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Need help with newtons laws. finding normal force.

  1. Oct 5, 2011 #1
    1. The problem statement, all variables and given/known data
    It includes a diagram so i just took a screenshot of the question.


    I don't undertand how to do this problem
    Last edited: Oct 5, 2011
  2. jcsd
  3. Oct 5, 2011 #2
    What does the normal force equal?

    EDIT: you should probably draw a free body diagram to start also.
  4. Oct 5, 2011 #3
    you can compute the normal force knowing two things: trigonometry and the laws of natural states of motion.
  5. Oct 5, 2011 #4
    Yes i know but how?
  6. Oct 5, 2011 #5
    [itex]F_{N}=mgcos(\theta)[/itex] is what you would use. But really, just draw a FBD and figure out this yourself with the trig.
  7. Oct 5, 2011 #6
    Yeah i got 28.411 but it is wrong...
  8. Oct 5, 2011 #7
  9. Oct 5, 2011 #8
    It's hard to see how you can get the normal force vector, but you can see it if you draw vectors large enough and spend enough time until you finally see the solution. The natural states of motion are rest and uniform speed which makes the net force 0.
  10. Oct 5, 2011 #9
    yeah its wrong.
  11. Oct 5, 2011 #10
    28N is probably the answer, if not then I don't know why it's marking it wrong and for the other 1 you just do [itex]F_{g}sin\theta[/itex]
  12. Oct 5, 2011 #11
    the other one is the sintheta version because in static equalibrium the action is going to be equal to the reaction and in this case its the gravity in the x direction of the block. so, 0=Fpush-mgsin45.
  13. Oct 5, 2011 #12
    The correct answers were
    56.824 N (I multiplyed 2 x 28.412 in a desperate attempt to guess)
    40.180 N (4.1 x 9.8 and also 28.412^2 + 28.412^2 = 40.180^2)

    I have no idea why or how these worked lmao
  14. Oct 14, 2011 #13
    N=mg(cosθ) +Fcosθ.......(1)

    i.e,F=mg=4.1 x 9.8=40.18

    from (1),N=mgcosθ +mgcosθ [F=mg]
    i.e,N=2mgcosθ=2 x 4.1 x 9.8 x cos 45°=56.83
    so, N=56.83
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