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Using Newton's Laws - A problem involving 3 masses and an incline

  1. Mar 15, 2013 #1
    This is my first time posting. I see there is a protocol for making proper posts, so I am sorry if I make any errors.

    1. The problem statement, all variables and given/known data

    I have a problem involving a system with an incline and three masses. A picture of the system is in the attachment labeled 15.1. I am to calculate the tension in each string and the acceleration of the system.

    2. Relevant equations



    3. The attempt at a solution

    I have an idea for solving this problem and would like to know if I'm on the correct track. I divided the system up into two individual systems seen in attachments 15.2 and 15.3. My reasoning is that I should be able to calculate the first string tension (T1) in system one and then calculate the second string tension (T2) in system two. I would then use the difference between T1 and T2 to derive a net force on the entire system. With a net force I could then calculate the acceleration of the system as a whole.
     

    Attached Files:

  2. jcsd
  3. Mar 15, 2013 #2

    berkeman

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    Staff: Mentor

    Welcome to the PF. Your first post is exemplary! :smile:

    Your approach may well work, but I'd be inclined to draw 3 FBDs (one for each mass), and solve the resulting equations simultaneously.
     
  4. Mar 15, 2013 #3

    gneill

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    Staff: Mentor

    What's the coefficient of friction between mass 2 and the inclined surface?

    I believe that you'll have difficulty finding the individual tensions by your method, since those tensions are going to depend upon the net acceleration of the system which, in turn, depends upon all three masses.
     
  5. Mar 15, 2013 #4
    Calculations

    I'm sorry, I made an error with attachment 15.3 as I forgot to add the force vector T1 to my diagram. I've corrected it in attachment 15.4. Attachment 15.5 shows how I've designated the masses and the two string tensions.

    The problem states:

    Three objects are connected on an inclined table as shown in the diagram (Attachment 15.1). The objects have masses 8.0 kg, 4.0 kg and 2.0 kg as shown, and the pulleys are frictionless. The tabletop is rough, with a coefficient of kinetic friction of 0.47, and makes an angle of 20.0° with the horizontal.

    My calculations are as follows:

    m1 = 8.0 kg, m2 = 4.0 kg, m3 = 2.0 kg, uk = 0.47, θ = 20 degrees

    System 1:

    m2g = Fg2 =(4.0 kg)(9.8 m/s^2) = 39.2 N
    Fg2x = (39.2 N)(sin(20)) = 13.4 N
    Fg2y = (39.2 N)(cos(20)) = 36.84 N

    Fg2y = Fn = 36.84 N

    Ffr = ukFn = (0.47)(36.84N) = 17.31 N

    m1g = (8.0 kg)(9.8 m/s^2) = 78.4 N

    Fnet = ma

    m1g - T1 = m1a

    T1 = m1g - m1a

    T1 - Ffr - Fg2x = m2a

    T1 = m2a + Ffr + Fg2x

    m1g - m1a = m2a + Ffr + Fg2x

    m1g - Ffr - Fg2x = m2a + m1a

    (m1g - Ffr - Fg2x) / (m2 + m1) = a

    (78.4 N - 17.31 N - 13.4 N) / (4.0 kg + 8.0 kg) = 3.97 m/s^2

    T1 = m1g - m1a

    m1a = (8.0 kg)(3.97 m/s^2) = 31.76 N

    T1 = 78.4 N - 31.76N = 46.64 N

    T1 = 46.64 N

    System 2

    m3g = (2.0 kg)(9.8 m/s^2) = 19.6 N

    Ffr = 17.31 N (As calculated in system 1)

    T1 + Ffr - Fg2x - T2 = m2a

    T2 = T1 + Ffr - Fg2x - m2a

    m3g - T2 = m3a

    T2 = m3g - m3a

    m3g - m3a = T1 + Ffr - Fg2x - m2a

    (m3g - T1 - Ffr + Fg2x) / (m3 + m2) = a

    (19.6 N - 46.64 N - 17.31 N + 13.4 N) / (2.0 kg + 4.0 kg) = -5.16 m/s^2

    T2 = m3g - m3a

    T2 = 19.6 N - (-10.32 N)

    T2 = 29.92 N

    Net Force : T1 - T2

    46.64 N - 29.92 N = 16.72 N

    Mass of entire system: m1 + m2 + m3

    8.0 kg + 4.0 kg + 2.0 kg = 14 kg = M

    Acceleration of system:

    a = Fnet / M

    a = 16.72 N / 14 kg =1.19 m/s^2

    a = 1.19 m/s^2 towards the 8.0 kg mass.
     
  6. Mar 16, 2013 #5

    gneill

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    Staff: Mentor

    Nope. Doesn't work. You're system 1 goes off the rails when you write:

    T1 - Ffr - Fg2x = m2a

    This equation cannot be true since it does not include the effect of T2 on the acceleration of m2. As a result your value for T1 is incorrect. That error propagates through the subsequent calculations.
     
  7. Mar 16, 2013 #6
    With this attempt, I've determined a new equation for m2 which accounts for both T1 and T2 and have incorporated all three masses into determining an acceleration for the entire system.

    m2g = Fg2 =(4.0 kg)(9.8 m/s^2) = 39.2 N
    Fg2x = (39.2 N)(sin(20)) = 13.4 N
    Fg2y = (39.2 N)(cos(20)) = 36.84 N

    Fg2y = Fn = 36.84 N

    Ffr = ukFn = (0.47)(36.84N) = 17.31 N

    m1g = (8.0 kg)(9.8 m/s^2) = 78.4 N

    T1 -T2 - Ffr - Fg2x = m2a

    m1g - T1 = m1a

    T1 = m1g - m1a

    T2 -m3g = m3a

    T2 = m3g + m3a

    (m1g - m1a) - (m3g + m3a) - Ffr - Fg2x = m2a

    m1g - m1a - m3g - m3a - Ffr - Fg2x = m2a

    ((m1 - m3)g - Ffr - Fg2x) / (m1 + m2 + m3) = a

    (6 kg (9.8 m/s^2)) - 17.296 N - 13.41 N / 14 kg = 2.0 m/s^2

    Acceleration of the system: 2.0 m/s^2 towards the 8.0 kg mass.
     
  8. Mar 16, 2013 #7

    gneill

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    Staff: Mentor

    Your acceleration value looks good :approve:
     
  9. Mar 16, 2013 #8
    Awesome! I'm very glad.

    Thank you both for the assistance and for the welcome to this forum. Much obliged. :)
     
  10. Mar 9, 2014 #9
    Im confused with the calculation for T1:

    "m1g - T1 = m1a

    T1 = m1g - m1a"

    why in this case is the tension subtracted from the gravity force and not vice versa? Thanks!
     
  11. Mar 10, 2014 #10
    also, how do you calculate which direction the system is moving?
     
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