Friction Fulton County Schools Home by nikeborome

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									           Physics 152
            Fall 2004
         Tuesday, Sep 14


Lecture #7 - Chapter 5.1: FRICTION
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Properties of Friction
          1. fs between any two surfaces
            in contact is opposite the
            applied force that is parallel
            to the surface and equal in
            magnitude

                       fs< µsN
               µs - coefficient of static friction
               N - Normal force

          2. When the object is on the
            verge of slipping
                fs,max = µsN
Properties of Friction
           3. fk is opposite the applied
             force that is parallel to the
             surface and is


                        fk = µkN

              µk - coefficient of kinetic friction
              N - Normal force

           For constant velocity:
                      F = fk
Properties of Friction
           4. µs and µk depend on the
             nature of the surface.
             However:


                       µk < µs

             µs - coefficient of static friction
             µk - coefficient of kinetic friction

           5. fs and fk do not depend on
             area because weight and µ
             do not depend on area.
           (empirical finding)
                               Reading Quiz
(1) In three experiments, three different horizontal forces are applied to
    the same block lying on the same countertop. The force magnitudes
    are F1= 12 N, F2= 4 N and F3= 8 N. In each experiment, the block
    remains stationary inspite of the applied force.
(a) Rank the forces according to the magnitude fs of the static frictional
    force on the block from the countertop, greatest first.
        (1) F2, F 3, F 1
        (2) F1, F 3, F 2
        (3) F3, F 2, F 1
(b) Rank the forces according to the maximum value fs,max of that force,
    greatest first.
        (1) F2, F 3, F 1
        (2) F1, F 2, F 3
        (3) F3, F 2, F 1
        (4) all tie
(2) In the figure (a), a 'Batman' thermos is sent sliding
    leftwards across a long plastic tray. What are the
    directions of kinetic frictional forces on...
(a) the thermos
       (a) downward
       (b) upward
       (c) leftward
       (d) rightward
(2) In the figure (a), a 'Batman' thermos is sent sliding
    leftwards across a long plastic tray. What are the
    directions of kinetic frictional forces on...
b) the tray (from each other)
       1. upward
       2. leftward
       3. rightward
       4. downward
(2) In the figure (a), a 'Batman' thermos is sent sliding
    leftwards across a long plastic tray.
(c) Does the kinetic frictional forces on the thermos increase
    or decrease the speed of the thermos relative to the
    floor?
       1. decrease
       2. increase
2. In figure (b), the tray is now sent sliding leftward beneath
    the thermos. What now are the directions of the kinetic
    frictional forces on...
(d) the thermos
       1. upward
       2. rightward
       3. downward
       4. leftward
2. In figure (b), the tray is now sent sliding leftward beneath
    the thermos. What now are the directions of the kinetic
    frictional forces on...
(e) the tray (from each other)
       1. rightward
       2. downward
       3. leftward
       4. upward
2 (f) Does the kinetic frictional forces on the thermos
increase or decrease the speed of the thermos relative to
the floor?
decrease
increase

(g) Do the kinetic frictional forces always slow the
objects?
yes
no
3) If you press an apple crate against a wall so hard that the
    crate cannot slide down the wall, what is the direction of...
    (a) the static frictional force fs on the crate from the wall?
           1. upward
           2. leftward
           3. rightward
           4. downward
(b) the normal force N on the crate from the wall?
           1. horizontal, left of you
           2. horizontal, towards you
           3. horizontal, right of you
           4. horizontal, towards wall
If you increase your push, what happens to...
(c) Fs
           1. no change
           2. decreases
           3. increases

(d) N
             1. decreases
             2. no change
             3. increases
(e) fs,max
             1. no change
             2. increases
             3. decreases
         Example: Static Friction
A box of mass m =10.21 kg is at rest on a floor. The
 coefficient of static friction between the floor and the box is
 µs = 0.40 . A rope is attached to the box and pulled at an
 angle of q = 30o above horizontal with tension T = 40 N.

Does the box move?

                                                     T

                                            q
 static friction (= 0.40 )    m
*Select axes & draw FBD of box:
                                           y

 *Apply FNET = ma
                                                   x
 y: N + T sin q - mg = maY = 0
                                      N
    N = mg - T sin q = 80 N
                                                       T
 x: T cos q - fs = maX
                            fs    m            q

 The box will move if:
    T cos q - fs > 0                  mg
 y: N = 80 N                                        y

 x: T cos q - fs = maX
                                                        x
 The box will move if: T cos q - fs > 0
                                                   N
      T cos q = 34.6 N
                                                            T
      fMAX = N = (0.40)(80N)
        = 32 N                     fMAX = N
                                               m        q



So T cos q > fMAX and the box does move            mg
        Example: Static Friction
A detached block of rock with m =1.8 x 107 kg is at rest on a
 “bedding plane fracture”. The coefficient of static friction
 between the rock and the fracture µs = 0.63 . The bedding
 plane fracture is inclined q = 24o above horizontal.
Does the block slide?


          y
                 fs   x
                                             q
                q

               W
     y
         N
             fs      x
                                        q
             q

          W

Fy = 0=N - Wcosq = N - mgcosq
                 N = mg cosq = 160 x 106 Newtons
     y
          N
              fs      x
                                            q
              q

          W

Fy = 0=N - Wcosq = N - mgcosq
                  N = mg cosq = 160 x 106
Newtons
fs = µs N = 102 x 106 Newtons
Wx= mg sinq = 72 x 106 Newtons

              Block will not slide because Wx < fs !
       y
           N
               fs   x
                                         q
               q

            W

Alternative Method:
    From example 5-2 in textbook:
           µs = tan q max           Arctan(µs) = q max
                                    q max= 32   o


                                    Will not slide because
                                    q = 24 o < q max
       Example: Dynamic Friction
A box of mass m1 = 1.5 kg is being pulled by a
  horizontal string having tension T = 90 N. It slides
  with friction (µk = 0.51) on top of a second box
  having mass m2 = 3.0 kg, which in turn slides on a
  frictionless floor.

What is the acceleration of the second box ?




   T                          m1    slides with friction (=0.51 )

   a=?                   m2              slides without friction
    *First draw FBD of the top box:
                                      N1,2
                N1,2


T             m1                  T          f k= N1 = m1g
                        fk


                                       W1=m1g
                   W1



          (Note: a1 ~ 55 m/s2)
*Newton’s 3rd law says the force box 2 exerts on box 1 is
equal and opposite to the force box 1 exerts on box 2.


   Action-Reaction Pair from Frictional Forces:


                                                  f1,2
                                      m1

                  f2,1               m2


   Action-Reaction Pair from Contact Forces:
                                        N1,2

                                    m1

                              m2
                                          N2,1
*Now consider the FBD of box 2:
                                                 N2,
                                                   g
                    N2,
                      g




       f2,1
                                  f2,1 = km1g
               m2

                                                   N2,1
                      N2,1                           =m1g
                                    W2=m2g
              W2
     *Finally, solve F = ma in the horizontal direction:



              N2,
                g                m1g = m2a




                                a = 2.5 m/s2
f2,1 = m1g


                N2,1
                  =m1g
  W2=m2g

								
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