Chapter 5 Applications of Newton's Laws

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					Chapter 5       Applications of Newton’s Laws


We shall apply Newton’s laws of motion to a variety
of situations in which several types of forces are
involved:
• Gravitational force
• Tension
• Contact Forces (including friction)


                                        (5-1)
                           Gravitational force Fg
                           Also known as the “weight” of
   m                Fg     an object
               g

       earth

All objects near the surface of the earth are uniformly
accelerated towards the center of the earth. The
magnitude of the acceleration g = 9.8 m/s 2
From Newton’s second law we get that Fg = ma = mg
                                       (5-2)
A practical method
of measuring Fg Use
a spring scale as
shown in the figure.                    y-axis
The spring has a
pointer attached to it
so that we can
determine its
elongation

System = grapes. There are two forces acting on the grapes:
The gravitational force mg and the spring force Fs which is
given by: Fs = k|y|. Here y is the elongation of the spring
The net force Fynet = Fs - mg = may = 0 because the grapes are
stationary. Thus mg = Fs = k|y|                (5.3)
                 T                    Tension
        crate     rope
                           It is the force exerted by a rope or a
         floor
                           cable

Tension has the following characteristics:
• Tension always pulls (never pushes)
• The direction of T is along the rope
                                     pulley
• The magnitude |T| is constant                        T
even when T changes direction            T

(as in the case of the pulley)      crate                  pull
(5-4)
Normal Force (contact force)
Characteristics of the normal force FN
• FN is perpendicular to the contact surface
• FN always pushes away from the contact surface
       (never pulls)



                  FN
    crate
                 Fg
        floor
                         (5-5)
The normal force and Newton’s third law
              FN = Fcf     System = crate
crate                      The normal force on the crate
                           exerted by the floor is labeled Fcf
    floor
                           System = floor
crate
                           The normal force on the floor
                           exerted by the floor is labeled F fc
   floor
              Ffc

Newton’s third law: Fcf + F fc = 0 → Ffc = - Fcf = - FN
The crate exerts a force F fc on the floor which is equal and
opposite to FN                          (5-6)
Q: How can we
measure F N ?
A: With a bathroom
scale!
The bathroom scale
measures the normal
force (and not
necessarily mg unless
the object on the scale
does not accelerate)

                          (5-7)
                             Fas is the normal force exerted
                             on the apple by the scale.
apple            FN = Fas    Fsa is the normal force exerted
                             on the scale by the apple
  scale
                             These two forces are equal and
                  FN = Fsa   opposite by virtue of Newton’s
                             third law
y-axis                       System = apple
                             Fynet = FN - mg
                 FN          Fynet = may = 0       →

                 mg          FN - mg = 0       →
         (5-8)               Indeed: FN = mg
                         What does the scale reads when the
y-axis
              elevator   elevator accelerates upwards ?
                         System = man
         ..              Fynet = FN - mg
                         Fynet = ma   (Newton’s second law)

                  FN     →    FN - mg = ma
                         →    FN = mg + ma

a                 mg          FN > mg
                         The scale in this case gives a reading
                         which is higher than the actual weight
          scale
                         mg                         (5-9)
                         What does the scale reads when the
y-axis
              elevator   elevator accelerates downwards ?
                         System = man
         ..              Fynet = FN - mg
                         Fynet = - ma   (Newton’s second law)

                  FN     →    FN - mg = - ma
                         →    FN = mg - ma

a                 mg          FN < mg
                         The scale in this case gives a reading
                         which is lower than the actual weight
          scale
                         mg                         (5-10)
                           Atwood’s Machine
        a
                    It consists of two masses m1 and
                a   m2 (m2 > m 1) connected with a
                  y
                    rope that goes around a pulley.
                    The rope and the pulley have
         m2         negligible masses.
              m1

The two masses are released and move under gravity which
acts downwards. The heavier mass m 2 will move down and
the lighter mass m1 will move up. Determine the
acceleration a of the two masses
                                          (5-11)
                         y                                          a
a                                                           T
                              T    a
           a
               y                                        y

                                  m1                            m2
                         m1g                             m2g
 m2
         m1
    System = m1                        System = m2
    Fynet = T − m1 g = m1a →           Fynet = T − m2 g = − m2a →
    T = m1g + m1a     eqs.1            T = m2 g − m2a       eqs.2

We combine eqs.1 and eqs.2 → m1 g + m1a = m2 g − m2 a Solve for a
                        ( m2 − m1 )
→                   a=g
                        ( m2 + m1 )                             (5-12)
Friction   Friction opposes motion or impending motion




                                               (5-13)
(5-14)
         motion               Friction was first studied by
                  FN          Leonardo da Vinci
   crate               T
                              He discovered that friction has the
        f              rope   following characteristics:
         floor    mg
• Friction opposes motion or impending motion
• Static and kinetic friction are independent of contact area
• The static and kinetic friction are proportional to FN
 Static friction fs                  0 < f S < µ S FN
µs is a constant known as “coefficient of static friction”
Kinetic friction fk                  f k = µ k FN
µk is a constant known as “coefficient of kinetic friction”
Leonardo da Vinci
(self portrait)
(1452-1519)
   motion                 Consider the following experiment:
             FN
                   T      We start pulling a heavy crate using
crate                     a rope with a force T that increases
     f            rope
                          with time. At some instant tm the
   floor     mg           crate will start to move

                                      f             µk < µs
We then plot the frictional force f
as function of time. The static                      µ s FN
friction increases with time. At
tm the static friction becomes                                    µ k FN
maximum (µs FN) and the crate             Static kinetic
                                          friction friction
starts to move. The kinetic                    tm
friction has a smaller constant                Motion         t
value of µk FN .           (5-15)              starts
                                  Example (5-6),page 121
                     y
                                  A mass m is at rest on a
                          x       horizontal ramp. The ramp is
                                  slowly raised. Find the critical
                                  angle θc at which the mass will
                                  start to slide

Fynet = FN - mg cos θ = ma y = 0 → FN = mg cos θ (eqs.1)
Fxnet = mg sin θ - f S = max = 0 → mg sin θ = f S = µ S FN (eqs.2)
From eqs.1 substitute FN into eqs.2 →
mg sin θ = µ S mg cos θ       →      tan θ C = µ S

                                                     (5-16)
Example (5-8) page 123. An automobile has a mass
m = 1000 kg. The static friction with the road is µs = 0.8
Determine the maximum forward acceleration (without
spinning the wheels)
Fynet = FN - mg = may = 0    →     FN = mg     (eqs.1)
Fxnet = fs = ma   →    a = fs /m
amax = (fs)max /m = µs FN /m (eqs.2)
From eqs.1 substitute FN into eqs.2          → amax = µsmg /m
amax = µs g = 0.8×9.8 = 7.8 m/s2
                                        When the rope breaks
                                        the book does not stay
                                        on its circular path but
                                        instead flies along the
                                        tangent to the orbit




Conclusion: For an object to be able to move on a circular
orbit a force that points towards the center of the circle is
required. This force is known as the “centripetal” force.
When the centripetal force stops acting the object cannot stay
on its circular orbit
                                                    (5-18)
(5-19)
         m                    Consider an object of mass m
                     m        moving on a circle of radius r
                              with uniform speed v. The
                              acceleration points towards the
                              center of the orbit and has
                              magnitude a = v2/r = ω2r
         m
  From Newton’s second law a force F = ma must act on the
  object . Thus the centripetal force F is given by the
  expression:
                       mv 2
                    F=          or     F = mω 2r
                        r
  The magnitude of F is constant, its direction is not. It
  always points towards the center of the circular orbit.
                           Recipe for problems that involve
                           uniform circular motion of an
         C   .   y
                  R        object of mass m on a circular
                           orbit of radius R with speed v


                      x
         m
                 v
• Draw the force diagram for the object
• Choose one of the coordinate axes (the y-axis in this
diagram) to point towards the orbit center C
• Determine Fynet
• Set Fynet = mv2/R or    Fynet = mω2R        (5-20)
                                   Example (5-11) page 128
                                   A small dice sits 0.15 m from
                                   the center of a turntable. If µs
                                   between the turntable and the
   (5-21)                          dice is 0.55 what is the largest
                       y
                                   possible angular speed such
                 x
                                   that the dice will not slide off


Fynet = FN - mg = may = 0 →          FN = mg     (eqs.1)
Fxnet = fs = mω2r (eqs.2) → ω2 = fs /mr
ω2max = (fs )max /mr       From eqs.1 we have:    (fs)max = µs FN
Thus    ω2max = µs mg/mr → ωmax = [µs g/r]1/2 = 6 rad/s
                         Example (5-12) page 129
                         A space station in the form of a hollow
                         circular tube rotates around its axis. If
                         the distance from the station outer wall
                         to the center is 50 m what should be
           y             the speed v of the outer wall so that a
                         scale will read what it would read on
                         earth
Fynet = FN = mv2/R      (eqs.1)
Also       FN = mg      (eqs.2)
If we combine the two equations we get: mv2/R = mg
→      v = [Rg]1/2 = [50×9.8]1/2 = 22.1 m/s
                                                      (5-22)