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Newton's Third Law problem

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[SOLVED] Newton's Third Law problem

Homework Statement


Three boxes, A, B, and C, are positioned next to each other on a horizontal, frictionless surface. An applied force acting on box A causes all the boxes to accelerate at [tex]1.5m/s^{2}[/tex]
. Calculate the force that box B exerts on box A.

http://img70.imageshack.us/img70/8836/thirdlaw2cf8.png [Broken]​
[/URL]

Homework Equations


[tex]F_{A on B}[/tex]= [tex]-F_{B on A}[/tex]
F=ma

The Attempt at a Solution


What seems to be the issue is how my teacher solved this problem. I have only realized this problem, so I can't have asked her personally. What she did is isolate box B and find the [tex]F_{net}[/tex] acting on it. This is the calculation she did:

[tex]F_{net}[/tex]= [tex]F_{A on B}[/tex]-[tex]F_{C on B}[/tex]
[tex]F_{net}[/tex]= [tex]m_{B}a[/tex] -(-7.5 N)
[tex]F_{net}[/tex]= (10 Kg)([tex]1.5m/s^{2}[/tex]) + 7.5 N
[tex]F_{net}[/tex]= 23 N


However, what confused me is the two negative signs she put infront of the 7.5 N. I think only one would suffice to take into consideration that [tex]F_{C on B}[/tex] is a negative value, being that its direction is left.

I solved this problem taking a different approach. I found [tex]F_{net}[/tex] of box B and C, and added them together. Here are my calculations:

[tex]F_{net}[/tex]= [tex]m_{B}a[/tex] + [tex]m_{C}a[/tex]
[tex]F_{net}[/tex]= (10 Kg)([tex]1.5m/s^{2}[/tex]) + (5.0 Kg)([tex]1.5m/s^{2}[/tex])
[tex]F_{net}[/tex]= 23 N


Could someone please confirm whether my method of solving the problem is correct or not?​
 
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Answers and Replies

  • #2
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Anyone?

Is their something wrong or confusing with my post that nobody has replied yet?
 
  • #3
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...

...anyone please?
 

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