CPO FP2nd Edition

1
(No Transcript)
2
Chapter 6 Forces and Equilibrium

• 6.1 Mass, Weight and Gravity
• 6.2 Friction
• 6.3 Equilibrium of Forces and Hookes Law

3
Chapter 6 Objectives

• Calculate the weight of an object using the
  strength of gravity (g) and mass.
• Describe the difference between mass and weight.
• Describe at least three processes that cause
  friction.
• Calculate the force of friction on an object when
  given the coefficient of friction and normal
  force.
• Calculate the acceleration of an object including
  the effect of friction.
• Draw a free-body diagram and solve
  one-dimensional equilibrium force problems.
• Calculate the force or deformation of a spring
  when given the spring constant and either of the
  other two variables.

4
Chapter 6 Vocabulary

• ball bearings
• coefficient of friction
• coefficient of static friction
• compressed
• deformation
• dimensions

• lubricant
• normal force
• prototype
• restoring force
• rolling friction
• sliding friction

• engineering
• engineering cycle
• extended
• free-body diagram
• g forces
• Hookes law

• spring
• spring constant
• static friction
• subscript
• viscous friction
• weightless

5
Inv 6.1 Mass versus Weight

• Investigation Key Question
• How are mass and weight related on Earth?

6
6.1 Mass, Weight, and Gravity

• Mass is a measure of matter.
• Mass is constant.
• Weight is a force.
• Weight is not constant.

7
6.1 Mass, Weight, and Gravity

• The weight of an object depends on the strength
  of gravity wherever the object is.
• The mass always stays the same.

8
6.1 Calculating weight with mass and gravity

• The weight of an object depends on its mass and
  the strength of gravity.
• The formula gives the weight (Fw) in terms of the
  mass of an object, m, and the strength of
  gravity, g.

9
6.1 Two meanings for g

• The symbol g stands for the acceleration of
  gravity in free fall, which is 9.8 m/s2.
• Another meaning for g is the strength of gravity,
  which is 9.8 N/kg.
• Sometimes it is more natural to discuss gravity
  in N/kg instead of m/s2 because objects may not
  be in motion but they still have weight.
• The two meanings for g are equivalent since a
  force of 9.8 N acting on a mass of 1 kg produces
  an acceleration of 9.8 m/s2.

10
6.1 Gravity, acceleration and weightlessness

• An object is weightless when it experiences no
  net force from gravity.
• If an elevator is accelerating downward at 9.8
  m/sec2, the scale in the elevator shows no force
  because it is falling away from your feet at the
  same rate you are falling.

11
6.1 Gravity, acceleration and weightlessness

• Airplane pilots and race car drivers often
  describe forces they feel from acceleration as g
  forces.
• These g forces are not really forces at all, but
  are created by inertia.
• Remember, inertia is resistance to being
  accelerated.

12
6.1 Using weight in physics problems

• Like other forces, weight is measured in newtons
  or pounds.
• Very often, weight problems involve equilibrium
  where forces are balanced.
• The other common type of weight problem involves
  other planets, or high altitudes, where the
  strength of gravity (g) is not the same as on
  Earths surface.

13
Calculating force required to hold up an object

• A 10-kilogram ball is supported at the end of a
  rope. How much force (tension) is in the rope?

• You are asked to find force.
• You are given a mass of 10 kilograms.
• The force of the weight is Fw mg and g 9.8
  N/kg.
• The word supported means the ball is hanging
  motionless at the end of the rope. That means the
  tension force in the rope is equal and opposite
  to the weight of the ball.
• Fw (10 kg) (9.8 N/kg) 98 N.
• The tension force in the rope is 98 newtons.

14
Calculating weight on Jupiter
How much would a person who weighs 490 N (110
lbs) on Earth weigh on Jupiter? Since Jupiter may
not have a surface, on means at the top of the
atmosphere. The value of g at the top of
Jupiters atmosphere is 23 N/kg.

• You are asked for the weight.
• You are given the weight on Earth and the
  strength of gravity on Jupiter.
• Use Fw mg.
• First, find the persons mass from weight on
  Earth
• m (490 N) (9.8 N/kg) 50 kg.
• Next, find the weight on Jupiter
• Fw (50 kg) (23 N/kg) 1,150 N (259 lbs)

15
Chapter 6 Forces and Equilibrium

• 6.1 Mass, Weight and Gravity
• 6.2 Friction
• 6.3 Equilibrium of Forces and Hookes Law

16
Inv 6.2 Friction

• Investigation Key Question
• What happens to the force of sliding friction as
  you add mass to a sled?

17
6.2 Friction

• Friction results from relative motion between
  objects.
• Friction is a resistive force.
• Describing friction as resistive means that it
  always works against the motion that produces it.

18
6.2 Types of Friction

• Static friction
• Sliding friction
• Rolling friction

19
6.2 Types of Friction

• Air friction
• Viscous friction

20
(No Transcript)
21
6.2 A model for friction

• No single model or formula can accurately
  describe the many processes that create friction.
  
• Some of the factors that affect friction include
  the type of material, the degree of roughness,
  and the presence of dirt or oil.
• Even friction between two identical surfaces
  changes as the surfaces are polished by sliding
  across each other.

22
6.2 A model for friction

• The coefficient of friction is a ratio of the
  strength of sliding friction between two surfaces
  compared to the force holding the surfaces
  together, called the normal force.
• The coefficient of friction is most often a
  number between zero and one.

23
6.2 Dry sliding friction
Normal force (N)
Ff m Fn
Friction force (N)
Coefficient of friction

• The symbol for coefficient of friction is the
  Greek letter µ.
• A coefficient of one means the force of friction
  is equal to the normal force.
• A coefficient of zero means there is no friction
  no matter how much force is applied to squeeze
  the surfaces together.

24
Calculate force of friction
A 10-N force pushes down on a box that weighs 100
N. As the box is pushed horizontally, the
coefficient of sliding friction is 0.25.
Determine the force of friction resisting the
motion.

• You are asked for the force of friction Ff.
• You are given weight Fw, applied force F, and
  coefficient of sliding friction µ.
• The normal force is the sum of forces pushing
  down on the floor, so use Ff µFn.
• First, find the normal force Fn 100 N 10
  N 110 N
• Use Ff µFn and substitute values Ff
  (0.25)(110 N) 27.5 N

25
6.2 Calculating the force of friction

• The normal force is the force perpendicular to
  two surfaces which are moving relative to each
  other.
• In many problems, the normal force is the
  reaction in an action-reaction pair.

26
6.2 Static friction

• It takes a certain minimum amount of force to
  make an object start sliding.
• The maximum net force that can be applied before
  an object starts sliding is called the force of
  static friction.

27
6.2 Static Friction
Normal force (N)
Ff msFn
Friction force (N)
Coefficient of sliding friction

• The coefficient of static friction ( µs) relates
  the maximum force of static friction to the
  normal force.
• It takes more force to break two surfaces loose
  than it does to keep them sliding once they are
  already moving.

28
6.2 Table of friction coefficients
29
Calculate the force of static friction
A steel pot with a weight of 50 sits on a steel
countertop. How much force does it take to start
to slide the pot?

1. You are asked for the force to overcome static
   friction Ff
2. You are given the weight Fw. Both surfaces are
   steel.
3. Use F f µs Fn
4. Substitute values F f (0.74) (50 N) 37
   N

30
6.2 Friction and motion

• When calculating the acceleration of an object,
  the F that appears in Newtons second law
  represents the net force.
• Since the net force includes all of the forces
  acting on an object, it also includes the force
  of friction.
• The real world is never friction-free, so any
  useful physics must incorporate friction into
  practical models of motion.

31
Calculating the acceleration of acar including
friction
The engine applies a forward force of 1,000
newtons to a 500-kilogram car. Find the
acceleration of the car if the coefficient of
rolling friction is 0.07.

1. You are asked for the acceleration a.
2. You are given the applied force F, the mass m,
   and the coefficient of rolling friction µ.
3. Use a F m, Ff µFn, Fw mg and g 9.8
   N/kg.

32
Calculating the acceleration of acar including
friction

• The normal force equals the weight of the car
• Fn mg (500 kg)(9.8 N/kg) 4,900 N.
• The friction force is Ff (0.07)(4,900 N) 343
  N.
• The acceleration is the net force divided by the
  mass
• a (1,000 N 343 N) 500 kg 657 N 500 kg
• a 1.31 m/s2

33
6.2 Reducing the force of friction

• Friction cannot be completely eliminated but it
  can be reduced.
• A fluid used to reduce friction is called a
  lubricant.
• In systems where there are axles, pulleys, and
  rotating objects, ball bearings are used to
  reduce friction.
• Another method of reducing friction is to
  separate two surfaces with a cushion of air.

34
6.2 Using friction

• There are many applications where friction is
  both useful and necessary.
• Friction between brake pads and the rim slows
  down a bicycle.
• All-weather tires have treads, patterns of deep
  grooves to channel water away from the road-tire
  contact point.
• Friction keeps nails and screws in place.
• Cleats greatly increase the friction between the
  sports shoe and the ground.

35
Chapter 6 Forces and Equilibrium

• 6.1 Mass, Weight and Gravity
• 6.2 Friction
• 6.3 Equilibrium of Forces and Hookes Law

36
Inv 6.3 Equilibrium of Forces and Hookes Law

• Investigation Key Question
• How do you predict the force on a spring?

37
6.3 Equilibrium and Hooke's Law

• When the net force acting on an object is zero,
  the forces on the object are balanced.
• We call this condition equilibrium.

38
6.3 Equilibrium and Hooke's Law

• A moving object continues to move with the same
  speed and direction.
• Newtons second law states that for an object to
  be in equilibrium, the net force, or the sum of
  the forces, has to be zero.

39
6.3 Equilibrium and Hooke's Law

• Acceleration results from a net force that is not
  equal to zero.

40
(No Transcript)
41
Calculating the net force from four forces
Four people are pulling on the same 200 kg box
with the forces shown. Calculate the acceleration
of the box.

• You are asked for acceleration.
• You are given mass and force.
• Use a F m.
• First add the forces to find the net force.
• F - 75N - 25N 45N 55N 0 N, so a 0

42
6.3 Free-body diagrams

• To keep track of the number and direction of all
  the forces in a system, it is useful to draw a
  free-body diagram.
• A free-body diagram makes it possible to focus on
  all forces and where they act

43
6.3 Free-body diagrams

• Forces due to weight or acceleration may be
  assumed to act directly on an object, often at
  its center.
• A reaction force is usually present at any point
  an object is in contact with another object or
  the floor.
• If a force comes out negative, it means the
  opposes another force.

44
6.3 Applications of equilibrium

• If an object is not moving, then you know it is
  in equilibrium and the net force must be zero.
• You know the total upward force from the cables
  must equal the downward force of the signs
  weight because the sign is in equilibrium.

What is the upward force in each cable?
45
Using equilibrium to find anunknown force
Two chains are used to lift a small boat. One of
the chains has a force of 600 newtons. Find the
force on the other chain if the mass of the boat
is 150 kilograms.

• You are asked for the force on one chain.
• You are given 2 forces and the mass
• Use net force zero, Fw mg and g 9.8 N/kg.
• Substitute values Fw mg (150 kg)(9.8 N/kg)
  1,470 N.
• Let F be the force in the other chain,
  equilibrium requires
• F (600 N) 1,470 N F 1,470 N 600 N
• So F 870 N.

46
6.3 Applications of equilibrium

• Real objects can move in three directions
  up-down, right-left, and front-back.
• The three directions are called three dimensions
  and usually given the names x, y, and z.
• When an object is in equilibrium, forces must
  balance separately in each of the x, y, and z
  dimensions.

47
6.3 The force from a spring

• A spring is a device designed to expand or
  contract, and thereby make forces in a controlled
  way.
• Springs are used in many devices to create force.
  
• There are springs holding up the wheels in a car,
  springs to close doors, and a spring in a toaster
  that pops up the toast.

48
6.3 The force from a spring

• The most common type of spring is a coil of metal
  or plastic that creates a force when it is
  extended (stretched) or compressed (squeezed).

49
6.3 The force from a spring

• The force from a spring has two important
  characteristics
• The force always acts in a direction that tries
  to return the spring to its unstretched shape.
• The strength of the force is proportional to the
  amount of extension or compression in the spring.

50
6.3 Restoring force and Hookes Law

• The force created by an extended or compressed
  spring is called a restoring force because it
  always acts in a direction to restore the spring
  to its natural length.
• The change a natural, unstretched length from
  extension or compression is called deformation.
• The relationship between the restoring force and
  deformation of a spring is given by the spring
  constant (k).

51
6.3 Restoring force and Hookes Law

• The relationship between force, spring constant,
  and deformation is called Hookes law.
• The spring constant has units of newtons per
  meter, abbreviated N/m.

52
6.3 Hooke's Law
Deformation (m)
F - k x
Force (N)
Spring constant N/m

• The negative sign indicates that positive
  deformation, or extension, creates a restoring
  force in the opposite direction.

53
Calculate the force from a spring
A spring with k 250 N/m is extended by one
centimeter. How much force does the spring exert?

1. You are asked for force.
2. You are given k and x.
3. Use F - kx
4. Substitute values F - (250 N/m)(0.01 m) F
   - 2.5 N

54
6.3 More about action-reaction and normal forces

• The restoring force from a wall is always exactly
  equal and opposite to the force you apply,
  because it is caused by the deformation resulting
  from the force you apply.

55
Calculate the restoring force
The spring constant for a piece of solid wood is
1 108 N/m. Use Hookes law to calculate
the deformation when a force of 500 N (112 lbs)
is applied.

1. You are asked for the deformation, x.
2. You are given force, F and spring constant, k.
3. Use F - kx, so x - F k
4. Substitute values x - (500 N/m) (1 108
   N/m)
5. x - 5 10-6 meters (a very small deformation)

56
The Design of Structures

• We are surrounded by structures.
• To design a structure well, you first need to
  know what forces act and how, and where the
  forces are applied.
• Engineering is the application of science to
  solving real-life problems, such as designing a
  bridge.