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Another Double Check ! FBD String Tension

  1. Feb 3, 2013 #1
    Calculate the tension in the cable connecting the two masses. Assume all surfaces are frictionless.

    FBD
    Physics Question 4 U1-C.gif

    m1 = 5 kg
    θ1 = 60°
    m2 = 6 kg
    θ2 = 70°

    Equation 1 ~

    FT - Fg = ma

    FT - mgSinθ = ma

    FT = 5a + 5(9.8)Sin60

    FT = 5a + 42.44

    Equation 2 ~

    Fg - FT = ma

    mgSinθ - FT = ma

    6(9.8)Sin70 - 6a = FT

    FT = 55.25 - 6a

    ~Set Equations Equal to Each Other ~

    5a + 42.44 = 55.25 - 6a

    11a = 12.81

    a = 1.16 m/s2

    ~Sub into Eq to find Force Tension on da String Son !~

    5(1.16) + 5(9.8)Sin60 = FT

    FT = 48.2 N
     
  2. jcsd
  3. Feb 3, 2013 #2

    SammyS

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    The results looks correct.

    Comment: If I were grading this, I would be concerned that you are using the same variable names for mass 1 quantities and mass 2 quantities, particularly Fg and m .
     
  4. Feb 3, 2013 #3
    Thanks for the input, I will make amends in future questions.
     
  5. Feb 3, 2013 #4

    SammyS

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    To be more specific: FT and a are the same in magnitude for both masses, so it's fine to use the same variable name for them.
     
    Last edited: Feb 3, 2013
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