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Forces acting on an inclined object

  1. Mar 15, 2009 #1
    1. The problem statement, all variables and given/known data
    There is a box on a frictionless surface that forms an angle of 37 with the horizontal. the box is pushed by a horizontal force (acting on the left top corner) such that it moves with a constant speed. If the mass of the object is 66.2 kg, calculate the magnitude of the external force.

    2. Relevant equations
    Fnet = ma
    ** Note my coordinate system is lined up so that the x-axis is parallel to the angled surface.

    3. The attempt at a solution
    I thought i knew what i was doing, but i can seem to get the right answer,

    for the system Fnet = 0 *constant velocity, means acceleration= 0
    Fnet = Fext - (x-component of gravity)

    Fext = (x-component of gravity)
    (x-component of Fg) = m*g*sin37 = 390 N

    Could someone please explain where im going wrong? Thanks
  2. jcsd
  3. Mar 15, 2009 #2


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    If the force is acting horizontally what is its component || along the incline that is balancing the m*g*sinθ ?
  4. Mar 15, 2009 #3
    would the || component be: m*g / sinθ ?
  5. Mar 15, 2009 #4


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    No. Sounds like guessing to me.

    The component of the horizontal force F is what you are looking for. It is acting up the incline to balance the m*g*sinθ that's acting down the incline.

    Make a drawing.
  6. Mar 15, 2009 #5
    i dont think i understand then... isnt the only force opposing the component of the external force= m*g*sinθ ? so wouldnt the external force just equal that in magnitude?
    Last edited: Mar 15, 2009
  7. Mar 15, 2009 #6


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    Half right. The horizontal force is ... well ... horizontal. So what part of that acts up the incline to balance the downward effect of gravity.
  8. Mar 15, 2009 #7
    so can i use a triangle that has the external force horizontal, a vertical force which would be mg and the component of external force that is inclined?
    and then the angle between the incline and the external force would be θ?

    please correct me as im more than often wrong when i try to draw the right-angle triangles to use, this is usually what messes me up
  9. Mar 15, 2009 #8


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    Well if it's horizontal then there is no vertical component. And that's the other force anyway - gravity.

    No, I think you want to resolve the F into the force components that are ⊥ and || to the incline.

    So which one is F*Sinθ and which is F*Cosθ ?

    The one you want is the one that is || to the incline, because that's the one that equals your balance to gravity.
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