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Newton's 3rd

  1. Oct 28, 2007 #1
    1. The problem statement, all variables and given/known data

    Blocks with masses of 1 kg, 2 kg, and 3 kg are lined up in a row on a frictionless table. All three are pushed forward by a 12 N force applied to the 1 kg block. How much force does the 2 kg block exert on the 3 kg block? How much force does the 2 kg block exert on the 1 kg block?

    2. Relevant equations

    Newtons third law
    F=m(a) ?

    3. The attempt at a solution

    This isn't my first time posting on this forum and I know that I really should include at least an attempt at the solution but what I'm asking for is just a general quick idea to help me get moving along, as I am stuck.

    Thank you very much.
  2. jcsd
  3. Oct 28, 2007 #2


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    Draw Free body diagrams of each block and of all three blocks together. Identify all forces acting in each diagram. Then use newton 2 for each. (It's easiest to start by looking at all three blocks together so you can quickly determine the acceleration of the system, which is the same as the acceleration of each block).
  4. Oct 28, 2007 #3
    Well finding the acceleration of the system was easy enough/

    I drew the fbd however I'm having trouble differentiating between the different blocks.

    For instance,
    Block 1- 1 kg acting on block 2- 2 kg...
    When I use the formula f=ma clearly I will arrive at two different products when substituting either 1 or 2 for mass. The thing that has me confused however is that I know that every force exerted on an object exerts an equal force back. Through this logic it makes me believe that possibly each block is exerting exactly 12 newtons on each other.

    I'm obviously not understanding a certain concept.
  5. Oct 28, 2007 #4
    Would the force exerted between 2-3 =

    x=6 N

    Force between 1-2
    x= 2(2)
    x= 4 N

    Seems incredibly simple so I'm sure there must be some sort of problem somewhere along the way.
  6. Oct 28, 2007 #5


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    You want to look at forces in the horizontal direction. Start by looking at block 3. There is a horizontal contact force from block 2 acting on block 3. Calculate it using F_net = ma.
  7. Oct 28, 2007 #6
    Let the magnitude of contact force between 1kg block and 2kg block be F1 and that between 2kg block and 3kg block be F2. (Note that I have already applied Newton's 3rd Law.)
    Now, do NOT do in mind.. use a piece of paper and pen. Draw free body diagram of each. Start from block 1kg. For example 1kg block will have 12N force and F1 in opposite direction; 3kg block will have only F2 in the direction of overall acceleration.
    Comeon.. now do it!
  8. Oct 28, 2007 #7
    The problem is indeed simple. There is no problem coming in the way. The acc of the system which you have got is correct. Not just try to solve the problem by the FBD. You will get it.
  9. Oct 28, 2007 #8
    Do not confuse, you can start from any block. It is just a matter of choice, I guess.
  10. Oct 28, 2007 #9
    I think, I got it what is it that you are not understanding.

    Please note that in F = m*a, F stands for net force on the particle of mass, m, whose net acceleration is a.

    When you wrote:
    "Force between 1-2
    x= 2(2)
    x= 4 N"

    You wrote 2nd law for 2kg mass: "x= 2(2)". This "x" is the net force on 2kg block. And, net force on 2kg block is sum of forces acted on it by 1kg block and 3kg block. {Note that you have already calculated force on 3kg block by 2kg block.}
  11. Oct 28, 2007 #10
    I reiterate, pease refer to free body diagram!
  12. Oct 28, 2007 #11
    Errr... This is painful...
    Thank you to the repliers, especially saket.
    You told me that I did arrive to the correct force between 2-3, 6 N.
    Seeing as the net force equals 12 like you said, would that have to lead me to believe that 12-6 equals Fnet-F2= F1= 6?

    Thank you for being so patient.
  13. Oct 28, 2007 #12
    2nd law on 2kg mass: (Fnet = ) F1 - F2 = 2kg*(2m/s^2) where F2 is 6N.
  14. Oct 28, 2007 #13
    Note that F1 and F2 are in opposite direction on 2kg mass. (Apply Newton's 3rd Law carefully.)
    Thus, F1 = 10N.
    Also, proceeding in a similar fashion, please verify yourself that Newton's second law also holds for the 1kg block.
  15. Oct 28, 2007 #14
    Ahhh yes!
    Applied force - F2 = 1 kg (2m/s^2)

    I believe that I finally understand what you were trying to explain. Thank you so much for your help and patience... It's been a long weekend.
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