Friction

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					                                                Chapter 4


      4.1




                       sN



                 Frictional
                 force




                                                     Fmax          Applied force



4.2         Since the block is in equilibrium under three forces, the three forces must pass through
            the same point. Thus the normal reaction will be at the point where the arrow showing
            the weight meets the inclined plane. This is shown below.

                               f
                                                     N




                                      mg

                                                               θ




4.3         The free body diagram of the block is as follows
                                                              N

                                                F


                                            
                             f

                                                           mg


Balancing the horizontal forces gives
                                                f  F sin 
Balancing the vertical forces gives
                                        N  mg  F cos
Since the maximum frictional force f m ax  N , for equilibrium we should have

                                                                        mg
                       F sin    mg  F cos   F 
                                                                  sin    cos


4.4     We consider two different situations when the weight on the table is about to move to
        the left or to the right. When it is about to move to the left, its free body diagram will
        look as follows

                                                       N




                            10g
                                                                            mg

                                                                   f



                                                     50g

By equilibrium conditions, we have
                                                 N = 50g
                                            f + mg = 10g
Since
                                               f  N
We have
                               f  10g  mg  0.1 50g  5  m
Thus the minimum value of m is 5kg when the frictional force is at its maximum pointing to the
right. As m is increased above 5kg, frictional force becomes less and less, eventually changing
direction and attaining its maximum value pointing left. In that situation, the free body diagram
of the block on the table is as follows.

                                                     N




                             10g
                                                                      mg


                                   f


                                                 50g

In this situation, the equation for horizontal force balance is
                                            f + 10g = mg
This coupled with f  N leads to
                               f  mg  10g  0.1 50g  m  15
Thus
                                             5  m  15


4.5 Taking the x axis along the plane and the y axis perpendicular to the plane, the free-body
diagram of the block looks as follows.
                                                               Y
                                    N
                                                                          X


                F         30º


                                                    30º
                    F

                                                 100g


The equations describing equilibrium in the X and the Y directions are

                          F    x    0  F cos30  F   100g sin 30  0

                          F    y    0  N  F sin 30  100g cos30  0

The first equation implies that
                                            F    F cos30  100g sin 30
                  2
Taking g = 9.8ms , the value of F for different values of F is
                                            F  600N  F   29.6 N
                                            F  500N  F   57.0 N
                                            F  100N  F   403.4 N

4.6   Free body diagram of the box when it is about to move (i.e. the frictional force is at its
      maximum) is shown below


                                                           N
                    F


                                        h                          b

                                                          a

                         N
                                                   mg
When the box is about to move, the friction is at its maximum and is equal to N. The force F
also equals N at this point. This creates a couple that is counterbalanced by the couple formed
by the weight of the box mg and N (=mg). This is the reason that N shifts towards the direction
                                                                                                  a
of the push. However, the maximum couple moment that can be created by mg and N is                  mg .
                                                                                                  2
Thus for the box not to topple, the couple created by F and the friction should remain less than
a
  mg . Thus implies
2
                                                         a                       a
                                        h  mg           mg        h
                                                         2                      2


4.7 suppose each break show makes an angle  at the centre as shown below



                                                          



The force F is assumed distributed uniformly over the shoe. Then the torque due to the
frictional force will be
                      b
                    
                               Fr
                                              r dr 
                                                                
                                                         2F b 3  a 3 
                                                                       
                                                                                     
                                                                         2F a 2  ab  b 2   
                      a
                           
                             b   2
                                       a2                     
                                                          3 b2  a2      3      a  b 
                           2
With two shoes therefore, the torque would be

                                                 
                                                           
                                                      4 F a 2  ab  b 2   
                                                        3      a  b 

4.8 It is given that mass M is balanced by mass m. The contact angle is π. Since each time the
    string is wound once more around the rod, the mass M that can be balanced by m becomes
    twice as large, we have
                               M  m exp( )    
                                                 
                               2 M  m exp(3 )   2  exp(2  )
                               4 M  m exp(5 )
   This gives  = 0.11


                                                                                       L1
4.9 Neglecting the length of the rope passing over the pulley, we have mass                 M on one
                                                                                    L1  L2

                                                     L2
    side of the pulley that is balanced by mass           M on the other side. Thus we have
                                                  L1  L2

                    L1                     L2
                         Mg  exp(  )         Mg                 L1  exp(  ) L2
                 L1  L2                L1  L2


4.10 As the weight is put, it has a tendency to move down. Hence the frictional force will be in
    the counterclockwise direction. Thus if the tension in the rope on the spring balance side is
    T1 and that on the weight side is T2 then
                                               
                                  T2  T1 exp   
                                               2


     Now it is given that T1 = 5g and T2 = mg. Thus we get
                                  m  5 exp(0.2   / 2)  6.85kg
     An interesting possibility exists if a person had pulled the weight down and then slowly
     brought it to equilibrium. In that case the tension will work in the other direction and
                                 m  5 exp(0.2   / 2)  3.65kg
     However we have not considered this possibility.


4.11 There is a range of M2 that exists because frictional force can act with its maximum value
    in one direction to the maximum in the other direction.         Largest value of M2 is when the
    mass M1 is about to slide up the plane. The free body diagram of M1 in that case is as
    follows
                            N
                                                          T




                                            f

                                
                                         M1g


When the mass M1 is about to slide up, we have
                  T  M 1 g sin   1 M1 g cos  M1 g (sin   1 cos )
                                                         
The contact angle between the rope and the pulley is    
                                                     2    
Since the rope has a tendency to move clockwise, the frictional force due to the pulley will be
acting counterclockwise. Thus we have
                                                                                 
       M 2 g  T exp 2            M 2 g  M 1 g sin   1 cos  exp 2    
                     2                                                     2       
       Thus
                                               
       M 2  M 1 sin   1 cos  exp 2    
                                        2       
The other extreme is when the mass M1 is about to slide down the plane. In that case the free
body diagram of M1 is


                            N
                                                          T




                                            f

                                
                                         M1g


Thus we have
                      T  M 1 g sin   1 M 1 g cos  M 1 g (sin   1 cos )
Now the rope has a tendency to move counterclockwise, the frictional force due to the pulley
will be acting clockwise. Thus we have
                                                                                     
      M 2 g exp 2      T           M 2 g  M 1 g sin   1 cos  exp  2    
                2                                                                 2    
      Thus
                                                
      M 2  M 1 sin   1 cos  exp  2    
                                            2    


4.12Free body diagram of the tire when it is loaded and is about to roll is as follows




                                                                       F


                                      f
                                                  W       N


Balancing the vertical forces gives
                                                                W
                                      N  abP  W       a
                                                                bP


Balancing the horizontal forces gives F = f


Balancing torque about the centre of the wheel gives
                                          a                    W2
                                      FR  W           F
                                          2                   2bPR

				
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posted:7/13/2012
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