Transcript Document

FORCE
A force is any influence that can change the velocity of a
body. Forces can act either through the physical contact
of two objects (contact forces: push or pull) or at a
distance (field forces: magnetic force, gravitational
force).
FREE BODY DIAGRAMS
In all but the simplest
problems that involve forces, it
is helpful to draw a free body
diagram (FBD) of the situation.
This is a vector diagram that
shows all the forces that act
on the body whose motion is
being studied. Forces that the
body exerts on anything else
should not be included, since
such forces do not affect the
body's motion.
4.1 Complete the free body diagram
showing all of the forces acting on the
mass M. Be sure to show the direction of
each force as an arrow and label each force
clearly!
FT force due to tension: a rope or
cord
FN force normal: acting
perpendicular to
a surface
Ff force of friction: opposes motion
Fg gravitational force or weight,
always
downward
Fa applied force: push or pull
F force exerted by a spring
Example:
FN
Fa
Ff
Fg
Fs
Fg
FN1
FN2
Fa
Ff
FT
Fg1
FT
Fg2
FN
Fa
Ff
Fg
FN
Fa
Ff
Fg
FN2
FN1
FT
Ff
Fg1
Ff
Fa
Fg1+ Fg2
FIRST LAW OF MOTION
According to Newton's
First Law of Motion:
Isaac Newton
(1642-1727)
" If no net force acts on it, a
body at rest remains at rest and
a body in motion remains in
motion at constant speed in a
straight line."
MASS
The property a body has of resisting any change in its
state of rest or of uniform motion is called inertia. The
inertia of a body is related to the amount of matter it
contains. A quantitative measure of inertia is mass.
The SI unit of mass is the kilogram (kg).
Motion tends to continue
unchanged.
The elephant at rest tends
to remain at rest.
Tablecloth trick:
Too little force, too little time to
overcome "inertia" of tableware.
SECOND LAW OF MOTION
According to Newton's Second Law of Motion, the net
force acting on a body equals the product of the mass
and the acceleration of the body. The direction of the
force is the
same as that of the acceleration. In equation form:
F = ma
In the SI system, the unit for force is the newton (N): A
newton is that net force which, when applied to a 1-kg
mass, gives it an acceleration of 1 m/s2.
Net force is sometimes designated F.
The second law of motion is the key to understanding
the behavior of moving bodies since it links cause
(force) and effect (acceleration) in a definite way.
4.2 A force of 3000 N is applied to a 1500-kg car at rest.
a. What is its acceleration?
F = 3000 N
m = 1500 kg
vo = 0 m/s
F = ma
3000
F
= 2 m/s2

a
m 1500
b. What will its velocity be 5 s later?
vf = vo + at
= 2(5)
= 10 m/s
4.3 A 1000 kg car goes from 10 to 20 m/s in 5 s. What force is
acting on it?
m = 1000 kg
vo = 10 m/s
vf = 20 m/s
t=5s
v
 20  10 
F = ma  m
 1000
 = 2000 N
t
 5 
4.4 A 60-g tennis ball approaches a racket at 15 m/s, is in contact
with the racket for 0.005 s, and then rebounds at 20 m/s. Find the
average force exerted by the racket.
m = 0.06 kg
vo = 15 m/s
t = 0.005 s
vf = - 20 m/s
v
F = ma m  0.06  20  15  = - 420 N
t
 0.005 
4.5 The brakes of a 1000-kg car exert 3000 N.
a. How long will it take the car to come to a stop from a velocity of
30 m/s?
m = 1000 kg
F  3000
F = -3000 N
2
a
=
3
m/s

vo = 30 m/s
m 1000
vf = 0 m/s
v f  vo 0  30
= 10 s

t
3
a
b. How far will the car travel during this time?
x = vot+½at2
= 30(10)+ ½ (-3)(10)2
= 150 m
THIRD LAW OF MOTION
According to Newton's third law of motion, when one
body exerts a force on another body, the second body
exerts on the first an equal force in opposite direction.
The Third Law of Motion applies
to two different forces on two
different objects: "The action
force one object exerts on the
other, and the equal but
opposite reaction force the
second object exerts on the
first."
Action and reaction forces never
balance out because they act on
different objects.
Action-Reaction Pair Examples
4.6 A book rests on a table.
a. Show the forces acting on the table and the corresponding
reaction forces.
b. Why do the forces acting on the table not cause it to move?
ACTION: FTB
REACTION: FBT
The forces FN and Fg
are balanced.
WEIGHT
The weight of a body is the gravitational force with
which the Earth attracts the body. Weight (a vector
quantity) is different from mass (a scalar quantity). The
weight of a body varies with its location near the Earth
(or other astronomical body), whereas its mass is the
same everywhere in the universe. The weight of a body
is the force that causes it to be accelerated downward
with the acceleration of gravity g.
From the Second Law of Motion:
W = mg
Units: Newtons (N)
THE NORMAL FORCE
A normal force is a force exerted by one surface on
another in a direction perpendicular to the surface of
contact.
Note: The gravitational force and the normal force are
not an action-reaction pair.
4.7 A net horizontal force of 4000 N is applied to a car at rest whose
weight is 10,000 N. What will the car's speed be after 8 s?
Fa = 4000 N
Fg = 10,000 N
t = 8s
Fg
10000

m
9.8 = 1020.4 kg
g
F
4000
a
2

=
3.92
m/s
m 1020 .4
vf = vo + at
= 0 +3.92(8)
= 3.14 m/s
4.8 A 5.0-kg object is to be given an upward acceleration of 0.3
m/s2 by a rope pulling straight upward on it. What must be the
tension in the rope?
m = 5 kg
a = 0.3 m/s2
FT
a (+)
Fg
Fg = mg
= 5(9.8)
= 49.1 N
ΣFy = FT - Fg = ma
FT = ma + Fg
= 5(0.3) + 49.1
= 50.5 N
4.9 A 200-N wagon is to be pulled up a 30 incline at constant
speed. How large a force parallel to the incline is needed if friction
is negligible?
Fg = 200 N
FN
Fa
θ
Fg
Fgy
Fg
x
ΣFx = Fa – Fgx = ma = 0
Fa = Fgx = Fgsin 30
= 200(sin30)
= 100 N
4.10 A cord passing over a frictionless pulley has a 7.0 kg mass
hanging from one end and a 9.0-kg mass hanging from the other.
(This arrangement is called Atwood's machine).
a. Find the acceleration of the masses.
m = 7 kg
1
m2 = 9 kg
FT
FT
a (+)
a (+)
Fg1
Fg2
FT
FT
a (+)
a (+)
Fg1
a
Fg2
Fg 2  Fg1
mT
ΣF = FT – Fg1 -FT + Fg2 = mTa
9(9.8)  7(9.8)
= 1.22 m/s2

97
FT
FT
b. Find the tension of the
cord
a (+)
a (+)
Fg1
Fg2
Use either side of the
pulley:
you get the same answer!
FT – Fg1 = m1a
FT = m1a+ Fg1
= m1(a+g)
= 7(1.22+9.8)
= 77.1 N
4.11 A crate, which has a mass of m = 45.0 kg, is being pulled up a
frictionless inclined plane, at an angle of 35˚ by a rope.
a. What will be the magnitude of the normal force (FN) acting on the
crate?
m = 45.0 kg
FN
FT
Fgy
θ
Fg
Fg
x
Fg = mg
= 45(9.8)
= 441 N
ΣFy = FN – Fgy = 0
FN = Fgy=Fg(cos35º)
= 441cos 35º
= 361.2 N
b. What will be the magnitude of the tension force (FT) in the rope?
ΣFx = FT – Fgx = 0
FN
FT
Fgy
θ
Fg
Fg
x
FT = Fgx=Fg(sin35˚)
= 441(sin 35˚)
= 252.9 N
FRICTION: STATIC AND KINETIC FRICTION
Frictional forces act to oppose relative motion between
surfaces that are in contact. Such forces act parallel to
the surfaces.
Static friction occurs between surfaces at rest relative to
each other. When an increasing force is applied to a
book resting on a table, for instance, the force of static
friction at first increases as well to prevent motion. In a
given situation, static friction has a certain maximum
value called starting friction. When the force applied to
the book is greater than the starting friction, the book
begins to move across the table.
FRICTION: STATIC AND KINETIC FRICTION
The kinetic friction (or sliding friction) that occurs
afterward is usually less than the starting friction, so
less force is needed to keep the book moving than to
start it moving.
COEFFICIENT OF FRICTION
The frictional force between two surfaces depends on
the normal (perpendicular) force N pressing them
together and on the natures of the surfaces. The latter
factor is expressed quantitatively in the coefficient of
friction  (mu) whose value depends on the materials in
contact. The frictional force is experimentally found to
be:
Ff   s FN
Ff   k FN
Static friction
Kinetic friction
4.12 A horizontal force of 140 N is needed to pull a 60.0 kg box
across the horizontal floor at constant speed. What is the
coefficient of friction between floor and box?
FN
Fa = 140 N
m = 60 kg
Fa
Ff
Fg
ΣFy = FN – Fg = 0
FN = Fg =mg
= 60(9.8)
= 588 N
ΣFx = Fa – Ff = ma = 0
Fa = Ff = μFN
Fa 140


FN 588
= 0.24
4.13 A 400-g block with an initial speed of 80 cm/s slides along a
tabletop against a friction force of 0.70 N.
a. How far will it slide before stopping?
m = 400x10-3 kg
vo = 0.8 m/s
Ff = 0.7 N
FN
Ff
ΣFx = – Ff = ma
 Ff
 0 .7
2
a

=
1.75
m/s
400 x10 3
m
Fg
FN
Ff
v 2f  vo2  2ax
Fg
 v o2
 (0.8) 2
= 0.18 m
x

2(1.75)
2a
b. What is the coefficient of friction between the block and the
tabletop?
ΣFy = FN – Fg = 0
FN = Fg =mg
= 400x103(9.8)
= 3.92 N
ΣFx = – Ff = ma = 0
Ff = μFN
Ff
0 .7


= 0.18
FN 3.92
4.14 A 600-kg go-cart is moving on a level road at 30 m/s.
a. How large a retarding force is required to stop it in a distance of
70 m?
m = 600 kg
vo = 30 m/s
v 2f  vo2 0  (30) 2
2
=
6.42
m/s
x = 70 m
a

2x
2(70)
vf= 0 m/s
F = ma
= 600(-6.42)
= -3852 N
b. What is the minimum coefficient of friction between the tires and
the road?
ΣFy = FN – Fg = 0
FN = Fg =mg
= 600(9.8)
= 5880 N
Ff = μFN
Ff
3852


= 0.65
FN 5880
4.15 A 70-kg box is slid along the floor by a horizontal 400-N force.
Find the acceleration of the box if the value of the coefficient of
friction
between the box and the floor is 0.50.
m = 70 kg
ΣFy = FN – Fg = 0
Fa = 400 N
FN = Fg =mg
μ = 0.5
= 70(9.8)
= 686 N
ΣFx = Fa – Ff = ma
Ff = μFN
= (0.5)(686)
= 343 N
a
Fa  F f
m
400  343

70
= 0.81 m/s2
4.16 A 70-kg box is pulled by a rope with a 400-N force at an angle
of 30 to the horizontal. Find the acceleration of the box if the
coefficient of friction is 0.50
m = 70 kg
FN
Fa = 400 N, 30
Fa μ = 0.5
Ff
Fax = 400 cos 30 = 346.4
Fg
N
Fay = 400 sin 30 = 200 N
Ff = μFN
ΣFy = FN + Fay - Fg = 0
= (0.5)(486)
FN = Fg - Fay
= 243 N
= 70(9.8) - 200
= 486 N
FN
Fa
Ff
ΣFx = Fax – Ff = ma
Fg
a
Fax  F f
m
346 .4  243

= 1.47 m/s2
70
4.17 A force of 400 N pushes on a 25-kg box at an angle of 50.
Starting from rest, the box achieves a velocity of 2.0 m/s in a time
of 4.0 s. Find the coefficient of friction between the box and the
floor.
FN
Fa = 400 N, 50
m = 25 kg
Ff
vo = 0 m/s
vf = 2 m/s
Fa
t=4s
Fg
a
v f  v0
t
20
= 0.5

4 m/s2
Fax = 400 cos 50 = 257.1
N
Fay = 400 sin 50 = 306.4 N
FN
Ff
Fa
ΣFy = FN - Fay - Fg = 0
FN = Fay +Fg
= 306.4 - 25(9.8)
= 551.4 N
Fg
ΣFx = Fax - Ff = ma
Ff = Fax - ma
= 257.1 - 25 (0.5)
= 244.6 N
Ff = μFN

Ff
FN
244 .6

551 .4
= 0.44
4.18 A 20-kg box sits on an incline that makes an angle of 30˚ with
the horizontal. Find the acceleration of the box down the incline if
the coefficient of friction is 0.30.
m = 20 kg
θ = 30
μ = 0.3
FN
Ff
Fgy
θ
Fg
Fg
x
Fg = 20(9.8)
=196 N
Fgx = 196 cos 30 = 98 N
Fgy = 196 sin 30 = 169.7 N
FN
Ff
ΣFy = FN - Fgy = 0
FN = Fg = 169.7
N
θ
Fg
Ff = μFN
= (0.3)(169.7)
= 50.9 N
Fgy
Fg
x
ΣFx = Fgxx – Ff = ma
a
Fgx  F f
m
98  50.9 = 2.35 m/s2

20
4.19 Two blocks m1 (300 g) and m2 (500 g), are pushed by a force F.
If the coefficient of friction 0.40.
a. What must be the value of F if the blocks are to have an
acceleration of 200 cm/s2?
m1 = 0.3 kg
m2 = 0.5 kg
μ = 0.4
a = 2 m/s2
FN1
FN2
Ff1
Ff2
Fg1 Fg2
F
ΣFx = F - Ff1 - Ff2 = mTa
F = mTa + Ff1 + Ff2
= mTa + FNT μ
= mTa +mTg μ
= 0.8(2) + 0.8 (9.8)(0.4)
= 4.7 N
Fg1 = FN1
Fg2 = FN2
b. How large a force does m1 then exert on m2?
m2 alone
ΣFx = m2a
F12 - Ff2 = m2a
F12 = Ff2 + m2a
= 0.5 (9.8) (0.4) + 0.5 (2)
= 2.96 N
4.20 An object mA = 25 kg rests on a tabletop. A rope attached to it
passes over a light frictionless pulley and is attached to a mass
mB = 15 kg. If the coefficient of friction is 0.20 between the table
and block A, how far will block B drop in the first 3.0 s after the
system is released?
FN
Ff
FT
FgA
FT
FgB
mA = 25 kg
mB = 15 kg
μ = 0.2
t=3s
FN = FgA
= 25(9.8)
= 245 N
FfA = μFN
= 0.2 (245)
= 49 N
ΣF = FT - FfA + FgB - FT = mTa
FN
Ff
a
FT
FgB  F fA
mT
FT
= 2.45 m/s2
FgA
FgB
15(9.8)  49

25  15
y = ½at2
=½
(2.45)(3)2
= 11 m