a man with mass m1=52 g stands at the left end of a uniform boat with mass m2=17

Question

a man with mass m1=52 g stands at the left end of a uniform boat with mass m2=170 kg and a length L=3.4 m let the origin of our coordinate system be the mans original location x-0 there is no friction or drag betweenthe boat and water. 1). what is the location of the center of mass of the system? 2).if the man now walksto the right edge of the boat what is the location of the center of mass of the system? 3).after the man walks to the right edge of the boat what is the new location the center of the boat? 4).now the man walks to the very center of the boat at what location does the man end up?

Explanation / Answer

Since there is no external force to the man-boat system, the center of mass will remain the same

So initially we have ((1.65m)*170kg + 0m*52))/(10 + 52)kg = x cm = 4.5241m

Now let d be the distance the boat moves so the boat's cm will be (1.65 - d) and the man will move
(3.4 -d)

So ((1.65 -d)*161 + (3.4 - d)*52)/(170 + 52) = 4.5241

Solving for d we have d =m

SO the man will move 3.4 -d = m

solve the value of d and put into the last line

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