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Newton's 3rd Law

  1. Apr 29, 2007 #1
    The lower block in Figure P8.28 is pulled on by a rope with a tension force of F = 28 N. The coefficient of kinetic friction between the lower block and the surface is 0.40. The coefficient of kinetic friction between the lower block and the upper block is also 0.40. What is the acceleration of the 2.0 kg block?

    Relevant Equations
    f_k = uN
    Newton's 2nd and 3rd Laws

    What I Did
    Well I made two systems, one with the block on the bottom (block 1) and one with the block on the top (block 2). Now one thing that seemed confusing to me was that block 1 was being pulled with a force of 28 N, but in the picture the force is depicted going away from the wall. I'm not too sure where block 2 comes into play, other than enacting a frictional force on block 1. So this is how I have Newton's 2nd law set up:

    Force_1x = friction(2 on 1) + friction (floor on 1) - 28 N
    = .4 (1) + .4(2) - 28 N
    Force_1y = 0
    Force_2x = 28 N - friction(1 on 2)
    = 28 N - .4(2)
    Force_2y = 0

    But this gets me a somewhat high acceleration, that is not right.
    As you can see, I'm doing something seriously wrong here. Please help me with this problem!
  2. jcsd
  3. Apr 29, 2007 #2


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    Gold Member

    The total frictional force is 1*0.40 + 3*0.40.

    Both blocks are resting on the floor.
  4. Apr 29, 2007 #3
    OK so that gives me a force of -26.4 N in the x direction. I divided by the mass of the bottom block, 2 kg, to get an acceleration of -13.2 m/s^2. I tried dividing by the total mass to get -8.8 m/s^2, which didn't work either. Is there some way that I set up Newton's second law wrong other than that? Also, did I make the right assumption with the direction of the tension, is it a negative tension because it is being pulled left? Or is it a positive force that is directed right as in the picture? I appreciate all of your help.
    Last edited: Apr 29, 2007
  5. Apr 29, 2007 #4
    Can anyone offer their insight?
  6. Apr 30, 2007 #5


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    you are forgetting about the tension caused by the rope in your force diagram. Heres what I came up with

    [tex] \sum F_2 = m_2 a_2 = F_a - F_{f2g} -F_{f21} -T [/tex]
    [tex] \sum F_1 = m_1 a_1 = F_{f1} - T [/tex]
  7. Apr 30, 2007 #6
    hints: a1=-a2, the only unknown variable is T.....
  8. Apr 30, 2007 #7
    i've got one more stab at this question...so i have to make sure that i get it right...
    so what is F_a, is that the tension force? i'm not sure i understand. also, when you say the force of 2 on 1, you mean block 2 on 1 right? and the lower block is moving to the left, correct? i'm quite confused, i thought the tension force was 28 N...
    Last edited: Apr 30, 2007
  9. Apr 30, 2007 #8
    F2g= frictional force between 2 and ground (floor).
    F12= frictional force between 1 and 2.
  10. Apr 30, 2007 #9
    so when you say a_2, you say that is the acceleration of the block on the floor? I'm working it out right now...
  11. Apr 30, 2007 #10
    OK heres what I did. For the record I am not at all confident in my answer :(....yet the accelerations seem to agree well...
    Following the format above, I did this:
    28 - 7.84 - 3.92 - T = 2* a_2
    7.84 - T = 1* a_1
    I messed around with the equations to get T=10.64 N a_2 = 2.8 m/s^2 and a_1 = -2.8 m/s^2

    So I suppose this answers the question with -2.8 m/s^2...
    So nervous to put it in and see if it's right, does it seem logical?
    Last edited: Apr 30, 2007
  12. Apr 30, 2007 #11
    Since the question was "What is the acceleration of the 2.0 kg block?": the answer should be 2.8 m/s^2(if your calculations are correct). for a2=-a1...
    Last edited: Apr 30, 2007
  13. Apr 30, 2007 #12
    Yeah it's 2.8 m/s^2. Thank you for all of your help, I really appreciate everyone's help! The clarifications helped out greatly Ahmed. I understand how to do these types of problems now. Someday maybe I'll help all of you out ;) ....after I actually learn how to do all of these things that is....
    Last edited: Apr 30, 2007
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