Chapter 6: Forces and Equilibrium
 6.1 Mass, Weight and Gravity
 6.2 Friction
 6.3 Equilibrium of Forces and Hooke’s
Law
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.
Chapter 6 Vocabulary
 ball bearings
 engineering
 lubricant
 coefficient of
friction
 engineering
cycle
 normal force  spring
constant
 extended
 prototype
 static friction
 free-body
diagram
 restoring
force
 subscript
 g forces
 rolling
friction
 weightless
 coefficient of
static friction
 compressed
 deformation
 dimensions
 Hooke’s law
 sliding
friction
 spring
 viscous friction
Inv 6.1 Mass versus Weight
Investigation Key Question:
How are mass and weight related on Earth?
6.1 Mass, Weight, and Gravity
 Mass is a measure of
matter.
 Mass is constant.
 Weight is a force.
 Weight is not
constant.
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.
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.
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.
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.
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.
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
Earth’s surface.
Calculating force required
to hold up an object

1.
2.
3.
4.
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.
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 Jupiter’s atmosphere is 23 N/kg.
1.
2.
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 person’s mass from weight on Earth:
3.
4.



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)
Chapter 6: Forces and Equilibrium
 6.1 Mass, Weight and Gravity
 6.2 Friction
 6.3 Equilibrium of Forces and Hooke’s
Law
Inv 6.2 Friction
Investigation Key Question:
What happens to the force of sliding friction as you
add mass to a sled?
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.
6.2 Types of Friction
 Static friction
 Sliding friction
 Rolling friction
6.2 Types of Friction
 Air friction
 Viscous friction
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.
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.
6.2 Dry sliding friction
Friction force (N)
Ff = m Fn
Normal 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.
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.
1.
2.
You are asked for the force of friction Ff.
You are given weight Fw, applied force F, and coefficient of sliding
friction μ.
3.
The normal force is the sum of forces pushing down on the floor, so use
Ff = μFn.
4.
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
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.
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.
6.2 Static Friction
Friction force (N)
Ff = msFn
Normal 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.
6.2 Table of friction coefficients
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
6.2 Friction and motion
 When calculating the acceleration of an object,
the F that appears in Newton’s 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.
Calculating the acceleration of a
car 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.
2.
3.
You are asked for the acceleration a.
You are given the applied force F, the mass m, and the
coefficient of rolling friction μ.
Use: a = F ÷ m, Ff = μFn, Fw = mg and g = 9.8 N/kg.
Calculating the acceleration of a
car including friction
4.
5.
6.
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
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.
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.
Chapter 6: Forces and Equilibrium
 6.1 Mass, Weight and Gravity
 6.2 Friction
 6.3 Equilibrium of Forces and Hooke’s
Law
Inv 6.3 Equilibrium of Forces and
Hooke’s Law
Investigation Key Question:
How do you predict the force on a spring?
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.
6.3 Equilibrium and Hooke's Law
 A moving object continues to move with the same speed
and direction.
 Newton’s second law states that for an object to be in
equilibrium, the net force, or the sum of the forces, has
to be zero.
6.3 Equilibrium and Hooke's Law
 Acceleration results from a net force that is
not equal to zero.
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.
1. You are asked for acceleration.
2. You are given mass and force.
3. Use a = F ÷ m.
4. First add the forces to find the net force.
 F = - 75N - 25N + 45N + 55N = 0 N, so a = 0
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
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.
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.
What is the upward
force in each cable?
 You know the total upward
force from the cables must
equal the downward force of
the sign’s weight because
the sign is in equilibrium.
Using equilibrium to find an
unknown 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.
1.
2.
3.
4.
5.
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.
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.
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.
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).
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.
6.3 Restoring force and Hooke’s 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).
6.3 Restoring force and Hooke’s Law
 The relationship between force, spring constant,
and deformation is called Hooke’s law.
 The spring constant has units of newtons per
meter, abbreviated N/m.
6.3 Hooke's Law
Force (N)
F=-kx
Deformation (m)
Spring constant N/m
 The negative sign indicates that positive
deformation, or extension, creates a restoring
force in the opposite direction.
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
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.
Calculate the restoring force
The spring constant for a piece of solid wood is 1 × 108
N/m. Use Hooke’s 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)
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.