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CHAPTER 4:
FORCES
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4. FORCES
4.1 Basic of Forces and Free Body Diagram
4.2 Newtons Laws of Motion
2
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At the end of this chapter, students should be able to:
Identify the forces acting on a body in different situations.
Weight
Tension
Normal force
Friction
Determine weight, static friction and kinetic friction
Draw free body diagram
Determine the resultant force
Learning Outcome:
4.1 Basic of Forces and Free Body Diagram
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PHYSICS CHAPTER 4
4.1 Basic of Forces and Free Body Diagram
Weight, is defined as the force exerted on a body under gravitational
field.
It is a vector quantity.
It is dependant on where it is measured, because the value ofgvaries at different localities on the earths surface.
It always directed toward the centre of the earth or in the same
direction of acceleration due to gravity, g.
The S.I. unit is kg m s-2 or Newton (N).
Equation:
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Tension, T
The tension force is the force that is transmitted through a
string, rope, cable or wire when it is pulled tight by forcesacting from opposite ends. The tension force is directed
along the length of the wire and pulls equally on the objects
on the opposite ends of the wire.
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Figure 4.1
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Normal (reaction) force,
is defined as a reaction force that exerted by the surface to
an object interact with it and the direction always
perpendicular to the surface.
An object lies at rest on a flat horizontal surface as shown in
Figure 4.2.
RorN
N
gmW
0mgNFyThereforeFigure 4.2
Action: weight of an object is exerted on the
horizontal surface
Reaction: surface is exerted a force, Non theobject
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A free body diagram is defined as a diagramshowing the chosen body by itself, with vectors
drawn to show the magnitude and directions of allthe forces applied to the body by the other bodies
that interact with it.
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Friction
is defined as a force that resists the motion of one surface
relative to another with which it is in contact. is independent of the area of contact between the two surfaces..
is directly proportional to the reaction force.
OR
Coefficient of friction,
is defined as the ratio between frictional force to reactionforce.
OR
is dimensionless and depends on the nature of the surfaces.
Nf
where
N
f
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There are three types of frictional force :
Static,fs
(frictional force act on the object before its move)
Kinetic,fk (frictional force act on the object when its move)
Rolling,fr (frictional force act on the object when its rolling)
Caution:
The direction of the frictional force exerted by a surfaceon an object is always in the opposite direction of the
motion.
The frictional and the reaction forces are always
perpendicular.
Nf kk
Nf ss
Nf rr
skr fff where
thus skr
Can be ignored
Simulation 4.1
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Example 4.1:
A mass is resting on a flat surface which has a normal force of98N, with a coefficient of static friction of 0.35. What force
would it take to move the object?
9
Solution: N = 98N, s = 0.35
Nf ss
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Example 4.2:
A 15 kg piece of wood is placed on top of another piece ofwood. There is 35N of static friction measured between them.
Determine the coefficient of static friction between the two
pieces of wood.
Solution: N = mg = 15(9.81) = 147.15 N, Fs
= 35 N
N
fss
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PHYSICS CHAPTER 4
Example 4.3
A dock worker loading crates on a ship finds that a 15 kg crate,initially at rest on a horizontal surface, requires a 50 N
horizontal force to set it in motion. However, after the crate is in
motion, a horizontal force of 30 N is required to keep it moving
with a constant speed. The acceleration of gravity is 9.8 ms-2.
Find the coefficient of kinetic friction.
11
Solution:
Mass of crate = m = 15 kg
Force required to set the crate in motion = F1 = 50 N
Force required to keep the crate in moving at constant speed =
fk = 30 N
Acceleration of gravity = g = 9.81 ms-2
Normal force, N = mg = =
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Solution:
From
N
fkk
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Resultant force Is defined as a single force that represents the combined
effect of two or more forces
1313
The figure above shows three forces F1, F2 and F3 acted on a
particle O. Calculate the magnitude and direction of the
resultant force on particle O.
Example 4.4:y
30o
O
)N30(2F
)N10(1
F
30ox
)N40(3F
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30o
Solution :
O
y
x
3F
30o
y3F
321 FFFFFr
yxr FFF
xxxx FFFF 321
yyyy FFFF 321
xF2
1F2F
60o
yF2
x3F
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Vector x-component y-component
1F
3F
2F
N01 xF 11 FFy
N011 yF
60cos302 xF N152 xF
60sin302 yF N622 yF
30cos403 xFN34.63 xF
30sin403 yFN203 yF
Vector
sum
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y
xO
Solution :
The magnitude of the resultant force is
and
Its direction is 162 from positive x-axis OR 18 above negative x-
axis.
22 yxr FFF
x
y
F
F 1tan
yF
xF
162
rF
18
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1. Given three vectors P, Q andR as shown in Figure 4.3.
Calculate the resultant vector of P, Q andR.
ANS. : 49.4 m s2; 70.1 above + x-axis
Exercise 4.1:
Figure 4.3
y
x0
502sm10 R
2sm35 P
2sm24 Q
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At the end of this chapter, students should be able to:
State Newtons First Law
Define mass as a measure of inertia.
Define the equilibrium of a particle.
Apply Newtons First Law in equilibrium of forces
State and apply Newtons Second Law
State and apply Newtons Third Law.
Learning Outcome:
4.2 Newtons laws of motion
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4.2 Newtons laws of motion
4.2.1 Newtons first law of motion states an object at rest will remain at rest, or continues to
move with uniform velocity in a straight line unless it isacted upon by a external forces
OR
The first law gives the idea of inertia.
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4.2.2 InertiaInertia
is defined as the tendency of an object to resist any changein its state of rest or motion.
is a scalar quantity.
Mass, m
is defined as a measure of a bodys inertia. is a scalar quantity.
The S.I. unit of mass is kilogram (kg).
The value of mass is independent of location.
If the mass of a body increases then its inertia will increase.
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Figures 4.4a and 4.4b show the examples of real experience ofinertia.
Figure 4.4
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4.2.3 Equilibrium of a particle
is defined as the vector sum of all forces acting on a particle(point) must be zero
.
The equilibrium of a particle ensures the body in translationalequilibrium and its condition is given by
This is equivalent to the three independent scalar equationsalong the direction of the coordinate axes,
There are two types of equilibrium of a particle. It is
Static equilibrium (v=0) body remains at rest (stationary).
Dynamic equilibrium (a=0) body moving at a uniform(constant) velocity.
Newtons first
law of motion
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Problem solving strategies for equilibrium of a
particle
The following procedure is recommended when dealing with
problems involving the equilibrium of a particle:
Sketch a simple diagram of the system to help
conceptualize the problem.
Sketch a separate free body diagram for each body. Choose a convenient coordinate axes for each body and
construct a table to resolve the forces into their
components.
Apply the condition for equilibrium of a particle in
component form :
Solve the component equations for the unknowns. 0xF 0yFand
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A load of 250 kg is hung by a cranes cable. The load is pulled by a
horizontal force such that the cable makes a 30 angle to thevertical plane. If the load is in the equilibrium, calculate
a. the magnitude of the tension in the cable,
b. the magnitude of the horizontal force. (Given g=9.81 m s2)
Solution :
Example 4.5:
30
F F
Free body diagram of the load :
gm
T
yT3060
xT
kg250m
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Solution :
1st method :
a.
Since the load is in the equilibrium, then
Thus
b. By substituting eq. (2) into eq. (1), therefore
0xF 060cos
TF
kg250m
Force x-component (N) y-component (N)
gm
0 9.81250mg
F
F 0
T
60cosT 60sinT
2453
0F
(1)
(2) 0yF 0245360sin T
060cos2833
F
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30
Solution :
2nd method :
a. Since the load is in the equilibrium, then a closed triangle of
forces can be sketched as shown below.
b.
30sinT
F
kg250m
30cosT
mg
30sin2833 F
F
gm
T
From the closed triangle of forces, hence
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Calculate the magnitude and direction of a force that balance the
three forces acted at point A as shown in Figure 4.5.
Example 4.6:
N121
FN202F
N303F
30.055.0
45.0A
Figure 4.5
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Solution :
To find a force to balance the three forces means the system must
be in equilibrium hence
N30N;20N;12 321 FFF
Force x-component (N) y-component (N)
1F 55.0cos12
F
xF yF
6.88
55.0sin129.83
2F 30.0cos20
17.3
30.0sin2010.0
3F 45.0cos30
21.2
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Solution :
The magnitude of the force,
and its direction,
0 yF021.210.09.83 yF
222y
2
x
FFF 1.3731.6
x
y1
F
F tan
31.6
1.37tan 1
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30
A window washer pushes his scrub brush up a vertical window at
constant speed by applying a force Fas shown in Figure 4.6.The brush weighs 10.0 N and the coefficient of kinetic friction is
k= 0.125. Calculate
a. the magnitude of the force F,
b. the normal force exerted by the window on the brush.
Example 4.7:
F
50.0
Figure 4.6
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Solution :
a. The free body diagram of the brush :
The brush moves up at constant speed (a=0) so that
Thus
0.125;N10.0 kW
W
F
N
kf
constant
speed
Force x-component (N) y-component (N)
F
50.0cosF
kf
0Nk
50.0sinF
W
0 10.0
N
N 0
N0.125
0amF
50.0cosFN0 xF (1)
(2)10.00.12550.0sin NF
0 yF
50.0
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Solution :
a. By substituting eq. (1) into eq. (2), thus
b. Therefore the normal force exerted by the window on the brush
is given by
10.050.0cos0.12550.0sin FF
50.0cosFN
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Exercise 4.2:
Use gravitational acceleration, g= 9.81 m s2
1.
The system in Figure 5.8 is in equilibrium, with the string at the
centre exactly horizontal. Calculate
a. the tensions T1, T2 and T3.
b. the angle .
ANS. : 49 N, 28 N, 57 N; 29
Figure 4.7
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Exercise 4.2:
2.
A 20 kg ball is supported from the ceiling by a rope A. Rope B
pulls downward and to the side on the ball. If the angle of A tothe vertical is 20 and if B makes an angle of 50 to the vertical
as shown in Figure 4.8, Determine the tension in ropes A and B.
ANS. : 134 N; 300 N
Figure 4.8
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Exercise 4.2:
3.
A block of mass 3.00 kg is pushed up against a wall by a force
P that makes a 50.0 angle with the horizontal as show in
Figure 4.9. The coefficient of static friction between the blockand the wall is 0.250. Determine the possible values for the
magnitude of P that allow the block to remain stationary.
ANS. : 31.8 N; 48.6 N
Figure 4.9
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Newtons second law of motion
states the rate of change of linear momentum of a moving
body is proportional to the resultant force and is in thesame direction as the force acting on it
OR
its can be represented by
where
momentumlinearinchange:pd
intervaltime:dt
forceresultant:F
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From the Newtons 2nd law of motion, it also can be written as
Case 1: Object at rest or in motion with constant velocity but with
changing mass. For example : Rocket
and
0dt
vd
and
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Case 2:
Object at rest or in motion with constant velocity and constant
mass.
Thus
dt
vdm
dt
dmvF
Newtons 1st
law of motion
0dt
where and
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Case 3:
Object with constant mass but changing velocity.
The direction of the resultant force always in the same
direction of the motion or acceleration.
dt
vdm
dt
dmvF
0dt
dman
d
and
where
objectanofmass:m
onaccelerati:a
forceresultant:F
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Newtons 2nd law of motion restates that The acceleration of
an object is directly proportional to the nett force acting on
it and inversely proportional to its mass.OR
One newton(1 N) is defined as the amount of nett force that
gives an acceleration of one metre per second squared to abody with a mass of one kilogramme.
OR 1 N = 1 kg m s-2
Notes:
is a nett force or effective force or resultant force. The force which causes the motion of an object.
If the forces act on an object and the object moving at
uniform acceleration (not at rest or not in the
equilibrium) hence
amFFnett
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Newtons third law of motion
states every action force has a reaction force that is equal
in magnitude but opposite in direction. For example :
When the student push on the wall it will push back with thesame force. (refer to Figure 4.10)
A (hand) B (wall)
BAF
ABF
Figure 4.10
is a force by the hand on the wall (action)Where
is a force by the wall on the hand (reaction)BAFABF
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When a book is placed on the table. (refer to Figure 4.11)
If a car is accelerating forward, it is because its tyres arepushing backward on the road and the road is pushingforward on the tyres.
A rocket moves forward as a result of the push exerted on it
by the exhaust gases which the rocket has pushed out.
In all cases when two bodies interact, the action and reactionforces act on different bodies.
Figure 4.11
Force by the book on the table (action)
Force by the table on the book (reaction)
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Applications of Newtons 2nd law of motion
From the Newtons second law of motion, we arrived at equation
There are five steps in applying the equation above to solveproblems in mechanics:
Identify the object whose motion is considered.
Determine the forces exerted on the object.
Draw a free body diagram for each object. is defined as a diagram showing the chosen body by
itself, with vectors drawn to show the magnitude anddirections of all the forces applied to the body by theother bodies that interact with it.
Choose a system of coordinates so that calculations may besimplified.
Apply the equation above,
Along x-axis:
Along y-axis:
maFFnett
xx maF
yy maF
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Three wooden blocks connected by a rope of negligible mass are
being dragged by a horizontal force, Fin Figure 4.12.
Suppose thatF= 1000 N, m1 = 3 kg, m2 = 15 kg and m3 = 30 kg.Determine
a. the acceleration of blocks system.
b. the tension of the rope, T1 and T2.
Neglect the friction between the floor and the wooden blocks.
Example 4.8:
Figure 4.12
1T
m1 m2 m32T
F
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Solution :
a. For the block, m1 = 3 kg
For the block, m2 = 15 kg
For the block, m3 = 30 kg
a
amTFF 11x
(1)
amTTF 221x
(2)
1T
m1
m2
m3
2T
F
aTF 1x 31000
1T
a
aTTF 21x 15
2T
a
amTF 32x(3)
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Solution :
a. By substituting eq. (3) into eq. (2) thus
Eq. (1)(4) :
b. By substituting the value of acceleration into equations (4) and
(3), therefore
045 aT1 (4)
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Two objects of masses m1 = 10 kg and m2 = 15 kg are connected
by a light string which passes over a smooth pulley as shown inFigure 4.13. Calculate
a. the acceleration of the object of mass 10 kg.
b. the tension in the each string.
(Giveng= 9.81 m s2
)Solution :
a. For the object m1= 10 kg,
Example 4.9:
Figure 4.13
m1
m2
1T
gmW 11
amgmTF 111y
(1)agT 1010 a where TTT 21
Simulation 4.2
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Solution :
a. For the object m2= 15 kg,
Eq. (1) + (2) :
b. Substitute the value of acceleration into equation (1) thus
Therefore
2T
gmW 22
amTgmF 222y
(2)agT 1515 a
aTgFy 1515
1.96109.8110 T
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Two blocks, A of mass 10 kg and B of mass 30 kg, are side by side
and in contact with each another. They are pushed along a smoothfloor under the action of a constant forceFof magnitude 200 Napplied to A as shown in Figure 4.14. Determine
a. the acceleration of the blocks,
b. the force exerted by A on B.
Solution :
a. Let the acceleration of the blocks is a. Therefore
Example 4.10:
ammF BAx
N200kg;30kg;10 Fmm BA
Figure 4.14
A BF
ammF BA
Simulation 4.3
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Solution :
b. For the object A,
From the Newtons 3rd law, thus
OR
For the object B,
amFFFABAx
5.010200 BAFF
a
BAF
A
BABF
a
amFFBABx
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1. A block is dragged by forces,F1 andF2 of the magnitude
20 N and 30 N respectively as shown in Figure 4.15. The
frictional forcefexerted on the block is 5 N. If the weight ofthe block is 200 N and it is move horizontally, determine the
acceleration of the block.
(Giveng
= 9.81 m s2)
ANS. : 1.77 m s2
Exercise 4.3:
50a
1F
2F
f
20
Figure 4.15
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2. One 3.5 kg paint bucket is hanging by a massless cord from
another 3.5 kg paint bucket, also hanging by a massless cordas shown in Figure 4.16. If the two buckets are pulled upward
with an acceleration of 1.60 m s2 by the upper cord, calculate
the tension in each cord.
(Giveng= 9.81 m s2)
ANS. : 39.9 N; 79.8 N
Exercise 4.3:
Figure 4.16
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THE END
Next ChapterCHAPTER 5 :
Work, Energy and Power
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