of mass would be zero.

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