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Homework Help: Newton's Third Law problem

  1. Jan 3, 2008 #1
    [SOLVED] Newton's Third Law problem

    1. The problem statement, all variables and given/known data
    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]​

    2. Relevant equations
    [tex]F_{A on B}[/tex]= [tex]-F_{B on A}[/tex]

    3. 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?​
    Last edited by a moderator: May 3, 2017
  2. jcsd
  3. Jan 5, 2008 #2

    Is their something wrong or confusing with my post that nobody has replied yet?
  4. Jan 8, 2008 #3

    ...anyone please?
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