1. Foundations of Physics
CPO Science
Foundations of Physics
Chapter 9
Unit 2, Chapter 6

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
mass
weight
weightless
g-force
friction
static friction
sliding friction
rolling friction
viscous friction
air friction
normal force
extension
net force
free-body diagram
lubricant
equilibrium
ball bearing
dimension
spring
Hooke’s law
compression
spring constant
deformation
restoring force
coefficient of friction
engineering
design cycle
subscript
prototype
coefficient of static friction

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)
Fw = mg
Weight force (N)
Mass (kg)

8. 6.1 Free fall and weightlessness
An elevator is accelerating downward at 9.8 m/sec2.
The scale feels no force because it is falling away from your feet at the same rate you are falling.
As a result, you are weightless.

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 Jupiter’s atmosphere is 23 N/kg.
(Since Jupiter may not actually have a surface, “on” means at the top of the atmosphere.)
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:
Fw = (50 kg) × (23 N/kg)
= 1,150 N (259 lbs)

10. 6.1 Calculate force
1) You are asked to find force.
2) You are given a mass of 10 kilograms.
3) The force of the weight is: Fw = 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.
Fw = (10 kg) × (9.8 N/kg) = 98 N.
The tension force in the rope is 98 newtons.
A 10-kilogram ball is supported at the end of a rope. How much force (tension) is in the rope?

11. 6.1 Mass, Weight, and Gravity
Key Question:
What is speed and how is it measured?
*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. Coefficient of friction
Normal force (N)
Ff = m Fn
Friction force (N)
Coefficient of friction

16. 6.2 Calculate force of friction
1) You are asked for the force of friction (Ff).
2) You are given the weight (Fw), the applied force (F) and the coefficient of sliding friction (μ)
3) The normal force is the sum of forces pushing down on thefloor (Ff = μFn).
4) First, find the normal force:
Fn = 100 N + 10 N = 110 N
Use Ff = μFn to find the force of friction:
Ff = (0.25) x (110 N) = 27.5 N
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 Ff = msFn Normal force (N) Friction force (N)
Coefficient of
sliding friction

18. Table of friction coefficients

19. 6.2 Calculate using friction
A steel pot with a weight of 50 N sits on a steel countertop.
How much force does it take to start the pot sliding?
1) You are asked for the force to overcome static friction (Ff).
2) You are given the weight (Fw) and both surfaces are steel.
3) Ff = μsFn.
4) Ff = (0.74) x (50 N) = 37 N

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.
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, Ff = μFn, (Fw = 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: Ff = (0.07)(4,900 N) = 343 N.
The acceleration is the net force divided by the mass:
a = (1,000 N N) ÷ (500 kg) = (657 N) ÷ (500 kg) = 1.31 m/sec2

21. *Students read Section 6.2 AFTER Investigation 6.2
6.2 Friction
Key Question:
How can we describe and model friction?
*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.

26. 6.3 Calculate net force
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.
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.

31. 6.3 Hooke's Law F = - k x Deformation (m) Force (N)
Spring constant N/m

32. 6.3 Calculate force
A spring with k = 250 N/m is extended by one centimeter.
How much force does the spring exert?

33. 6.3 Equilibrium and Hooke's Law
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.

34. 6.3 Calculate using equilibrium
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 force.
2) You are given one of two forces and the mass.
3) Relationships that apply: net force = zero, Fw = mg and g = 9.8 N/kg.
4) The weight of the boat is Fw = 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.

35. 6.3 Equilibrium of Forces and Hooke's Law
Key Question:
How do you predict the force on a spring?
*Students read Section 6.3 AFTER Investigation 6.3

36. Application: The design of structures