A man with mass m 1 = 64 kg stands at the left end of a uniform boat with mass m
ID: 1286090 • Letter: A
Question
A man with mass m1 = 64 kg stands at the left end of a uniform boat with mass m2 = 174 kg and a length L = 3.3 m. Let the origin of our coordinate system be the mans original location as shown in the drawing. Assume there is no friction or drag between the boat and water.
1)
What is the location of the center of mass of the system?m
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2)
If the man now walks to the right edge of the boat, what is the location of the center of mass of the system?m
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3)
After walking to the right edge of the boat, how far has the man moved from his original location? (What is his new location?)m
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4)
After the man walks to the right edge of the boat, what is the new location the center of the boat?m
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5)
Now the man walks to the very center of the boat. At what location does the man end up?m
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Explanation / Answer
1)
The location of the center of mass is, as seen from the right side of the boat:
CM = (174*1.65 + 64*3.3)/(64+174) = 2.09 m
Cm = 3.3 - 2.09 = 1.21 m from the left side (1.21 m from the origin)
2) the location of the center of mass will not change, as long as there is no force on the system from the outside.
3) The displacement of the man is given by
?x = s/(1 + m(man)/m(boat)) = 3.3/(1+64/174) = 2.41 m to the right from the origin.
4) The center of mass of the boat has moved to the left by
?x = 3.3/(1 + m(boat)/man) = 3.3/(1+174/64)= 0.89 m to the left from its original position.
= 1.65 m - 0.89 = 0.76 right side from the origin. (This is the mass center of the boat alone, the mass center of the boat with man is still at the old distance from the origin = 1.21 m).
5)
?s = 1.65/(1+64/174) = 1.21 m to the left
the new position of man = 2.41 - 1.21 = 1.20 m to the right of the origin.
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