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Gravitational and normal force

  1. Mar 15, 2016 #1
    1. The problem statement, all variables and given/known data
    A 60 kg skier is beginning her descent down a 26 degree slope. The frictional force between the skis and the snow is 4 N
    a) What is the gravitational force the skier exerts on the slope?
    b) What is the normal force the slope exerts back on the skier?

    2. Relevant equations


    3. The attempt at a solution
    a) What is the gravitational force the skier exerts on the slope?
    60(kg)*9.8(m/s^2)*cos(26) = 528.5(N)

    b) What is the normal force the slope exerts back on the skier?
    not sure how to do this one
     
  2. jcsd
  3. Mar 15, 2016 #2
    First, sketch a free-body diagram. Use the slope as the X axis. Since there is no acceleration in the Y axis of this system, you can solve for the normal force.
     
  4. Mar 15, 2016 #3
    did i do part a correctly?
     
  5. Mar 15, 2016 #4
    As there is no acceleration in the Y axis, it means the force exerted by the skier on the slope and the one the slope exerts on the skier is the same. It's kind of the definition of the normal force, it is a force exerted by the surface on the object, it is a force of reaction to another one, which here is the gravitational force.
     
  6. Mar 15, 2016 #5

    haruspex

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    Is that the word-for-word statement of the question? It doesn't make any sense.
    The Earth exerts a gravitational force on the skier, and the skier exerts an equal and opposite gravitational force on the Earth; the skier exerts a contact force on the slope, which can be resolved into a combination of forces in different directions, like normal and tangential.
    I do not know what "the gravitational force exerted by the skier on the slope" means. Maybe it means the vertical component of the contact force?
     
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