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UNIT-07
Newton’s Three Laws of Motion
1. Learning Objectives:
1. Understand the three laws of motion, their proper areas of applicability and especially the
difference between the statements of the first and the third laws
2. Know the difference between inertial and non-inertial frames of reference.
3. Be able to draw free-body diagrams for a single or a system of several objects.
4. Be able to use F = Ma (with item #3 above) to solve problems for a system of one or
several objects.
5. Be familiar with concepts of static and kinetic friction.
6. Understand the concepts of tangential velocity and centripetal acceleration and be able to
solve problems involving circular motion
2. Uniform Motion
One of the central concepts in Newtonian dynamics is that of uniform motion. In Figure
1a-e are shown the positions of a particle at regular intervals of time. The motion of the
particle can be described as follows:
[a] The velocity V = constant = 0 , the particle is at rest.
[b] The velocity V = Vi, where V is constant .
[a]
X
[b]
X
[c] The speed is increasing, the direction remains unchanged.
Thus, velocity is not constant and the particle has an acceleration.
[d] The speed is constant, the direction of motion is changing.
Thus, velocity is not constant and the particle has an acceleration.
[c]
X
[e] The speed and the direction of motion are changing. Thus,
velocity is not constant and the particle has an acceleration.
Points to note:
[i] Motion with constant velocity implies motion in a straight
line with constant speed. When a particle moves with constant
velocity it is said to be in a state of uniform motion. Figure1a,b
are examples of uniform motion.
[d]
[e]
FIGURE 1
[ii] A particle not moving in a straight line is necessarily accelerating (Fig.1d,e).
3. Newton’s Laws
The first two laws are about change in the state of motion of a mass.
The First Law: A mass will continue in a state of uniform motion unless an external
agent exerts an influence on the mass.
The Second Law:
F = ma = m dV/dt
…[1]
Comments:
[a] The “external agent” mentioned in the first law is now identified as the force F.
[b] If F = 0, then dV = 0 ,which implies uniform motion. Thus the statement of
Newton’s
first law is contained in the second law.
[c] F in eq.[1] is the net force on m, F = F1+ F2+ F3…… , where F1, F2, F3, etc. are
the individual forces acting on m. To emphasize this point, sometimes the second law
(eq.[1]) is written as:
!
!
"
"
"
+
F
2
3 + .. = ma
!F = F + F
1
…[2]
[d] Remember F = ma is a vector equation which means it expresses three relations: ΣFx = max ; ΣFy = may ; ΣFz = maz These equations imply that the acceleration of an object in a given direction is governed only by the net force in that direction. When ΣFx = ΣFy = ΣFz = 0, the system is said to be in equilibrium and the equation, ΣFx = ΣFy = ΣFz = 0 is known as the first condition of equilibrium. [e] Newton had actually stated the second law as:
F = dp/dt = d(mV)/dt = m dV/dt + Vdm/dt
…[2]
Here p = mV is called the linear momentum. We will discuss the linear momentum in
a few weeks. If the mass of the system doesn’t change during motion, dm/dt = 0 and
eq.[2] reduces to eq.[1].
4. The Free-body Force Diagram.
To apply F = ma to an object you must first find out the net force acting on it. This is
done by drawing a free-body force diagram or simply free-body diagram (FBD) for
the object. Review all the FBD examples discussed in the lectures.
The Third Law
The third law deals with interaction between two objects (call them 1 and 2) and
states that the force exerted by 1 on 2 is equal in magnitude and opposite in direction
to the force exerted by 2 on (BE CAREFUL :Every time you come across a pair of
forces that are equal and opposite, don’t confuse it with the action and reaction forces
of the third Law.)
Try the following.
1. You are pulling a crate by applying a force on it. The crate exerts an equal and
opposite force on you. If these forces add up to zero according to the third Law
then what makes the crate move?
(Ans. The forces act on two different objects – you and the crate – and cannot be
added to yield zero force. The motion of the crate is determined by the force, F,
acting on the crate. Since there is a net force exerted on it (by you), it moves
according to F = ma.
2. A book of mass m is resting on the top of a table which in turn is resting on the
floor (Fig.a). The FBD for the book is shown in Fig.(b) below. The force exerted by
the Earth
on the book, Fbe = mg is equal and opposite to the normal reaction force N exerted by
the table on the book. Is the fact that Fbe – N = 0 and therefore the book is in
equilibrium an instance of Newton’s third law or Newton’s first Law?
Answer: Both forces are acting on the same object. Therefore this is an instance of
Newton’s first Law (and the second Law), but not of the third Law. The reaction to
Fbe is the force exerted by the book on the Earth Feb (see Fig.(c), it is NOT an FBD).
The fact that Feb = - Fbe is an instance of the third Law.
book
book
(a)
(b)
table
Fbe
floor (earth)
N
book
(C)
Fbe
Feb
EARTH
4.1 Using the FBD – some examples. (We may not discuss this topic due to time constraints) Problem 1:A block of mass m = 50.0kg is released from rest on a frictionless inclined plane of angle 370. What is the magnitude of the acceleration of the block? N
Α
A
370
500sin37ο
37ο
=300N
500cos37o=400N
500N
Step1. Draw the FBD for each part of the system. Make sure you label each force (or force component) and any angles involved. Remember you must draw all the forces acting ON the object of your interest. Do not include in the FBD any forces that the object exerts on other parts of the system or some external system. Improper or incomplete labeling can lead to an incorrect application of the second law. Step 2. Choose appropriate set of axes and decompose all forces along the two axes. In the example above we have chosen the x-­‐axis along the incline pointing upward and the y-­‐axis normal to the incline. Step-­‐3. Apply the first condition of equilibrium: ΣFx = max ; ΣFy = may to each FBD. In the present case we get the following equations for mass m: N – 400 = 0 -­‐-­‐-­‐-­‐-­‐-­‐-­‐-­‐-­‐[1] 500sin37o= 300 = 50a -­‐-­‐-­‐-­‐-­‐-­‐-­‐-­‐[2] Or a = 6.0m/s2 Problem 2 : A block of mass m = 50.0kg on a frictionless inclined plane of angle 370 is connected by a cord over a massless and frictionless pulley to a second mass M= 100.0 kg. [a] What is the magnitude of the acceleration of each block ? [b] What is the tension T in the cord? (Note: In this problem we have simply attached another block B to the existing block A.. Notice how the FBD of block A and the resulting F = ma equation changes.) T
N
T
A
A
B
B
500sin37ο
37ο
=300N
37o
1000N
500cos37o=400N
500N
Step1. Draw the FBD for each part of the system. Make sure you label each force (or force component) and any angles involved. Remember you must draw all the forces acting ON the object of your interest. Do not include in the FBD any forces that the object exerts on other parts of the system or some external system. Improper or incomplete labeling can lead to an incorrect application of the second law. Step 2. Choose appropriate set of axes and decompose all forces along the two axes. In the example above we have chosen the x-­‐axis along the incline pointing upward and the y-­‐axis normal to the incline. Step-­‐3. Apply the first condition of equilibrium: ΣFx = max ; ΣFy = may to each FBD. In the present case we get the following equations for mass m: N – 400 = 0 -­‐-­‐-­‐-­‐-­‐-­‐-­‐-­‐-­‐[1] T – 300 = 50a -­‐-­‐-­‐-­‐-­‐-­‐-­‐-­‐[2] (Note: we are assuming the mass would move up the incline. If our assumption is right the value of a would turn out to be positive, if we made the wrong assumption the value of a would be negative. The negative value of a does not render our solution wrong. The physical interpretation would then tell us that the mass would be moving downward.) From the FBD of M, we get 1000 –T = 100a -­‐-­‐-­‐-­‐-­‐-­‐-­‐[3] Step-­‐4. Solve the equations obtained in Step-­‐3. Thus we get, N = 400 N Adding [3] and [4] gives, 700 = 150a or a = 4.7m/s2. From [2] T = 300 +50a = 533.3 N. Step-­‐5. SMILE 5. Concept Questions 1. Which of the following statements is most correct? (a) It is possible for an object to have motion in the absence of forces on the object. (b) It is possible to have forces on an object in the absence of motion of the object. (c) Neither (a) nor (b) is correct. (d) Both (a) and (b) are correct. Answer: (d). Choice (a) is true. Newton’s first law tells us that motion requires no force: An object in motion continues to move at constant velocity in the absence of external forces. Choice (b) is also true: A stationary object can have several forces acting on it, but if the vector sum of all these external forces is zero, there is no net force and the object remains stationary. 2. You push an object, initially at rest, across a frictionless floor with a constant force for a time interval Δt, resulting in a final speed of v for the object. You repeat the experiment, but with a force that is twice as large. What time interval is now required to reach the same final speed v? (a) 4 Δt (b) 2 Δt (c) Δt (d) Δt/2 (e) Δ
t/4 Answer: (d). With twice the force, the object will experience twice the acceleration. Because the force is constant, the acceleration is constant, and the speed of the object, starting from rest, is given by v = at. With twice the acceleration, the object will arrive at speed v at half the time. 3. If a fly collides with the windshield of a fast-­‐moving bus, which experiences an impact force with a larger magnitude? (a) The fly does. (b) The bus does. (c) The same force is experienced by both. Which experiences the greater acceleration? (d) The fly does. (e) The bus does. (f) The same acceleration is experienced by both. Answer: (c), (d). In accordance with Newton’s third law, the fly and the bus experience forces that are equal in magnitude but opposite in direction. Because the fly has such a small mass, Newton’s second law tells us that it undergoes a very large acceleration. The huge mass of the bus means that it more effectively resists any change in its motion and exhibits a small acceleration. 4. Which of the following is the reaction force to the gravitational force acting on your body as you sit in your desk chair? (a) the normal force from the chair (b) the force you apply downward on the seat of the chair (c) neither of these forces Answer: (c). The reaction force to your weight is an upward gravitational force on the Earth caused by you. PROBLEMS 
1. A force F applied to an object of mass m1 produces an acceleration of 3.00 m/s2. The same force applied to a second object of mass m2 produces an acceleration of 1.00 m/s2. (a) What is the value of the ratio m1/m2? (b) If m1 and m2 
are combined, find their acceleration under the action of the force F . Solution: For the same force F, acting on different masses F = m1a1 and F = m2 a2 so m1a1 = m2 a2 . (a)
m1 a2
1
= =
m2 a1
3
m2 = 3m1
(b)
(
F = ( m1 + m2 ) a = 4m1a = m1 3.00 m s 2
)
a= 0.750 m s 2
2. (a) A car with a mass of 1000.0 kg is moving to the right with a constant speed of 1.8 m/s. What is the total force on the car? (b) What is the total force on the car if it is moving to the left? Solution: Since the car is moving with constant speed and in a straight line, the resultant force on it must be zero regardless of whether it is moving (a)
toward the right or
(b)
the left.
!
3. A 3.00-­‐‑kg object undergoes an acceleration given by a = (2.00 î + 5.00 ĵ) m/s2. Find the resultant force acting on it and the magnitude of the resultant force. Solution: m = 3.00 kg
!
a = 2.00 î + 5.00 ĵ m s 2
!
!
! F = ma = 6.00 î + 15.0 ĵ N
(
)
(
!
)
! F = ( 6.00 ) + (15.0 )
2
2
N = 16.2 N


4. Two forces F1 and F2 act on a 5.00-­‐‑kg object. If F1 = 20.0 N and F2 = 15.0 N, find the accelerations in (a) and (b) of Figure 4. Solution: (a)

F2
! ! !
! F = F1 + F2 = 20.0 î + 15.0 ĵ N
(
!
!
! F = ma :
)
!
20.0 î + 15.0 ĵ = 5.00 a
!
a = 4.00 î + 3.00 ĵ m s 2
(
)
or

F1

F2
a= 5.00 m s 2 at θ = 36.9°
(b)
F2 x = 15.0 cos 60.0° = 7.50 N
F2 y = 15.0 sin 60.0° = 13.0 N
!
F2 = 7.50 î + 13.0 ĵ N
! ! !
!
!
! F = F1 + F2 = 27.5 î + 13.0 ĵ N = ma = 5.00a
(
!
a=
(
)
( 5.50 î + 2.60 ĵ) m s
)
2
= 6.08 m s 2 at 25.3°

F1
FIG. 4a(top) and b !
!
5. Three forces, given by F1 = (!2.00 î + 2.00 ĵ) N, F2 = (5.00 î ! 3.00 ĵ) N, and !
F3 = (!45.0 î) N act on an object to give it an acceleration of magnitude 3.75 m/s2. (a) What is the direction of the acceleration? (b) What is the mass of the object? (c) If the object is initially at rest, what is its speed after 10.0 s? (d) What are the velocity components of the object after 10.0 s? Solution: 

∑ F = ma reads ( !2.00 î + 2.00 ĵ + 5.00 î ! 3.00 ĵ ! 45.0 î ) N = m ( 3.75 m s ) â 2

where a represents the direction of a ( !42.0 î ! 1.00 ĵ) N = m ( 3.75 m s ) â ! F = ( 42.0 ) + (1.00 )
! F = 42.0 N at 181° = m ( 3.75
2
!
2
2
" 1.00 %
N at tan !1 $
below # 42.0 '&
the –x-­‐axis !
)
m s 2 â . For the vectors to be equal, their magnitudes and their directions must be equal. (a)
! â is at 181° counterclockwise from the x-axis
(b)
m=
(d)
!
! !
!
v f = v i + at = 0 + 3.75 m s 2 at 181° 10.0 s so v f = 37.5 m s at 181°
42.0 N
= 11. 2 kg
3.75 m s 2
(
)
!
v f = 37.5 m s cos181° î + 37.5 m s sin181° ĵ so
!
v f = !37.5 î ! 0.893 ĵ m s
(
)
(c) 
v f = 37.52 + 0.8932 m s = 37.5 m s 6. If a man weighs 900 N on the Earth, what would he weigh on Jupiter, where the free-­‐‑fall acceleration is 25.9 m/s2? Solution: Fg = mg = 900 N , m =
900 N
= 91.8 kg 9.80 m s 2
(F )
(
)
= 91.8 kg 25.9 m s 2 = 2.38 kN g on Jupiter
7. An electron of mass 9.11 × 10–31 kg has an initial speed of 3.00 × 105 m/s. It travels in a straight line, and its speed increases to 7.00 × 105 m/s in a distance of 5.00 cm. Assuming that its acceleration is constant, (a) determine the net force exerted on the electron and (b) compare this force with the weight of the electron. Solution:
(a)
∑ F = ma and v 2f = v i2 + 2ax f or a =
v 2f − v i2
2x f
.
Therefore,
!F = m
(v
2
f
" vi2
)
2x f
! F = 9.11 # 10 "31
(b)
(
) (
)
$ 7.00 # 10 5 m s 2 2 " 3.00 # 10 5 m s 2 2 &
'=
kg %
3.64 # 10 "18 N .
2 ( 0.050 0 m )
The weight of the electron is
(
)(
)
Fg = mg = 9.11 ! 10 "31 kg 9.80 m s 2 = 8.93 ! 10 "30 N
The accelerating force is
4.08 ! 1011 times the weight of the electron.