I got the first one correct ! A man with mass m_1 = 56 kg stands at the left end

Question

I got the first one correct !

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

Explanation / Answer

2) See, when the person moves to the right, there's a leftward force on the boat applied by his foot(equal and opposite on him towards right)

So, Total force on the boat-man system = 0 (External forces = 0, internal forces cancel out)

F = macm

=> acm = 0

Initially, centre of mass was at rest and acm = 0, so it would remain there.

So, even now, location of centre of mass = 1.14 m

3) Centre of mass would be 1.14 m from the right side(from the man's new location, you can calculate this as done in part 1) ).

and centre of mass' new location = 1.14m from origin.

=> new location of man = 1.14 + 1.14 = 2.28 m

4) In the new case, the centre of mass is 1.14 m from the right edge of the boat, while centre of boat is 1.5 m from the right edge => centre of boat is 0.36 m(1.5 - 1.14) towards left from centre of mass of the boat

=> centre of boat location = 1.14 - 0.36 = 0.78 m.

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