PHY 105 (Module 2)
March 30 – April 13, 5 hours
Newton’s Laws (of motion)
Moment of a Force
Equilibrium of a particle
Newton’s Law of Universal Gravitation
Lecturer: Prof Joshua O. Ojo
Office: Room PY 122 White House
Now, your study of Physics begins!
• All you’ve learnt so far are very important
tools to enable us discuss Physics. But you
might as well have learnt those in a Maths
class.
• Physics deals with the study of the physical
universe with the objective of finding how
things work – the Laws behind it all.
(Note: we assume there are laws, and we try to state
these empirically)
Newton’s Laws (of motion)
-named after Sir Isaac Newton, British scientist.
• The first suite of Laws we’ll be studying deals with
motion. From the motion of a rocket headed for
the moon to that of blood flowing in your
capillaries.
• As is true for most topics in Physics, they are
really easy-to-grasp and “common-sensical”
• 90% of the rest of this Course is essentially the
application of what you’ll be learning today. So
better pay close attention!
Newton’s 1st Law
An object continues in a state of rest or of uniform motion along
a straight line unless a net force acts on it
• Frequently called the law of inertial, this law essentially
introduces the very important physical quantities:
Force and Inertial mass. It says:
• 1. A body at rest will continue to be at rest except
there is a net force acting on it.
• 2. A body in uniform motion will continue with its
uniform motion unless there is a net force acting on
it.
(Uniform motion => No change in either magnitude or
direction of the velocity vector. Obviously Statement 1 is
just a special case of Statement 2).
Summary: Inertial (mass) makes a physical object reluctant
to welcome changes. Net Force compels it !
Poser: How about timed bombs?
Practical Illustrations
Use of safety belts.
Here car stopped by the wall. Net force
compels motion of car to change
However, driver (when not restrained
to the car by safety belt) is not
affected by the force provided by the
wall. He continues in his motion along
a straight line!
Further illustrations below.
Later on in this and other modules, we’ll consider cases where
change in motion is essentially that of change in direction. The
nature of Force can be gravitational (this module), electric,
magnetic, nuclear weak and strong – PHY !06
Newton’s 2nd Law
the rate of change of momentum is directly
proportional to the net force
• This gives us an empirical/operational
definition of “Force”. (Most of the calculations
you will be doing in Physics is based directly
on this empirical definition we are about to
describe)
• The 2nd Law of motion introduces to us the
vector quantity Momentum.
2nd Law
the rate of change of momentum is directly
proportional to the net force
• It means the more the applied/net force, the
greater the rate of change of momentum
• Momentum is the product of Mass and Velocity
• Questions: What is the momentum of an object
of
• i. mass 50 kg moving with a velocity of 100 m/s?
• ii. Mass = 1000 kg, v = 0.001 cm/s?
• Iii. Mass = 10 g, v = 1,0000 m/s?
2nd Law
•
•
•
•
•
the rate of change of momentum is directly
proportional to the net force
It means the more the applied/net force, the greater
the rate of change of momentum
Change of Momentum means: Momentum after the
application of the net force less Momentum before the
application of the Force
i.e. (Final Mass x Final velocity) – (Initial Mass x Initial
velocity)
Example: i. What is the momentum of a tipper (mass
3,000 kg) when loaded with 2,000 kg of sand if it is
moving with a uniform velocity of 10 m/s?
Ii. What is the momentum of the same tipper if it
moves with the same velocity after discharging its
load? [Units?]
2nd Law
the rate of change of momentum is directly
proportional to the net force
• What is rate of change?
• It means how long did it take to effect the
change.
• Example: If momentum of a car changes
uniformly from 2500 kgms-1 to 1500 kgms-1
in 10 s, what is the rate of change of
momentum?
• Ii. What is this rate if the time involved were
20s?
2nd Law
the rate of change of momentum is directly
proportional to the net force
• Summary of the 2nd Law:
• For momentum to change, something called
force must be involved (a more general
statement of the 1st law).
• The larger the force involved, the greater the
rate of change of momentum
• Mathematically,
2nd Law
the rate of change of momentum is directly
proportional to the net force
• If we choose appropriate
unit system, we can equate
the constant of
proportionality to 1, hence
we can write
F = ma
This famous equation you will use repeatedly
in your study of Physics. But remember, it is
only a special form of the 2nd Law of motion,
WHEN THE MASS IS NOT CHANGING.
For instance in the rocket problem, mass is
not constant. (Neither for a running athlete –
sweating, exhaling etc implies loss of water,
not to talk of fat burned off!)
Assignment
Calculate the Forces involved for various
values of masses, velocities and times.
Note unit of Force is Kgms-2 which in SI
Unit is Newton (N)
• Last class, we found an operational way of quantifying force.
• The Home work was for you to play with different values of
the different related physical quantities. Values of force can
affect values of mass, velocity, or time of the process.
• Before we take more critical look at:
• types of forces: Tensile force, compressive force, friction, Air
resistance force, Normal force, weight, etc;
• Nature of Forces: Contact, gravitational, electromagnetic,
nuclear
• And derivative quantities: work,energy, power, action,
impulse, moment, torque, etc
• let’s consider the 3rd Law.
Newton’s 3rd Law of motion
To every action (force), there is an equal and
opposite reaction (force)
• Forces always exist in pairs. Whenever there is an
interaction between two objects, there is a force upon
each of the objects. When the interaction ceases, the
two objects no longer experience the force [motion
then continues uniformly from that point on]. Forces
only exist as a result of an interaction
• When an object exerts a force (the action force) on a
second object, the second object also in turn exerts a
reaction force on the first object.
• When two bodies interact, the forces on the bodies
from each other, are always equal (magnitude) and in
opposite directions.
Newton’s 3rd Law of motion
To every action (force), there is an equal and
opposite reaction (force)
• Some every-day examples:
• Birds flapping wings to soar.
Upthrust.
• Fishes push water with fins for
forward propulsion (same with
canoes, etc)
• Cars push back the road to move
forward. (consider car without grip
on the road – eg, smooth/muddy
road or threadless tire, or jacked-up
car!)
Newton’s 3rd Law of motion
To every action (force), there is an equal and
opposite reaction (force)
• One major implication of the third law is the very
important law of conservative of momentum.
• Both Action Force and the Reaction Force
produces changes of momentum in the two
interacting bodies.
• Since the two forces are equal and opposite, and
act over exactly the same time period, equal
changes in momenta are produced in the two
different bodies.
• The relative masses of the bodies will therefore
determine what relative velocities they get
Newton’s 3rd Law of motion
To every action (force), there is an equal and opposite reaction
(force)
• Examples:
• See, the car moves backward after the impact. The Wall
also moves backward, even if imperceptibly in this case
• Recoil of a gun during shooting
• Remember, it’s momentum (product of mass and velocity)
that is conserved. The greater mass has the less velocity
after the interaction
Some particular forces
• A Gravitational Force Fg on a body is a pull by
another body. In most situations that we shall be
dealing with in this course, the other body is the
Earth (or some other astronomical body). Here,
• the force is directed down toward the centre of
the earth, and has
• magnitude Fg = mg, [From F = ma]
• where g here is the acceleration due to gravity
(more on this later)
Some particular forces….
• Weight W of a body is the magnitude of the
upward force needed to balance the
gravitational force on the body due to Earth
(or some other astronomical body). It is
related to the body’s mass by
• W= mg
Some particular forces……
• Frictional Force
• Is the force on a body when the body slides or
attempts to slide along a surface. The force is
always parallel to the surface and directed so as
to oppose the motion of the body. On a
frictionless surface, the frictional force is
negligible.
• When a body moves through air, friction is
termed air-resistance. In fluid, friction is
reflected as viscosity.
Illustration of Gravitation Force, Weight,
Air resistance/Friction
g is the same, but with air resistance net force is reduced. More later…..
See text: 6-4
Normal Force...
• Normal Force N is the force on a body from a
surface against which the body presses. The
normal force is always perpendicular (Normal)
to the surface
• Illustrate
j
N
F
ma
fF
mg
i
Tension
• Tension: When a cord is under tension, it pulls
on a body at each of its ends. The pull is
directed along the cord, away from the point
of attachment to each body. For a massless
cord (a cord with negligible mass), the pulls at
both ends of the cord have the same
magnitude T, even if the cord runs around a
massless, frictionless pulley (a pulley with
negligible mass and negligible friction on its
axle to oppose its rotation).
Applying Newton’s Laws to solve
various problems.
• Navigate to Answers for:
• Questions #1-7
Questions #8-#36
Questions #37-#46
Questions #47-#60
Examples for Discussion
• 8a. Orobo has a mass of 100 kg on the earth. What is Orobo's mass on the
moon where the force of gravity is approximately 1/6-th that of Earth's?
________ Explain or show your work.
• 8b. Sis Lepa weighs 360 N on Earth. What is Sis Lepa's mass on the moon
where the force of gravity is approximately 1/6-th that of Earth's?
________ Explain or show your work.
• 9. TRUE or FALSE:
• An object which is moving rightward has a rightward force acting upon it.
• 10. The amount of net force required to keep a 5-kg object moving
rightward with a constant velocity of 2 m/s is ____.
• a. 0 N
• b. 0.4 N
• c. 2 N
• d. 2.5 N
• e. 5 N
Examples for Discussion
• 12. Which one(s) of the following force
diagrams depict an object moving to the right
with a constant speed? List all that apply.
• .
Examples for Discussion
• 13. According to Newton's third law, every force is
accompanied by an equal and opposite reaction force.
The reason that these forces do not cancel each other
is ____.
• a. the action force acts for a longer time period
• b. the two forces are not always in the same direction
• c. one of the two forces is greater than the other
• d. the two forces act upon different objects; only forces
on the same object can balance each other.
• e. ... nonsense! They do cancel each other. Objects
accelerate because of the presence of a third force
Examples for Discussion
• 48. A 72-kg skydiver is falling from 5000 m. At an instant
during the fall, the skydiver encounters an air resistance
force of 540 Newtons. Determine the acceleration of
the skydiver at this instant. PSYW
• 50. A 5.20-N force is applied to a 1.05-kg object to
accelerate it rightwards across a friction-free surface.
Determine the acceleration of the object. (Neglect air
resistance.)PSYW
• 52. Determine the applied force required to accelerate a
3.25-kg object rightward with a constant acceleration of
1.20 m/s/s if the force of friction opposing the motion is
18.2 N. (Neglect air resistance.)
Examples for Discussion
A 921-kg sports car is moving rightward with a speed of 29.0 m/s.
The driver suddenly slams on the brakes and the car skids to a
stop over the course of 3.20 seconds with the wheels locked.
Determine the average resistive force acting upon the car.
[F = d(mv)/dt]
A 1250-kg small aircraft decelerates along the runway from 36.6
m/s to 6.80 m/s in 5.10 seconds. Determine the average
resistive force acting upon the plane. (Assume that its
engine/propeller makes no contributes to its forward motion).
[F = d(mv)/dt]
A tow truck exerts a 18300-N force upon a 1210 car to drag it out
of a mud puddle onto the shoulder of a road. A 17900 N force
opposes the car's motion. The plane of motion of the car is
horizontal. Determine the time required to drag the car a
distance of 6.90 meters from its rest position.
[F = d(mv)/dt; v2=u2+2ax; v-u+at]
Examples for Discussion
A 4.44-kg bucket suspended by a rope is accelerated upwards
from an initial rest position. If the tension in the rope is a
constant value of 83.1 Newtons, then determine the speed
(in m/s) of the bucket after 1.59 seconds. v=u+at
A 22.6-N horizontal force is applied to a 0.0710-kg hockey
puck to accelerate it across the ice from an initial rest
position. Ignore friction and determine the final speed (in
m/s) of the puck after being pushed for a time of .0721
seconds. [F = d(mv)/dt]
A train has a mass of 6.32x104 kg and is moving with a speed
of 94.3 km/hr. The engineer applies the brakes which
results in a net backward force of 2.43x105Newtons on the
train. The brakes are held for 3.40 seconds. How far (in
meters) does the train travel during this time?
More on Friction, Normal reaction,
Gravitation (lesson 3)
• We want to look a little more closely at some
of the very important Forces we come across
in using F = ma.
• friction - resistance encountered by an object
as it interacts with its surroundings
• o static friction - when friction stops an object
from moving, static friction opposite & equal
to the force being exerted on the object
• Fs ≤ µsN
(where the dimensionless constant µs is called
the coefficient of static friction )
• And N is the Normal Reaction.
More on Friction
• See, the Frictional force depends on the
Normal Reaction
• Force of friction acts to oppose motion:
– Parallel to surface.
– Perpendicular to Normal force.
– Since fF = mN , the force of friction does not
depend on the area of the surfaces in contact
j
N
F
i
ma = 0
fF
mg
See text: 5-2
Static Friction...
• mS is discovered by increasing F until the block
starts to slide:
i:
j:
FMAX - mSN = 0
N = mg
mS = FMAX / mg
N
FMAX
mSmg
j
i
mg
Magnitude of force of Friction
depends on Normal reaction
More on Friction
• kinetic friction - friction force for an object in
motion
• Fk = µkN
µk - dimensionless constant called the
coefficient of kinetic friction
–.
j
N
F
ma
Fk
See figure 5-11
mg
i
Coeff. Of static friction µs increases to
a max. µk is smaller than µs
See text: 6-4
Coeff. Of static friction µs increases to
a max. µk is smaller than µs
• Graph of Frictional force vs Applied force:
fF = mSN
fF = mKN
fF
fF = FA
FA
See text: 6-4
Problem: Box on Truck
• A box with mass m sits in the back of a truck.
The static coefficient of friction between box
and truck is mS
– What is the maximum acceleration a that the
truck can have without the box slipping?
m
a
mS N
See text: 6-4
Problem: Box on Truck
• Draw Free Body Diagram for box:
– Consider case where fF is max...
(i.e. if the acceleration were any
larger, the box would slip).
N
j
i
f F = m SN
mg
See text: 6-4
Problem: Box on Truck
• Use FTOT = ma for both i and j components
–i
–j
mSN = maMAX
N = mg
N
aMAX = mS g
j
aMAX
i
f F = m SN
mg
See text: 6-4
Problem: Putting on the brakes
• Anti-lock brakes work by regulating making
sure the wheels roll without slipping. This
maximizes the frictional force slowing the car
since mS > mK .
• The driver of a car moving with speed vo slams
on the brakes. The coefficient of static friction
between wheels and the road is mS . What is
the stopping distance D?
ab
See example Stopping a car
vo
v=0
D
See text: 6-4
Problem: Putting on the brakes


D = v0 t - 1/2 abt2
0 = v0 - abt
In our problem:
v 02
D=
2 ab
Solving for D:
v 02
D=
2 ms g
Putting in ab = mSg
(F=ma=mg=umg)
ab
vo
v=0
D
See example : Stopping a car
=> t=vo/a
Incline planes….
Fgrav acts vertically downwards
Fnorm acts normally from the plane
Ffric acts parallel to the plane
Here, we replace the force of gravity by two
components. Then solve for the net force and
acceleration the usual way
• Example
• A 100-kg crate is sliding down an
inclined plane.
• The plane is inclined at an angle of
30o
• The coefficient of friction
between the crate and the incline
is 0.3.
• Determine the net force and
acceleration of the crate.
• First, find the force of gravity
acting upon the crate and the
components of this force parallel
and perpendicular to the incline.
• The force of gravity is 980 N (mg)
• Fparallel = 490 N (980 N • sin 30
degrees) and
• Fperpendicular = 849 N (980 N • cos30
degrees).
• Example
• A 100-kg crate is sliding down an
inclined plane.
• The plane is inclined at an angle of
30o
• The coefficient of friction
between the crate and the incline
is 0.3.
• Determine the net force and
acceleration of the crate.
•
Now the normal force can be
determined to be 849 N (it must
balance the perpendicular
component of the weight vector).
• The force of friction can be
determined from the value of the
normal force and the coefficient of
friction; Ffrict is 255 N (Ffrict =
"mu"*Fnorm= 0.3 • 849 N).
• Example
• A 100-kg crate is sliding down an
inclined plane.
• The plane is inclined at an angle of
30o
• The coefficient of friction between
the crate and the incline is 0.3.
• Determine the net force and
acceleration of the crate.
• The net force is the vector sum of all
the forces. The forces directed
perpendicular to the incline balance;
the forces directed parallel to the
incline do not balance.
• The net force is 235 N (490 N - 255
N).
• The acceleration is 2.35 m/s/s
(Fnet/m = 235 N/100 kg).
see, we can repeat the last problem, substituting
‘car’ for ‘crate.’ It’s exactly the same problem.
Gravitation
LAW OF UNIVERSAL GRAVITATION
• NEWTON’S
Every particle in the Universe attracts every other particle
•
with a force that is proportional to the product of their
masses and inversely proportional to the square of the
distance between them. [Irrespective of intervening medium]
•
• Where G = 6.673 x 10-11 N.m2/kg2 (a very weak
force!)
Example: Tolu, Tunde, Titi and Tope are located at the four
corners of a square auditorium (length 15m). If each of these
students has a mass of 60 kg, what is the net gravitational force
on one of them due to the other 3?
• F1=F12+F13+F14
R12
R14
• R13 2 = R12 2 + R14 2
Moment (Torque) of a Force
• Several derivatives of Force are used in
Physics: eg Work /Energy (Force and distance
moved by force)
• Power (Work/Time)
• Impulse (Force * Time)
• Moment (Force and distance from a
fulcrum/axis)
• etc
The Torque is measure of the tendency
of a force to rotate an object about some
axis.
It is a vector quantity τ = F x d
The turning effect, when only the
perpendicular distance between the
Force and the turning axis is used, is
called Moment
Moment = Force X perpendicular
distance from the turning point/axis
examples
F2
d1
F1
d2
Conditions for Equilibrium
1. No net force
•  Sum of forces along x axis = 0 (ΣFx = 0)
Sum of forces along y axis = 0. (ΣFy = 0)
[Sum of forces along z axis = 0. (ΣFz = 0)]
2. No net moment
3. Please check textbooks for several
examples.
Homework: The leg and cast in the figure weigh 220N
(w1). Determine the weight w2 and the angle α needed
so that no force is exerted on the hip joint by the leg
plus cast.
(note, if all the forces did not act on the same point as
in this case, there would be turning effects also to
consider).
Homework: Two packing crates of masses 10.0 kg and
5.00 kg are connected by a light string that passes over a
frictionless pulley as in the figure. The 5.00 kg crate lies
on a smooth incline of angle 40.0o. Find the acceleration
of the 5.00 kg crate and the tension in the string.
END OF MODULE 2
THANK YOU FOR YOUR ATTENTION
Tutorial Questions and e-copies of these
teaching slides will be made available on the
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Tutorial Classes being arranged by the Physics
Department. Details will be provided by the
Course Secretary/Teaching Assistants.
WISHING YOU ALL THE BEST!