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Homework Help: Newton's Third Law problem

  1. Jan 9, 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}= -F_{B on A}[/tex]
    F = ma

    3. The attempt at a solution
    What seems to be the issue is how my teacher solved this problem. 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}= F_{A on B} - F_{C on B}[/tex]
    [tex]F_{net}= m_{B}a - (-m_{C}a)[/tex]
    [tex]F_{net}= (10 Kg)(1.5m/s^{2}) + (5.0 Kg)(1.5m/s^{2})[/tex]
    [tex]F_{net}= 23 N[/tex]

    Therefore, she says that [tex]F_{B on A} = 23 N

    However, what confused me is the two negative signs she put infront of [tex]m_{C}a[/tex]. 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}= m_{B}a + m_{C}a[/tex]
    [tex]F_{net}= (10 Kg)(1.5m/s^{2}) + (5.0 Kg)(1.5m/s^{2})[/tex]
    [tex]F_{net}= 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 9, 2008 #2

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    You can find the total force. You know the mass of A. So, you know the force B applies back on A.
  4. Jan 9, 2008 #3
    I found the total force that box B and C exerted on box A.
  5. Jan 9, 2008 #4

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    Why does C have to come into the picture?

    F_A = F_net - F_B_on_A =>

    M_A*a = M_tot*a - F_B_on_A.

    I think this is conceptually the simplest. Ultimately, after algebra, all the methods are the same.
  6. Jan 9, 2008 #5

    your teacher has [tex]F_{B on A}[/tex] pointing in the opposite direction as [tex]F_{A on B}[/tex]
    This is rather confusing. First you need a minus sign in the first formula, because the two forces acting on B now point in opposite directions, and then you need another minus sign in [tex] F_{C on B}=-m_{C}a[/tex]. It's Easier to have positive=rightwards for all forces and accelerations.
    What's worse is that between the first and the second line of her calculations, The meaning of [tex]F_{net}[/tex] changes, at first it's the net force on B, but then it's the net force on B and C as you had. It seems she changed her mind in mid-calculation.

    Your calculation is ok.​
  7. Jan 9, 2008 #6

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    Regardless of the value you have got, why are you calling the sum of the forces on B and C as Fnet(left). Fnet is the force which is making all the boxes go.​
  8. Jan 9, 2008 #7


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    Boxes A, B, and C have total mass 23 kg. If they are acceletated at 1.5 m/s2, the force must F= ma= (23)(1.5)= 34.5 Newtons- and it is exerted upon A directly. A alone has mass 8 kg. In order to accelerate it alone at 1.5m /s2, it must be pushed by a net force of Fnet= (8)(1.5)= 12 Newtons. What force must B be exerting on A?
  9. Jan 9, 2008 #8

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    Hi Halls,

    That's exactly what I've given in post #4 (without the littersome numericals, of course).
  10. Jan 9, 2008 #9
    So my mistake was solving for [tex]F_{net}[/tex]? What if instead I had [tex]F_{A on B} = 23N
    [/tex], and according to Newton's Third Law, [tex]F_{B on A} = 23N
  11. Jan 9, 2008 #10

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    Would you do it leaving C out of it? Or you just can't give up your method? It happens to all of us.
  12. Jan 9, 2008 #11
    I suppose your method of solving the problem is easier. It makes me realize just how many ways one can go about solving such problems. Thanks for the help.
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