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Normal force

  1. Sep 18, 2006 #1
    Susie has a box that weighs 75N that is lying on top of a horizontal floor that is not frictionless. She pushes on the box with a force of 40N directed at an angle 41 degrees below the horizontal. Calculate the normal force that the floor exerts on the box.
    We have not learned anything about friction yet, and it's not in the chapter, so I have no idea if there being friction matters or not for this problem. But ignoring that, it seems easy simple enough.
    I break the force exerted into the x and y components.
    40*cos41 = 30.1 N
    40*sin41 = 26.2 N
    x is pointed in the positive direction (+30.1N), and y is pointed in the negative direction (-26.2N). Since the box weighs 75N, it has a force of 75N downward, and that means the normal force is currently 75N upwards. Now, my question is, if everything I did so far is correct, is the new normal force just the total downward force (75N + 26.2N) = 101.2N, or is it equal to the sum of all the forces? Sqrt((75N+26.2N)^2+(30.1N)^2) = 105N?
  2. jcsd
  3. Sep 18, 2006 #2


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    The normal force from the ground must balance out the other forces acting upon the box in the normal direction.
    (I.e, the box experience zero acceleration in the normal direction).
  4. Sep 18, 2006 #3


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    You have the right instinct. The total normal force opposes the weight plus the vertical component of the pushing force. Good job. Welcome to PF, BTW.
  5. Sep 18, 2006 #4
    It's just 75+26.2 N.
  6. Sep 18, 2006 #5
    Aren't we all excited to see a problem already worked out? ;)
  7. Sep 18, 2006 #6


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    Yep! :biggrin:
  8. Sep 19, 2006 #7
    Okay, so I think I am getting normal force now from that example. So if I apply what is in this problem for my inclined problem, this should be the same I believe.
    A 30000kg truck traveling at 31.3 m/s goes up a 100m frictionless inclined plane. What angle must the inclined plane be above the horizontal for the truck to stop before falling off.

    V(i) = 31.3 m/s
    V(f) = 0 m/s
    Since it is constant, a = 0, and there is no initial force. So normal force cancels out just the mass, and I just need to find the the downward or parallel force.

    First thing I did:
    Find the acceleration needed to stop in that distance:
    V(f)^2 – v(i)^2 = 2a(x(f)-x(i))
    31.3 m/s ^2 = 2a(100m)
    a =-4.9 m/s^2

    I use this equation: a = g * sin O
    -4.9m/s^2/-9.8m/s^2 = sin O = O = 30 degrees

    Normal force means mass never matters on its own?
    Last edited: Sep 19, 2006
  9. Sep 20, 2006 #8
    Here is a question
    A force,F, is acting against the wall on a book of mass,m,. Find the minimum mu so that there is equilibrium.
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