Explains Newtons Laws of Motion **More good stuff available at: www.wsautter.com and http://www.youtube.com/results?search_query=wnsautter&aq=f

1. Copyright Sautter 2015

2. 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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3. “Objects in motion tend to remain in motion, at the same rate,
And in the same direction, unless acted on by an outside force”
Outside
Force
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4. Objects tend to move in the same
direction and at the same rate
unless acted on by an outside force
Constant velocity
Outside
Force
Constant velocity stops
Acceleration occurs 4

5. 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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6. Force
Force
Large masses with the same force applied
results in small accelerations
Small masses with the same force applied
results in large accelerations
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7. 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.
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8. FORCE
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10. 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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11. “For every action there must be an equal but
opposite reaction”
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12. For every action there must
always be an equal and
opposite reaction.
Pushes and pulls occur in pairs.
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13. 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 = (Vi
2 – Vo
2) / 2 S (displacement, velocity &
acceleration, time not required)
• Vave = (V1 + V2) / 2 (average velocity)
• Vave = S / t (average velocity)
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14. 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) 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)
a = 30,000 m/s2
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15. 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! 15

16. scale
150 lbs
scale
25.6 lbs
scale
406 lbs
g = 9.81 m/s2 g = 1.67 m/s2 g = 26.6 m/s2
Mass is the
same in
all cases
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17. YOU COULD LOSE WEIGHT BY MOVING
TO ANOTHER PLANET BUT YOUR MASS
WOULD BE THE SAME AND YOU WOULD
STILL LOOK THE SAME!
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18. 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?
• (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.
20 mph
10 mph
time = 5 sec
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19. 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.
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