Download 4.1 The Concepts of Force and Mass

Chapter 4
Forces and Newton’s
Laws of Motion
In 1D
4.1 The Concepts of Force and Mass
A force is a push or a pull.
Contact forces arise from physical
contact .
Action-at-a-distance forces do not
require contact and include gravity
and electrical forces.
4.1 The Concepts of Force and Mass
Arrows are used to represent forces. The length of the arrow
is proportional to the magnitude of the force.
15 N
5N
4.1 The Concepts of Force and Mass
Mass is a measure of the amount
of “stuff” contained in an object.
4.2 Newton’s First Law of Motion
Newton’s First Law
An object continues in a state of rest
or in a state of motion at a constant
speed along a straight line, unless
compelled to change that state by a
net force.
The net force is the vector sum of all
of the forces acting on an object.
4.2 Newton’s First Law of Motion
The net force on an object is the vector sum of
all forces acting on that object.
The SI unit of force is the Newton (N).
Individual Forces
4N
10 N
Net Force
6N
4.2 Newton’s First Law of Motion
Individual Forces
Net Force
5N
64
3N
4N
4.2 Newton’s First Law of Motion
Inertia is the natural tendency of an
object to remain at rest in motion at
a constant speed along a straight line.
The mass of an object is a quantitative
measure of inertia.
SI Unit of Mass: kilogram (kg)
4.2 Newton’s First Law of Motion
An inertial reference frame is one in
which Newton’s law of inertia is valid.
All accelerating reference frames are
noninertial.
4.3 Newton’s Second Law of Motion
Mathematically, the net force is
written as

Fnet = ∑ F
where the Greek letter sigma
denotes the vector sum.
The net force acting on an object is the vector
sum of all the individual forces acting on it.
4.3 Newton’s Second Law of Motion
Newton’s Second Law
When a net external force acts on an object
of mass m, the acceleration that results is
directly proportional to the net force and has
a magnitude that is inversely proportional to
the mass. The direction of the acceleration is
the same as the direction of the net force.

a=
∑

F
m
Fnet =
∑


F = ma
4.3 Newton’s Second Law of Motion
SI Unit for Force
 m  kg ⋅ m
( kg )  2  = 2
s
s 
This combination of units is called a newton (N).
4.3 Newton’s Second Law of Motion
4.3 Newton’s Second Law of Motion
A free-body-diagram is a diagram that
represents the object and the forces that
act on it. Find the net force acting on
the car.
4.3 Newton’s Second Law of Motion
The net force in this case is:
275 N + 395 N – 560 N = +110 N or 110N@right
and is directed along the + x axis of the coordinate system.
4.3 Newton’s Second Law of Motion
If the mass of the car is 1850 kg then, by
Newton’s second law, the acceleration is
a=
∑
F
+ 110 N
2
=
= + 0.059 m s
m
1850 kg
a = 0.059 m s @ right
2
4.4 The Vector Nature of Newton’s Second Law
The direction of force and acceleration vectors
can be taken into account by using x and y
components.
∑


F = ma
is equivalent to
Fnety =
Fnetx =
∑
∑
Fy = ma y
Fx = ma x
BUT we will study examples in the next PART. We will focus no on
Simple 1D examples.
4.5 Newton’s Third Law of Motion
Newton’s Third Law of Motion
Whenever one body exerts a force on a
second body, the second body exerts an
oppositely directed force of equal
magnitude on the first body.
4.5 Newton’s Third Law of Motion
Suppose that the magnitude of the force is 36 N. If the mass
of the spacecraft is 11,000 kg and the mass of the astronaut
is 92 kg, what are the accelerations?
4.5 Newton’s Third Law of Motion
 
F = P.


F = − P.
∑
On the astronaut ∑
On the spacecraft


P
+ 36 N
2
as =
=
= + 0.0033 m s
ms 11,000 kg


− P − 36 N
2
aA =
=
= − 0.39 m s
mA
92 kg
SOME SIMPLE APPLICATION OF NEWTON's 2nd LAW
Source: Physics of everyday phenomena from Griffith
McGrawHill
Frictional force fk opposes the motion on an object.
Always long the surface, opposing the motion
Normal force FN is always perpendicular to the surface.
It is the reaction force from the ground. FN
Weight W is straight down to center of Earth
FN
fk
We will study them later
In details. In addition you can
Have drive force (push or pull)
Thrust, air resistance.
W
Air resistance slows down the
Sky diver. It decreases with
The speed of the diver.
Once the air resistance = weight
The terminal speed is reached.
Find the net force
Find the acceleration at that time.
5N
10 N
25 N
5kg
Find the acceleration
Find the net force
5N
10 N
25 N
5kg
Find the acceleration
Find the net force
The right side is winning : sum(right) – sum(left) = mass x acceleration
25+5 – 10 = 5 a
20/5 = a = 4m/s/s
Fnet = mass x acceleration = 5 x 4 = 20N
Or Fnet = by how much right is winning
Fnet = 30 – 10 = 20N
At a given instant time, a 4kg rock that has been dropped from a high cliff
experiences a force of air resistance of 15N.
What are the magnitude and direction of the acceleration of the rock/
(Do not forget the gravitational force = weight of the rock)
down-up=mass x acceleration
40 – 15 = 4 a
a=6.25m/s/s@down
A 0.4kg book rests on a table. A downwards force of 6N is exerted on top of the
book by a hand pushing down on the book,
A)What is the magnitude of the gravitational force acting upon the book ? (weight)
B)What is the magnitude of the upward (normal) force exerted by the table
on the book/ (the book is at rest).
First do a sketch
Then trace a free body diagram.
Attach all the forces to the origin.
Then :
All up = all down
FN
W=mg
F
sum(up) – Sum(down) = mass x acceleration
FN – mg – F = ma plug it in and solve for a
An upward force of 18N is applied via a string to lift a ball with a mass of 1.5 kg.
A)What is the net force acting upon the ball?
B)What is the acceleration of the ball ?
Draw a free body diagram
Then sum(up) – sum(down) = net force = Fnet
And Fnet = mass x acceleration
An upward force of 18N is applied via a string to lift a ball with a mass of 1.5 kg.
A)What is the net force acting upon the ball?
B)What is the acceleration of the ball ?
Y
Tension=T
Fnet= Up – down
Fnet=tension – weight
Fnet= T – mg
Fnet = 18 – 15 =3N
W=mg
a=Fnet/m = .......
A 60kg woman in an elevator is accelerating upward at a rate of 1.2m/s/s
A)What is the net force acting upon the woman ? (use Fnet= ma )
B)What is the gravitational force acting upon the woman / (weight = mg)
C)What is the normal force pushing upward on the woman feet
Hint: trace a free-body . Then up – right = m a = Fnet
D)If the woman were standing on a scale, the scale would read this normal force.
A scale reads what ever force you are exerting on it.
This is called her apparent weight. What is her apparent weight?
A 60kg woman in an elevator is accelerating upward at a rate of 1.2m/s/s
A)What is the net force acting upon the woman ? (use Fnet= ma )
Fnet = ma = 72 N
B)What is the gravitational force acting upon the woman / (weight = mg)
Weight = 600N
C)What is the normal force pushing upward on the woman feet
Hint: trace a free-body . Then up – right = m a = Fnet
FN
mg
FN – mg = Fnet
FN – 600 = 72
FN = 672 N
D)If the woman were standing on a scale, the scale would read this normal force.
A scale reads what ever force you are exerting on it.
This is called her apparent weight. What is her apparent weight?
672N instead of 600N on Earth or in an elevator moving at a constant speed.
A rope exerts a constant force of 250N to pull a 60kg crate across the floor.
The velocity of the crate is observed to increase form 1m/s to 3m/s in a time of
2 seconds under the influence of this force and the frictional force exerted by
the floor on the crate.
A)What is the acceleration of the crate (kinematics)
B)What is the total force acting upon the crate ? (Fnet=ma)
C)What is the magnitude of the frictional force acting on the crate?
(right – left = mass x acceleration)
D)What force would have to be applied to the crate by the rope for the crate
to move with a constant velocity / explain./
A rope exerts a constant force of 250N to pull a 60kg crate across the floor.
The velocity of the crate is observed to increase form 1m/s to 3m/s in a time of
2 seconds under the influence of this force and the frictional force exerted by
the floor on the crate.
A)What is the acceleration of the crate (kinematics)
A=3-1 / 2 = 1m/s/s
B)What is the total force acting upon the crate ? (Fnet=ma)
Net=60x1=60N
C)What is the magnitude of the frictional force acting on the crate?
(right – left = mass x acceleration)
fk
F
Right – left = m a = Fnet
250 – fk = 60 so fk = 190N
D)What force would have to be applied to the crate by the rope for the crate
to move with a constant velocity / 190N so right = left no acceleration
Two blocks tied together by a horizontal string are being pulled across the table
by a 30N as shown. The 2kg block has a 6N frictional force exerted on it by the table ,
and the 4kg block has an 8N frictional force acting on it.
A)What is the net force acting on the entire two-block system?
(right – left) = Fnet
B)What is the acceleration of this system / (Fnet=ma)
C)What force is exerted on the 2kg block by the connecting string /
(consider only the forces acting on this block. Its acceleration is the same as that
of the entire system. Draw a free-body diagram for this block alone.
D)Find the net force acting on the 4 block and calculate its acceleration.
How does this value compare to that found in part B ? focus on the block alone.
Trace the free-body diagram.
2kg
6N
4kg
8N
30N
Two blocks tied together by a horizontal string are being pulled across the table
by a 30N as shown. The 2kg block has a 6N frictional force exerted on it by the table ,
and the 4kg block has an 8N frictional force acting on it.
A)What is the net force acting on the entire two-block system?
(right – left) = Fnet = 30 -6-8=16N
B)What is the acceleration of this system / a=16/6 = 2.7m/s/s
C)What force is exerted on the 2kg block by the connecting string /
(consider only the forces acting on this block. Its acceleration is the same as that
of the entire system. Draw a free-body diagram for this block alone.
fk
T
2kg
4kg
T
30N
Free-body diagram along
X-axis . T – fk = m a
T – 6 = 2 (2.7)
T=11.4N
6N
8N
4kg pulls on 2kg with a tension T so 2kg
Pulls back on 4gk with a tension T
D)Find the net force acting on the 4 block and calculate its acceleration.
How does this value compare to that found in part B ? focus on the block alone.
Trace the free-body diagram.
D)Find the net force acting on the 4 block and calculate its acceleration.
How does this value compare to that found in part B ? focus on the block alone.
Trace the free-body diagram.
2kg
4kg
T
30N
6N
fk
8N
4kg pulls on 2kg with a tension T so 2kg
Pulls back on 4gk with a tension T
F
T
Free-body diagram along
X-axis . Fnet = right – left
Fnet = 30 -8 -11.4 = 10.6
Fnet = mass x acceleration
10.6 = 4 a so a = 2.7m/s/s @ right same as before
A 60kg crate is lowered from a loading dock to the floor using a rope passing
over a fixed support. The rope exerts a constant upward force on the crate of 500N.
A)Will the crate accelerate / Explain.
B)What are the magnitude and direction of the acceleration of the crate /
C)How long will it take for the crate to reach the floor if the height of the loading
doc is 1.4m above the floor /
D)how fast is the crate traveling when it hits the floor ?
A 60kg crate is lowered from a loading dock to the floor using a rope passing
over a fixed support. The rope exerts a constant upward force on the crate of 500N.
A)Will the crate accelerate / 2 forces act on the crate: tension @ up = 500N
And the weight =600N @ down. Down > up so it is accelerating @ down
B)What are the magnitude and direction of the acceleration of the crate /
T
Down – up = 600 – 500 =100 = 60 a
So a 1.67 ms/s
W
C)How long will it take for the crate to reach the floor if the height of the loading
doc is 1.4m above the floor / use y=0.5 a t2 1.4 = 0.5 (1.67)t2 solve for t
D)how fast is the crate traveling when it hits the floor ?
Use V = a t