A man with mass m_1 = 56 kg stands at the left end of a uniform boat with mass m

Question

A man with mass m_1 = 56 kg stands at the left end of a uniform boat with mass m_2 = 178 kg and a length L = 3 m. Let the origin of our coordinate system be the man's original location as shown in the drawing. Assume there is no friction or drag between the boat and water. What is the location of the center of mass of the system? If the man now walks to the right edge of the boat, what is the location of the center of mass of the system? After walking to the right edge of the boat, how far has the man moved from his original location? (What is his new location?) After the man walks to the right edge of the boat, what is the new location the center of the boat? Now the man walks to the very center of the boat. At what location does the man end up?

Explanation / Answer

1)

so intital condition is as such:-
m1 is at x=0
m2 is at x=L/2
so center of mass position=(m1*0+m2*(L/2))/(m1+m2)=1.141 m

2)

If the man now walks to the right edge of the boat, the location of the center of mass of the system remains unchanged as there is no external force.

3)

The displacement of the man is given by
?x = s/(1 + m(man)/m(boat)) = 3/(1+56/178) = 2.282 m to the right from the origin.

4)

let the boat shifts x
178*3/2 + 67 *0=56*(3-x) + 178*(3/2 -x)
so x=0.72 m
so centre=(3/2-x) = 0.78 m from origin

5)

when the man walks to the very center of the boat,then that position be x=d1
then [m1*d1+m2*d1]/(m1+m2)=1.141
or d1=1.141 m
so he ends up at the center of the boat.

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