Asked by Burrito
A 70 kg man stands at the back of a 3.0 meter long boat that has mass 2X kg. The front of the boat is
touching a dock. The man walks to the front of the boat. How far from the dock is he? Assume there is
no friction between the boat and the water.
touching a dock. The man walks to the front of the boat. How far from the dock is he? Assume there is
no friction between the boat and the water.
Answers
Answered by
Damon
Boats do not have backs. They have sterns or transoms.
They also have bows.
Anyway:
Find original center of mass from the pier (not a dock but that term is actually often used these days for a pier). I will assume that the center of mass of the boat is 1.5 meters from the pier.
70*3 + 2X*1.5 - (70+2X)Xcg
so
Xcg = (210+ 3X)/(70+2X)
from dock
The Xcg from dock will not change if there are no external forces.
call distance from bow to dock d
[70*d + 2X*(1.5+d)]/(70+2X) = same old Xcg
so
210 + 3X = (70+3X)d + 3 X
d = 210/(70+3X)
They also have bows.
Anyway:
Find original center of mass from the pier (not a dock but that term is actually often used these days for a pier). I will assume that the center of mass of the boat is 1.5 meters from the pier.
70*3 + 2X*1.5 - (70+2X)Xcg
so
Xcg = (210+ 3X)/(70+2X)
from dock
The Xcg from dock will not change if there are no external forces.
call distance from bow to dock d
[70*d + 2X*(1.5+d)]/(70+2X) = same old Xcg
so
210 + 3X = (70+3X)d + 3 X
d = 210/(70+3X)
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