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Applied force in horizontal direction on block on a inclined plane.

  1. Jan 2, 2014 #1
    1. The problem statement, all variables and given/known data
    10kg block on a inclined plane (32degrees) (frictionless) and an applied force 120N horizontally. What is acceleration?


    2. Relevant equations

    Fnet = ma

    3. The attempt at a solution

    mgsin32 = 51.9N (from mg)
    From applied force, in the same plane, we have xcos32=120 therefore, x = 141.5
    141.5-51.9=ma
    a=9m/s^2 but the answer is 5.0m/s^2
     
  2. jcsd
  3. Jan 2, 2014 #2
    Explain how you got xcos32=120
     
  4. Jan 2, 2014 #3
    because the horizontal force is 120N, so I need to find the hypotenuse component. the angle in that triangle is also 32, so xcos32 = 120
     
  5. Jan 2, 2014 #4
    Think carefully.

    The horizontal force of 120N has a component along the plane and another component perpendicular to the plane and you need to find the component along the plane.
     
  6. Jan 2, 2014 #5
    Let me think...
     
  7. Jan 2, 2014 #6
    Remember that when a force is resolved into components, the force itself will be what you are calling the 'hypotenuse'.
     
  8. Jan 2, 2014 #7
    Oh I get it now! Thank you!
     
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