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Homework Help: Why can't I show Newton's Third Law by different means?

  1. Oct 3, 2015 #1
    1. The problem statement, all variables and given/known data
    A small mass (1 kg) sits next to a larger mass (3 kg) on a table. A force of 5 newtons pushes from left to right on the system while a force of 3 newtons pushes from right to left on the system. Am I justified to conclude that the net force on the larger block has magnitude 2 newtons?

    2. Relevant equations
    Why can't I get the same answer when I solve separately the force of block two onto block one (the larger onto the smaller)? I get the answer when I solve for F_12 and then state that F_21 is - F_12 by Newton's Third Law. But, why can't I do the reverse, F_21 solved first? Why do I get two separate answers?

    3. The attempt at a solution
    I correctly answer that the answer is no. In fact, the force on block two (the larger block) is actually 1.5 newtons. First, I treat the whole as a single system to get the acceleration, which is

    a = (F_right - F_left) / (m1 + m2) = 0.5 m/s^2

    I then solve for the force of block one onto block two to get

    F_12 = m2*a = [ m2 / (m1 + m2) ] * (F_right - F_left) = 1.5 newtons (this is the answer)

    To get the force of block 2 onto block one, I simply chant that because they are an interaction pair,

    F_12 = - F_21.

    [Onion]. BUT, if I start the problem solving for the force of block two onto block one, F21, I cannot get the answer. Actually, I get that the F_21 = 0.5 newtons. This does not make any sense.

    F_21 = m1*a = [ m1 / (m1 + m2) ] * (F_right - F_left) = 0.5 newtons ??? (why isn't it 1.5 N?)

    My point is that I only get the right answer when I do it one way. When I try different ways to reinforce that I understand, being mathematically poetic, I quickly hit the skids.

    Thank you and looking forward to see where I went wrong.


    Is it correct for me to write that the force on block one is

    F_1 = F_right - F_21 = m1*a ?

    and that that force on block two is

    F_2 = F_12 - F_left = m2*a ?
    Last edited: Oct 3, 2015
  2. jcsd
  3. Oct 3, 2015 #2


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    Staff: Mentor

    Which mass is left, which one is right?
    This is not correct, as F_12 is not the only force acting on mass 2. There is another one.
    The other derivation has the same issue, so you get two wrong answers.

    That is a better approach and it should lead to a correct answer.
  4. Oct 3, 2015 #3

    Block one is the left block and block two sits to its right. Do you mean that this is wrong, too? 'F_12 = m2*a = [ m2 / (m1 + m2) ] * (F_right - F_left) = 1.5 newtons. ' 1.5 newtons is the solution at the back of the book.

  5. Oct 3, 2015 #4


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    1.5 N is certainly not the force between the blocks.
    An easy cross-check: the right, heavy block has a force of 3 N towards the left side and the force |F_12| in the opposite direction. It is accelerating to the right. Clearly |F_12| has to be larger than 3 N.
  6. Oct 3, 2015 #5
    Yes, but there is also a force of 5 newtons pushing to the right (on block one side). The 3 newton force is pushing leftward on block two.
  7. Oct 4, 2015 #6


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    Staff: Mentor

    Sure, the 5 N force makes sure that the block accelerates to the right. Details are not relevant for the argument that the mutual force between the blocks has to exceed 3 N.
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