How Does the Center of Mass Affect Distance When a Dog Walks on a Boat?

[Twigs]

The problem is a dog weighing 10lb is standing on a flatboat so that he is 20ft from the shore. He walks 8 ft on the boat toward shoare and then halts. The boat weighs 40 lb, and one can assume there is no friction between it and the water. How far is the dog from the shore at the end of this time?

Alright, since there is no friction i know the center of mass between the dog and boat stays the same. So i tried setting the center of mass in the beginning equal to the center of mass at the end. However for the equation i need to know the mass of each which i have and their distance from some point(i chose the shore). Using the shore i could get the distance the dog started at 20 ft. Not sure on the rest though. Any help is appreciated.

 

[vsage]

why not set the center of mass of the boat to be 20 feet from shore in the beginning as well? Just pick a place for it to reside and it'll turn out right.

 

[wais]

Hi there,

Thank you for reaching out for help with your center of mass problem. I can definitely provide some guidance to help you solve it.

To start, let's define the center of mass as the point where the total mass of the system is equally distributed on both sides. In this case, the system consists of the dog and the boat.

Since the center of mass remains constant, we can use the equation for center of mass:

x_cm = (m1x1 + m2x2) / (m1 + m2)

Where x_cm is the distance of the center of mass from a reference point, m1 and m2 are the masses of the dog and boat respectively, and x1 and x2 are their respective distances from the reference point.

In this problem, our reference point can be the shore, which means the initial distance of the dog from the shore is 20 ft. Now, let's determine the initial center of mass of the system.

x_cm = (10lb * 20ft + 40lb * 0ft) / (10lb + 40lb)

x_cm = 8ft

This tells us that the initial center of mass is 8 ft from the shore. Now, let's determine the final center of mass after the dog walks 8 ft towards the shore.

x_cm = (10lb * x1 + 40lb * x2) / (10lb + 40lb)

We know that the dog walked 8 ft, so x1 = 20ft - 8ft = 12ft. And since the boat is stationary, x2 remains at 0ft.

x_cm = (10lb * 12ft + 40lb * 0ft) / (10lb + 40lb)

x_cm = 12/5 ft = 2.4 ft

This means that the final center of mass is 2.4 ft from the shore. Since the center of mass remains constant, this is also the final distance of the dog from the shore.

I hope this helps you solve the problem. Let me know if you have any further questions. Good luck!