Using Newton's Second Law for a System of Particles

  • Thread starter blue5t1053
  • Start date
  • #1
23
1
The Problem:
A 4 kg dog stands on a 16 kg flatboat at distance 12 m from the shore. The dog walks 5 m along the boat toward the shore and then stops. Assuming no friction between the boat and the water, find how far the dog is then from the shore.
The dog's displacement is towards the shore. The boat's displacement is away from the shore.

My work:
I set the shore as the origin.
Then, ((4 kg x 12 m) + (16 kg x 12 m))/(4 kg + 16 kg) = 12 m for original center of mass. ((4 kg x 7 m) + (16 kg x 12 m))/(4 kg + 16 kg) = 11 m.
So I get a 1 m difference.

My question: does it mean that the dog is currently 7 m + 1 m away from the shore, or does it mean that the dog is 12 m - 1 m away from the shore?
 

Answers and Replies

  • #2
382
2
i think the answer will be 8.25 m since when the dog walks towards the shore. the boat will move away from the shore according to the principle of momentum conservation.
 
  • #3
kdv
348
6
The Problem:
A 4 kg dog stands on a 16 kg flatboat at distance 12 m from the shore. The dog walks 5 m along the boat toward the shore and then stops. Assuming no friction between the boat and the water, find how far the dog is then from the shore.
The dog's displacement is towards the shore. The boat's displacement is away from the shore.

My work:
I set the shore as the origin.
Then, ((4 kg x 12 m) + (16 kg x 12 m))/(4 kg + 16 kg) = 12 m for original center of mass.
Is the dog initially in the middle of the boat? You are using this here.
((4 kg x 7 m) + (16 kg x 12 m))/(4 kg + 16 kg) = 11 m.
So I get a 1 m difference.

My question: does it mean that the dog is currently 7 m + 1 m away from the shore, or does it mean that the dog is 12 m - 1 m away from the shore?

The last calculation is wrong. How di dyou get 7 meters?? And you are assuming that the boat has not moved! No, you must call the new position of the dog "x". It's an unknown. The center of the boat has moved but since the dog walked 5 meters relative to th eboat, the center of the boat is now at x+5 meters from the shore.
Use that the center of mass has not moved and solve for x.
 
  • #4
23
1
Is the dog initially in the middle of the boat? You are using this here.


The last calculation is wrong. How di dyou get 7 meters?? And you are assuming that the boat has not moved! No, you must call the new position of the dog "x". It's an unknown. The center of the boat has moved but since the dog walked 5 meters relative to th eboat, the center of the boat is now at x+5 meters from the shore.
Use that the center of mass has not moved and solve for x.

If I set the new position of the dog as:

((4 kg x (X+5) m) + (16 kg x 12 m))/(4 kg + 16 kg) = ? m

Then solve for X?
 
  • #5
kdv
348
6
If I set the new position of the dog as:

((4 kg x (X+5) m) + (16 kg x 12 m))/(4 kg + 16 kg) = ? m

Then solve for X?

No. You have again treated the boat as if it had not moved!

As for "x", use it to represent the position of the dog. Then the position of the center of the boat is x+5 .
 
  • #6
23
1
No. You have again treated the boat as if it had not moved!

As for "x", use it to represent the position of the dog. Then the position of the center of the boat is x+5 .

Alright, I think I have my answer.

Total distance of the dog moving toward the shore is: 5 m - [tex]\Delta[/tex]X_boat;
The final distance of the dog is: X_dog_initial - 5 m + [tex]\Delta[/tex]X_boat;
X_dog_intial = 12 m
After some algebra: [tex]\Delta[/tex]X_boat = [M_dog/(M_dog + M_boat)] x (5 m);

[tex]\Delta[/tex]X_boat = [(4 kg)/(4 kg + 16 kg)] x (5 m) = 1 m;

X_dog_initial - 5 m + [tex]\Delta[/tex]X_boat;
12 m - 5 m + 1 m = 8m
 
  • #7
kdv
348
6
Alright, I think I have my answer.

Total distance of the dog moving toward the shore is: 5 m - [tex]\Delta[/tex]X_boat;
The final distance of the dog is: X_dog_initial - 5 m + [tex]\Delta[/tex]X_boat;
X_dog_intial = 12 m
After some algebra: [tex]\Delta[/tex]X_boat = [M_dog/(M_dog + M_boat)] x (5 m);

[tex]\Delta[/tex]X_boat = [(4 kg)/(4 kg + 16 kg)] x (5 m) = 1 m;

X_dog_initial - 5 m + [tex]\Delta[/tex]X_boat;
12 m - 5 m + 1 m = 8m

And you can double check your answer by calculating the final position of the center of mass ans see that it is the same as before.

(and note that the dog has moved with respect to the shore by 4 times more distance than the boat has moved which makes sense since the boat is 4 times more massive)
 

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