An 80-kg person stands at one end of a 130-kg boat. He then walks to the...

An 80-kg person stands at one end of a 130-kg boat. He then walks to the other end of the boat so that the boat moves 80 cm with respect to the bottom of the lake.

A. What is the length of the boat?

B. How much did the center of mass of the person-boat system move when the person walked from one end to the other?

Expert Solution

Let length of boat = L cm

Suppose we have a coordinate system relative to the bottom of the lake with:
Boat's position initially from x=0 cm to x=L cm
Suppose person (S) initially at x= L cm (right hand end of boat)
Let Boat's centre of mass at x = C cm

The overall (system's) initial centre of mass is at X1 where (taking moments about x=0):
(80+130)X1 = 130C + 80L (i)
After S walks left to the other end of the boat, the boat moves right 80cm.

Boat's position is now from x=80 cm to x=(L+ 80) cm
S's position is x = 80 (left end of boat)
Boat's centre of mass is at C+80

The overall (system's) final centre of mass is at X2 where (taking moments about x=0):
(80+130)X2 = (130)(C+80) + 80*80 (ii)
Since no external forces acted on the system, the system's centre of mass has not moved, so X1=X2. This means the left hand sides of both equations (i)and (ii) are equal. So we can equate the right hand sides giving:
130C + 80L = (130)(C+80) + 80*80
. 80L=16800 cm

L=210 cm=2.1 m

Length of boat=2.1 m

(B)center of mass of the person-boat system did not move when the person walked from one end to the other because there was no external force.

genius_generous answered 1 year ago

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