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Copyright Sautter 2003
THE LAWS OF MOTION
• Isaac Newton, in the 1600s, proposed three fundamental laws of
motion which are found to be correct even today!
• Newton’s First Law of Motion – Inertia – “Objects in motion
tend to remain in motion at the same rate (speed) and the in
same direction, unless acted on by an outside force”
• This law says essentially then, that objects keep doing what they
have been doing, unless they are forced to change by an external
factor.
• This law explains why for example a car skids on ice when the
brakes are applied (the lack of the outside force of friction on
the tires) or why it is difficult to negotiate a tight curve at a high
rate of speed (the direction of motion tends to remain straight
due to inertia).
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Outside
Force
“Objects in motion tend to remain in motion, at the same rate,
And in the same direction, unless acted on by an outside force”
Objects tend to move in the same
direction and at the same rate
unless acted on by an outside force
Outside
Force
Constant velocity stops
Constant velocity
Acceleration
occurs
THE LAWS OF MOTION
• Newton’s Second Law of Motion – “the acceleration of
an object is directly proportional to the force exerted on
it and inversely proportional to its mass’.
• Force means a push or a pull and the second law then
says that the harder you push or pull an object, the
more rapidly it speeds up or slows down. Cars with
large engines (more available force) accelerate more
quickly and those with smaller engines!
• The second law also tells us that large, massive objects
are harder to speed up or slow down that small objects.
Mass then is the inertial property of matter which
means it determines how readily an object maintains its
state of motion. As an example, it is much easier to
accelerate a sports car than a 10 ton truck!
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Force
Large masses with the same force applied
results in small accelerations
Force
Small masses with the same force applied
results in large accelerations
THE LAWS OF MOTION
• Newton’s Second Law of Motion in abbreviated mathematical
form states, F = MA, (force equals mass times acceleration).
• The units used in expressing force, mass and acceleration vary
depending on the measurement system which is used.
• Three systems are available and the one chosen depends on the
units cited in the problem to be solved.
• (1) MKS – metric units involving meters as displacement units,
kilograms as mass units and seconds as time units. MKS force
units are newtons
• (2) CGS – metric units involving centimeters as displacement
units, grams as mass units and seconds as time units. CGS force
units are dynes.
• (3) English Units– involving feet as displacement units, slugs as
mass units and seconds as time units. English force units are
pounds.
FORCE
THE LAWS OF MOTION
• Newton’s Third Law of Motion tells us that “for every action
there must always be an equal and opposite reaction”
• What the Third Law says then is if you push on something it
must push back equally hard or nothing will happen.
• As an example, pretend that you are at an ice skating rink,
standing on skates at the center of the rink and you attempt to
move by pushing on the surrounding air with your hands. No
motion occurs because the air cannot push back sufficiently!
You do not move.
• Now, pretend that a friend in next to you on skates and you
chose to push on him. You move one way and your friend
moves the other. Equal and opposite pushes result in motion
occurring!
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“For every action there must be an equal but
opposite reaction”
For every action there must
always be an equal and
opposite reaction.
Pushes and pulls occur in pairs.
Review of Kinematic Equations
• S = displacement, t = time
Vo = original velocity, a = acceleration
Vi = instantaneous velocity, Vave = average velocity
• Equations
• S = Vot + ½ at2 (displacement vs. time)
• Vi = Vo + at (velocity, acceleration & time)
• a = (Vi2 – Vo2) / 2 S (displacement, velocity &
acceleration, time not required)
• Vave = (V1 + V2) / 2 (average velocity)
• Vave = S / t (average velocity)
Solving Force Problems
A 5 Kg rifle fires a 9 gram bullet with an acceleration of 30,000
m/s2. (a) What force acts on the bullet? (b) What force acts on the
rifle? ( c) If the bullet is in the rifle for 0.007 seconds, what is its
muzzle velocity?
a = 30,000 m/s2
(a) F = ma, F = 0.009 kg x 30,000 m/s2 = 270 nt (grams must be
converted to kilograms in order to use the MKS system and
get newtons as force unit answer)
(b) Newton’s 3rd Law (Action / Reaction) – the rifle is “shot” in
the opposite direction with the same force as the bullet or
– 270 nts
(c) (muzzle velocity means the velocity at which the bullet leaves
the rifle barrel) Using the equation Vi = Vo + at, Vi = o +
30,000 x 0.007 = 210 m/s (Vo = 0 since the bullet is not moving
before it is fired)
Force, Weight & Gravity
• Weight and mass are related but they are not the same!
Weight requires gravity while mass exists independent of
gravity. Your weight in outer space would be zero, your
mass would not be zero but the same as it is on earth or
anywhere for that matter.
• As the mass of an object increases, its weight increases
proportionally if gravity is present. Greater mass gives
greater weight.
• Although, weight in Europe is measured in kilograms, it is
technically incorrect! Weight being a force should be
measured in newtons. The exact meaning of weight
measured in kilograms is “kilograms of force” meaning the
mass of an object times Earth’s gravity.
• In the English system, pounds in the correct force unit and
weight values given in pound units are correct. English mass
units are slugs!
Mass is the
same in
all cases
scale
scale
150 lbs
g = 9.81 m/s2
scale
25.6 lbs
g = 1.67 m/s2
406 lbs
g = 26.6 m/s2
YOU COULD LOSE WEIGHT BY MOVING
TO ANOTHER PLANET BUT YOUR MASS
WOULD BE THE SAME AND YOU WOULD
STILL LOOK THE SAME!
Solving Force Problems
(a) How much force is needed to reduce the velocity of a 6400 lb
truck from 20 mph to 10 mph in 5 seconds? (b) What is its stopping
distance?
time = 5 sec
20 mph
10 mph
• (a) F = ma, we must first find mass. The weight is 6400 lbs. Wt
= mass x gravity therefore, m = w/g,
mass = 6400 lbs /32 ft/s2 = 200 slugs.
• Now, to find acceleration, Vi = Vo + at or a = (Vi – Vo) / t
• Velocities are given in mph and we need ft / sec. To convert mph
to ft / sec we multiply by 5280 ft / mile and divide by 3600
seconds in an hour to get Vo = (20 x 5280) / 3600 = 29.4 ft /sec
and Vi = (10 x 5280) / 3600 = 14.9 ft/sec.
• a = (14.9 – 29.4)/ 5 = - 2.98 ft/sec2 and F = 200 x (-2.98) = - 596 nt
(negative means the force is opposing the motion)
• (b) Using S = Vot + ½ at2 we get S = (29.4 x 5) + ½ (-2.98) 52 =
110 feet is the distance traveled during stopping.
Solving Force Problems
• Solving problems in physics involving forces often
uses the idea of net force. The net force on a object is
the vector sum of all forces acting on that object and
the net force gives the object its acceleration.
• As an example, if I push a chair across the floor with a
force of 25 newtons and the force of friction opposing
the motion is 10 newtons, the net force accelerating
the chair is 25 + (-10) or 15 newtons in the direction
that I am pushing the chair.
• When the applied forces are acting at angles other
than straight line motion, vector addition methods
must be used to determine the net force on the object.
Frictional Forces
• Friction is a force that is commonly encountered in physics
problems. FRICTION is a force which ALWAYS OPPOSES THE
MOTION OF AN OBJECT
• Friction is related to two factors. (1) the force holding the surfaces
in contact with each other (normal force) and (2) the nature of the
surfaces (different materials cause different frictional forces)
• The normal force holding surfaces in contact is simply the weight
of an object on the horizontal. As the the angle of inclination
changes the normal force is reduced until at 90 degrees none of
the weight holds the surfaces in contact and the normal force is
zero. The trig function which describes this relationship is the
cosine, Normal force = weight x cosine of the angle of inclination.
(Fnormal = w x cos  and since weight is mass x gravity,
Fnormal = mg cos ).
• The nature of the surfaces in contact is described by the
coefficient of friction with a symbol called “mu” ().
• Force of friction = coefficient of friction x normal force
• Ffriction =  w cos  =  m g cos 
Solving Force Problems
A force of 20 nts will just set a 50 kg box in motion. What is the
coefficient of friction between the box and the floor?
200 nts
50 kg
friction
• Ffriction =  w cos  =  m g cos 
•  = Ffriction / (m g cos ) since the box just moves the
force acting on the box and the force of friction are
just about equal, therefore Ffriction = 200 nts
•  = 200 / (50 x 9.8 x cos 00) = 0.410
• The coefficient of friction is a ratio which has no
units!
Solving Force Problems
• A common problem and laboratory experiment
involving forces uses the Atwood’s Machine. It is
simply a pulley with two mass suspended on each side.
When released the masses accelerate at different rates
depending on the masses used.
• The experiment may be done in two different ways. In
one it can be used to predict and verify the
acceleration of the system or it may be use to verify
the acceleration due to gravity.
• The calculations shown on the next slide are used to
predict the acceleration of the system. An important
factor to remember when doing an Atwood’s Machine
problem is that both masses are accelerated do to the
net force acting on the system.
pulley
3 Kg
2 Kg
F3 = 3 x 9.8 = 29.4 nt
F2 = 2 x 9.8 = 19.6 nt
Acceleration
1.96 m/s2
Fnet = F3 – F2 = 29.4 – 19.6 = 9.8 nt
Fnet = ma, mass = 2 + 3 = 5 kg
9.8 nt = 5 kg x a, a = 1.96 m/s2
Solving Force Problems
• Additional calculations may be applied to the
Atwood’s Machine problem, that is to tend the
tension in the cord connecting the masses. Doing
a problem such as this requires the use of the
Free Body Diagram concept.
• Using the Free Body Diagram requires us to
select a point within the system and examine it
with only the forces acting at that specific point.
• Free Body Diagrams are used to simplify
otherwise complex systems dealing with one
point within the system at a time!
Tension of
The cord
Free Body Diagram
a = 1.96 m/s2
Point to be
examined
Weight
2 x -9.8 = -19.6 nt
3 Kg
2 Kg
Force net = mass x acceleration
Tension + weight = mass x acceleration
T + (-19.6) = 2 x (+1.96) (positive means upwards)
T = 3.92 + 19.6 = 23.52 nts (tension in cord)
Solving Force Problems
• Inclined plane are often solved using forces. The forces
acting on a body resting on an incline are that of gravity
(the weight of the object), the normal force of the plane
supporting the object on its surface and the the force of
friction.
• A free body diagram may be constructed for inclined
plane situations using the center of mass of the object as
the point of force interaction.
• Center of mass refers to a point within the object around
which the mass of the body is equally distributed . It is a
simplifying technique by which the entire mass of the
body can be considered to be present at a single point
rather than working with each and every point of mass
within the object.
• For a symmetric body of uniform density, the center of
mass is the geometric center of the body.
Center of Mass
Upward force of plane
Friction
force
Force parallel
Normal force

P
weight
W

N
Vector diagram for Inclined Plane
W = weight vector
P = parallel force vector
N = normal force vector
P = W sin 
N = W cos 
Solving an Inclined Plane Problem
A box slides down an plane 8 meters long inclined at 300 with a
coefficient of friction of 0.25. (a) what is the acceleration of the box?
(b) what is its velocity at the bottom of the incline?
FParallel = W sin 
Center of Mass
Friction
Upward force of plane
FNormal = W cos 
force
FParallel = w sin 300
Ffriction =  Fnormal
Force
parallel
Normal force
Ffriction = 0.25 x w cos 300
Fnet = P – Ffriction

Fnet = 0.5 w – 0.17 w
weight
• (a) Fnet = ma, w = mg , m = w /g , Fnet = (w/g) a
• 0.5 w – 0.17 w = (w/9.8) a , a = 2.77 m/s2
• (b) Vo = 0 (the box starts at rest), a = (Vi2 – Vo2) / 2 S
• Vi =( 2 S a + Vo)1/2 , Vi = ( 2 x 8 x 2.77 + 02)1/2 , Vi = 6.7 m/s
Solving Force Problems
• In addition to Atwood’s Machine problems and
inclined plane problems another common force
problem involves elevators.
• When a person stands on a scale in a motionless
environment (acceleration is zero) the scale reading
gives his normal weight (the effect of gravity on his
mass that is mass x gravity - remember w = mg)
• When a person stands on a scale in a moving
environment the scale still gives his normal weight if
the motion is at a constant velocity that is acceleration
is still zero.
• If, however, the system is accelerated upwards or
downwards, the scale reading changes and his normal
weight is not the scale reading !
Solving Force Problems
• If the system is accelerating upwards, the scale must
support the weight of the individual and also provide
additional force to accelerate the person upwards. The
scale reading therefore is greater than the normal weight
of the person.
• If the system is accelerating downwards, the scale must
provide a reading less than the normal weight since if is
provided an upward force equal to the weight the net force
would be zero and no acceleration would occur ! The
extreme example of this situation would be when both the
person and the scale are in free fall. In this case, no
upward force is supplied by the scale and its reading is
zero. The acceleration of both the person and scale then is
that of gravity (- 9.8 m/s2)
Upward force
of scale
Upward force
of scale
Upward force
of scale
Downward force
of weight
Downward force
of weight
Downward force
of weight
Net force = 0
Acceleration = 0
Net force = up
Acceleration = +
Net force = down
Acceleration = -
Scale reads normal weight
Scale reads > normal weight Scale reads < normal weight
GOING UP
Acceleration = + 8 ft/s2
T = tension
in elevator cable
W = weight
Net force = MA
T + W = MA
Scale
Reading
100 lbs
Scale
Reading
125 lbs
GOING DOWN
Acceleration = - 8 ft/s2
Scale
Reading
75 lbs
Solving Force Problems
An 800 nt man stands on a scale in an elevator. What is the scale
reading when (a) it is ascending a constant velocity of 3 m/s (b)
descending at the same rate ( c) ascending with acceleration of 0.8
m/s2 (d) descending with acceleration of 0.8 m/s2 ?
Questions
(a) and (b)
Upward force
of scale
Downward force
of weight
Net force = 0
Acceleration = 0
Scale reads normal weight
In any unaccelerated
system (constant velocity)
the scale reading equals normal
weight, therefore in parts (a)
and (b) the scale reads 800 nts
Elevator Problem (cont’d)
Part ( c)
Ascending
a = 0.8 m/s2
Upward force
of scale (P)
Force net = upward push of scale (+) + weight downward (-)
Force net = P + (-800)
Force net = mass x acceleration
Weight = mass x gravity
-800 = m (-9.8), mass = (-800 / - 9.8)
P + (- 800) = (- 800 / -9.8) (0.80)
P = ( 81.6) (0.80) + 800
P = + 865 nt
Downward force
of weight
Net force > 0 (up)
Acceleration = +
Scale reads > normal weight
The scale reading is greater than
the normal weight reading of 800 nts
and upward acceleration occurs
Elevator Problem (cont’d)
Part ( c)
descending
a = - 0.8 m/s2
Upward force
of scale (P)
Force net = upward push of scale (+) + weight downward (-)
Force net = P + (-800)
Force net = mass x acceleration
Weight = mass x gravity
-800 = m (-9.8), mass = (-800 / - 9.8)
P + (- 800) = (- 800 / -9.8) ( - 0.80)
P = ( 81.6) (- 0.80) + 800
P = + 735 nt
Downward force
of weight
Net force < 0 (down)
Acceleration = Scale reads < normal weight
The scale reading is less than
the normal weight reading of 800 nts
and downward acceleration occurs
A force which gives a 2.0 kg object an acceleration of 1.6 m/s2
would give an 8.0 kg object what acceleration ?
(A) 0.2 m/s2 (B) 0.4 m/s2 (C) 1.6 m/s2 (D) 6.4 m/s2
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A car slows form 50 ft/s to 15 ft/s in 10 seconds when the brakes
exert a force of 200 pounds. What is the weight of the car ?
(A) 57 lbs (B) 560 lbs (C) 1830 lbs (D) 22,400 lbs
The coefficient of friction between the tires of a car and the road is 0.80.
If the car is to stop in 3.0 seconds what can be its maximum speed?
(A) 2.4 m/s (B) 2.6 m/s (C) 7.8 m/s (D) 23.5 m/s
A skier stands on a 5 degree hill The coefficient of friction is 0.1
Does the skier slide down the hill ?
(A) yes (B) no (C) it is impossible to tell
A cable supports an elevator which is 2000 Kg. The tension in the cable
lifting the elevator is 25 KN. What is the acceleration ?
(A) 2.7 m/s2 (B) 0.4 m/s2 (C) 1.6 m/s2 (D) 8.4 m/s2