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Does force normal change when you push an object at an angle?

  1. Nov 12, 2012 #1
    1. The problem statement, all variables and given/known data

    You exert a force of 131 N [30° below horizontal] onto a 12.6 kg lawn mower, What is the force of gravity and normal force on the lawn mower?


    2. Relevant equations
    Fg= m × g
    FA (y component) = Sin30° × FA


    3. The attempt at a solution
    The force of gravity is equal to the force of normal,
    ∴ FN = m × g
    = 12.6 kg × 9.8 m/s2
    = 123.48 [up]​

    But, the normal force gets added on by the force opposing the y component of the applied force right?

    So, FA (y component)Sin30° × FA
    = 65.5 N [down]​
    So by newtons 3rd law there would be a force acting up on the lawn mower of 65.5 N[up].


    I'm just wondering if these two forces are added to each other for the total normal force which is 123.48 N + 65.5 N = 188.98 N [up]?
     
  2. jcsd
  3. Nov 12, 2012 #2

    Delphi51

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    Homework Helper

    Yes, all correct. The vertical component of the push adds to the weight to make the normal force pressing the mower down on the ground.
     
  4. Nov 12, 2012 #3
    Thanks, but I have a question. Why doesn't the force of gravity on the lawn mower increase when you apply your force on it? Aren't you essentially making the lawn mower heavier?
     
  5. Nov 12, 2012 #4

    Delphi51

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    Homework Helper

    The force of gravity depends on the MASS of the mower. You probably know it is
    Fg = GMm/d² where M is the mass of the Earth and m the mass of the mower.
     
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