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					       FN




  Ff                  FA



       Fg



An Unbalanced Force

      Fnet
   a
       m
 What is an unbalanced force?
• According to Newton’s Second Law of
  Motion:
  – an unbalanced force is one that causes an
    object to accelerate.
    • Positively or negatively (increasing or decreasing
      velocity)
    or
    • Change direction. (a force that causes an object to
      move in a circular path)
 Strategies for solving force related problems.
1.       Draw a free-body diagram that summarizes all the forces
         that act on the object in the vertical (y) and horizontal (x)
         directions.        F    N

                                                         FA = applied force
                                                         Ff = frictional force
                        Ff                          FA   FN = normal force
                                                         Fg = force due to gravity
                                                         or weight
                                Fg

2.       Write a mathematical expression that summarizes these
         forces in each direction.
     –      Fnet(x) = FA – Ff
           •   Since the length of the vector for FA is longer than the one for Ff, the
               block should be accelerating on the positive x-direction.
     –      Fnet(y) = FN – Fg
           •   Since the length of the vectors for FN and Fg are the same length,
               the net force in the vertical direction should be zero as we would
               expect. This also means that FN = Fg.
 Strategies for solving force related problems.
3.       For horizontal surfaces:
     •      FN = Fg = mg.
4.       For inclined surfaces, the force due to gravity (weight)
         must be broken into two vectors; one that is
         perpendicular to the incline, and the other that is parallel
         to the incline.                      F  N


     –      FN = Fg() = Fgcos
     –      Fg(||) = Fgsin
                                                                Ff


                                              Fg(||)
                                                            
                                                       

                                                                     Fg( | )

                                                           Fg
Forces Acting on Objects in the Horizontal Direction
                                  Does the applied
Is there an                                                v <> 0
                    Yes           force cause the                         Yes
applied force?
                                  object to move?
                                          v=0

      No                                                            Does the object
                                           No       v = constant
                                                                    accelerate?
                                          a=0

                                        Fnet = 0
Is the object in
                     Yes         v = 0 m/s: FA = Ff(static)               Yes
motion?
                               v = constant: FA = Ff(kinetic)
                     a≠0                                                  a≠0



      No           Fnet = Ff                                          Fnet = FA - Ff
      a=0


                                                                      If no friction
    Fnet = 0               Fnet = ma                                     Fnet = FA
        No Forces in the Horizontal Direction
                                       FN




                                       Fg

If there is no horizontally applied force, then the object will be:
    • stationary (v = 0 m/s)
    • or in motion, sliding along a frictionless surface at a constant velocity
     (v = constant).
        •Under both circumstances, Fnet = 0 N since there is no acceleration.




                                                                    Return to Flow Chart
  Friction in the Absence of an Applied Force
                                       FN



                                                                  v
                        Ff



                                       Fg


Under the circumstance where the object is in motion and being acted
  upon by a only a frictional force, the object will experience an
  unbalanced force due to friction (Fnet = Ff).
   •   For example, assume the block has a mass of 10 kg, and undergoes
       an acceleration of -5.0 m/s2 as it slides from the left to the right.
        1. Determine the frictional force.
        2. Determine the normal force.
        3. Determine the coefficient of friction.
 Friction in the Absence of an Applied Force
                                     FN



                                                                        v
                        Ff



                                     Fg
1. As per our free-body diagram, the only force acting in the x-direction is friction.
   Therefore:
               Fnet = -Ff
               Ff = ma
               Ff = (10kg)(-5m/s2) = -50N
2. On a horizontal surface, the normal force equals the weight.
               FN = Fg
               FN = mg
               FN = (10kg)(9.81m/s2) = 98.1N
                          F   50 N
3. Since Ff = FN,   f           0.51.
                          FN 98.1N                                Return to Flow Chart
           An Applied Force with Friction

                                      FN




                        Ff                                    FA



                                      Fg

When multiple forces act on an object, they need to be summed up to
  determine the net force (Fnet = FA - Ff).
   •   Assume that there is an applied force of 120N acting on a 10kg block
       to the right that causes it to go through an acceleration of 4m/s2.
        1. Determine the net force.
        2. Determine the frictional force.
        3. Determine the coefficient of friction.
            An Applied Force with Friction
                                            FN




                         Ff                                             FA



                                           Fg
1. The net force is simply determined by multiplying the object’s mass by its
   acceleration.
              Fnet = ma
              Fnet = (10kg)(4m/s2) = 40N
2. From the free-body diagram, the net force is the sum of the forces acting in the x-
   direction. We will then solve the relationship for the frictional force.
              Fnet = FA - Ff
              Ff = FA - Fnet
              Ff = 120N – 40N = 80N
                         Ff        80N
3. Since Ff = FN,                    0.82.
                         FN       98.1N                          Return to Flow Chart
         An Applied Force with No Friction
                                              FN




                                                                 FA



                                              Fg

When the surfaces are considered frictionless, the applied force will
  equal the net force (Fnet = FA).
    •   For example: Assume that there is an applied force of 120N acting on
        a 10kg block to the right.
         1. Determine the acceleration of the block.
             • Since Fnet = FA, a = FA/m
                       Fapplied       120 N
                  a                        12 m / s 2
                         m            10 kg                Return to Flow Chart
  Stationary or Moving at a Constant Velocity
                                        FN




                Ff                                              FA



                                        Fg

When objects move at a constant velocity (same speed in a linear direction)
  then the net force will be zero. Similarly, if the object is not moving, then the
  net force must be zero as well. Why?
    •   In both cases, the acceleration is zero, hence (Fnet = ma = 0)
    •   And if the net force is zero, then the applied force must equal the frictional force
        (FA = Ff).
    •   Example: A wooden 10kg block is placed on a wooden floor.
           Case #1: the block is stationary.
           Case #2: the block is moving at a constant velocity.
  Stationary or Moving at a Constant Velocity
                                        FN




                Ff                                             FA


                                                                     Return to Flow Chart
                                        Fg
 Case #1: Determine the maximum force that can be applied to the block without
  causing it to move.
 Case #2: Determine the applied force required to cause the block to move at a
  constant velocity.
 In both cases, you will need to refer to your reference table to find the appropriate
  values for the coefficient of friction for wood on wood.

Case #1: Stationary                                Case #2: Constant Velocity
s = 0.42                                          k = 0.30
FA = Ff                                            FA = Ff
FA = FN                                           FA = FN
FA = mg                                           FA = mg
FA = (0.42)(10kg)(9.81m/s2) = 41.2N                FA = (0.30)(10kg)(9.81m/s2) = 29.4N
          Special Case – The Elevator Question
1.       When talking about elevators, they
         spend a short period of time at the                 FN
         beginning and end of their ascent
         accelerating and decelerating,
         respectively.
2.       All motion and forces occur in the
         vertical direction.
3.       Write a mathematical expression that
         summarizes these forces.
     –     Fnet(y) = FN – Fg
           •   Note: FN will be greater than Fg if the
               acceleration is in the upward direction and
               vice-versa if the acceleration is in the
               downward direction.
Special Case – The Elevator Question




Note: You are weightless when in free-fall!
Unbalanced Forces and Uniform Circular Motion

• Whenever an object moves in a circular path, it
  experiences an unbalanced force.
• The unbalanced force always acts perpendicular
  to the direction of motion and towards the center
  of the circular path.
• A centripetal force is an unbalanced force, which
  is also a net force except that the motion is
  circular instead of linear.
                        mv 2
          Fnet    Fc 
                         r
Uniform Circular Motion – A Horizontal Path(FBD’s)
    Car on Road          Ball on String           The Rotor

                    Ff       FT
                                                              Ff
                                          v
                                              v


                                                    FN
                               FT
               Ff


                                                             Fg


     Fc = Ff               Fc = FT                 Fc = FN
 Uniform Circular Motion – A Vertical Path(FBD’s)
         Roller Coaster        Ball on String        Ferris Wheel
                           v
                                  FT                                  FN
Top
            FN                          Fg
                 Fg
                                                                      Fg


          Fc = F N + F g        Fc = FT + F g         Fc = FN - Fg

                                                                      FN


Bottom           FN                     FT
                                                                      Fg
                                                 v
                 Fg
                                        Fg
          Fc = FN - Fg          F c = FT - F g        Fc = FN - F g