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Homework Help: Newton's Law and a rope

  1. Jul 19, 2007 #1
    1. The problem statement, all variables and given/known data
    The two blocks are connected by a heavy uniform rope with a mass of 4.00kg. An upward force of 200N is applied as shown.
    a.) What is the acceleration of the system?
    b.) What is the tension at the top of the heavy rope?
    c.) What is the tension at the midpoint of the rope?
    The right is my own drawing of FBD and the sketch on the left side is from the book
    2. Relevant equations
    [tex] w = mg[/tex]
    [tex] F = ma[/tex]

    3. The attempt at a solution
    b.)Acceleration of the system:
    [tex] \sum{F_{net}} = F + (-W_{m1}) + (-W_{rope}) + (-W_{m2})[/tex]
    Fnet = 200N - (6.00kg)(9.80m/s^2) - 4.00kg(9.80m/s^2) - (5.00kg)(9.80m/s^2)
    Fnet = 53.0N
    m_total = 4 + 5 + 6
    m_total = 15kg
    a = (Fnet)/(mtotal)
    a = 53N / 15kg
    a = 3.53m/s^2

    b.) tension at the top of the heavy rope?
    i dont know why the answer is 120N

    c.) midpoint of rope,....
    same here 93.3N
    Last edited: Jul 19, 2007
  2. jcsd
  3. Jul 19, 2007 #2


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    Your acceleration looks good for the system.... but when you take your FBD's of the parts, you are forgetting to include the rope tension and/or rope mass. For sample, isolate the top block just at the top of the rope, and you have the 200N force up, the block weight down, AND the tension at the top of the cord down...
  4. Jul 19, 2007 #3
    so 200N is the upward force
    and you need to find the total downward force of the 6.00kg block which is the tension of
    the upper rope?

    [tex] F_{y} = F_{applied} - W_{6kgblock} - F_{6kg block}[/tex] <<< is the force of 6 kg block negative?
    F_{y} = 200 - (6.00kg)(9.8m/s^2) - (6.00kg)(3.53m/s^2)
    = 120N

    still confused in part C
    i think i did this:
    T = 5kg(3.53m/s^2) + 5kg(9.8m/s^2)
    T = 66.65N

    For computing the midpoint of the tension of the rope:
    Average Force = (120N + 66.65N) / 2
    = 93.3N
    is the acceleration of the system negative?
    Last edited: Jul 19, 2007
  5. Jul 20, 2007 #4
    Are my answers correct? they agree at the back of my book but im not sure if my solutions/assumptions here are correct..

  6. Jul 20, 2007 #5
    Last question

    Befor i finalize my answer:
    is my FBD correct?

  7. Jul 20, 2007 #6


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    You're getting the right answers, but you're means of arriving at these answers is going to mess you up in the long run. And tension in ropes should not be considered as a 'force ' of the block.

    Let's look at (b) first. To get the tension in the top of the rope, you draw an FBD of the top block by circling a 'cloud' around it. Wherever that 'cloud' cuts thru a contact force, there will be forces acting, which are in addition to the weight force on the block which is always there. So, in the FBD of the top block, you have 3 forces acting:

    The applied force of 200N acting up (positive),
    the weight of the block, 58.8N, acting down (negative)
    the tension at the top of the rope, T_top, acting down (negative).
    (Note that tension forces always pull away from the isolated object).
    (Also note that I have chosen up as the positive direction consistent with the positive upward direction of the acceleration).

    Thus, the NET force acting on the top block is 200 - 58.8 - T_top.

    F_net = ma
    200 -58.8 - T_top = (6)(3.53)
    200 - 58.8 - T_top = 21.2
    T_top = 120N.

    Now in part (c), you have first calculated the tension in the bottom of the rope correctly, but please be consistent in your FBD. Isolating the bottom block, you have the tension at the bottom of the cord acting up (away from the object, positive), and the block weight, 49N, acting down (negative). Thus,
    F_net = T_bot -49

    F_net = ma
    T_bot - 49 = 5(3.53)
    T_bot = 66.6N

    So then you took the average to find T_midpoint, which is OK in this problem due to the linearly varying tension, but in failing to recognize that, it is perhaps better instead to take a free body 'cloud' of the lower block that cuts through the mid-point of the rope, then apply newton 2 again noting that you must include 1/2 of the rope's weight and mass:
    F_net = ma
    T_mid - 49 - 2(9.8) = (5 +2)(3.53)
    T_mid = 93.3N
  8. Jul 20, 2007 #7
    thanks phantomjay

    thank you so much
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