A man with mass m1 = 52 kg stands at the left end of a uniform boat with mass m2 = 171 kg and a length L = 3.5 m.?

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A man with mass m1 = 52 kg stands at the left end of a uniform boat with mass m2 = 171 kg and a length L = 3.5 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.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 boat, what is the location of the center of mass of the system?m 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 4)After the man walks to the right edge of the boat, what is the new location the center of the boat?m 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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  • NCS
    Lv 7
    1 month ago
    Favorite Answer

    1) X_cm = 171kg * 3.5m/2 / (171+52)kg = 1.342 m

    2) Still 1.342 m. No external forces have moved the CoM.

    3) The man is now at x, and so the CoM of the boat is at x-1.75 m.

    1.342 m = [52kg*x + 171kg*(x-1.75m)] / 223kg

    solves to

    x = 2.684 m

    4) x' = 0.934 m

    5) That would be 1.342 m.

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