Newtons Third Law

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Homework Statement


Two blocks are lined up in a row. Block A a mass of 1kg, and Block B with a mass of 2kg. If block a is A is pushed forward by 12N, how much force does Block B exert on Block A. Friction can be nelected.




Homework Equations


Newtons Third law
F(A on B)=-F(B on A)
For the system of A:
[tex]F_{Force on A}-F_{B on A}= m_{a}a_{a}[/tex]
For the system of B:
[tex]F_{A on B}=m_{B}a_{b}[/tex]




The Attempt at a Solution


I know something is wrong because when I plug in values I get [tex]F_{B on A}=0[/tex]

I think it may be just me but I've had some trouble with Newtons Third law, I can't stand the book we are using: Physics for engineers and scientists by knight, any suggestions?
 
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Answers and Replies

  • #2
Doc Al
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Homework Equations


Newtons Third law
F(A on B)=-F(B on A)
Good.
For the system of A:
[tex]F_{Force on A}-F_{B on A}= m_{a}a_{a}[/tex]
I'd write that as:
[tex]F_{Force on A}+F_{B on A}= m_{a}a_{a}[/tex]
For the system of B:
[tex]F_{A on B}=m_{B}a_{b}[/tex]
Good.

Now what can you say about their accelerations?
 
  • #3
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Well I know that the Acceleration of B is going to be less than A. Since A acts on B's mass of 2kg, B will act on A less than 12N.
Right?
Good.

I'd write that as:
[tex]F_{Force on A}+F_{B on A}= m_{a}a_{a}[/tex]
But isn't [tex]F_{B on A}[/tex] acting in the other direction to make it [tex]-F_{B on A}[/tex]
 
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  • #4
Doc Al
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Well I know that the Acceleration of B is going to be less than A.
Oh really? So you're saying that A will pass through B? :wink:
Since A acts on B with 12N, B will have to act on A less than 12N because there is a Net Force.
12 N is the applied force on A, not the force that A and B exert on each other.

But isn't [tex]F_{B on A}[/tex] acting in the other direction to make it [tex]-F_{B on A}[/tex]
By definition, the force on A due to B is [tex]F_{B on A}[/tex]. So the net force on A will be the applied force (12N) plus the force of B on A.

Let's get specific. Assume A is on the left and B on the right. And let's take to the right as the positive direction. Let's call the magnitude of the force of B on A (and A on B) to be Fn. So [tex]F_{B on A}[/tex] = -Fn, since that force points to the left. And [tex]F_{A on B}[/tex] = +Fn, since that force points to the right.

It's certainly true that [tex]F_{B on A} = -F_{A on B}[/tex]. Make sense?
 
  • #5
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By definition, the force on A due to B is [tex]F_{B on A}[/tex]. So the net force on A will be the applied force (12N) plus the force of B on A.

Let's get specific. Assume A is on the left and B on the right. And let's take to the right as the positive direction. Let's call the magnitude of the force of B on A (and A on B) to be Fn. So [tex]F_{B on A}[/tex] = -Fn, since that force points to the left. And [tex]F_{A on B}[/tex] = +Fn, since that force points to the right.

It's certainly true that [tex]F_{B on A} = -F_{A on B}[/tex]. Make sense?
Got yah on that.

Ok, so by definition. of [tex]F_{B on A} = -F_{A on B}[/tex]
The agent acts on A with 12N, then A must act on the agent with 12N, right(neglecting friction)? The force that A imposes on B:
[tex]F_{AonB}=m_{a}a_{acceleration of a}[/tex]
I get it conceptually, but I guess I have trouble representing it mathematically.
 
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  • #6
Doc Al
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Ok, so by definition. of [tex]F_{B on A} = -F_{A on B}[/tex]
That's true not by definition, but by Newton's 3rd law.

The agent acts on A with 12N, then A must act on the agent with 12N, right(neglecting friction)?
That's certainly true, but we don't really care about what happens to the agent of that force in this problem.
The force that A imposes on B:
[tex]F_{AonB}=m_{a}a_{acceleration of a}[/tex]
I get it conceptually, but I guess I have trouble representing it mathematically.
Careful: you mixed up your A's and B's a bit.

Since the only force acting on block B is the force from A, then [tex]F_{A on B}[/tex] is the net force on B. Then [tex]F_{A on B} = m_b a_b[/tex] follows from Newton's 2nd law.

But return to the accelerations for a minute. How does the acceleration of A relate to the acceleration of B? Hint: Use common sense!
 
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  • #7
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The accelerations are going to be the same?
 
  • #8
Doc Al
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The accelerations are going to be the same?
You got it.
 
  • #9
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So.
Acceleration is going to be [tex]a=F/m=12N\_div1Kg=12m/s^2[/tex]
[tex]F_{A on B} = m_b a_b= (2kg)(12m\s^2)=12N[/tex]
then this equals [tex]-F_{B on A}[/tex] ?
 
  • #10
Doc Al
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So.
Acceleration is going to be [tex]a=F/m=12N\_div1Kg=12m/s^2[/tex]
That acceleration is incorrect. When you apply Newton's 2nd law to block A, you must use the net force on block A. You just used the applied force of 12 N.

(Note that you can also treat block A and B together as a single system. What's the net force on that system? What's its mass?)
 
  • #11
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ok..
[tex]F_{AB}=(m_{a}+m_{a})a_{AB}[/tex]
The force on A in contact with B is 12N so the net force is 12N.
[tex]a_{AB}=F_{net}divided(m_{a}+m_{b})=4m/s^2[/tex]
 
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  • #12
Doc Al
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Yes, that's the correct value for the acceleration of the blocks. Now you can solve for the force they exert on each other.
 
  • #13
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Thanks Doc...did I really make this problem harder than it really was?
 
  • #14
Doc Al
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did I really make this problem harder than it really was?
Problems often appear much easier after you've figured them out! :smile:
 

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