Unit 1.4: Newton's Laws

Notes

These are the topic areas covered in the notes for this unit.

Law of Inertia         Action / Reaction Forces        Newton's Second law

Newton's laws and Motion           Uniform Circular motion









 

Law of Inertia          (Qs and Ps for this section)       (back to the top)

Initially proposed by Galileo Galilei (1632) as a means of explaining the natural motion of objects as opposed to Aristotles theory (360 BC).  Aristotle believed that objects naturally came to rest unless some force caused them to keep going.  Galileo disagreed, he thought, and proved, just the opposite.  Isaac Newton (1665) thought it was cool also, so he adopted it as his first law.

Inertia - the tendency of an object to maintain its current state of motion, whether at rest or in
             motion.



Because of inertia, only external forces can alter an objects state of motion; internal forces cannot.

         Internal force - forces generated by one part of a system of objects on another part of
                                     the same system.

                           EX. Your left hand pushes on your right hand.  The engine of a car
                                  pushes on the transmission which pushes on the wheels.


         External force - forces generated by something outside the system of objects.

                          EX. A tennis racket hits a tennis ball. Your foot kicks a chair.
 


The key point to remember is that no change of motion can occur unless a force from outside the object causes it to change.

A simple enough concept, but sometimes hard to remember in practice.

The amount of inertia an object possesses is indicated by its mass.

            Mass - A measure of an objects resistance to change in motion.

The larger the mass of an object, the more it resists a change in its motion.  Smaller masses have less resistance to change in motion.  This is an operational definition of mass, it requires direct measurement of an objects change in motion to determine its value.


Remember, the basic unit of mass in the MKS system is the Kilogram, Kg  (1000 g).
 


 

Action / Reaction Forces           (Qs and Ps for this section)       (back to the top)

Action and reaction forces were introduced in Unit 1.3 (Forces).  The following comments are a review of that information.

Simply put, forces are social beings, they always operate in pairs.  It is not possible to have an isolated, or single force.  Isaac Newton expressed it a bit more formally:

If an object expresses a force upon another object, then the latter of
the two exerts a force of equal magnitude back upon the first.


We call these action and reaction forces.  It doesnt make any difference which you call the action and which the reaction since they happen concurrently and are reversible.

EX.  Your body pushes on the chair, the chair pushes on your body.
        A tennis racket hits the ball, the ball hits the racket.

Point to remember:  Although these two forces are equal, but oppositely directed, they do
                                   not cancel each other out.  They operate on different objects so they
                                   cannot be added together.

If you kick a chair, your foot exerts a force on the chair.  The chair exerts a force on your foot.

See, one force is on the chair, the other force is on your foot; two different objects.

The only force which can cancel any applied force is another force on the same object.  The only force which can cancel your foots force on the chair is another force on the chair, like friction between the chair and the floor.


Remember inertia?  How an internal force cannot cause a change in motion?  Heres how it works in the real world.

Stand up and jump straight up into the air.  If you think about what you did you bent your knees, pushed on the floor, and went up into the air.  We say I jumped.

Actually, you cannot do anything to yourself to make you go into the air from rest on the floor.  There has to be a force on your from outside of you to make you go into the air.

Why do you push on the floor?  To get the floor to push on you!  Thats the action/reaction part of the operation.  The floors force on you is whats needed to make you go up.  Pretty tricky, huh?  Thats why it took you a while to learn to walk and jump.



 

Newton's Second Law        (Qs and Ps for this section)       (back to the top)

As laws of our physical universe go, Newtons second law is right up there with the biggest.

His first law, also Galileos law of inertia, concerned itself with what happens to objects when the total, external force on them is zero - they dont change their motion.  The second law is for the cases when the external force is not zero.

Newton did not state this law the same way we learn it in school, but the gist of it is:

If an object experiences an external force, not equal to zero, it will accelerate proportional to the magnitude of the force, but inversely proportional to its own mass.
In equation form:  FT  = M a      where Force (F) is in Newtons,
                                                          mass (M) is in Kg,
                                                          and acceleration (a) in m/s2

This means that the acceleration of an object depends on the ratio of the total force to the mass of the object.  The direction of the acceleration is the same as that of the total force which caused it.

Newtons second law is the work horse of our mechanical world; it connects the changes in motion of all objects with the forces needed to make that change.   Its a simple equation to work, but its application can become complicated as you will see in the next section.

Some examples:

a)  If a total force of 10 N is placed on  2 Kg ball, what is its acceleration?


b)  If you push with a force of 25 N on a 2 Kg box to slide it across the floor,
     what is the acceleration of the box if the frictional force is 10 N?



 

Newton's Laws and Motion       (Qs and Ps for this section)       (back to the top)
 

First of all, many students consider this the hardest part of the whole year.  It requires time and patience and a clear eye on what you are doing.  You only get good at it if you practice.


Newtons second law (FT  = M a) is not a difficult equation to solve; any algebra I student can do it.  The difficulty arises in getting to the equation or in using the answer when you are done.

Like any problem situation, the key is to correctly identify what you know at the beginning and what you want at the end.  This law of Newton connects the force information with the motion information.

Think of forces and motion as two separate areas in the physical world.  Newton's second law connects them like a bridge.  If you are going to cross a bridge in your town, you first have to get to it.  You aren't usually right where the bridge is located.  Also, once you cross the bridge you usually want to go somewhere besides the end of the bridge.  Solving dynamics problems is much the same.

The only place in the area of forces you can be to use Newton's bridge is the total force ( FT ).  Some of the easier problems start you here, so you can use the bridge right away.  Most of the time you have to use the procedures of handling forces to calculate the total force first, then you can use the bridge.

On the motion side, the only value which accesses the bridge is acceleration (a).  Sometimes you are given the acceleration initially, but usually you have to solve a motion equation to find it.

Many times a problem is solved in a three step sequence.  You could be given information about the forces on an object and asked to find something specific about its motion.

The problem could go the other way; starting with some motion information and asking you to find a particular force.  Your procedure would be the reverse of the one above.


The key to success with these problems is patience.  Determine what procedure you need to follow, then keep track of your progress as you work through the problem.
 

The easier stuff

Sometimes, but rarely, do you have the information to simply plug in two numbers into the equation and solve for the third, but it does happen.

EX.  What is the total force needed to accelerate an 800 Kg car at 3 m/s2 ?

FT = M a   so   FT = 800 Kg (3 m/s2 ) = 2400 N

Simple, right?          Remember, mass is always in Kg,
                                                force is in Newtons,  and
                                               acceleration is in m/s2 .
 

The real stuff

Ususlly you are asked to find one of the applied forces rather than the total force or you are asked to find some aspect of an objects motion rather than its acceleration.  This complicates the procedure a bit.  You might find it helpful to make yourself a plan for solving the problem to avoid getting lost.

EX.  If the average force applied to a 10 g bullet in a rifle is 1000 N, how fast will it be
         moving at the end of the 0.6 m long barrel?

1.  Identify what you know and what you are looking for.

 FT  - The average force (1000 N) is the total force since no other forces are involved.
 M - the mass of the bullet is 10 g = 0.010 Kg
 d - the bullet will travel 0.6 m
 VF  - you are looking for the final velocity.
2.  Make a plan.

The only way to get VF  is from a motion equation.  You need to get the acceleration to have enough information to use a motion equation.  You can use Ns 2nd law directly to find the accel.

 so...
           a.  use Newtons 2nd law to find the acceleration.
           b.  use one of the motion equations to find the final velocity.

 a.  FT  = M a     a = FT / M = 1000 N / 0.010 Kg = 1 E5 m/s2

 b.  Vo  = 0     d = 0.6 m     a = 1 E5 m/s2      VF  = ?

                VF 2  = Vo 2  + 2 a d = 0 + 2 (1E5) (0.6) = 1.2 E5
                VF  = sqrt(1.2 E5) = 350 m/s
 

Another one

You are throwing your 20 Kg cousin straight up in the air (dont ask why).  During the throwing motion you move your cousin straight up through a distance of 0.8 m and he/she leaves your hands at a speed of 7 m/s.  How much force do you have to use to get the job done?

1.  Identify the variables.

 M - your cousin is 20 Kg
 d - 0.8 m during the throw
 VF  - 7 m/s at the end of the throw
 FY  - how much you have to push


2.  Make a plan

Your push is not the total force, but one of the forces applied to your cousin.  You have to take Fg  into account.  You know about the throwing motion, so use a motion equation to find the acceleration then Ns 2nd law to find the total force, then use a force diagram to find your force.

 a.  find accel. from a motion equation
 b.  use Ns 2nd law to find the total force.
 c.  calculate Fg  for your cousin.
 d.  use a force diagram to find your force.


 a.  Vo  = 0        VF  = 7 m/s         d = 0.8 m      a = ?

                      VF 2  = Vo 2  + 2  a  d
                         49 = 0 + 2 (a) (0.8)

                             a = 31.6 m/s2  [up]

 b.  FT  = M a     FT  = 20 Kg (31.6 m/s2 [up])
                              = 632 N [up]
                        notice the direction of FT  is the same as the acceleration

 c.  Fg  = M g = 20 Kg (9.8 m/s2 ) = 196 N

 d. 
 
 


 

Uniform Circular Motion       (Qs and Ps for this section)       (back to the top)

A special type of accelerated motion is when the object has a constant speed, but its direction changes at a uniform rate.  An object in uniform circular motion is an example of this type of accelerated motion.

Things do not turn on their own (remember the law of inertia?).  If an object changes direction, there has to be an unbalanced force causing it to turn.

 
 
 

If this force is perpendicular to the motion of the object, the speed of the object will not change, but it will change direction.
 
 
 
 
 
 
 
 
 
 

Uniform Circular Motion (UCM) - an object following a curved path of constant
                                                          radius at a constant speed.


Swinging a ball around on a string, an elephant running around the center ring of the circus, a race car going around the curve, and a rider on a Ferris wheel are some examples of UCM.

Since this is accelerated motion Newton's second law must apply.  The total force on the object causes the mass to accelerate, but the speed doesn't change.  This is different than you have seen before.

As before, the direction of the acceleration is the same direction as the force which causes it.  Since the force is always perpendicular to the velocity of the object they both must be directed toward the center of the circular motion.

These are called centripetal force and centripetal acceleration.

Centripetal - toward the center
The value of the acceleration can be calculated from the following formula.

ac = V2 / r

In this formula V is the speed of the object in circular motion and r is the radius of the motion.  This formula is good to use for objects which are moving around curves, or partial circles.  You may know the speed of these objects, like a car going around a curve in the road.

EX.  What is the centripetal acceleration of a car going around a curve of 50 m radius at a
        speed of 20 m/s?

ac = V2 / r = (20 m/s)2 / 50 m = 8.0 m/s2

To find the force required to create this acceleration use Newton's second law.  For the example above suppose the mass of the car is 600 Kg.  What is the centripetal force needed on the car?

Fc = M ac = 600 Kg (8 m/s2) = 4800 N

The direction for the acceleration and force are already stated by the term centripetal, its toward the center of the circular path.  This force is the total force on the object.  It does not indicate the source of the force, that is dependent on the circumstances of the individual problem.  The source might be friction, tension, or a push from a person, but the total force has to be toward the center of the circle for UCM to occur.

For some cases the motion is around a circle repetitively, like the tip of a fan blade or a rider in a spinning amusement ride.  Usually the speed of the rider is not known, but other information is.

Consider that the speed of any object is the distance travelled divided by the time for the motion.  For an object going in a circular path the distance is the circumference of the circle and the time is the period (time for one revolution) of the motion.

By making the appropriate substitutions the speed and acceleration formulas can be rewritten.  These would be more useful for spinning objects rather that for objects going around curves.

Another circular motion quantity which is often used is the frequency of the motion.  Frequency is the number of revolutions in one second.  The frequency and the period are reciprocals of each other.

If an object makes 2 revolutions in one second then the period is 1/2 second for a revolution.  The unit for the period is the same as for any time; sec, etc.  The unit for frequency would be 1/sec (sec-1), or the Hertz (Hz).  A frequency of 15 Hz would be 15 revolutions in one second.

The centripetal force equations can be rewritten using these new circular motion variables



EX.  There is a merry-go-round at the local playground.  If the ride has a 4 m radius and
         takes 3.5 sec to make a revolution, what force would a 50 Kg person have to exert to
         stay on at the edge?


Fc = ?        r = 4 m         T = 3.5 s         m = 50 Kg

Fc = m 4 p2 r / T2  =  50 Kg (4) (3.14)2 ( 4 m) / ( 3.5 s)2  = 644 N

The fact that it is a force directed toward the center of the circle which causes the circular motion is a difficult concept for most students to accept.  Based on your experience it "feels" like there is a force pushing you toward the outside (centrifugal) of the motion.  Think of going around a curve in a car at a fast speed, riding in a circular amusement park ride and being pressed to the outside of the ride, mud being "thrown off" a bicycle wheel.

In all of these cases there is no force which is pulling, or pushing the object to the outside, just the centripetal force pushing it toward the inside.  But why do you feel like there is?  The answer has to do with inertia.
 


 
 

Consider the car going around the curve.  According to the law of inertia you (and everything else) would travel in a straight line unless you are forced to change by an outside force.  As the car approaches the curve you are moving in a direction tangent to the curve.
 
 
 
 
 
 
 


 
 

The car would follow this path also, but the friction with the road pushes it toward the center of the circle and it changes its direction.  You would keep going straight, but the friction with the seat or the contact with the door pushes you toward the center of the circle.
 
 








You want to go straight, but the seat or door keep pushing you around the curve.  You keep running into the door.  If the door were to suddenly open you would follow the path tangent to the circle in accordance with the law of inertia.  See?  No force pushing you out.

Swing a ball on a string in a circle and you can see this for yourself.  You have to pull on the string which pulls on the ball.  These forces are toward the center of the circle.  If you let go the ball will go in a straight line which is tangent to the circle at the point it is released.

In information and examples given so far the motions have been in the horizontal plane.  The force of gravity on the object has not been considered since it is perpendicular to the motion and does not directly affect the motion.  But what about objects which move in circles along a vertical plane?

In these cases the force of gravity must be taken into account.

For the following examples it is desired that the circular motion of the object be uniform; the speed of the object remains constant.  If this is the case then the Fc has to be constant.  Keep this in mind as you go through the cases.
 


 
 

Suppose you swing a ball on a string in a vertical circle at a constant speed.  The source of the centripetal force is the tension in the string.  Notice that at the top of the circular motion the tension and the weight of the object are in the same direction, while at the bottom of the circle they are in opposite directions.  The Fc must always be directed toward the center.
 
 
 
 
 
 
 

This means the tension does not have to be as much at the top of the motion since the weight is helping it provide the centripetal force, while at the bottom the tension needs to be greater since it has to work against the weight to provide the same centripetal force.
 

EX.  You tie a 0.5 Kg pumpkin to a 0.8 m rope and swing it in a vertical circle.  If the pumpkin
        takes 1.2 sec to complete the circular path, what are the tensions in the rope at the top and
        bottom of the motion?


You can see that the tension is less at the top of the swing than at the bottom.  This is also the case for riders in rollercoasters which have vertical loops.  In these rides the Normal force from the seat replaces the tension by the string.
 


 
 

If the rider goes fast enough there is a normal force pushing them toward the center of the circle.  The rider "feels" pushed into the seat as a result.  Depending on the amount of Normal force needed they could feel lighter, or normal at the top - even though they are upside down.
 
 
 
 
 
 
 
 
 
 
 
 


 
 

If the ride does not go fast enough then the weight of the rider is greater than the centripetal force needed for the motion.  Something must provide an additional upward force, like a shoulder harness or lap belt.  The rider feels pushed into these and has the feeling of falling out of the ride (which they would do if the restraints are not there)
 
 
 
 
 
 
 

At the bottom of the ride the situation is just the opposite.  If the ride were not moving the rider would feel their normal weight, but when in motion the Normal force must provide the additional force needed to create the proper centripetal force.  This means that the rider will always "feel" pushed into the seat with more than their normal weight.

There would exist a speed of the rider which would require a centripetal force equal to their weight.  At the top of the ride there would be no need of any force from the seat, so the rider would not feel pushed into the seat, not into the harness.  They would feel "weightless", like floating through the top part of the ride.  To calculate this minimum speed:

You can see that the mass of the individual rider is not needed.  Every rider would have the same relative experience.  Truly an equal opportunity experience.  If the ride went faster than this speed the rider would feel pressed into the seat at the top of the ride, no lap bar or shoulder harness would be needed.  If the ride went slower than this speed the riders would need to be held in the seat or they would fall out.  As a safety factor most rides are designed to move at 150% of the minimum speed through the top of the loop.

EX.  What is the minimum speed needed to pass through the top of the first loop in the Demon
         at Great America if the radius of the loop is 5 m?

V = sqrt (g r) = sqrt (9.8 m/s2 x 5 m) = 7 m/s

The actual speed is close to 10 m/s.
 
 

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