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PHYSICS CHAPTER 3
CHAPTER 3: CHAPTER 3: Force, Momentum and ImpulseForce, Momentum and Impulse
(5 Hours)(5 Hours)
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PHYSICS CHAPTER 3
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3.0 Force is defined as something capable of changing state of something capable of changing state of
motion or size or dimension of a bodymotion or size or dimension of a body . There are four types of fundamental forces in nature:
Gravitational forces Gravitational forces (refer to figures 3.1 and 3.2) The forces involve attraction between massive
body. is a long-range forces. the weakest forces in nature.
Figure 3.1Figure 3.1 Figure 3.2Figure 3.2
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Electromagnetic forces Electromagnetic forces (refer to figures 3.3) The attractive and repulsive forces between electric
charges. is a long-range forces.
Strong nuclear forces Strong nuclear forces (refer to figures 3.4) The attractive forces bonding neutron and proton in
atomic nucleus. is a short-range forces and the strongest forces in
nature.
Figure 3.3Figure 3.3 Figure 3.4Figure 3.4
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Weak nuclear forces Weak nuclear forces (refer to figures 3.5) cause the unstable condition for atomic nucleus and is
responsible for the radioactive decay. is a short-range forces and 12 times weak compare with
electromagnetic forces.
is a vector quantity. The dimension of the force is given by
The S.I. unit of force, F is kg m skg m s-2-2 or newton (N)newton (N)
amF 2MLTF
Figure 3.5Figure 3.5
PHYSICS CHAPTER 3
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At the end of this chapter, students should be able to: At the end of this chapter, students should be able to: Explain Explain Newton’s First Law and the concept of mass and Newton’s First Law and the concept of mass and
inertia.inertia. DefinitionDefinition of inertia and mass.of inertia and mass.
ExplainExplain and use Newton’s Second Law and use Newton’s Second Law
ExplainExplain Newton’s Third Law. Newton’s Third Law.
Learning Outcome:
3.1 Newton’s laws of motion (2 hours)
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vm
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mvmv
tF
d
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PHYSICS CHAPTER 3
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3.1 Newton’s laws of motion
3.1.1 Newton’s first law of motion states “an object at rest will remain at rest, or continues to an object at rest will remain at rest, or continues to
move with uniform velocity in a straight line unless it is move with uniform velocity in a straight line unless it is acted upon by a external forcesacted upon by a external forces”
OR
The first law gives the idea of inertia.Inertia Inertia is defined as the tendency of an object to resist any change the tendency of an object to resist any change
in its state of rest or motionin its state of rest or motion. is a scalar quantity.
0FFnett
PHYSICS CHAPTER 3
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Figures 3.6a and 3.6b show the examples of real experience of inertia.
Figure 3.6bFigure 3.6b
Figure 3.6aFigure 3.6a
PHYSICS CHAPTER 3
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Mass, Mass, mm is defined as a measure of a body’s inertia.a measure of a body’s inertia. is a scalar quantityscalar quantity. The S.I. unit of mass is kilogram (kg)kilogram (kg). The value of mass is independent of locationindependent of location. If the mass of a body increases then its inertia will increase.
Weight,Weight, is defined as the force exerted on a body under gravitational the force exerted on a body under gravitational
field.field. It is a vector quantityvector quantity. It is dependant on where it is measureddependant on where it is measured, because the value value
of of gg varies at different localities on the earth’s surface varies at different localities on the earth’s surface.
inertiamass
W
PHYSICS CHAPTER 3
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It always directed toward the centre of the earthdirected toward the centre of the earth or in the
same direction of acceleration due to gravity, same direction of acceleration due to gravity, gg. The S.I. unit is kg m skg m s-2-2 or newton (N)newton (N). Equation:
3.1.2 Newton’s second law of motion states “the rate of change of linear momentum of a moving the rate of change of linear momentum of a moving
body is proportional to the resultant force and is in the body is proportional to the resultant force and is in the same direction as the force acting on itsame direction as the force acting on it”
OR
its can be represented by
gmW
dt
where
momentumlinear in change : pd
interval time:dt
forceresultant : F
PHYSICS CHAPTER 3
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From the Newton’s 2nd law of motion, it also can be written as
Case 1:Case 1: Object at restrest or in motion with constant velocityconstant velocity but with
changing masschanging mass. For example : Rocket
dt
vdm
dt
dmvF
dt
dt
vmdF
dt
vdm
dt
dmvF
mvp and
0dt
vd
dt
dmvF
and
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Case 2:Case 2: Object at restrest or in motion with constant velocityconstant velocity and
constant massconstant mass.
Thus
dt
vdm
dt
dmvF
Newton’s 1Newton’s 1stst law of motion law of motion
0 dt
constantp
0dt
dm 0
dt
vd
0F
where and
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Case 3:Case 3: Object with constant massconstant mass but changing velocitychanging velocity.
The directiondirection of the resultant forceresultant force always in the same same direction of the motiondirection of the motion or accelerationacceleration.
dt
vdm
dt
dmvF
0dt
dmand
amF
dt
vdmF
dt
vda
and
where
objectan of mass : monaccelerati :a
forceresultant : F
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Newton’s 2nd law of motion restates that “The acceleration of The acceleration of an object is directly proportional to the nett force acting on an object is directly proportional to the nett force acting on it and inversely proportional to its massit and inversely proportional to its mass”.
OR
One newton(1 N) is defined as the amount of net force that the amount of net force that gives an acceleration of one metre per second squared to a gives an acceleration of one metre per second squared to a body with a mass of one kilogrammebody with a mass of one kilogramme.
OR 1 N = 1 kg m s1 N = 1 kg m s-2-2
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 uniform acceleration (not at rest or not in the equilibrium)equilibrium) hence amFFnett
m
Fa
F
PHYSICS CHAPTER 3
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3.1.3 Newton’s third law of motion states “every action force has a reaction force that is equal every action force has a reaction force that is equal
in magnitude but opposite in directionin magnitude but opposite in direction”. For example :
When the student push on the wall it will push back with the same force. (refer to figure 3.7)
BAAB FF
A (hand)
B (wall)
BAF
ABF
Figure 3.7Figure 3.7
is a force by the hand on the wall (action)(action)Where
is a force by the wall on the hand (reaction)(reaction)BAFABF
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When a book is placed on the table. (refer to figure 3.8)
If a car is accelerating forward, it is because its tyres are pushing backward on the road and the road is pushing forward 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 reaction action and reaction forces act on different bodiesforces act on different bodies.
Figure 3.8Figure 3.8
Force by the book on the table (action)(action)
Force by the table on the book (reaction)(reaction)
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3.1.4 Applications of Newton’s 2nd law of motion From the Newton’s second law of motion, we arrived at equation
There are five steps in applying the equation above to solve problems in mechanics: Identify the object whose motion is considered. Determine the forces exerted on the object. Draw a free body diagram free body diagram for each object.
is defined as a diagram showing the chosen body by a diagram showing the chosen body by itself, with vectors drawn to show the magnitude and itself, with vectors drawn to show the magnitude and directions of all the forces applied to the body by the directions of all the forces applied to the body by the other bodies that interact with itother bodies that interact with it.
Choose a system of coordinates so that calculations may be simplified.
Apply the equation above, Along x-axis:
Along y-axis:
maFF nett
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, F in figure 3.9.
Suppose that F = 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 1 :
Figure 3.9Figure 3.9
1T
m1 m2m3
2T
F
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Solution :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
aTT 21 15 (2)
1T
m1
m2
m3
2T
F
aTF 1x 3100010003 aT1
1T a
aTTF 21x 15
2T a
amTF 32x
aT2 30 (3)
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Solution :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)
48
1000a
2s m 20.8 a
N 9361T
N 6242T
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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 in figure 3.10. Calculate
a. the acceleration of the object of mass 10 kg.
b. the tension in the each string.
(Given g = 9.81 m s2)
Solution :Solution :
a. For the object m1= 10 kg,
Example 2 :
Figure 3.10Figure 3.10
m1
m2
1T
gmW 11
amgmTF 111y
(1)agT 1010 a where TTT 21
Simulation 3.1
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Solution :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
25
9.815
25
5
ga
2s m 1.96 a
118NT 1.96109.8110 T
N 118 TTT 21
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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 smooth
floor under the action of a constant force F of magnitude 200 N applied to A as shown in figure 3.11. Determine
a. the acceleration of the blocks,
b. the force exerted by A on B.
Solution :Solution :
a. Let the acceleration of the blocks is a. Therefore
Example 3 :
ammF BAx
2s m 5.0 a
N 200 kg; 30 kg; 10 Fmm BA
Figure 3.11Figure 3.11
A BF
ammF BA a3010200
Simulation 3.2
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Solution :Solution :
b. For the object A,
From the Newton’s 3rd law, thus
OR
For the object B,
N 150BAF
amFFF ABAx
5.010200 BAF
N 150 BAAB FF
F
a
BAF
A
BABF
a
amFF BABx
5.030ABF
N 150ABF
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1. A block is dragged by forces, F1 and F2 of the magnitude
20 N and 30 N respectively as shown in figure 3.12. The
frictional force f exerted on the block is 5 N. If the weight of the block is 200 N and it is move horizontally, determine the acceleration of the block.
(Given g = 9.81 m s2)
ANS. : 1.77 m sANS. : 1.77 m s22
Exercise 3.1 :
50a
1F
2F
f
20
Figure 3.12Figure 3.12
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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 cord as shown in figure 3.13. 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.
(Given g = 9.81 m s2)
ANS. : 39.9 N; 79.8 NANS. : 39.9 N; 79.8 N
Exercise 3.1 :
Figure 3.13Figure 3.13
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At the end of this chapter, students should be able to: At the end of this chapter, students should be able to: StateState the principle of conservation of linear momentum. Explain and applyExplain and apply the principle of conservation of
momentum in elastic and inelastic collisions Define and useDefine and use the coefficient of restitution, e
to determine the types of collisions. DefineDefine impulse J = Ft and use F-t graph to determine
impulse
Learning Outcome:3.2 Conservation of linear momentum and impulse (2 hours)
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uu
vve
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3.2 Conservation of linear momentum and impulse 3.2.1 Linear momentum, is defined as the product between mass and velocitythe product between mass and velocity. is a vector quantity. Equation :
The S.I. unit of linear momentum is kg m skg m s-1-1. The direction of the momentumdirection of the momentum is the samesame as the direction direction
of the velocityof the velocity. It can be resolve into vertical (y) component and horizontal (x)
component.
p
vmp
xp
pyp
θmvθppx coscos θmvθppy sinsin
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3.2.2 Principle of conservation of linear momentum states “In an isolated (closed) system, the total momentum In an isolated (closed) system, the total momentum
of that system is constantof that system is constant.”
OR
“When the net external force on a system is zero, the total When the net external force on a system is zero, the total momentum of that system is constantmomentum of that system is constant.”
In a Closed system,
From the Newton’s second law, thus
0 dt
0F
0pd
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According to the principle of conservation of linear momentum, we obtain
OR
The total of initial momentum = the total of final momentumThe total of initial momentum = the total of final momentum
fi pp
constantp
constant xp
constant yp
Therefore then
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Linear momentum in one dimension collisionLinear momentum in one dimension collision
Example 4 :
Figure 3.14 shows an object A of mass 200 g collides head-on with object B of mass 100 g. After the collision, B moves at a speed of 2 m s-1 to the left. Determine the velocity of A after Collision.
SolutionSolution: :
1s m 6 Au
AB
1s m 3 Bu
Figure 3.14Figure 3.14
fi pp
BBAABBAA vmvmumum 20.1000.20030.10060.200 Av
1s m 3.5 Av
1s m 6 ;kg 0.100 ;kg 0.200 ABA umm11 s m 2 ;s m 3 BB vu
to the leftto the left
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Linear momentum in two dimension collisionLinear momentum in two dimension collision
Example 5 :
A tennis ball of mass m1 moving with initial velocity u1 collides with a soccer ball of mass m2 initially at rest. After the collision, the
tennis ball is deflected 50 from its initial direction with a velocity v1 as shown in figure 3.15. Suppose that m1 = 250 g, m2 = 900 g,
u1 = 20 m s1 and v1 = 4 m s1. Calculate the magnitude and direction of soccer ball after the collision.
Figure 3.15Figure 3.15
1u
Before collision After collision
m1 m2
m1 1v
50
Simulation 3.3
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Solution :Solution :
From the principle of conservation of linear momentum,
The x-component of linear momentum,
fi pp
x22x11x22x11 vmvmumum
1s m 4.84 x2v
;s m 20 ;kg 0.900 ;kg 0.250 1 121 umm0 ;s m 4 ;0 1 5θvu 112
fxix pp
x211 vθv 0.900cos0.2500200.250
x2v0.90050cos40.2505
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Solution :Solution :
The y-component of linear momentum,
Magnitude of the soccer ball,
Direction of the soccer ball,
y22y11 vmvm 0
1s m 0.851 y2v
fyiy pp
y2v0.90050sin40.2500
2y22
x22 vvv
1s m 4.910.8514.84 222v
4.84
0.851tantan 1
x2
y212 v
vθ
9.972θ from positive x-axis anticlockwisefrom positive x-axis anticlockwise
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1. An object P of mass 4 kg moving with a velocity 4 m s 1 collides elastically with another object Q of mass 2 kg moving with a velocity 3 m s1 towards it.
a. Determine the total momentum before collision.
b. If P immediately stop after the collision, calculate the final
velocity of Q.
c. If the two objects stick together after the collision, calculate
the final velocity of both objects.
ANS. : 10 kg m sANS. : 10 kg m s11; 5 m s; 5 m s11 to the right; 1.7 m s to the right; 1.7 m s11 to the right to the right
2. A marksman holds a rifle of mass mr = 3.00 kg loosely in his hands, so as to let it recoil freely when fired. He fires a bullet of mass mb = 5.00 g horizontally with a velocity 300 m s-1. Determine
a. the recoil velocity of the rifle,
b. the final momentum of the system.
ANS. : ANS. : 0.5 m s0.5 m s11; U think.; U think.
Exercise 3.2.1 :
PHYSICS CHAPTER 3
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3.
In figures 3.16 show a 3.50 g bullet is fired horizontally at two blocks at rest on a frictionless tabletop. The bullet passes through the first block, with mass 1.20 kg, and embeds itself in the second block, with mass 1.80 kg. Speeds of 0.630 m s1 and 1.40 m s-1, respectively, are thereby given to the blocks. Neglecting the mass removed from the first block by the bullet, determine
a. the speed of the bullet immediately after it emerges from the first block and
b. the initial speed of the bullet.
ANS. : 721 m sANS. : 721 m s11; 937.4 m s; 937.4 m s11
Figure 3.16Figure 3.16
1.20 kg 1.80 kg
0.630 m s-1 1.40 m s-1
Before
After
PHYSICS CHAPTER 3
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Figure 3.17Figure 3.17
4. A ball moving with a speed of 17 m s1 strikes an identical ball that is initially at rest. After the collision, the incoming ball has been deviated by 45 from its original direction, and the struck ball moves off at 30 from the original direction as shown in figure 3.17. Calculate the speed of each ball after the collision.
ANS. : 8.80 m sANS. : 8.80 m s 1 1; 12.4 m s; 12.4 m s11
Exercise 3.2.1 :
PHYSICS CHAPTER 3
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3.2.3 Collision is defined as an isolated event in which two or more bodies an isolated event in which two or more bodies
(the colliding bodies) exert relatively strong forces on each (the colliding bodies) exert relatively strong forces on each other for a relatively short timeother for a relatively short time.
From the Newton’s Law of impact, the coefficient of restitutioncoefficient of restitution is defined as the ratio of the relative velocity after collision the ratio of the relative velocity after collision to the relative velocity before collisionto the relative velocity before collision.
OR
12
12
uu
vve
Where
collisionafter velocity relative :12 vv nrestitutio oft coefficien :e
collision before velocity relative :12 uu
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The coefficient of restitution, e is used to measure the measure the elasticity of the colliding bodieselasticity of the colliding bodies where its value always positive (0positive (0 e e 1) 1).
The coefficient of restitution, e is dimensionless (no unit)dimensionless (no unit). Table 3.1 shows the type of collision based on the value of e.
Table 3.1Table 3.1
Coefficient of
restitution, eType of collision
1
<1
0
Elastic
Inelastic
Completely inelastic
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Elastic collisionElastic collision is defined as one in which the total kinetic energy (as well as one in which the total kinetic energy (as well as
total momentum) of the system is the same before and after total momentum) of the system is the same before and after the collisionthe collision.
Figure 3.18 shows the head-on collision of two billiard balls.
11 22
Before collision
At collision
After collision
11 2222um11um
11 2222vm11vm
Figure 3.18Figure 3.18
Simulation 3.4
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The properties of elastic collision are
a. The coefficient of restitution, ee = 1 = 1
b. The total momentum is conservedtotal momentum is conserved.
c. The total kinetic energy is conservedtotal kinetic energy is conserved.
OR
fi pp
fi KK
222
211
222
211 vmvmumum
2
1
2
1
2
1
2
1
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Inelastic (non-elastic) collisionInelastic (non-elastic) collision is defined as one in which the total kinetic energy of the one in which the total kinetic energy of the
system is not the same before and after the collision (even system is not the same before and after the collision (even though the total momentum of the system is conserved)though the total momentum of the system is conserved).
Figure 3.19 shows the model of a completely inelastic completely inelastic collisioncollision of two billiard balls.
11 22At collision
After collision (stick together)
11 22v
Figure 3.19Figure 3.19
Before collision 11 2211um 02u
2m
Simulation 3.5
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Caution: Not allNot all the inelastic collision is stick togetherstick together. In fact, inelastic collisions include many situationsmany situations in which
the bodies do not stickbodies do not stick. The properties of inelastic collision are
a. The coefficient of restitution, 0 0 ee < 1 < 1b. The total momentum is conservedtotal momentum is conserved.
c. The total kinetic energy is not conservedtotal kinetic energy is not conserved because some of the energy is converted to internal energyinternal energy and some of it is transferred away by means of sound or heatsound or heat. But the total total energy is conservedenergy is conserved.
OR
fi pp
fi EE energy losses fi KK
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Two titanium spheres approach each other head-on with the same speed and collide elastically. After the collision, one of the spheres, whose mass is 500 g, remains at rest. Calculate the mass of the other sphere.
Solution :Solution :
By using the principle of conservation of linear momentum, thus
Example 6 :
0 ; kg; 0.500 1211 vuuum
Before collision 11 22uu
After collision 11 22
?2v01v
?2m
fi pp
22112211 vmvmumum 2221 vmumum (1)
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Solution :Solution :
Since the collision is elastic then e = 1, thus
By substituting eq. (2) into eq. (1), therefore
12
12
uu
vve
uu
v2 01
uv2 2 (2)
0.5003
1
3
1 12 mm
kg 0.1672m
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A ball is dropped from a height of 2.00 m above a tile floor and rebounds to a height of 1.30 m.
a. Determine the ball’s speed just before and after strike the floor.
b. State the type of the collision between ball and floor. Give
reason. (Given g = 9.81 m s2)
Solution :Solution :
a. i. Before collision,
Thus
Example 7 :
m 1.30 m; 2.00 10 hh
1
1
Floor (2)
1 1
m 2.00
1v
m 1.30
0 v
'1v
0u m 2.00 0y hs
y22
1 gsuv 2 2.009.8120 2
1v1s m 6.26 1v
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Solution :Solution :
a. ii. After collision,
Thus
b. The initial and final velocities of the floor are zero.
By using equation of Newton’s law of restitution,
Therefore the collision between ball and floor is inelastic inelastic.
m 1.30 1y hs
y2
12 gsvv 2'
1s m 5.05' 1v
1.309.812' 21v0
6.260
5.050'
12
12
vu
vve
0.807e
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3.2.4 Impulse, Let a single constant force, constant force, FF acts on an object in a short time
interval (collision), thus the Newton’s 2nd law can be written as
is defined as the product of a force, the product of a force, F F and the time, and the time, tt OR the change of momentumthe change of momentum.
is a vector quantityvector quantity whose directiondirection is the samesame as the constant forceconstant force on the object.
J
constant dt
pdFF
12 pppddtFJ
momentum final: 2p
where
momentum initial: 1p
force impulsive :F
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The S.I. unit of impulse is N sN s or kg m skg m s11. If the forceforce acts on the object is not constantnot constant then
Since impulse and momentum are both vector quantities, then it is often easiest to use them in component form :
dtFdtFJ av
t
t
2
1
where force impulsive average :avF
xxx1x2xavx uvmppdtFJ
yyy1y2yavy uvmppdtFJ
zzz1z2zavz uvmppdtFJ
consider 2-D consider 2-D collision onlycollision only
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When two objects in collision, the impulsive force, F against
time, t graph is given by the figure 3.20.
1t 2tFigure 3.20Figure 3.20 t0
F
Shaded area under the Ft graph = impulse
Picture 3.1
Picture 3.2
Picture 3.3
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A 0.20 kg tennis ball strikes the wall horizontally with a speed of 100 m s1 and it bounces off with a speed of 70 m s1 in the opposite direction.
a. Calculate the magnitude of impulse delivered to the ball by the
wall,
b. If the ball is in contact with the wall for 10 ms, determine the
magnitude of average force exerted by the wall on the ball.
Solution :Solution :
Example 8 :
Wall (2)11
1s m 100 1u
111s m 70 1v
0 22 uv
kg 0.201 m
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Solution :Solution :
a. From the equation of impulse that the force is constant,
Therefore the magnitude of the impulse is 34 N s34 N s.
b. Given the contact time,
12 ppdpJ 111 uvmJ
100700.20 J
s N 34J
s 1010 3dtdtFJ av 3101034 avF
N 3400avF
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An estimated force-time curve for a tennis ball of mass 60.0 g struck by a racket is shown in figure 3.21. Determine
a. the impulse delivered to the ball,
b. the speed of the ball after being struck, assuming the ball is
being served so it is nearly at rest initially.
Example 9 :
0.2 1.8 mst0
kNF
1.0
18
Figure 3.21Figure 3.21
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Solution :Solution :
a. From the force-time graph,
b. Given the ball’s initial speed,
graph under the area tFJ
33 1018100.21.82
1 J
s N 14.4J0u
uvmdpJ
01060.014.4 3 v1s m 240 v
kg 1060.0 3m
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1. A steel ball with mass 40.0 g is dropped from a height of 2.00 m onto a horizontal steel slab. The ball rebounds to a height of 1.60 m.
a. Calculate the impulse delivered to the ball during impact.
b. If the ball is in contact with the slab for 2.00 ms, determine
the average force on the ball during impact.
ANS. : 0.47 N s; 237. 1 NANS. : 0.47 N s; 237. 1 N
2. A golf ball (m = 46.0 g) is struck with a force that makes an angle of 45 with the horizontal. The ball lands 200 m away on a flat fairway. If the golf club and ball are in contact for 7.00 ms, calculate the average force of impact. (neglect the air resistance.)
ANS. : 293 NANS. : 293 N
Exercise 3.2.2 :
PHYSICS CHAPTER 3
55
Figure 3.22Figure 3.22
3.
A tennis ball of mass, m = 0.060 kg and a speed, v = 28 m s1 strikes a wall at a 45 angle and rebounds with the same speed at 45 as shown in figure 3.22. Calculate the impulse given by the wall.
ANS. : 2.4 N s to the left or ANS. : 2.4 N s to the left or 2.4 N s2.4 N s
Exercise 3.2.2 :
PHYSICS CHAPTER 3
56
At the end of this chapter, students should be able to: At the end of this chapter, students should be able to: UseUse Newton’s Third Law to Newton’s Third Law to explainexplain the concept of normal the concept of normal
reaction force.reaction force. State and useState and use equation for frictional force and equation for frictional force and
distinguishdistinguish between static friction, between static friction,
and kinetic (dynamic) friction, and kinetic (dynamic) friction,
Learning Outcome:
3.3 Reaction and frictional forces (1 hour)
ww
w.k
mp
h.m
atri
k.ed
u.m
y/p
hys
ics
ww
w.k
mp
h.m
atri
k.ed
u.m
y/p
hys
ics
Nf ss
Nf kk
PHYSICS CHAPTER 3
57
3.3 Reaction and frictional forces
3.3.1 Reaction (normal) force, is defined as a reaction force that exerted by the surface to a reaction force that exerted by the surface to
an object interact with it and the direction always an object interact with it and the direction always perpendicular to the surfaceperpendicular to the surface.
Case 1: Horizontal surfaceCase 1: Horizontal surface An object lies at rest on a flat horizontal surface as shown in
figure 3.23.
RN
or
N
gmW
0mgNFy
mgN ThereforeFigure 3.23Figure 3.23
Action: weight of an object is exerted on weight of an object is exerted on the horizontal surfacethe horizontal surface
Reaction: surface is exerted a force, surface is exerted a force, NN on on the object . the object .
PHYSICS CHAPTER 3
58
Figure 3.24Figure 3.24
yW
Case 2 : Inclined planeCase 2 : Inclined plane An object lies at rest on a rough inclined plane as shown in
figure 3.24.
gmW
θmgWx sin
N
xW
θmgWy cos
0yy WNF
Component of the weight :
xy
ThereforecosmgN
Action: y-component of the object’s y-component of the object’s weight is exerted on the inclined weight is exerted on the inclined surface.surface.
Reaction: surface is exerted a force, surface is exerted a force, NN on on the object. the object.
PHYSICS CHAPTER 3
59
Case 3 : Motion of a liftCase 3 : Motion of a lift Consider a person standing inside a lift as shown in figures
3.25a, 3.25b and 3.25c.
a. Lift moving upward at a uniform velocity
gmW
N
Since the lift moving at a uniform velocity, thus
0yaTherefore
0yF0 mgN
mgN Figure 3.25aFigure 3.25a
PHYSICS CHAPTER 3
60
a
b. Lift moving upwards at a constant acceleration, a
gmW
N By applying the newton’s 2nd law
of motion, thus
yy maF
mamgN
gamN Figure 3.25bFigure 3.25b
PHYSICS CHAPTER 3
61
a
c. Lift moving downwards at a constant acceleration, a
Caution : N is also known as apparent weightapparent weight and W is true true weight weight.
gmW
N
By applying the newton’s 2nd law of motion, thus
yy maF
maNmg
agmN Figure 3.25cFigure 3.25c
PHYSICS CHAPTER 3
62
3.3.2 Frictional force, is defined as a force that resists the motion of one surface a force that resists the motion of one surface
relative to another with which it is in contactrelative 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, Coefficient of friction, is defined as the ratio between frictional force to reaction
force.OR
is dimensionless and depends on the nature of the surfaces.
f
Nf
Nf force frictional:f
friction oft coefficien : μforcereaction : N
where
N
f
PHYSICS CHAPTER 3
63
There are three types of frictional force :
Static, fs (frictional force act on the object before its before its
movemove)
Kinetic, fk (frictional force act on the object when its movewhen its move)
Rolling, fr (frictional force act on the object when its rollingwhen its rolling)
Caution: The direction of the frictional forcedirection of the frictional force exerted by a surface on
an object is always in the opposite direction of the motionopposite direction of the motion. The frictional and the reaction forcesfrictional and the reaction forces are always
perpendicularperpendicular.
Nf kk
Nf ss
Nf rr
skr fff where
thus skr
Can be ignored
Simulation 3.6
PHYSICS CHAPTER 3
64
Case 1 : Horizontal surfaceCase 1 : Horizontal surface Consider a box of mass m is pulled along a horizontal surface
by a horizontal force, F as shown in figures 3.26.
x-component :
y-component :
Figure 3.26Figure 3.26
maFF nettx
F
a
gm
N
f
mafF
0yF
mgN
PHYSICS CHAPTER 3
65
Case 2 : Inclined planeCase 2 : Inclined plane Consider a box of mass m is pulled along an inclined plane by a
force, F as shown in figures 3.27.
x-component
(parallel to the inclined
plane) :
y-component
(perpendicular to the inclined plane:
a
Figure 3.27Figure 3.27
N
gmW
xy
yW
xW
F
f
0yF0 yWN
θmgN cos
maFx
mafWF x fθmgmaF sin
Simulation 3.7
PHYSICS CHAPTER 3
66
A box of mass 20 kg is on a rough horizontal plane. The box is
pulled by a force, F which is applied at an angle of 30 above horizontal as shown in figure 3.28. If the coefficient of static friction between the box and the plane is 0.3 and the box moves at a constant speed, calculate
a. the normal reaction force,
b. the applied force F,
c. the static friction force.
(Given g = 9.81 m s-2)
Example 10 :
Figure 3.28Figure 3.28
30
F
PHYSICS CHAPTER 3
67
Solution :Solution :
a. Since the box moves at constant speed thus
x-component :
0.3 kg; 20 sμm
30
Fconstant speed
N
gm
sf 30cosF
30sinF
0xF
030cos sfF
0a
30cos
0.3NF
030cos NμF s
(1)
PHYSICS CHAPTER 3
68
Solution :Solution :
y-component :
By substituting eq. (1) into eq. (2), hence
b. Therefore the applied force is given by
c. The static friction force is
0yF
030sin mgFN
N 167N
(2)
9.812030sin FN
N 57.9
30cos
1670.3 F
19630sin FN
19630sin30cos
0.3
NN
Nμf ss N 50.11670.3 sf
PHYSICS CHAPTER 3
69
A block of mass 200 kg is pulled along an inclined plane of 30 by
a force, F = 2 kN as shown in figure 3.29. The coefficient of kinetic friction of the plane is 0.4. Determine
a. the normal force,
b. the nett force,
c. the acceleration of the block,
d. the time taken for the block to travel 30 m from rest.
(Given g = 9.81 m s-2)
Example 11 :
Figure 3.29Figure 3.29
F
30
20
PHYSICS CHAPTER 3
70
30
Solution :Solution :
a. y-component :
0.4 N; 2000 kg; 200 kμFm
0yF
030cos20sin mgFN
N 1015N
F
xy
30
20cosF
N
20sinF20
gm
kf 30cosmg
30sinmg
a
030cos9.8120020sin2000 N
PHYSICS CHAPTER 3
71
Solution :Solution :
b. The nett force is directed along the inclined plane surface.
x-component :
c.
d. Given
xnett FF
knett fmgFF 30sin20cosNμmgFF knett 30sin20cos
N 492nettF 10150.430sin9.8120020cos2000
nettF
maFnett a200492 2s m 2.46 a
0 m; 30 us2
2
1atuts 22.46
2
130 t0
s 4.94t
PHYSICS CHAPTER 3
72
Figure 3.30Figure 3.30
Exercise 3.3 :1.
A 5.00 kg object placed on a frictionless, horizontal table is connected to a string that passes over a pulley and then is fastened to a hanging 9.00 kg object as in figure 3.30.
a. Sketch free body diagrams of both objects,
b. Calculate the acceleration of the two objects and the
tension in the string.
(Given g = 9.81 m s2)
ANS. : 6.30 m sANS. : 6.30 m s22; 31.5 N; 31.5 N
PHYSICS CHAPTER 3
73
Figure 3.31Figure 3.31
2. Two object are connected by a light string that passes over a frictionless pulley as in figure 3.31.
The coefficient of kinetic friction of
the plane is 0.3 and m1 = 2.00 kg,
m2 = 6.00 kg and = 55.
a. Sketch free body diagrams of
both objects.
b. Determine
i. the accelerations of the objects,
ii. the tension in the string
iii. the speed of each object 2.00 s after being released from
rest. (Given g = 9.81 m s2)
ANS. : 2.31 m sANS. : 2.31 m s22; 24.2 N; 4.62 m s; 24.2 N; 4.62 m s11
Exercise 3.3 :
PHYSICS CHAPTER 3
74
3. A 5.00 g bullet is fired horizontally into a 1.20 kg wooden block resting on a horizontal surface. The coefficient of kinetic friction between block and surface is 0.20. The bullet remains embedded in the block, which is observed to slide 0.230 m along the surface before stopping. Calculate the initial speed of the bullet.
(Given g = 9.81 m s2)
Tips : Use Newton’s second law of motion involving
acceleration. Principle of conservation of linear momentum. Equation of motion for linear motion.
ANS. : 229 m sANS. : 229 m s11
Exercise 3.3 :
PHYSICS CHAPTER 3
75
Figure 3.32Figure 3.32
4. The block shown in figure 3.32,
has mass, m =7.0 kg and lies on a smooth frictionless plane tilted
at an angle, = 22.0 to the horizontal.
a. Determine the acceleration of
the block as it slides down the
plane.
b. If the block starts from rest
12.0 m up the plane from its
base, calculate the block’s
speed when it reaches the
bottom of the incline plane.
(Given g = 9.81 m s2)
ANS. : 3.68 m sANS. : 3.68 m s22; 9.40 m s; 9.40 m s11
Exercise 3.3 :
76
PHYSICS CHAPTER 3
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CHAPTER 4 :Work, Energy and Power