FORCE

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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).
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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.
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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 Fs force
exerted by a spring
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Example
FN
Ff
Fa
Fg
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Fs
Fg
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FN1
FN2
Ff
Fa
FT
FT
Fg1
Fg2
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FN
Fa
Ff
Fg
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FN
Fa
Ff
Fg
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FN2
FN1
FT
Ff
Fa
Ff
Fg1
Fg1 Fg2
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FIRST LAW OF MOTION   According to Newton's
First Law of Motion " 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."
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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).
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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.
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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.
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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
2 m/s2
b. What will its velocity be 5 s later?
vf vo at 2(5) 10 m/s
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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 5 s
F ma
2000 N
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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
F ma
- 420 N
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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 N vo 30 m/s vf 0 m/s
- 3 m/s2
10 s
b. How far will the car travel during this time?
x vot½at2 30(10) ½ (-3)(10)2 150 m
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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.
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Action-Reaction Pair Examples
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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.
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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)
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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.
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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
1020.4 kg
3.92 m/s2
vf vo at 0 3.92(8) 3.14 m/s
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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
Fg mg 5(9.8) 49.1 N SFy FT -
Fg ma FT ma Fg 5(0.3) 49.1
50.5 N
FT
a ()
Fg
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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
SFx Fa Fgx ma 0 Fa Fgx Fgsin 30?
200(sin30?) 100 N
Fgy
?
Fg
Fgx
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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.
m1 7 kg m2 9 kg
FT
FT
a ()
a ()
Fg2
Fg1
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FT
FT
a ()
a ()
SF FT Fg1 -FT Fg2 mTa
Fg1
Fg2
1.22 m/s2
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FT
FT
b. Find the tension of the cord
a ()
a ()
Use either side of the pulley you get the same
answer! FT Fg1 m1a FT m1a Fg1
m1(ag) 7(1.229.8) 77.1 N
Fg1
Fg2
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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
Fg mg 45(9.8) 441 N   SFy FN
Fgy 0 FN FgyFg(cos35º)
441cos 35º 361.2 N
FT
Fgy
?
Fg
Fgx
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b. What will be the magnitude of the tension
force (FT) in the rope?
SFx FT Fgx 0 FT FgxFg(sin35)
441(sin 35) 252.9 N
FN
FT
Fgy
?
Fg
Fgx
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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.
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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.
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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 Static friction   Kinetic
friction
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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
SFy FN Fg 0 FN Fg mg 60(9.8)
588 N
     
Fg
SFx Fa Ff ma 0 Fa Ff µFN
0.24
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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?
FN
m 400x10-3 kg vo 0.8 m/s Ff 0.7 N
Ff
Fg
 
SFx Ff ma
- 1.75 m/s2
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FN
Ff
Fg
0.18 m
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b. What is the coefficient of friction between
the block and the tabletop?
SFy FN Fg 0 FN Fg mg
400x10-3(9.8) 3.92 N
SFx Ff ma 0 Ff µFN
0.18
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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 x 70 m vf 0 m/s
- 6.42 m/s2
F ma 600(-6.42) -3852 N
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b. What is the minimum coefficient of friction
between the tires and the road?
SFy FN Fg 0 FN Fg mg
600(9.8) 5880 N
Ff µFN
0.65
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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 Fa 400 N µ 0.5
SFy FN Fg 0 FN Fg mg
70(9.8) 686 N
SFx Fa Ff ma Ff µFN
(0.5)(686) 343 N
0.81 m/s2
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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 Fa 400 N, 30? µ 0.5
FN
Fa
Ff
Fax 400 cos 30? 346.4 N Fay 400 sin 30?
200 N
Fg
Ff µFN (0.5)(486) 243 N
SFy FN Fay - Fg 0 FN Fg - Fay
70(9.8) - 200
486 N
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FN
Fa
Ff
SFx Fax Ff ma
       
Fg
1.47 m/s2
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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 vo 0 m/s vf 2 m/s t
4 s
Ff
Fa
Fg
Fax 400 cos 50? 257.1 N Fay 400 sin 50?
306.4 N
0.5 m/s2
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FN
SFy FN - Fay - Fg 0 FN Fay Fg
306.4 - 25(9.8) 551.4 N
Ff
Fa
Fg
Ff µFN
SFx Fax - Ff ma Ff Fax - ma
257.1 - 25 (0.5) 244.6 N
0.44
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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 ? 30? µ 0.3
FN
Ff
Fg 20(9.8) 196 N
Fgy
?
Fg
Fgx 196 cos 30? 98 N Fgy 196 sin 30?
169.7 N
Fgx
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FN
Ff
SFy FN - Fgy 0 FN Fg 169.7 N
?
Ff µFN (0.3)(169.7) 50.9 N
Fg
Fgy
Fgx
SFx Fgxx Ff ma
2.35 m/s2
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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?
FN1
FN2
Fg1 FN1 Fg2 FN2
m1 0.3 kg m2 0.5 kg µ 0.4 a 2 m/s2
Ff1
F
Ff2
Fg1
Fg2
SFx 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
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b. How large a force does m1 then exert on m2?
m2 alone SFx m2a F12 - Ff2 m2a F12 Ff2
m2a 0.5 (9.8) (0.4) 0.5 (2)
2.96 N
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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
FT
FgA
FN FgA 25(9.8) 245 N FfA µFN
0.2 (245) 49 N
FgB
mA 25 kg mB 15 kg µ 0.2 t 3 s
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SF FT - FfA FgB - FT mTa
FN
Ff
FT
FT
2.45 m/s2
FgA
y ½at2 ½ (2.45)(3)2 11 m
FgB
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