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					AP homework on Newton’s Laws
          WS #1 - #1 Ch.5 4th ed. - Newton's Laws
                       The Problem
1.        A force, F, applied to an object of mass m1
          produces an acceleration of 3.0 m/s². The
          same force applied to a second object of
          mass m2 produces an acceleration of 1.0
          m/s².
     A.     What is the value of the ratio m1/m2?

                                                    F
      Draw the                              M1
      diagram
                                            M2
                                                    F
    WS #1 - #1 Ch.5 4th ed. - Newton's Laws
                 The Solution

             a1  3.0 m         a2  1.0 m
                          s²                 s²
                 m1 a2         1.0 m
                           s²  .33
                 m2 a1 3.0 m
                             s²
If m1 and m2 are combined, find their acceleration under the
                   action of the force F.
                    a2 m1  m2     m
                               1 1
                    a12   m2       m2
                    
         a12  a2   1 
                    
                          m1 
                                        
                               1.0 m s²  1  .33 
                          m2 
                           a12  .75 m
                                       s²
     WS #2 - #26 Ch.5 4th ed - Newton's Laws

2.   Find the tension in each cord for the systems
     described below. (Neglect the mass of the
     cords.)

         40.°= 1           = 50.°    60.°
                            2
     T                           T2
      1                               T1

                                                       90.°
                    T3                                T2
                                              T3



                    5.0 kg
                                              10 kg
                     (a)                       (b)
         WS #2 - #26 Ch.5 4th ed - Newton's Laws
                   Free body Diagrams




T1                    T2           T
                                       1
  = 40.°         = 50.°            1= 60.°    2 = 0.0°   T2
     1            2

            T3                              T3
WS #2 - #26 Ch.5 4th ed - Newton's Laws
        The Working Equations

             T3  mg ˆ j
                   ˆ
    T1  T1 cos1 i  T1 sin1 ˆ
                                j
                     ˆ
     T2  T2 cos  2 i  T2 sin  2 ˆ
                                    j

             F    x    ma x
      T1 cos1  T2 cos 2  0

         T1 cos1  T2 cos 2
                     cos1
             T2  T1
                     cos 2
WS #2 - #26 Ch.5 4th ed - Newton's Laws
        The Working Equations

             F   y    ma y

     T1 sin 1  T2 sin  2  T3  0
                 cos1
   T1 sin1  T1        sin 2  T3
                 cos 2
                cos1 sin  2 
   T1  sin 1                  T3
                  cos 2 

    sin1 cos 2  cos1 sin  2 
T1                                 T3
              cos 2             
WS #2 - #26 Ch.5 4th ed - Newton's Laws
        The Working Equations

              T3 cos 2
 T1 
      sin1 cos 2  cos1 sin 2

               T3 cos 2
         T1 
              sin 1   2 

         From Symmetry
               T3 cos 1
         T2 
              sin  2  1 
         WS #2 - #26 Ch.5 4th ed - Newton's Laws
                   The Solution Part a
        1  40     2  50     m  5.0 kg


                         
   T3  mg   5.0 kg  9.80 m
                                     s²   49 N
      T3 cos 2         49 N  cos  50 
T1                                            31.5 N
     sin 1   2       sin  40  50 

       T3 cos1          49 N  cos  40 
 T2                                            37.5 N
      sin  2  1       sin  50  40 
       WS #2 - #26 Ch.5 4th ed - Newton's Laws
                 The Solution Part b
        1  60      2  0    m  10. kg


                        
   T3  mg   10 kg  9.80 m
                                    s²   98 N
      T3 cos 2         98 N  cos  0 
T1                                           110 N
     sin 1   2       sin  60  0 

      T3 cos1          49 N  cos  60 
T2                                            57 N
     sin  2  1        sin  0  60 
              WS #3 - #28 Ch.5 4th ed - Newton's Laws
                           The Problem
3.        A 225 N weight is tied to the middle of a strong rope,
          and two people pull at opposite ends of the rope in an
          attempt to lift the weight.
     a.     What is the magnitude F of the force that each person must
            apply in order to suspend the weight as shown in the Figure
            (P5.28)?
     b.     Can they pull in such a way as to make the rope horizontal?
            Explain.

                            T1                    T2

                            10.°= 1            2= 10.°
                                       T3



                                       225 N
WS #3 - #28 Ch.5 4th ed - Newton's Laws
             The Diagram



  T1                        T2

  10.°= 1              2= 10.°
              T3
                     225 N
     WS #3 - #28 Ch.5 4th ed - Newton's Laws
The Solution - The equations come from problem 26
         1  10     2  10   T3  225 N

            T3 cos 2         225 N  cos10
      T1                  
           sin 1   2     sin  10  10 

                    T1  650 N

            T3 cos1          225 N  cos10
      T2                  
           sin  2  1     sin  10  10 

                    T2  650 N
        WS #4 - #31 Ch.5 4th ed - Newton's Laws
                     The Problem
4.   A bag of cement hangs from three wires as
     shown on page 134 figure p5.31. Two of the
     wires make angels of q1 and q2 with the
     horizontal. If the system is in equilibrium.

                             w cos  2
           Show that T2 
                          sin 1   2 

                 See problem 26
         WS #4 - #31 Ch.5 4th ed - Newton's Laws
                      The Problem
4.   Given that w = 325 N, q1 = 10°,and q2 = 25°, find the tensions
     T1, T2, and T3 in the wires.

             1  10        2  25       T3  325 N

               T3 cos 2         325 N  cos 25
         T1                  
              sin 1   2     sin  10  25 

                         T1  510 N
                T3 cos1          325 N  cos10
          T2                  
               sin  2  1     sin  10  25 
                          T2  560 N
           WS #5 - #17 Ch.5 3th ed - Newton's Laws
                        The Problem
5.     Two forces F1 = (3i - 5j) N and F2 = (2i + j) N
       act on a 1.50 kg body.
     1. What are the magnitude and direction of the
         acceleration?
     2. If the body is initially resting at the origin,
         what are the coordinates of its position at t =
         4.00 s?
        WS #5 - #17 Ch.5 3th ed - Newton's Laws
                     The Solution



     ˆ
F1  3i  5 ˆ N
            j             ˆ j
                      F2  2i  ˆ N              m  1.50kg

                       F  ma
a
   F       ˆ     j  
             3i  5 ˆ N  2i  ˆ N
                           ˆ j                  
                                                       ˆ     
                                                       5i  4 ˆ N
                                                              j
    m                 1.50kg                            1.50kg

                  
              a  3.33i  2.67 ˆ
                      ˆ        j       m
                                            s2

               a  4.27 m s2 at -38.7
          WS #5 - #17 Ch.5 3th ed - Newton's Laws
              The Problem and The Solution
  B.     If the body is initially resting at the origin, what
         are the coordinates of its position at t = 4.00 s?


   
a  3.33i  2.67 ˆ
        ˆ        j           m
                                  s2
                                       vi  0   so  0                t  4.00s
                       s  so  vi t  at   1
                                            2
                                                2



       s  0   0 t    1
                          2          ˆ
                                  3.33i  2.67 ˆ
                                               j       m
                                                            s2    4.00s 
                                                                             2




                          
                    s  26.6i  21.4 ˆ m
                            ˆ        j          
         WS #6 - #18 Ch.5 4th ed - Newton's Laws
              The Problem and Diagrams
6.   Two forces F1 and F2 act on a 5.0 kg mass. If F1 = 20. N and F2
     = 15 N, Find the acceleration in both of the following figures.
WS #6 - #18 Ch.5 4th ed - Newton's Laws
       The Solution to diagram 1



      m  5.0kg                 ˆ
                        F1  20.iN       F2  15 ˆN
                                                 j

                a
                   F  F        1  F2
                         m         m

                a
                           ˆ        
                         20.i  15 ˆ N
                                   j
                            5.0kg

                    
                a  4.0i  3.0 ˆ
                       ˆ       j        m
                                             s2
WS #6 - #18 Ch.5 4th ed - Newton's Laws
       The Solution to diagram 2


                 m  5.0kg                ˆ
                                  F1  20.iN

                 
            F2  15cos 60 i  15sin 60 ˆ N
                           ˆ             j           
                     a
                        F  F      1    F2
                            m           m

         a
             20.iˆ  15cos 60iˆ  15sin 60 ˆj  N
                                5.0kg

                        
                     a  5.5i  2.6 ˆ
                            ˆ       j      m
                                                s2
          WS #7 - #67 Ch.5 3rd ed - Newton's Laws
                       The Problem
7.   The largest caliber antiaircraft gun
     operated by the Luftwaffe during World
     War II was the 12.8-cm Flak 40. This
     weapon fired a 25.8 kg shell with a
     muzzle velocity of 880 m/s. What
     propulsive force was necessary to attain
     the muzzle velocity within the 6.0 m
     barrel? (Assume constant acceleration
     and neglect the earth’s gravitational
     effect.) (Physics for Scientists and Engineers by Serway, 3rd Edition, Ch 5# 67)
WS #7 - #67 Ch.5 3rd ed - Newton's Laws
             The Diagram




             6.0 m
           25.8 kg              F=?
           V = 880 m/s
WS #7 - #67 Ch.5 3rd ed - Newton's Laws
             The Solution
        d  6.0 m         v f  880 m/s
     m  25.8 kg           vo  0    F ?
                     F  ma
                   v 2  vi2  2ad
                     f

                         v 2  vi2
                    a
                           f

                           2d
       v 2  vi2                  880 m/s   0
                                            2

F m                 25.8 kg 
         f

         2d                         2  6.0 m 
               F  1.7  106 N
             WS #8 - #43 Ch.5 4th ed - Newton's Laws
                          The Problem
8.        A 72.0-kg man stands on a spring scale in an
          elevator. Starting from rest, the elevator
          ascends, attaining its maximum speed of 1.20
          m/s in 0.800s. It travels with this constant
          speed for the next 5.00 s. The elevator then
          undergoes a uniform acceleration in the
          negative y direction for 1.50 s and comes to
          rest. What does the spring scale register:
     a.     before the elevator starts to move?
     b.     during the first 0.80 s.?
     c.     while the elevator is traveling at constant speed?
     d.     during the time it is slowing down?
WS #8 - #43 Ch.5 4th ed - Newton's Laws
             The Diagram

        Elevator


                   Normal
           72 kg Force of the scale
                   is your weight
                                      1.2m/s

                     Force
                      of
                    Gravity
          Scale
          WS #8 - #43 Ch.5 4th ed - Newton's Laws
                       The Solution
          a.   What does the spring scale register: before the
               elevator starts to move?




     m  72.0kg            g  9.80 m s 2          ay  0

                      F     y    ma y
                       N  Fg  0
N  Fg   mg    72.0kg   9.8 0 m s 2   706 N
           WS #8 - #43 Ch.5 4th ed - Newton's Laws
                        The Solution
            b.   What does the spring scale register: during the first
                 0.80 s.?

   m  72.0kg     g  9.80 m s2          v f  1.20 m
                                                    s
                                                        t  0.80 s   vi  0
                                    v f  v i  at
                         v f  vi     1.20 m  0
                  a                      s
                                                  1.50 m s2
                            t           0.80s
                                     Fy  ma y
                                N  Fg  ma y
                                N  Fg  ma y

                                   
N  mg  ma y  m g  a y   72.0kg   9.80 m s2  1.50 m s2   814N
           WS #8 - #43 Ch.5 4th ed - Newton's Laws
                        The Solution
      c.     What does the spring scale register: while the
             elevator is traveling at constant speed

   m  72.0kg              g  9.80 m s2           ay  0

                       F     y    ma y
                        N  Fg  0
N  Fg  mg   72.0kg   9.80 m s2   706N
                 WS #8 - #43 Ch.5 4th ed - Newton's Laws
                              The Solution
               d.    What does the spring scale register: during the time
                     it is slowing down?

    m  72.0kg       g  9.80 m s 2          vf  0      t  1.50s   v i  1.2 0 m
                                                                                 s

                                        v f  v i  at
                         v f  vi     0  1.20 m
                    a                        s
                                                   0.800 m s 2
                             t          1.50s
                                     F y  ma y
                                    N  Fg  ma y
                                    N  Fg  ma y

                                   
N  mg  ma y  m g  a y   72.0kg   9.80 m s 2  0.8 00 m s 2   648 N
              WS #9 - #45 Ch.5 4th ed - Newton's Laws
                    The Problem and Solution
 9.      A net horizontal force F = A + Bt3 acts on a 3.50-kg object, where A
         = 8.60 N and B = 2.50 N/s3. What is the horizontal speed of this
         object 3.00 s after it starts from rest?
                           F = A + Bt 3    A  8.60N          B  2.50 N s3
                          m  3.50kg      t  3.00s     vi  0       vf  ?
                                            F  ma
                                          A + Bt 3  ma
                                             A + Bt 3
                                         a
                                                 m
                        3.00       3.00               3.00       3.00
                                        A + Bt 3           A          Bt 3
                  v f   a  dt               dt        dt          dt
                         0          0
                                          m            0
                                                           m      0
                                                                      m
                           
                                                    2.50   3.00s         A         B   0  
                  3.00                                                  4                          4
     A    Bt 
              4
                              8.60N                      N
vf   t                          3.00s               s3
                                                                                0              
     m    4m  0
                           3.50kg                  4  3.50kg            m           4m  
                                                                                                 
                                 v f  7.37 m  14.5 m  21.8 m
                                            s        s        s
        WS #10 - #48 Ch.5 4th ed - Newton's Laws
                         The Problem
10.   A 25.0 kg block is initially at rest on a horizontal
      surface. A horizontal force of 75.0 N is required to set
      the block in motion. After it is in motion, a horizontal
      force of 60.0 N is required to keep the block moving
      with constant speed. Find the coefficients of static and
      kinetic friction from this information.
       WS #10 - #48 Ch.5 4th ed - Newton's Laws
                                The Solution
m  25kg     Fs  75N        Fk  60.N  s  ?   k  ?
                                                            N
                      Fy  ma y  0
                                                   f             F
             Fy  N  mg  0  N  mg                    25kg

                     F  x    ma x  0
             F  x    F  f 0 f  F                      mg
            fs  F       75N
     s        s                      .306
            N mg  25kg   9.80 m s2 
          fk   Fk      60N
     k                            .245
          N m g  25kg   9.80 s2 
                               m
           WS #11 - #72 Ch.5 4th ed - Newton's Laws
                          The Problem
11.     In the figure, the man and the
        platform together weigh 750
        N. Determine how hard the
        man would have to pull to hold
        himself off the ground. (Or is
        it impossible? If so, explain
        why.)


         F    y    ma  0                  T T

      T  T  T  750N  0
           T  750N                      750N      T down
                                                 on the man’s
                                                     arms
      WS #12 - #53 Ch.5 4th ed – Newton’s Law
                                       The Problem
11.        A boy drags his 60.0 N sled at constant speed up a 15°
           hill. He does so by pulling with a 25 N force on a rope
           attached to the sled. If the rope is inclined at 35° to the
           horizontal
      1.     what is the coefficient of kinetic friction between the sled and the
             snow?
      2.     At the top of the hill, he jumps on the sled and slides down the
             hill. What is the magnitude of his acceleration down the slope?
             (Physics for Scientists and Engineers by Serway, 4th Edition, Ch 5# 53)
WS #12 - #53 Ch.5 4th ed – Newton’s Law
              The Diagram
         WS #12 - #53 Ch.5 4th ed – Newton’s Law
                       The Solution


mg  60.N       F  25N
 =20°        15
F   y    ma y  0
N  F sin   mg cos   0
N  mg cos   F sin 
F   x    ma x  0
F cos   mg sin   F f  0
F cos   mg sin    N  0
  WS #12 - #53 Ch.5 4th ed – Newton’s Law
                   The Solution

F cos   mg sin     mg cos   F sin    0
   mg cos   F sin    mg sin   F cos 
  mg cos   F sin    F cos   mg sin 
   F cos   m g sin 

   mg cos   F sin 


    25N  cos 20°   60.N  sin 15  .16
    60.N  cos 15   25N  sin 20°
      WS #12 - #53 Ch.5 4th ed – Newton’s Law
                             The solution Part b
 1.    At the top of the hill, he jumps on the sled and slides down the
       hill. What is the magnitude of his acceleration down the slope?
       (Physics for Scientists and Engineers by Serway, 4th Edition, Ch 5# 53)




F     y    ma y  0
N  m g cos   0
  N  m g cos 
         WS #12 - #53 Ch.5 4th ed – Newton’s Law
                                 The solution Part b
    1.     At the top of the hill, he jumps on the sled and slides down the
           hill. What is the magnitude of his acceleration down the slope?
           (Physics for Scientists and Engineers by Serway, 4th Edition, Ch 5# 53)



                    F      x    ma x
             mg sin   F f  ma x
             mg sin    N  ma x
         mg sin    m g cos   ma x
a x  g sin    g cos   g  sin    cos  

  a x   9.80 m s 2   sin 15   .16  cos 15 
                     a x  1.0 m s 2
      WS #13 - #39 Ch.5 4th ed - Newton's Laws
                    The Problem
13.       A small bug is placed between two blocks of masses m1 and m2
          (m1 > m2) on a frictionless table. A horizontal forcec F, can be
          applied to either m1, as in the first figure below or m2 as in the
          second figure below. For which of these two cases does the bug
          have a greater chance to surviving? Show! (Physics for Scientists and
          Engineers by Serway, 4th Edition, Ch 5# 39)




      F            m1                                     m1                F
                             m2                                   m2
   WS #13 - #39 Ch.5 4th ed – Newton’s Laws
                 The Solution

                    a
                       F               F
                       m m            1  m2

  When force is on m1                      When force is on m1
m1 exerts a force F2 on m2             m2 exerts a force F1 on m1
       F2  m2 a                                  F1  m1a
           m2 F                                       m1 F
     F2                                        F1 
          m1  m2                                    m1  m2
                        if m2  m1
                             F2  F1
     WS #14 - #50 Ch.5 4th ed – Newton’s Laws
                   The Problem

14. What force must be exerted on Block A
    in order for Block B not to fall. The
    coefficient of static friction between
    blocks A and B is 0.55, and the horizontal
    surface is frictionless. (Physics for Scientists and
     Engineers by Serway, 4th Edition, Ch 5# 50)
WS #14 - #50 Ch.5 4th ed – Newton’s Laws
              The Diagram




                               Block B
              Block A          10.0 kg


              100. kg
WS #14 - #50 Ch.5 4th ed – Newton’s Laws
        The Free Body Diagrams
WS #14 - #50 Ch.5 4th ed – Newton’s Laws
       The Solution – 10 kg mass


 F    y   0             F    x    mb a
F f  mb g  0              N  mb a
 F f  mb g               mb g
                                     mb a
  N  mb g                 
                                     g
  N
       mb g                  a
                                    
   WS #14 - #50 Ch.5 4th ed – Newton’s Laws
      The Solution – both mass together
g  9.80 m s 2   .55 ma  100 kg m b  10.0 kg
                             g
                     a
                             
                 F   x    a m
                     g
               F          ma  mb 
                    
           9.80 m s 2
        F             100 kg  10.0 kg 
             .5 5
                 F  1960 N
           WS #15 - #71 Ch.5 4th ed – Newton’s Laws
                         The Problem
15.        Three baggage carts of masses m1, m2, and m3 are
           towed by a tractor of mass M along an airport apron.
           The wheels of the tractor exert a total frictional force F
           on the ground as shown. In the following, express your
           answers in terms of F, M, m1, m2, m3 and g.
      a.     What are the magnitude and direction of the horizontal force
             exerted on the tractor by the ground?
      b.     What is the smallest value of the coefficient of static friction
             that will prevent the wheels from slipping? Assume that each
             of the two drive wheels on the tractor bears 1/3 of the tractor’s
             weight.
      c.     What is the acceleration, a, of the system (tractor plus baggage
             carts)?
      d.     What are the tensions T1, T2, and T3 in the connecting cables?
      e.     What is the net force on the cart of mass m2?
                   (Physics for Scientists and Engineers by Serway, 4th Edition, Ch 5# 71)
WS #15 - #71 Ch.5 4th ed – Newton’s Laws
           The Solution Part a
a.    What are the magnitude and direction of the horizontal
      force exerted on the tractor by the ground?
     a.   From Newton’s Third Law since the force of the tractor on
          the ground is F, the magnitude of the force of the ground on
          the tractor must also be F but in the opposite direction.
          Therefore the force exerted on the tractor is F to the right.
         WS #15 - #71 Ch.5 4th ed – Newton’s Laws
                    The Solution Part b
            b.   What is the smallest value of the coefficient of static friction
                 that will prevent the wheels from slipping? Assume that each of
                 the two drive wheels on the tractor bears 1/3 of the tractor’s
                 weight.
                                                            N
F   y    ma y  0
N  2 mg  0
    3

 N  2 mg                                                                F
     3                      F  f friction   N                  2/3 mg
                               F     F
                                2
                               N 3 mg
                                3F
                            
                               2m g
WS #15 - #71 Ch.5 4th ed – Newton’s Laws
           The Solution Part c
   c.   What is the acceleration, a, of the system (tractor plus
        baggage carts)?




             a
                F                external

                 m
              F
    a
       M  m1  m2  m3
         WS #15 - #71 Ch.5 4th ed – Newton’s Laws
                    The Solution Part d
                d.   What are the tensions T1, T2, and T3 in the connecting
                     cables?




There is one force on m1          T2 is accelerating two masses m1 and m2
        F  1
                 m1a
                                                  F  m     1    m2  a
               F
T1  m1
        M  m1  m2  m3                      T2 
                                                         m2  m1  F
                                                     M  m1  m2  m3
             m1 F
 T1 
        M  m1  m2  m3
    WS #15 - #71 Ch.5 4th ed – Newton’s Laws
               The Solution Part d




T3 is accelerating three masses m1 and m2 and m3
             F  m   1    m2  m3  a

            T3 
                    m1  m2  m3  F
                   M  m1  m2  m3
WS #15 - #71 Ch.5 4th ed – Newton’s Laws
           The Solution Part e
   e.   What is the net force on the cart of mass m2?




                The net force is  F
                      F    2
                                 m2a
                          F
          F2  m2 M  m  m  m
                        1   2   3

                       m2F
            F2  M  m  m  m
                       1   2   3
     WS #16 - #37 Ch.5 4th ed – Newton’s Laws
                   The Problem
1.   In the system on the next slide, a
     horizontal force Fx acts on a 8.00 kg
     mass.
        a. For what values of Fx does the 2.00 kg mass
           accelerate upward?
        b. For what values of Fx is the tension in the cord
           zero?
        c. Plot the acceleration of the 8.00 kg mass versus
           Fx. Include values of Fx from -100 N to +100 N.
            (Physics for Scientists and Engineers by Serway, 4th Edition, Ch 5# 37)
WS #16 - #37 Ch.5 4th ed – Newton’s Laws
              The Diagram
WS #16 - #37 Ch.5 4th ed – Newton’s Laws
              The Diagram
                        y

                              x


         x

    y
WS #16 - #37 Ch.5 4th ed – Newton’s Laws
        The Free Body Diagram


        y                       y

                 x
                                           x
WS #16 - #37 Ch.5 4th ed – Newton’s Laws
           Find F such that a > 0

     m  2.0 kg        4m  8.0 kg
             F    x    ma x
        F  mg   m  4m  a
         a  0 if F  mg  0
                F  mg
  F   2.0 kg   9.80 m s2   19.6 N
    WS #16 - #37 Ch.5 4th ed – Newton’s Laws
               Find F such that T = 0


   For m                             For 4m

  F  ma                          F  ma
                                  F  T  4ma
T  mg  m a
                                  F  0  4ma
0  mg  ma
                     F  4m   g   4 2.0 kg  9.80 m s2 
   a  g
                                  F  78.4 N
WS #16 - #37 Ch.5 4th ed – Newton’s Laws
             Part C – The Plot
      WS #17 - #47 Ch.5 4th ed – Newton’s Laws
                    The Problem
17.   Two blocks of masses m and
      4m are connected by a light
      string passing over a fixed
      pulley as shown at the right.
      The 4m mass is on a smooth
      frictionless surface and is
      pushed toward the left by a
      constant horizontal force F.
             a.     What value of F will give the
                    system an acceleration of 0.400 g
             b.     Find the tension in the string
                    when F has the value determined
                    in (A).
             c.     What happens when F>4.00 mg?
                   (Physics for Scientists and Engineers
      by Serway, 3rd Edition, Ch 5# 47)
WS #17 - #47 Ch.5 4th ed – Newton’s Laws
              The Diagram
                        y

                             x


          x

     y
     WS #17 - #47 Ch.5 4th ed – Newton’s Laws
                   The Solution
 (A) What value of F will give the system an acceleration of
                           0.400 g


                      a system   
                                   F
                                   m
                              F  mg
                 0.400 g 
                              4m  m
             0.400 g  4m  m    F  mg
F  0.400 g  5m   mg  0.400 g  5m   mg  2 m g  mg
                          F  mg
        WS #17 - #47 Ch.5 4th ed – Newton’s Laws
                        The Solution
  (b) Find the tension in the string when F has the value
                    determined in (A).

a
   F                                         4m
                                                  y
   m
            F  T
0.400 g 
              4m                                            x
 F  T   0.400 g  4m
T  4m  0.400 g   F
T  4m  0.400 g   mg
T  .600mg
         WS #17 - #47 Ch.5 4th ed – Newton’s Laws
                       The Solution
           (c) What happens when F>4.00 mg?

         a
            F                             4m
            m                                 y
             F  T
        a
              4m
        F  T  4ma                                x
       T  4ma  F
T  4ma  4mg  4m  a  g 
    but amax  g  T  0
   The String goes slack
    WS #18 - #46 Ch.5 4th ed – Newton’s Laws
                  The Problem
18. Mass m1 on a smooth horizontal table is
    connected to mass m2 through a very
    light pulley P1 and a light fixed pulley P2
    as shown.
        a.   If a1 and a2 are the accelerations of m1 and m2,
             respectively, what is the relationship between these
             accelerations?
        b.   Express the tensions in the strings, and
        c.   the accelerations a1 and a2 in terms of the masses m1, m2,
             and g.
             (Physics for Scientists and Engineers by Serway, 4th Edition, Ch 5# 46)
WS #18 - #46 Ch.5 4th ed – Newton’s Laws
              The Diagram
        T’              P1
                              T     P2
          m1      T’


                                         T


                                        m2
                                  m2g
       WS #18 - #46 Ch.5 4th ed – Newton’s Laws
If a1 and a2 are the accelerations of m1 and m2, respectively,
      what is the relationship between these acceleration

A) Find a1 in terms of a 2 .
       2d1           2d 2
   a1 = 2        a2 = 2        But when m 2 moves a distance d 2 ,
        t             t
                               Pulley P1 moves a distance d 2 ,
                               which means that the string
                               between m1 and P1 gets a distance
                               d 2 shorter. This means that m1 moves 2d 2 .
     2d1 2  2d 2     2d
a1 = 2 =            = 2 2 2 = 2a 2
      t      t2         t
a1 = 2a 2
WS #18 - #46 Ch.5 4th ed – Newton’s Laws
 Express the (B) tensions in the strings,
                     For m1
               F     1   = m 1 a1
                 T1 = m1a1
                      T1
                 a1 =
                      m1
                     For P1
            F  p1    = m p1 ap1 = 0
              T2 - T1 - T1 = 0
                          T2
                     T1 =
                          2
WS #18 - #46 Ch.5 4th ed – Newton’s Laws
 Express the (B) tensions in the strings,
                      For m 2
                   F    2   = m 2a2
                                          T1
                                 a1            m1
      m 2 g - T2 = m 2 a 2 = m 2     = m2
                                 2             2
                                    T2
                        T1                  m 2 T2
     m 2 g - T2 = m 2        = m2 2 =
                      2m1           2m1     4m1
                              m 2 T2
                 m 2 g - T2 =
                              4m1
             4m1m 2 g - 4m1 T2 = m 2 T2
                         4m1m 2 g
                  T2 =
                         m 2 + 4m1
  WS #18 - #46 Ch.5 4th ed – Newton’s Laws
   Express the (B) tensions in the strings,



          4m1m 2 g
     T2   m 2 + 4m1   2m1m 2 g
T1 =    =           =
     2        2       m 2 + 4m1
     WS #18 - #46 Ch.5 4th ed – Newton’s Laws
(c) the accelerations a1 and a2 in terms of the masses
                    m1, m2, and g

             2m1m 2 g 
                       
  a1 =
       T1
          =  m 2 + 4m1  = 2m 2 g
       m1         m1       m 2 + 4m1
                2m 2 g
         a1   m 2 + 4m1      m2g
    a2 =    =           =
         2        2       m 2 + 4m1
          WS #19 - #85 Ch.5 4th ed - Newtons Laws
                       The Problem
19.   The three blocks to the right are connected by mass-
      less strings that pass over massless frictionless pulleys.
      The acceleration of the system is 2.35 m/s2 to the left
      and the surfaces are rough. Mass 1 is 10.0 kg, mass 2
      is 5.00 kg, and mass 3 is 3.00 kg. Find:
          a.   the tensions in the strings and
          b.   the coefficient of kinetic friction between blocks and surfaces.
               (Assume the same m for both blocks)

               (Physics for Scientists and Engineers by Serway, 4th Edition, Ch 5# 85)


                                             5.00
                                              kg
                                       T1
                                                                    T2       3.00
                                                                              kg
                             10.0
                              kg
WS #19 - #85 Ch.5 4th ed - Newtons Laws
             The Diagram
      WS #19 - #85 Ch.5 4th ed - Newtons Laws
              The Free body Diagram




They all acceleration as a unit. The tensions
                are internal.
          WS #19 - #85 Ch.5 4th ed - Newtons Laws
(b) the coefficient of kinetic friction between blocks and surfaces.


m1  10.0kg             m2  5.00kg               m3  3.00kg
              a  2.35 m s2   25.0

  For body 2                             For body 3
F    y     ma y  0                 F      y    ma y  0
N 2  m2 g  0                      N 3  m3 g cos   0
   N 2  m2 g                          N 3  m3 g cos 
          WS #19 - #85 Ch.5 4th ed - Newtons Laws
   (b) the coefficient of kinetic friction between blocks and surfaces.

                        F      x
                                      ma x
 F f 3  m3 g sin   F f 2  m1 g   m1  m2  m3  a
 N 3  m3 g sin    N 2  m1 g   m1  m2  m3  a
  N 2  N 3    m1  m2  m3  a  m1 g  m3 g sin 

      
          m1  m2  m3  a  m1 g  m3 g sin 
                                    N2  N3

      
          m1  m2  m3  a  m1 g  m3 g sin 
                           m3 g cos   m2 g
                   WS #19 - #85 Ch.5 4th ed - Newtons Laws
          (b) the coefficient of kinetic friction between blocks and surfaces.




               
                   m1  m2  m3  a  m1 g  m3 g sin 
                                       m3 g cos   m2 g

      10.0  5.00  3.00    2.35    10.0   9.80    3.00   9.80  sin 25.0   kg m s 2
                                                                                        
                        9.8 0 m s 2    3.00kg  cos 25.0    5.0 0kg  
                                               .57 2
WS #19 - #85 Ch.5 4th ed - Newtons Laws
           (a) the tensions in the strings


        Find T1 using body 1
                F  m a        1

            T1  m1 g  m1a
    T1  m1a  m1 g  m1  a  g 
T1   10.0kg   2.35 m s 2  9.80 m s 2 
                T1  74.5N
                WS #19 - #85 Ch.5 4th ed - Newtons Laws
                             (a) the tensions in the strings


                            Find T2 using body 3
                                   F  m a      3

                          m3 g sin   F f 3  T2  m3 a
                     T2   m3 a  m3 g sin    m3 g cos 
                      T2  m3   a  g  sin    cos   

                 
T2   3.00kg    2.35 m s 2   9.80 m s 2  sin 25.0   .572  cos 25.0    
                                    T2  34.7N
     WS #20 - #65 Ch.5 4th ed - Newtons Laws

20. A mass M is held in place by an applied
    force FA and a pulley system as shown at
    the right. The pulleys are mass-less and
    frictionless. Find the tension in each
    section of rope, T1, T2, T3, T4, and T5 and
    the applied force FA.
  WS #20 - #65 Ch.5 4th ed - Newtons Laws
             Starting solutions

Mass M
   y

         x
             On mass M
             F m    Ma  0
             T5  Mg  0
               T5  Mg
      WS #20 - #65 Ch.5 4th ed - Newtons Laws
                      The Solution - T2,T3

             On Pulley P2                    Pulley 2
            T2  T3  T5  0                     y
T2 and T3 are the same string  T2  T3
       T3  T3  T5  2T3  T5  0
                                                        x
                  T5 Mg
             T3     
                  2     2
                       Mg
             T3  T2 
                       2
         WS #20 - #65 Ch.5 4th ed - Newtons Laws
                         The Solution – T1,T4
                  On Pulley P1
              T1  T2  T3  T4  0                Pulley 1
T1 , T2 and T3 are the same string  T1  T2  T3       y
                         Mg
                   T1 
                          2                                    x
             Mg Mg Mg
                             T4  0
              2    2       2
                        3 Mg
                  T4 
                          2
                           Mg
                F  T1 
                             2
              Newton’s Law 1998 AP Question - Solution
                            From the 1998 Physics C Mech AP Test




1.   Block 1 of mass ml is placed
     on block 2 of mass m2, which
                                                                   m
                                                                   1


     is then placed on a table. A
     string connecting block 2 to a
                                                                   m
                                                                   2


     hanging mass M passes over
     a pulley attached to one end
     of the table, as shown to the
     right. The mass and friction
                                                                       M
     of the pulley are negligible.
     The coefficients of friction
     between blocks 1 and 2 and
     between block 2 and the
     tabletop are nonzero and are
     given in the following table.
                    Newton’s Law - Solution
                       From the 1998 Physics C Mech AP Test




                      Coefficient Between                       Coefficient Between
                        Blocks 1 and 2                        Block 2 and the Tabletop
      Static                         s1                                s2

      Kinetic                        k1                                k2




Express your answers in terms of the masses, coefficients of
      friction, and g, the acceleration due to gravity.
                              Newton’s Law - Solution
                                  From the 1998 Physics C Mech AP Test




(a)      Suppose that the value of M is small enough that                          m1

         the blocks remain at rest when released. For each                         m2

         of the following forces, determine the magnitude
         of the force and draw a vector on the block
         provided to indicate the direction of the force if it                          M
         is nonzero.
i.       The normal force N1 exerted on block 1 by block 2



                                                                          F
      Forces
        on
        m1      N1
                                                N1                             y   =0
                                                                         N1 -m1g = 0
                m1g
                                      m         1
                                                                          N 1 = m 1g
                             Newton’s Law - Solution
                                 From the 1998 Physics C Mech AP Test




ii.       The friction force f1 exerted                                     m
                                                                            1


          on block 1 by block 2                                             m
                                                                            2




F    x   = max = m1 (0) = 0                                                    M

f1 = 0
                                                                 m      1



This is true since there is no motion and friction only opposes motion, there is
no horizontal force for the friction to oppose.
                          Newton’s Law - Solution
                             From the 1998 Physics C Mech AP Test




iii.    The force T exerted on block                                        m1

        2 by the string                                                     m2




                                                                                  M
   Forces
     on
     m2          N2
                                                                    F      = ma x = 0
        f2       T           m         2
                                                     T                  x

                                                                    Mg - T = 0
            N1
                 m2g
                                                                    T = Mg
                     Newton’s Law - Solution
                          From the 1998 Physics C Mech AP Test



                                                                 m
                                                                 1


iv.   The normal force N2 exerted                                m
                                                                 2
      on block 2 by the tabletop

Forces
  on
  m2          N2
                      F           y    =0                                M

      f2      T        N 2 - N1 - m 2 g = 0
         N1
              m2g      N 2 = N1 + m 2 g
                       N 2 = m1g + m 2 g                             N2

                       N 2 = g(m1 + m 2 )
                                                                     m
                                                                     2
                       Newton’s Law - Solution
                            From the 1998 Physics C Mech AP Test
                                                                            m
                                                                            1


                                                                            m
                                                                            2
v.    The friction force f2 exerted
      on block 2 by the tabletop
 Forces
   on
                                                    F             x   =0       M
   m2          N2
                                                    T - f2 = 0
      f2       T
          N1
               m2g
                                                    f 2 = T = Mg

                              f2                        m          2
                      Newton’s Law - Solution
                           From the 1998 Physics C Mech AP Test



                                                                     m  1


(b)   Determine the largest value
      of M for which the blocks
                                                                     m  2


      can remain at rest.
                From (a) v. above
                                                                                 M
                f 2max = Mg
                              f 2max              s2 N 2
                M max =                    =
                                 g                       g
                but from (a) iv. above N 2 = g(m1 + m 2 )
                            s2 g(m1 + m 2 )
                M max =                                           = s2 (m1 + m 2 )
                                              g
                          Newton’s Law - Solution                                m
                               From the 1998 Physics C Mech AP Test                  1


                                                                                 m   2

(c)     Now suppose that M is large
        enough that the hanging block Forces
        descends when the blocks are                                        Forces
                                        on                                    on              M
        released. Assume that blocks 1 m                              N2
                                          2                                   M
        and 2 are moving as a unit (no
        slippage). Determine the
                                            f2                        T
        magnitude a of their                 N1                                 T        Mg
        acceleration.                                                 m2g


  F    2x    a2 x  M
  Mg  f 2  a2 x  m1  m2  M                                 M  2k  m1  m2 
                                                        a2 x  g
          Mg  2 k g  m1  m2                                   m1  m2  M
 a2 x   
             m1  m2  M
                               Newton’s Law - Solution
                                  From the 1998 Physics C Mech AP Test

 (d) Now suppose that M is large enough that as the hanging block descends, block
            1 is slipping on block 2. Determine each of the following.

                                                                         m
                                                                         1



i.      The magnitude a1 of the                                          m
                                                                         2

        acceleration of block 1

     Forces
       on
       m1          N1                                  Fx  m1a x              M
                          f1                               f1 m1 g 1k
                                                      ax     
                    m1g                                    m1   m1
                                                      a x  g 1k
                                 Newton’s Law - Solution
                                      From the 1998 Physics C Mech AP Test
      (d) Now suppose that M is large enough that as the hanging block descends, block 1 is
                    slipping on block 2. Determine each of the following.

                                                                                         m  1


                                                                                         m  2

ii.        The magnitude a2 of the
           acceleration of block 2
 Forces                      Forces                                                                M
   on                          on
   m2              N2          M                                             N 2  m2 g  N 1
          f2        T                                                        N 2  m2 g  m1 g
           f1                       T               Mg
                                                                             N 2  g  m2  m1 
                  N1
                   m2g
                                  Newton’s Law - Solution
                                       From the 1998 Physics C Mech AP Test
       (d) Now suppose that M is large enough that as the hanging block descends, block 1 is
                     slipping on block 2. Determine each of the following.        m              1


                                                                                                 m
                                                                                                 2




ii.        The magnitude a2 of the                   Forces                           Forces
           acceleration of block 2                     on                               on                M
                                                       m2                     N2        M
      N 2  g  m2  m1                                        f2             T
                                                                 f1           N1             T       Mg
        F     x    ax  m                                                   m2g

ax    
        F    x
                   
                     Mg  f1  f 2
        m            m2  M
                                                  Mg  m1 g 1k  2 k g  m1  m2 
        Mg  m1 g 1k  2 k N 2             ax 
ax                                                          m2  M
              m2  M
                                                           g  M  m11k  2 k  m1  m2  
                                             ax 
                                                                                    m2  M

				
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