A man with mass m_1 = 50 kg stands at the left end of a uniform beat with mass m

Question

A man with mass m_1 = 50 kg stands at the left end of a uniform beat with mass m_2 = 166 kg and a length L = 3.1 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 beat and water. 1) What is the location of the center of mass of the system? m 2) If the man now walks to the right edge of the beat, what is the Location of the center of mass of the system? m 3) After walking to the right edge of the beat, how far has the man moved from his original location? (What is his new location?) m 4) After the man walks to the right edge of the beat, what is the new location the center of the beat? m 5) Now the man walks to the very center of the beat. At what location does the man end up? m

Explanation / Answer

1)
The location of the center of mass is, as seen from the right side of the boat:
CM = (166*1.55 + 50*3.1)/(50+166) = 1.908 m
Cm = 3.1-1.908= 1,191 m from the left side

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)) = 3.1/(1+50/166) = 2.38 m to the right from the origin.

4) The center of mass of the boat has moved to the left by
x = 3.1/(1 + m(boat)/man) = 3.1/(1+166/50)= 0.717 m to the left from its original position.
= 1.55 m - 0.717 = 0.832 right side from the origin.
5)
s = 1.55/(1+50/166) = 1.191 m to the left
the new position of man = 2.38 - 1.191 = 1.189 m to the right of the origin

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