Copyright Sautter 2003

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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).

“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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Objects tend to move in the samedirection and at the same rate

unless acted on by an outside force

Constant velocity

OutsideForce

Constant velocity stopsAcceleration 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!

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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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!

“For every action there must be an equal butopposite reaction”

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For every action there mustalways 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 – Vo

2) / 2 S (displacement, velocity & acceleration, time not required)

• Vave = (V1 + V2) / 2 (average velocity)

• Vave = S / t (average velocity)

Solving Force ProblemsA 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

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!

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

YOU COULD LOSE WEIGHT BY MOVINGTO ANOTHER PLANET BUT YOUR MASSWOULD 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?

• (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 mphtime = 5 sec

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 ProblemsA 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?

• 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!

50 kg200 nts friction

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.

2 Kg

3 Kg

Acceleration1.96 m/s2

pulley

F2 = 2 x 9.8 = 19.6 ntF3 = 3 x 9.8 = 29.4 nt

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!

Free Body Diagram

Weight 2 x -9.8 = -19.6 nt

Tension ofThe cord

Force net = mass x accelerationTension + 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)

2 Kg

3 Kg

Point to beexamined

a = 1.96 m/s2

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.

Normal force Force parallel

Upward force of planeFrictionforce

weight

W

N

P

Vector diagram for Inclined PlaneW = weight vector

P = parallel force vectorN = normal force vector

P = W sin N = W cos

Center of Mass

Solving an Inclined Plane ProblemA 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?

• (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 – Vo

2) / 2 S • Vi =( 2 S a + Vo)1/2 , Vi = ( 2 x 8 x 2.77 + 02)1/2 , Vi = 6.7 m/s

Normal force Force parallel

Upward force of planeFrictionforce

weight

Center of Mass

FParallel = w sin 300

Ffriction = Fnormal

Ffriction = 0.25 x w cos 300

Fnet = P – Ffriction

Fnet = 0.5 w – 0.17 w

FParallel = W sin

FNormal = W cos

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 forceof scale

Downward forceof weight

Net force = 0 Acceleration = 0

Scale reads normal weight

Upward forceof scale

Downward forceof weight

Net force = up Acceleration = +

Scale reads > normal weight

Upward forceof scale

Downward forceof weight

Net force = down Acceleration = -

Scale reads < normal weight

ScaleReading100 lbs

ScaleReading125 lbs

GOING UPAcceleration = + 8 ft/s2

ScaleReading

75 lbs

GOING DOWNAcceleration = - 8 ft/s2

Net force = MAT + W = MA

T = tension in elevator cable

W = weight

Solving Force ProblemsAn 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 ?

Upward forceof scale

Downward forceof weight

Net force = 0 Acceleration = 0

Scale reads normal weight

Questions(a) and (b)

In any unacceleratedsystem (constant velocity)

the scale reading equals normalweight, therefore in parts (a)

and (b) the scale reads 800 nts

Elevator Problem (cont’d)

Part ( c)Ascendinga = 0.8 m/s2

Upward forceof scale (P)

Downward forceof weight

Net force > 0 (up) Acceleration = +

Scale reads > normal weight

Force net = upward push of scale (+) + weight downward (-)Force net = P + (-800)

Force net = mass x accelerationWeight = mass x gravity

-800 = m (-9.8), mass = (-800 / - 9.8)P + (- 800) = (- 800 / -9.8) (0.80)

P = ( 81.6) (0.80) + 800P = + 865 nt

The scale reading is greater thanthe normal weight reading of 800 nts

and upward acceleration occurs

Elevator Problem (cont’d)

Part ( c)descending

a = - 0.8 m/s2

Upward forceof scale (P)

Downward forceof weight

Net force < 0 (down) Acceleration = -

Scale reads < normal weight

Force net = upward push of scale (+) + weight downward (-)Force net = P + (-800)

Force net = mass x accelerationWeight = mass x gravity

-800 = m (-9.8), mass = (-800 / - 9.8)P + (- 800) = (- 800 / -9.8) ( - 0.80)

P = ( 81.6) (- 0.80) + 800P = + 735 nt

The scale reading is less thanthe normal weight reading of 800 ntsand 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

A car slows form 50 ft/s to 15 ft/s in 10 seconds when the brakesexert 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 0.1Does 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 cablelifting 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

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