Applications of Newtons Laws Walker, Chapter 6

1
Applications of Newtons LawsWalker, Chapter 6

• Phys 111
• Prof. Charles E. Hyde-Wright
• Fall 2005

2
Manipulating Fractions
r
y
q
x
3
Macroscopic Forces

• Static and Kinetic Friction
• Force perpendicular (Normal) to a surface
• String Tension
• Spring (harmonic) forces
• Viscous Friction

4
Friction

• As a block slides on the table, the force from
  the surface of the table acting on the bottom of
  the block has components both perpendicular to
  the surface, and parallel to the surface.
• The component perpendicular to the surface we
  call the Normal force, N.
• The component parallel to the surface is friction.

5
Approximate Model of Force of Kinetic (Sliding)
Friction (Fk)

• Force of friction Fk points in direction opposite
  to velocity v (parallel to surface).
• Force of friction Fk acting on object m is
  proportional to Normal force N also acting on m
  at the same surface Fk mk N
• N is sometimes but often not equal in magnitude
  to force of gravity
• Fk independent of area of contact and
  velocity v

• Coefficient mk
• Depends on materials, surface conditions
• Independent of v or N

6
Motion on an Incline Kinetic Friction
y
x

• Cart rolls down incline
• Friction in bearings is equivalent to kinetic
  friction of block sliding on incline.
• If the angle of the incline is q10º and the
  coefficient of kinetic friction is mk 0.05,
• Find the acceleration of the cart as it rolls up
  the incline.
• Find the acceleration of the cart as it rolls
  down the incline.
• Notice N is perpendicular to incline and Fk is
  parallel to incline.

q
Free Body Diagram
v
Fk
N
mg
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Motion on an Incline Kinetic Friction, Sliding
uphill
Fk
N

• Draw all force vectors with a common origin.
• Find all components of vectors
• Nx 0, NyN
• Wx mg cos(270-q) -mg sinq lt 0
• Wy mg sin(270-q) -mg cosq lt 0
• Fk,x - mk N, Fk,y 0
• Apply Fma to each component
• Y N mg sin(270-q) m ay 0
• . N -mg sin(270-q) mg cosq
• X - mk N- mg sinq m ax
• Combine solve
• - mk mg cosq - mg sinq m ax
• ax - mk g cosq g sinq

mg
y
N
x
q
Fk
q
Wmg
Bigger than without friction
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Motion on an Incline Kinetic Friction, Sliding
downhill
v
Fk
N

• Find all components of vectors
• Nx 0, NyN
• Wx mg cos(270-q) -mg sinq lt 0
• Wy mg sin(270-q) -mg cosq lt 0
• Fk,x mk N, Fk,y 0
• Apply Fma to each component
• Y N - mg sinq m ay 0
• . N mg cosq
• X mk N- mg sinq m ax
• Combine solve
• mk mg cosq - mg sinq m ax
• ax - g sinq - mk g cosq

mg
y
N
x
Fk
q
q
Wmg
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Motion down an inclineKinetic Friction

• ax - g sinq - mk g cosq - g sinq - mk
  cosq
• Friction slows the downward acceleration for
  downward motion.
• q 10, mk 0.05
• ax - g sin 10 - 0.05 cos 10 - g 0.174 -
  0.0492
• ax - g0.124
• What if we make mk larger?
• mk sinq /cosq tan q
• ax 0. Constant downward velocity
• mk gt sinq /cosq
• ax gt 0 (as v lt 0), Motion slows to stop, even
  though started downhill.

10
Cart rolling on inclined track, magnetic bounce
at end
x(t)
vx(t) positive means downhill
gsinq mk cosq
ax(t)
g sinq - mk cosq
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Static Friction (v0)
P

• In spite of the push force P, the block does not
  move (v0, a0)
• Static friction does not have a fixed magnitude
  or direction (but in the absence of glue, static
  friction must be parallel to surface).
• Static friction takes whatever value necessary to
  keep a0 via Fnet 0
• But static friction has a maximum value Fs ? mS N
• Due to a natural welding action, generally mS gt
  mk .

Fs
mg
N
12
Friction Driving

• In ordinary driving, is the friction between tire
  and road static or kinetic?
• In a skid, is the friction between tire and road
  static or kinetic?
• How can anti-lock brakes stop a car in less
  distance than just slamming on the brakes?

13
Friction and Driving

• In ordinary driving
• The friction between tire and road is static
  friction.
• The friction between tire and road is kinetic
  friction.

14
Friction and Driving

• In a skid
• The friction between tire and road is static
  friction.
• The friction between tire and road is kinetic
  friction.

15
Friction Driving

• How can anti-lock brakes stop a car in less
  distance than just slamming on the brakes?
• They dont, its just a scam to make you feel
  safe.
• Anti-lock brakes are more powerful than your foot
• Since ms gt mk, anti-lock brakes make sure you
  have the maximum braking force without actually
  skidding.

16
Strings(application of Action-Reaction)

• A String is floppy, it cannot push either
  straight or sideways. It can only pull in
  tension.
• A String is a chain of tiny masses
• Consider Fma for the middle segment.
• T1 and T2 are the forces on the mass segment from
  the adjacent segments to left and right.
• If the mass m of each segment is small enough,
  then ma0 even if a?0. Therefore Fnet 0, and
  T1T2.
• By action-reaction, the segment pulls back on its
  neighbors with equal and opposite forces.

T1
T2
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Tension
T1

• Free Body Diagram for the Box.
• The Earth pulls down with force mg
• The string pulls up with tension T1
• If a0, then T1-mg0.
• By Action-Reaction,
• Tension T2 T1
• T3 T2 T1
• The rope pulls down on the upper hook with
  tension T T1
• If a string pulls at one end with tension T, it
  pulls (in the opposite direction) at the other
  end with the same tension T.

mg
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Ideal Pulley
Force on person

• Pulley spins freely without friction, neglect
  (rotational) inertia (mass) of pulley.
• Pulley changes direction, not magnitude, of
  tension.

Force On Box
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Each Spring-Scale reads 9.81 N !!!
Each Scale has two forces of 9.81 N acting in
opposite directions on its two ends.
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Tension, and Forces on Pulleys

• Free body diagram of pulley.

Roof holds mg
Roof holds 2mg
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Connected Objects

• Tension replaces contact force from example in
  chapter 5 notes.
• Box 2 Tm2a
• Box 1 F-Tm1a (same acceleration)
• Add the equations F0(m2m1)a a F/(m2m1)
• Find the tension T F m2/(m2m1)

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Atwood Machine
Find the acceleration a, Tension, 2T, in support
rope.
2T
Expect a gt 0, if m2 gt m1 Expect a ltlt g,
if m2 ? m1
T
a
m1
m2
a
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Atwood Machine,Free Body Diagrams

• T - m1g m1a
• T - m2g - m2a (note sign)
• Get rid of T by subtracting
• T-m1g - T - m2g (m1m2)a
• (m2 - m1)g (m1m2)a
• a g (m2 - m1) / (m1m2) lt g
• a lt 0 if (m2 - m1)lt0
• Plug back in to get T
• T m1(ga) 2g m1 m2 / (m1m2)
• Note if m1 m2 ,
• then a 0 and T m1g

T
T
a
a
m1g
m2g
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Walker Problem 4, pg. 165
When you push a 1.80 kg book resting on a
tabletop it takes 2.25 N to start the book
sliding. Once it is sliding, however, it takes
only 1.50 N to keep the book moving with constant
speed. What are the coefficients of static and
kinetic friction between the book and the
tabletop?
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Two blocks stacked on the floor.
M1

• Fill in the table describing which external
  object is responsible for each force acting on
  mass M2.
• (hint, gravity is not caused by the floor).

M2
M1
M1g
N
M1
M2
N1
Earth
M2

• Which force acting on mass M1 is the
  action-reaction partner of the force N1 acting on
  M2?

M2g
N
N2
Phys111 Fall 2003
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(Ideal) Springs

• An ideal spring exerts a force proportional to
  the springs displacement from its equilibrium
  position. The magnitude of the force is given by
  Hookes Law
• Fx ?kx
• k is the spring constant (units of
  Force/length N/mkgm/s2)
• The direction of the force is toward the
  equilibrium position. The force F is referred to,
  therefore, as a restoring force.

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Ideal Non-ideal Springs

• Hookes law F- k x is a good approximation, not
  just for springs, but for the stretch of
  non-ideal wire in tension, the bending of a steel
  beam, atomic, molecular, and nuclear compression
  as well.
• For many springs we see in real life, however,
• F- k x for xgt0,
• but F is almost infinite if xlt0. (stretches,
  but does not compress)
• In this case, just stretch the spring by an
  amount x0
• F F0 - k (x-x0) F0 - k x0 F
• Now we have an ideal Hookes law spring for
  displacements around the point x0.

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Walker Problem 23, pg. 166
F-kxkx F-Nmax0 FS-mg may
0 mN-mg0 mF-mg0 mkx-mg0 xmg/(mk) x (0.27
kg)(9.8 m/s2) /0.46(120 N/m) x 0.048 N/ (N/m)
0.048 m
x
As drawn, block will slip
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Translational Equilibrium
When the sum of the forces on an object is zero,
it is in translational equilibrium. It must
have zero acceleration.
Example A 5.0 kg mass is raised at a constant
speed of 6 m/s using a rope. What is the tension
in the rope? Notice that Fnet is not a new
force. It is just the sum of the existing
forces. ma is also not a force, it is a
consequence of the forces. The 5kg mass is raised
with a constant acceleration of 1.0 m/s2. What
is the tension in the rope?
Fnet T - mg Fnet ma 0 Tmg T
(5.0kg)(9.8m/s2) T 49 N
T
Fnet T-mg Fnet ma gt 0 maT - mg
Tmgma T (5.0kg)(9.8m/s2 1.0m/ s2) T 54 N
mg
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Walker Problem 30, pg. 167
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Walker Problem 40, pg. 169
A 3.50-kg block on a smooth (frictionless)
tabletop is attached by a string to a hanging
block of mass 2.80 kg, as shown in the Figure.
Which statement is true?

• The 2.80 kg block accelerates downward and the
  3.50 kg block remains motionless
• Both blocks remain motionless
• Both blocks move at equal constant speed
• The 3.50 kg block accelerates to the right and
  the 2.80 kg block moves with equal downward
  acceleration.

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Walker Problem 40, pg. 169
A 3.50-kg block on a smooth (frictionless)
tabletop is attached by a string to a hanging
block of mass 2.80 kg, as shown in the Figure.
The blocks are released from rest and allowed to
move freely.

• Is the tension in the string
• greater than,
• less than, or
• equal to the weight of the hanging mass?
• Hint is the net force on the mass up, down, or
  zero?

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Walker Problem 40, pg. 169

• A 3.50-kg block on a smooth (frictionless)
  tabletop is attached by a string to a hanging
  block of mass 2.80 kg, as shown in the Figure.
  The blocks are released from rest and allowed to
  move freely.
• Find the acceleration of the blocks and the
  tension in the string.

TMa N-Mg0
-Tmgma
mgmaT(mM)a a mg/(mM)
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Circular Motion
Consider an object moving at constant speed in a
circle. The direction of motion is changing, so
the velocity is changing (even though speed is
constant). Therefore, the object is
accelerating. The direction of the acceleration
is toward the center of the circle and so we call
it centripetal acceleration. The magnitude of
the acceleration is
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Centripetal Force

• A string cannot push sideways or lengthwise.
• A string in tension only pulls.
• The string pulls the ball inward toward the
  center of the circle
• Centripetal force

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Centripetal Acceleration

• The best estimate of the acceleration at P is
  found by calculating the average acceleration for
  the symmetric interval 1?2.

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Centripetal Force

• If there is a centripetal acceleration, then the
  net force must also be a centripetal force

• What is the centripetal force responsible for the
  following circular motion?
• A car driving in a circle
• A ball twirled in a circle on the end of a string
• Clothes moving in a circle during the spin cycle
  on a washing machine

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Walker Problem 50, pg. 169
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Walker Problem 71, pg. 171
40
Banked CurvesIf cars travel around a banked
curve at speed v, what is the optimum angle such
that cars will not skid even if rain/ice reduces
friction almost to zero?
?v
av2/r
q
y
x
N
mg
Free body diagram Without friction
Independent of mass of car