Determine the angle of incline

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

    Determine the angle of incline on this frictionless air track.

    Acceleration = 0.147 m/s2.

    The answer is supposed to be 4.10°.

    2. Relevant equations

    Fnet = ma

    3. The attempt at a solution

    Mass 1 (FBD):

    Perpendicular (y):

    Fny = +Fn
    Ty = 0
    ay = 0
    Fgy = -m1gcosθ
    = (428)(-9.80)cosθ
    = -4194.4cosθ
    Fn + Fgy = 0
    Fn = +4194.4cosθ

    Parallel (x):

    Fnx = 0
    = -m1gxsinθ
    = -4194.4sinθ
    Fnet = m1a
    Fn + Fgx + T = m1a

    +4194.4cosθ – (-4194.4sinθ) + T
    = (428)(0.147)
    = 62.9

    Mass 2 (FBD):

    Fg2 = (20.0)(9.80)
    = 196
    Fg2 + T = m2a

    196 + T = (20.0)(0.147)
    = 2.94
    T = 2.94 – 196
    T = -490

    Final calculations:

    +4194.4cosθ – (-4194.4sinθ) + T = 62.9
    (-) T = -490
    ----------------
    +4194.4cosθ – (-4194.4sinθ) = 552

    Not only do I not know how to solve that final equation, I'm also sure the answer will be wrong. I really don't get how to properly go about this question.
     

    Attached Files:

  2. jcsd
  3. kuruman

    kuruman 3,449
    Homework Helper
    Gold Member

    Stop!!!. This equation is wrong. The normal force is along y (as you correctly point out) and does not belong in the x-equation. Furthermore, the tension is up the incline and the component of the weight is down the incline, so they must have opposite signs.
     
  4. Thank you! I omitted Fn and changed the equation to this:

    Code (Text):
    -Fgx + T = -m1a
    -4194.4sinθ + T
       = (428)(-0.147)
       = -62.9
    So that the final calculations were:

    Code (Text):
        T = -62.9 + 4194.4sinθ
    (-) T = -490
    ---------------------------
        0 = 427.1 + 4194.4sinθ
    -427.1 = 4194.4sinθ
    -427.1/sinθ = 4194.4
    sinθ = -427.1/4194.4
       θ = sin-1(-427.1/4194.4)
       θ = 5.84°
    Is this just a big percentage error, or am I still doing something wrong?
     
  5. kuruman

    kuruman 3,449
    Homework Helper
    Gold Member

    The equation from the second FBD is also incorrect. The weight and the tension are in opposite directions. When you fix it, make sure that the direction of the acceleration is consistent between FBD1 and FBD2.
     
  6. I just realized I had no idea how I got 2.94 – 196 = -490 or other glaring errors. Is this correct?:

    Code (Text):
    Fg2 - T = -m2a

    196 - T = (20.0)(-0.147)
            = -2.94
    -T = -2.94 – 196
    T = 198.94

    Final calculations:

        T = -62.9 + 4194.4sinθ
    (-) T = 198.94
    ---------------------------
        0 = -260.94 + 4194.4sinθ
    260.94 = 4194.4sinθ
    260.94/sinθ = 4194.4
    sinθ = 260.94/4194.4
        θ = sin-1(260.94/4194.4)
        θ = 3.57°
     
  7. kuruman

    kuruman 3,449
    Homework Helper
    Gold Member

    That looks about right considering round offs. I got 3.55o.

    *** On edit ***
    Note that according to your drawing, arcsin(0.162/2.28) = 4.09o. If that's not close enough, you may have to explain the discrepancy.
     
  8. Awesome, thank you so much!
     
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