Newton’s Laws
Apologia Physics, Module 5
Newton’s First Law
 An object in motion (or at rest) will tend to stay in motion (or at rest)
until it is acted upon by an outside force.
 This is also called the law of inertia
 Friction – A force that opposes motion, resulting from the contact of
two surfaces.
Inertia Lab Stations
Newton’s Second Law
 When an object is acted on by one or more outside forces, the
vector sum of those forces is equal to the mass of the object times
the resulting vector.
 ∑F = ma where both the sum of the forces and the acceleration are
vectors.
 A force is a push or pull exerted on an object in an effort to change that
objects velocity.
 The unit for force is the Newton (N) which is defined as
𝑘𝑔∙𝑚
𝑠2
OYO 5.3
A man is pushing a car that has run out of gas. If
he pushes with a force of 15.0N west, and the car
has a mass of 1.23x106 grams, what will the car’s
acceleration be?
OYO 5.3
 A man is pushing a car that has run out of gas. If he pushes with a force
of 15.0N west, and the car has a mass of 1.23x106 grams, what will the
car’s acceleration be?
 Our first step is always to make sure our units match. A Newton is
have to convert our grams to kg.
 ∑F = ma
15.0N = (1,230kg)a
a = 0.012 m/s2
𝑘𝑔∙𝑚
𝑠2
so we
Mass and Weight
 Mass- The amount of matter in an object (Scaler)
 Units- slugs, grams (kg,mg,cg,etc)
 Weight- The measure of the gravity pulling on an object (Vector)
 Weight is a vector quantity because it is a measure of force (gravity)
 w = mg (mass x gravity)
 Units- Newtons
𝑘𝑔∙𝑚
), dynes, pounds
𝑠2
(
OYO- 5.5
 What is the weight (on earth) of a 13.5 gram object?
 First we see that we have a unit problem. We are looking for Newtons but we
are given grams instead of kg. So 13.5g = 0.0135kg.
 w=mg
 w= (0.0135kg)(9.8 m/sec2) = 0.13 N
OYO- 5.6
 A Martian rock weights 15 N on Mars. How much will it weigh on
earth?
 This is a two part problem. First we need to find out the mass of the rock by
removing the gravity on Mars.
 15N = m (3.7 m/s2) so m= 4.1kg
 Now we can use the mass to find the weight on Earth.
 w = (4.1kg)(9.8 m/s2) = 40 N or 4.0x101 N (correct scientific notation)
The Normal Force
 Definition- A force that results from the contact of two bodies and is
perpendicular to the surface of contact.
 Gravity pulls down on a man so the force of gravity is ∑f = (mass of the man) (9.8m/s2)
 Because the man does not fall through the earth, there must be a force that is
pushing up on the man opposite of the force of gravity. ∑f = (mass of the man)
(9.8m/s2 (force of the floor)). This is the normal force and it is opposite of the
gravitational force.
 To calculate the normal force, we calculate the gravitational force and change the
sign.
Friction
 Friction is a result of the molecules of two surfaces being attracted to each other.
 The more the attraction (the closer they can get to each other) the greater the friction.
 The more the friction, the more the resistance to motion.
 There are two types of friction
 Kinetic friction – The friction that opposed motion once the motion has already started
 Static friction – The friction that opposes the initiation of motion
Lab 5.2
Lab 5.2 – Key Points
 Static Friction can vary to counteract the force that is attempting to move an object, but
there is a maximum value for static friction. Once the force applied is greater than the
maximum value of the static friction, the object will move.
 We saw this in the experiment during step 8.
 The Kinetic Frictional Force is always less the maximum static frictional force.
 We saw this because the rubber band was stretched less when the tub was moving than when
the motion started.
 The frictional force is dependent of the molecules that make up the surfaces that are in
contact.
 This point was shown when we used the soap and the aluminum foil on the board.
An Equation for the Frictional Force
 Like almost everything else we have studied in Physics, frictional force has an
equation.
 We know that amount of friction depends on the surface of the object and the normal force.
 When you put those together you get the equation – f
= μFn
 f represents the magnitude of the frictional force, μ is the coefficient of friction and Fn is the normal force.
 You will notice that even though the frictional force and the normal force are vectors, we only have the
magnitude in the equation. We already know the direction because the frictional force always opposes
motion.
 The coefficient of friction μ changes depending on if we are talking about static friction or kinetic friction.
 We can only know the coefficient of friction if we determine it experimentally or if we are given it in a problem.
We do know that the coefficient of static friction is greater than the coefficient of kinetic friction for any given
object.
Newton’s 2nd Law Problems
 The main thing to remember when solving Newton’s 2nd Law Problems is to be sure to
account for all the forces involved. (Draw a picture)
 To find the frictional force  First we have to account for gravity and determine the weight of the object being moved. W=mg
 Once we have the weight, we can determine the normal force because is the opposite of the
weight.
 Now we can use f = μFn to determine the frictional force. In order to know which coefficient to
use, we need to determine what kind of friction we are calculating.
 To find the total force The equation for total force is F-f = ma. The total force minus the frictional force is equal to
the mass x the acceleration.
OYO 5.8

A man is pushing on a 12.0 slug rock. What is the maximum
frictional force that will keep the rock from moving? (μs = 0.65 and
μk = 0.33)
OYO 5.8

A man is pushing on a 12.0 slug rock. What is the maximum frictional force that will keep
the rock from moving? (μs = 0.65 and μk = 0.33)
 To determine the frictional force, we need to know the normal force. We
determine the normal force by finding the weight.
 W= (12.o slugs)(32 ft/s2) = 380 pounds
 Now that we have the weight, we have the normal force and can find our
frictional force. Remember to determine which frictional force we are
talking about so that we know which coefficient to use. In this case, we
are getting the object to move so we are talking about static friction
 f = (o.65)(380 pounds) = 250 pounds
OYO- 5.10
 A man pushes a 6,675N rock with a force of 95N. If the rock is
already moving, what will the acceleration be? (μs = 0.15 and μk
= 0.011)
OYO- 5.10
 A man pushes a 6,675N rock with a force of 95N. If the rock is already moving, what will
the acceleration be? (μs = 0.15 and μk = 0.011)
 Looking at the problem, we can see that they gave us the weight so we
know that we have the normal force. Because the rock is already moving
we will use μk.
 First we need to find the frictional force (f).
 f = (0.011)(6675N)= 73N
 Next we need to find the mass of the rock.(w=mg)
 6,675N = m (9.8 m/s2) so m=680kg
 Now we have enough information to use our force equation. F-f=ma
 95N- 73N = (680kg) a so a= 0.032 m/s2
OYO- 5.11
 A child pushes a 2.3kg toy along the floor at a constant velocity.
What force is the child exerting? (μs = 0.25 and μk = 0.13)
OYO- 5.11

A child pushes a 2.3kg toy along the floor at a constant velocity. What force is the child exerting? (μs = 0.25 and μk
= 0.13)
 The first thing we need to realize is that anything at constant velocity has no
acceleration. So Newton’s 2nd Law says that the sum of the forces is equal to the
mass times the acceleration. If the acceleration is 0 than the sum of the forces
has to be zero.
 Also, the object is already in motion so we are talking about the kinetic friction.
 To get the normal force, we need the weight of the toy.
 W= (2.3kg)(9.8 m/s2) = 23N
 Now we can find the frictional force.
 f = (0.13)(23N) = 3.0N
 Finally we can use F-f = ma
 F-3.0N = 0
 F= 3.0N
Newton’s Third Law
 For every action, there is an equal and opposite reaction.
 Just because the forces are equal does not mean that the masses are
equal.
 The opposite but equal forces are on two objects.
If it was just one object there would be no motion