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Find the Normal Force

  1. Oct 7, 2008 #1
    1. The problem statement, all variables and given/known data
    A 30 kg box is on a frictionless floor. The box is being pulled with a force of 120 N at an angle of 15 degrees. Find the normal force. Why is the normal force not equal to simply the gravitational force. What would make the person pull the box vertically into the air?


    2. Relevant equations
    Fg = mg


    3. The attempt at a solution
    I solved for the horizontal component of the applied force, added it to the normal force, and set it equal to 0 plus the gravitational force (30 * 9.8 = 294 N). I think I have done the first part correctly, but I am not sure how to answer the second two parts.
     
  2. jcsd
  3. Oct 8, 2008 #2
    Okay, I'm a little confused with what you've done. Firstly, you solved for the horizontal component of the applied force? Why is this? You should have solved for the vertical component.

    You then said that you added it to the normal force, however as defined by the question; the normal force is what we're solving for. So how exactly did you manage that?

    You should solve for the vertical component of the applied force and subtract it from the magnitude of mg. This should give you the normal force.

    You can see that it's not simply equal to mg because there are more forces in the y plane than simply mg.

    I don't completely understand the wording of the third question, if it's asking for a numerical answer or not, but you can see that the box would leave the ground if the vertical component of applied force exceeds that of gravity.
     
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