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Newton’s 3 Laws
of Motion
1. If
∑ F=
0 → No change in motion

2. F=
ma → Change in motion
net
3.


F1 on 2 = − F2 on 1
Newton’s First Law
(Law of Inertia)
An object will remain at rest or in
a constant state of motion unless
acted upon by external net forces.
Statics Problem
+y
Find the tensions in the wires.
+x
T2
T1
43
55
W
Newton’s 2nd Law
∑ F = ma
Fnet
a=
m
The acceleration of object is directly
related to the net forces acting on it and
inversely proportional to its mass.
Problem
Starting from rest, Sally pulls Billy on the
sled (total mass = 60kg) with a total force
of 100 N at an angle of 40 degrees above
the horizontal, as shown. After 5 seconds,
how fast is the sled moving and how far
has it traveled from where it started?
∑F
x
= max
F cos θ = max
F cos θ
v=
v0 + at
f
ax =
m
F cos θ
100 N cos 40
v f = v0 + at =
t=
5s = 6.38m / s
m
60kg
Frictional Forces
Friction always opposes the applied force and is in
the opposite direction of motion. The greater the
normal force the greater the frictional force.
f s = µs N
fk = µk N
Statice vs Kinetic Friction
fs > fk
Fig. 5.16, p.131
Problem
The magnitude of F1 is 75.0N and F2 is 50.0N. The
coefficient of friction between the block and the floor is 0.04.
What is the acceleration of the block?
a) First of all, which way is the box going to go? You have to figure that out first. To do that
compare the x-component of F2 to the x-component of F1. Which is larger?
F1x = F1 cos θ = 75 N cos 65= 31.7 N < 50 N = F2
So the box is going to end up moving to the left!!! So I’m going to make LEFT the positive x
direction in my FBD. Also, since F1 is pushing down, the normal force is going to be larger than
the weight!!! F2 should be longer than F1x too.
Apply Newton’s Second Law:
=
ax
F
∑
=
F2 − F1x 50 N − 31.7 N
=
= 3.66m / s 2
5kg
m
x
m
b) Now there is friction which will slow it down. You have to know which direction the
box is going without friction first because friction ALWAYS acts in the opposite direction to slow
it down. You can draw the friction vector at the end of F1x or from the center, your choice. The
frictional force is proportional to the normal force. To find that, apply Newton’s second law in the
y direction:
∑F
y
=0 → N =mg + F1 y =(5kg )(9.8)m / s 2 + 75 N sin 65 =117 N, f =µ N =0.04(117 N ) =4.68 N
=
ax
F
∑
=
x
m
F2 − F1x − f 50 N − 31.7 N − 4.68
=
= 2.72m / s 2
m
5kg
At an instant when a 4.0-kg object
has an acceleration equal to (5i + 3j)
m/s2, one of the two forces acting on
the object is known to be (12i + 22j)
N. Determine the magnitude of the
other force acting on the object.
a. 2.0 N
b. 13 N
c.
18 N
d. 1.7 N
e.
20 N
Newton’s 3rd Law
Fhand on wall = − Fwall on hand
To every force there is an equal but
opposite reaction force.
Newton’s 3rd Law
Fhand on wall = − Fwall on hand
You can’t TOUCH without being
TOUCHED back!!
Newton’s 3rd Law
Fhand on wall = − Fwall on hand
To every force there is an equal but
opposite reaction force.
Newton’s 3rd Law
Fhand on wall = − Fwall on hand
This is an INTERACTIVE Universe.
Gravity is an Interaction
FEarth on Rock = − FRock on Earth
Gravity is an Interaction
The Earth pulls on you, you pull on the Earth.
You fall to the Earth, the Earth Falls to you.
You accelerate towards the Earth with g =9.8m/s2.
With what acceleration is the Earth falling towards you?
FEarth on You = − FYou on Earth
mg = − M E aE
mg
aE =
ME
2
(65kg )(9.8m / s )
2
−22
1.1x10 m / s
=
24
5.98 x10 kg
This is your weight:
aE
Force is not Acceleration
Force is the Same!
Acceleration is NOT!
FEarth on You = − FYou on Earth
aEarth to You = −aYou to Earth
An interaction requires a pair of
forces acting on two objects.
kick
Gun Pushes Bullet out.
Bullet Pushes back on Gun (& Man)
Action Reaction Pairs
kick
Gun Pushes Bullet out.
Bullet Pushes back on Gun (& Man)
Rocket Thrust
Rocket Pushes Gas Out.
Gas Pushes Back on Rocket.
rd
3
Newton’s
Law
Exploding Systems
Reading Question 7.3
The propulsion force on a car is due to
A.
B.
C.
D.
Static friction.
Kinetic friction.
The car engine.
Elastic energy.
© 2013 Pearson Education, Inc.
Slide 7-14
Reading Question 7.3
The propulsion force on a car is due to
A.
B.
C.
D.
Static friction.
Kinetic friction.
The car engine.
Elastic energy.
© 2013 Pearson Education, Inc.
Slide 7-15
In order to get an object moving,
you must push harder on it than it
pushes back on you.
A) True
B) False
Question
You push a heavy car by hand. The car, in turn,
pushes back with an opposite but equal force on
you. Doesn’t this mean the forces cancel one
another, making acceleration impossible?
How is it that the car moves?
The System
Action-Reaction forces act on different objects.
For F = ma, the forces must act on ONE object: the system.
Interacting Objects
 If object A exerts a force on object B, then object
B exerts a force on object A.
 The pair of forces, as shown, is called an
action/reaction pair.
Slide 7-23
QuickCheck 7.11
Block A is accelerated across a frictionless table. The string is
massless, and the pulley is both massless and frictionless.
Which is true?
A. Block A accelerates faster in case a than in case b.
B. Block A has the same acceleration in case a and case b.
C. Block A accelerates slower in case a than in case b.
© 2013 Pearson Education, Inc.
Slide 7-80
QuickCheck 7.11
Block A is accelerated across a frictionless table. The string is
massless, and the pulley is both massless and frictionless.
Which is true?
A. Block A accelerates faster in case a than in case b.
B. Block A has the same acceleration in case a and case b.
C. Block A accelerates slower in case a than in case b.
© 2013 Pearson Education, Inc.
Slide 7-81
Problem
=
m1 2=
kg , m2 3=
kg , m3 5kg
The three blocks are pushed across a rough
surface by a 40-N force. If the coefficient of
kinetic friction between each of the blocks and
the surface is 0.20, determine the magnitude of
the force exerted by m2 on m3.
a)
b)
c)
d)
e)
20 N
30 N
10 N
15 N
25 N
QuickCheck 7.6
Boxes A and B are sliding to the right on a frictionless surface.
Hand H is slowing them. Box A has a larger mass than B.
Considering only the horizontal forces:
A.
FB on H = FH on B = FA on B = FB on A
B.
FB on H = FH on B > FA on B = FB on A
C.
FB on H = FH on B < FA on B = FB on A
D.
FH on B = FH on A > FA on B
Slide 7-67
QuickCheck 7.6
Boxes A and B are sliding to the right on a frictionless surface.
Hand H is slowing them. Box A has a larger mass than B.
Considering only the horizontal forces:
A.
FB on H = FH on B = FA on B = FB on A
B.
FB on H = FH on B > FA on B = FB on A
C.
FB on H = FH on B < FA on B = FB on A
D.
FH on B = FH on A > FA on B
Slide 7-67
Reading Question 7.4
Is the tension in rope 2 greater than, less
than, or equal to the tension in rope 1?
A. Greater than rope 2.
B. Less than rope 2.
C. Equal to rope 2.
© 2013 Pearson Education, Inc.
Slide 7-16
Reading Question 7.4
Is the tension in rope 2 greater than, less
than, or equal to the tension in rope 1?
A. Greater than rope 2.
B. Less than rope 2.
C. Equal to rope 2.
© 2013 Pearson Education, Inc.
Slide 7-17
QuickCheck 7.8
The two masses are at rest. The pulleys are frictionless.
The scale is in kg. The scale reads
A.
B.
C.
0 kg.
5 kg.
10 kg.
Slide 7-73
QuickCheck 7.8
The two masses are at rest. The pulleys are frictionless.
The scale is in kg. The scale reads
A.
B.
C.
0 kg.
5 kg.
10 kg.
Slide 7-74
Tension Forces
Tension forces are transmitted undiminished through the rope.
Different T
Same T
QuickCheck 7.7
All three 50-kg blocks are at rest.
The tension in rope 2 is
A.
greater than the tension in rope 1.
B.
equal to the tension in rope 1.
C.
less than the tension in rope 1.
© 2013 Pearson Education, Inc.
Slide 7-71
QuickCheck 7.7
All three 50-kg blocks are at
rest. The tension in rope 2 is
A.
greater than the tension in rope 1.
B.
equal to the tension in rope 1.
C.
less than the tension in rope 1.
Each block is in static equilibrium, with
© 2013 Pearson Education, Inc.
.
Slide 7-72
Find the acceleration and
tension of the system.
© 2013 Pearson Education, Inc.
Slide 7-75
QuickCheck 7.5
Boxes A and B are being pulled to the right on a
frictionless surface. Box A has a larger mass than B.
How do the two tension forces compare?
A.
B.
C.
D.
T1 > T 2
T1 = T 2
T1 < T 2
Not enough information to tell.
© 2013 Pearson Education, Inc.
Slide 7-61
QuickCheck 7.5
Boxes A and B are being pulled to the right on a
frictionless surface. Box A has a larger mass than B.
How do the two tension forces compare?
A.
B.
C.
D.
T1 > T 2
T1 = T 2
T1 < T2
Not enough information to tell.
© 2013 Pearson Education, Inc.
Slide 7-62
Pulleys
 Block B drags block A across a frictionless table as it
falls.
 The string and the pulley are both massless.
 There is no friction where the pulley turns on its axle.
 Therefore, TA on S = TB on S.
© 2013 Pearson Education, Inc.
Slide 7-69
Pulleys
 Since TA on B = TB on A, we can draw the simplified freebody diagram on the right, below.
 Forces
and
act as if they are in an
action/reaction pair, even though they are not
opposite in direction because the tension force gets
“turned” by the pulley.
© 2013 Pearson Education, Inc.
Slide 7-70
Pulleys, Masses, Strings
What is the acceleration of the system?
(If they are connected, it is the same for both masses!)
What is the tension in the string?
1. If it falls from rest
2. If it is dragged to the left
3. If the string is cut
FIRST: Draw free-body diagrams for each mass!!!
© 2013 Pearson Education, Inc.
Problem
A force F = 40 N pulls the two masses. If the table is
frictionless, find the tension in the string.
=
m1 3=
kg , m2 1.5kg
a)
b)
c)
d)
e)
13 N
36 N
23 N
15 N
28 N
QuickCheck 7.10
The top block is accelerated across a frictionless table
by the falling mass m. The string is massless, and the
pulley is both massless and frictionless. The tension in
the string is
A. T < mg.
B. T = mg.
C. T > mg.
© 2013 Pearson Education, Inc.
Slide 7-78
QuickCheck 7.10
The top block is accelerated across a frictionless table
by the falling mass m. The string is massless, and the
pulley is both massless and frictionless. The tension in
the string is
A. T < mg.
B. T = mg.
C. T > mg
© 2013 Pearson Education, Inc.
Tension has to be
less than mg for
the block to have
a downward
acceleration.
Slide 7-79
Force Vector Diagrams
Draw free-body diagrams for every object!
Note: T and a are the same! Ropes connected by ideal pulleys
have the same tension everywhere!
Pulleys, Masses, Strings
What is the acceleration of the system?
(If they are connected, it is the same for both masses!)
What is the tension in the string?
1. If it falls from rest
2. If it is dragged to the left
3. If the string is cut
FIRST: Draw free-body diagrams for each mass!!!
© 2013 Pearson Education, Inc.
Problem
A constant force F pulls the system as
shown. The pulleys are frictionless.
The coefficient of kinetic friction
between the block and the table is µ.
a) Draw free body diagrams for both
masses.
b) Find an expression for the
acceleration in terms of the given
variables.
HW Problem
In the figure shown, the coefficient of kinetic friction
between the block and the incline is 0.29. What is the
magnitude of the acceleration of the suspended block
as it falls? Disregard any pulley mass or friction in the
pulley. Draw the free body diagrams for each mass.
Derive a general solution for the acceleration in terms
of M, and g, box it, then put the numbers in and get a
numerical value – then box that too. Then find a
numerical value for the tension in the string. Box that.
Show all your work and make it pretty! Use 3
significant figures.
2M
30
M
Tension
Two 10 N weights are pulling on the spring scale as shown
(the right side is attached to a hook, the left side is attached to the
body of the scale) What does the scale read?
a) 0 N
b) 10 N
c) 20 N
Force Vector Diagrams
Align axes to simplify the problem!
QuickCheck 7.1
A mosquito runs head-on into a truck. Splat! Which is true
during the collision?
A.
B.
C.
D.
E.
The mosquito exerts more force on the truck than the truck
exerts on the mosquito.
The truck exerts more force on the mosquito than the
mosquito exerts on the truck.
The mosquito exerts the same force on the truck as the
truck exerts on the mosquito.
The truck exerts a force on the mosquito but the mosquito
does not exert a force on the truck.
The mosquito exerts a force on the truck but the truck does
not exert a force on the mosquito.
Slide 7-39
QuickCheck 7.1
A mosquito runs head-on into a truck. Splat! Which is true
during the collision?
A.
B.
C.
D.
E.
The mosquito exerts more force on the truck than the truck
exerts on the mosquito.
The truck exerts more force on the mosquito than the
mosquito exerts on the truck.
The mosquito exerts the same force on the truck as the
truck exerts on the mosquito.
The truck exerts a force on the mosquito but the mosquito
does not exert a force on the truck.
The mosquito exerts a force on the truck but the truck does
not exert a force on the mosquito.
Slide 7-40
QuickCheck 7.2
A mosquito runs head-on into a truck. Which is true during
the collision?
A.
B.
C.
D.
E.
The magnitude of the mosquito’s acceleration is larger
than that of the truck.
The magnitude of the truck’s acceleration is larger than that
of the mosquito.
The magnitude of the mosquito’s acceleration is the same as
that of the truck.
The truck accelerates but the mosquito does not.
The mosquito accelerates but the truck does not.
Slide 7-41
QuickCheck 7.2
A mosquito runs head-on into a truck. Which is true during
the collision?
A.
B.
C.
D.
E.
The magnitude of the mosquito’s acceleration is larger
than that of the truck.
The magnitude of the truck’s acceleration is larger than that
of the mosquito.
The magnitude of the mosquito’s acceleration is the same as
that of the truck.
The truck accelerates but the mosquito does not.
The mosquito accelerates but the truck does not.
Newton’s second law:
Don’t confuse cause and effect! The same force can have very different effects.
Slide 7-42