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Frictional Force & Normal forces

  1. Jan 30, 2010 #1
    1. The problem statement, all variables and given/known data

    A car (m = 1840 kg) is parked on a road that rises 18.7 ° above the horizontal. What are the magnitudes of (a) the normal force and (b) the static frictional force that the ground exerts on the tires?

    2. Relevant equations

    F=ma Fg=mg Ff=uN

    3. The attempt at a solution

    errr.... what forces cancels out? is it FN=Fg=Ff or is it FN+Fg+Ff=0? i don't remember...
    1. The problem statement, all variables and given/known data



    2. Relevant equations



    3. The attempt at a solution
     
  2. jcsd
  3. Jan 30, 2010 #2

    rock.freak667

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    On your free body diagram, at the angle of θ=18.7°, what are the forces acting? (After you split the weight of the car into components parallel to the road and perpendicular to the road)
     
  4. Jan 30, 2010 #3
    frictional force, force of gravity, and normal force.
     
  5. Jan 30, 2010 #4

    rock.freak667

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    So if there is no net movement perpendicular to the road, that is the magnitude of the normal force?
     
  6. Jan 30, 2010 #5
    the same as the gravitational force? 1 840 * 9.8 = 18 032 N?
     
  7. Jan 30, 2010 #6

    rock.freak667

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    No, the component of the weight perpendicular to the plane is equal to the normal force.
     
  8. Jan 30, 2010 #7
    so that's equal to the y-component of the gravitational force?

    18032 cos(18.7deg ) = 17080.10 N

    and so frictional force is 18032sin18.7=5781.29 N ?
     
  9. Jan 30, 2010 #8

    rock.freak667

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    Yes but only because the net force in the x and y directions are zero.
     
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