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Newton laws object on incline in equlibrium

  1. Sep 27, 2008 #1
    1. The problem statement, all variables and given/known data

    A block of mass m=2.0kg is held in equilibrium on an incline of angle=60 degrees by the horizontal force F.
    a)determine the value of F
    b)determine the normal force exerted by the incline on the block (ignore friction).

    2. Relevant equations
    Fnet = ma
    Fnetx = 0
    Fnety = 0

    3. The attempt at a solution

    So, first I thought about it
    If it is held in equilibrum, doesn't it means the object is at rest, at least, shouldn't Fnet = 0 since a = 0?

    I investigated my Fnetx and Fnet y.
    Fnetx = 0, then I have F - FgSin60 = 0
    Fnety = 0, then I have Fg - Fgcos60 = 0

    But it doesn't make sense because fg doesn't equal to Fgcos60

    the answer for a) 34N and b) 39

    please correct my approach, where have I gone wrong??
  2. jcsd
  3. Sep 27, 2008 #2
    anyone would like to give me a hand?
  4. Sep 27, 2008 #3


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    Homework Helper

    I don't think you are interpreting the horizontal force correctly. It is not acting along the incline (so not parallel to it). It is at an angle with respect to the incline. You will need to break it up into its components, and put them into your equations. Don't forget about the normal force either, it is there too. Does that help at all? See what you can come up with.

    Yes, a = 0, for both directions.
  5. Sep 27, 2008 #4
    then shouldn't F means Fgsinx?
    you know how, Fg|| or Fg |_ like fgsin and fgcos
    i was stuck at breaking it
    by the way, i really thank you your reply
  6. Sep 27, 2008 #5


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    Homework Helper

    I'm not sure what you mean. F and Fg are not the same thing. If you take your y direction to be perpendicular to the incline, and the x direction to be parallel to the incline then you need to figure out how to make each of the forces "fit" into those directions. This is where the trig comes in. Draw a diagram, and label all the forces (there are three). Then break them up into the components so you can sum them up in those two directions. The horizontal force being applied to the block is just one of the forces you need to do this to.
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