2. Unit 2: Motion and Force in
One Dimension
Chapter 6: Forces and Equilibrium
6.1 Mass, Weight and Gravity
6.2 Friction
6.3 Equilibrium of Forces and Hooke’s
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 Terms
mas s n o rmal c o m p r e s s io n
w e ig h t force s p r in g
w e ig h t l e s s e x t e n s io n c o n s tan t
g -f o r c e net force d e f o r m a t io n
f r ic t io n f r e e -b o d y r e s t o r in g
d ia g r a m force
s t a t ic
f r ic t io n l u b r ic a n t c o e f f ic ie n t
s l id in g e q u il ib r iu m o f f r ic t io n
f r ic t io n b a l l b e a r in g e n g in e e r in g
r o l l in g d im e n s io n d e s ig n c y c l e
f r ic t io n s p r in g s u b s c r ip t
v is c o u s prototype
5. 6.1 Mass, Weight, and Gravity
Mass is a measure of
matter.
Mass is constant.
Weight is a force.
Weight is not constant.
6. 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.
7. 6.1 Weight
Gravity (9.8 m/sec2)
Weight force (N) Fw = mg
Mass (kg)
8. 6.1 Free fall and weightlessness
An e l e v a t o r is a c c e l e r a t in g
d o w n w a r d a t 9 .8 m /s e c 2 .
T h e s c a l e f e e l s n o f o r c e b e c a u s e it is
f a l l in g a w a y f r o m y o u r f e e t a t t h e
s a m e r a t e y o u a r e f a l l in g .
As a r e s u l t , y o u a r e w e ig h t l e s s .
9. 6.1 Calculate weight
How much would a person
who weighs 490 N ( 110
lbs) on Earth weigh on
Jupiter?
The value of g at the top of
J upiter’s atmosphere is 23
N kg.
/
(Since J upiter may not
actually have a surface,
“on” means at the top of
the atmosphere.)
10. 6.1 Calculate force
A 10-kilogram ball is supported at
the end of a rope. H much force
ow
(tension) is in the rope?
11. 6.1 Mass, Weight, and Gravity
Key Question:
W h a t is s p e e d a n d h o w is it
meas ured ?
*Students read Section 6.1 BEFORE Investigation 6.1
12. 6.2 Friction
Friction results from relative motion
between objects.
Frictional forces are forces that resist or
oppose motion.
13. 6.2 Types of Friction
Static friction
Sliding friction
Rolling friction
14. 6.2 Types of Friction
Air friction
Viscous friction
15. 6.2 Friction
Normal force (N)
Friction force (N) Ff = µ Fn
Coefficient of friction
16. 6.2 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.
17. 6.2 Sliding Friction
Normal force (N)
Friction force (N) Ff = µ sFn
Coefficient of
sliding friction
19. 6.2 Calculate using friction
A s t e e l p o t w it h a w e ig h t o f 5 0 N s it s
on a s teel countertop.
H o w m u c h f o r c e d o e s it t a k e t o s t a r t
t h e p o t s l id in g ?
20. 6.2 Calculate using friction
The engine applies a forward force of
1,000 newtons to a 500-kg car.
Find the acceleration of the car if the
coefficient of rolling friction is 0.07.
21. 6.2 Friction
Key Question:
H o w c a n w e d e s c r ib e a n d m o d e l
f r ic t io n ?
*Students read Section 6.2 AFTER Investigation 6.2
22. 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.
23. 6.3 Equilibrium and Hooke's Law
Newton’s second law simply requires that for an object to be in
equilibrium, the net force, or the sum of the forces, has to be
zero.
24. 6.3 Equilibrium and Hooke's Law
Many problems have more than one force applied to an
object in more than one place.
25.
26. 6.3 Calculate net force
Four people are pulling on the same 200 kg box
with the forces shown.
Calculate the acceleration of the box.
27. 6.3 Calculate force using equilibrium
Two chains are used to lift a
small boat. One of the chains has
a force of 600 newtons.
Find the force in the other chain if
the mass of the boat is 150
kilograms.
28. 6.3 Equilibrium and Hooke's Law
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) .
29. 6.3 Equilibrium and Hooke's Law
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.
30.
31. 6.3 Hooke's Law
Deformation (m)
Force (N) F=-kx
Spring constant N/m
32. 6.3 Calculate force
A spring with k = 250 N m is extended by one
/
centimeter.
H much force does the spring exert?
ow
33. 6.3 Equilibrium and Hooke's Law
T h e r e s t o r in g
force from a
w a l l is a l w a y s
exactly eq ual
a n d o p p o s it e t o
th e fo rce yo u
a p p l y , b e c a u s e it
is c a u s e d b y t h e
d e f o r m a t io n
r e s u l t in g f r o m
34. 6.3 Calculate using equilibrium
The spring constant for a piece of solid wood is 1 × 10 8
N/m.
Use Hooke’s law to calculate the deformation when a
force of 500 N ( 112 lbs) is applied.
35. 6.3 Equilibrium of Forces and Hooke's Law
Key Question:
Ho w d o yo u
p r e d ic t t h e
force on a
s p r in g ?
*Students read Section 6.3 AFTER Investigation 6.3
1) You are asked for the weight. 2) You are given the weight on Earth and the strength of gravity on Jupiter. 3) Fw = mg. 4) First, find the person’s mass from the weight and strength of gravity on Earth: m = (490 N) ÷ (9.8 N/kg) = 50 kg Next, find the weight on Jupiter: F w = (50 kg) × (23 N/kg) = 1,150 N (259 lbs)
1) You are asked to find force. 2) You are given a mass of 10 kilograms. 3) The force of the weight is: F w = mg and g = 9.8 N/kg. 4) 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. F w = (10 kg) × (9.8 N/kg) = 98 N. The tension force in the rope is 98 newtons.
1) You are asked for the force of friction (F f ). 2) You are given the weight (F w ), the applied force (F) and the coefficient of sliding friction (μ) 3) The normal force is the sum of forces pushing down on thefloor (F f = μFn). 4) First, find the normal force: F n = 100 N + 10 N = 110 N Use F f = μFn to find the force of friction: F f = (0.25) x (110 N) = 27.5 N
1) You are asked for the force to overcome static friction (F f ). 2) You are given the weight (F w ) and both surfaces are steel. 3) F f = μsF n . 4) F f = (0.74) x (50 N) = 37 N
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) Relationships that apply: a = F ÷ m, F f = μF n , (F w = mg and g = 9.8 N/kg). 4) The normal force equals the weight of the car: Fn = mg = (500 kg)(9.8 N/kg) = 4,900 N. The friction force is: F f = (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) = 1.31 m/sec 2
1) You are asked for acceleration. 2) You are given mass and force. 3) a = F ÷ m. 4) F = -75N - 25N +45N +55N = 0 N, so a = 0.
1) You are asked for the force. 2) You are given one of two forces and the mass. 3) Relationships that apply: net force = zero, F w = mg and g = 9.8 N/kg. 4) The weight of the boat is F w = mg = (150 kg) (9.8 N/kg) = 1,470 N. Let F be the force in the other chain, The condition of equilibrium requires that: F + (600 N) = 1,470 N, therefore F = 870 N.