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Newton's Third Law - Interacting Systems

  1. Oct 17, 2007 #1
    1. The problem statement, all variables and given/known data

    The 1.0 kg block in the figure (attached) is tied to the wall with a rope. It sits on top of the 2.0 kg block. The lower block is pulled to the right with a tension force of 20 N. The coefficient of kinetic friction at both the lower and upper surfaces of the 2.0 kg block is mu_k = 0.500.

    What is the tension in the rope holding the 1.0 kg block to the wall?
    What is the acceleration of the 2.0 kg block?

    2. Relevant equations

    F_a on b = -F_b on a
    a_ax = - a_bx

    3. The attempt at a solution

    I started by drawing force diagrams for both boxes and then calculating the net force in the x-direction for each box.

    For box 1:
    Fnet x = -T_1 - (mu_k)(n) = -1T_1 - (0.5)(9.8)(1) = -1T_1 - 4.9

    For box 2:
    Fnet x = T_2 - 2(mu_k)(n) = 20 - (2)(0.5)(20.6) = -0.6

    At this point, I've tried to find the acceleration by setting the Fnet = ma, but get the wrong answer for the acceleration and solving for T_1. Anyone know where I may be making the mistake?
     

    Attached Files:

  2. jcsd
  3. Oct 17, 2007 #2

    Doc Al

    User Avatar

    Staff: Mentor

    The first equation's true but the second is not: the top box doesn't accelerate.

    (1) Careful with signs: What direction does the friction act?
    (2) What does the net force equal?

    Recompute the friction forces. Realize that there are two friction forces acting on box 2, but they are not the same; write each one separately.
     
  4. Oct 17, 2007 #3
    The first equation's true but the second is not: the top box doesn't accelerate.

    Thanks! I rewrote the Fnet as you suggested:

    Box 1: Fnetx = (mu_k)(n) - T1 = 4.9 - T1
    Box 2: Fnetx = T2 - (mu_k)(9.8)(1) - (mu_k)(9.8)(2) = 5.3

    I set Box 1 Fnet = 0 and was able to solve for the first part of the problem.

    For the second part of the problem (finding the acceleration of box 2), I tried to use:
    Fnet = (m)(a) = 5.3 = (2)(a)
    which gave me 2.65 m/s for acceleration, which was incorrect.

    I'm assuming I have to use the Fnet for box 2 because it does have a net force, unlike box 1. Did I calculate Fnetx for box 2 incorrectly?
     
  5. Oct 17, 2007 #4

    Doc Al

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    Staff: Mentor

    That second friction force is incorrect. Hint: What's the total force that box 2 exerts on the floor?
     
  6. Oct 17, 2007 #5
    Thanks so much! I had just realized that I had to change the mass somewhere when you replied. Thanks again for pointing me in the right direction!
     
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