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Center of Mass (Boxcar water example)

  • Thread starter BMcC
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  • #1
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A boxcar of length 8.00 m and height 2.40 m is at rest on frictionless rails.

The empty boxcar has a mass of 3750 kg. Inside the boxcar, located at the left end, is a tank containing 2450 kg of water. The tank is 2.00 m long and 2.40 m tall. The tank starts to leak, and the water fills the floor of the boxcar uniformly. Assume that all the water stays in the box car.

1) What is the displacement of the boxcar 14 s after the water has settled in the bottom?




SO my original thinking was that the final velocity of the boxcar will be zero after all of the water has leaked out. What I've been trying to do is find the Center of Mass of the boxcar initially and subtracting it from 4m, which I believe to be the final center of mass once the water has settled. Taking the lower left corner of the boxcar as the origin:

I went Xcmi = (4m)(3750kg boxcar) + (1m)(2450kg tank), and dividing this by the total mass (3750kg + 2450kg) to get 2.9m. Then when I go 4m - 2.9m, I get 1.19m which I thought would be the displacement, but I was wrong.

Could anybody steer me in the right direction? I know I'm close, I'm just not quite sure what to do from here.
 

Answers and Replies

  • #2
ko2
1
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you're right, you just forgot to indicate the magnitude. The boxcar moves to the left so displacement is -1.19m
 
  • #3
32
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Not according to my online assignment!
 
  • #4
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Anyone?
 
  • #5
199
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I went Xcmi = (4m)(3750kg boxcar) + (1m)(2450kg tank), and dividing this by the total mass (3750kg + 2450kg) to get 2.9m. Then when I go 4m - 2.9m, I get 1.19m which I thought would be the displacement, but I was wrong.
this is the center of mass that you have calculated. if the box car doesn't move when water spreads evenly on the floor, then the center of mass would be at 4 meters from the left end of the car. But that is not so, boxcar has to move so that center of mass doesn't change, so it has to move such that its center lies 2.81 meters from the initial position of left end of the car. Now you tell me in which direction boxcar has to move!!!:wink:

the above is an intuitive method, you can also do it by equating final and initial center of mass.

P.S. - in the equation you wrote, you forgot to write M term on the left side of the equation. I assume this a typo because rest of the calculation seems correct apart from rounding 2.81 to 2.9.
 
  • #6
32
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Okay so it would lie 2.81 meters to the left of its initial position? And yeah that was a typo with the 2.9 - oops!
 
  • #7
199
15
Okay so it would lie 2.81 meters to the left of its initial position? And yeah that was a typo with the 2.9 - oops!
I guess it wasn't clear enough what I wrote :

it has to move such that its center lies 2.81 meters from the initial position of left end of the car
the center of mass doesn't change. so the center car of the car moves to the center of mass calculated but initially it was at 4 meters, that is the boxcar moves to the left by 1.19 meters or a total displacement of -1.19.

The intuitive method is not easy to explain, it develops with practice.

Try the regular method:

##MX_{cmi}= (4m)(3750kg) + (1m)(2450kg)= x(3750kg + 2450kg)##
##MX_{cmf}= x(3750kg + 2450kg)##
##MX_{cmf}=MX_{cmi}##
Solving this for x gives final position of the boxcar.

Displacement will be final minus initial position.
 

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