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An object of mass M can slide on a horizontal table. Two weights of masses M1 and M2 are attached to it on both sides of a table with weightless thread as on this picture.
Assuming mass M1 is greater than M2, the object will start sliding to the left under a pull of a bigger weight.
Tensions T1 and T2, as well as friction act on an object on a table.
Assume that this object, as a result of actions of these forces, moves with acceleration a.
What is the friction coefficient μ and magnitudes of tensions T1 and T2?
Hint:
Use the Second Newton's Law for each of three objects (one on a table and two weights) participating in the motion.
Solution:
For object on a table:
T1 − T2 − μ·M·g = M·a
For weight on the left:
M1·g − T1 = M1·a
For weight on the right:
T2 − M2·g = M2·a
Got linear system of three equations with three unknowns: friction coefficient μ and magnitudes of tensions T1 and T2.
There is a ramp of mass M and angle φ to horizon, lying, but not fixed, on a horizontal surface (it's like a triangular prism lying on a side).
An object of mass m slides down this ramp. As it slides, the ramp also moves along the horizontal surface it's lying on.
What is the acceleration of a ramp relative to a horizontal surface, as
object slides down, if the coefficient of friction between an object and
a ramp is μ0, and the coefficient of friction between a ramp and a horizontal surface is μ1?
The free fall acceleration is g, so the weight of an object of mass m is m·g.
Hint:
(1) Review Problem B from Mechanics - Dynamics - Superposition of Forces - Inclined Plane of this course.
(2) Take into account that, according to the Third Newton's Law,
friction between an object and a ramp is, on one hand, a force exhorted
by a ramp that pulls object uphill against its sliding downhill and, on
the other hand, is an action of an object on the ramp in an opposite
direction, slowing its horizontal movement.
There is an object of mass m on an inclined plane that makes angle φ with horizon.
To move this object up on this inclined plane with constant speed you need to apply a force Fup directed uphill.
The coefficient of friction is unknown.
Let's assume that, left by itself, the object would slide down the
inclined plane under its own weight. The free fall acceleration is g, so the weight of an object of mass m is m·g.
What is the acceleration a of an object when it moves downhill under its own weight?
Hint
Fup = m·g·sin(φ)+μ·m·g·cos(φ)
a = g·sin(φ)−μ·g·cos(φ)
Answer:
a = 2·g·sin(φ) − Fup/m
Problem B
An object is lying on a horizontal platform that moves with acceleration a=10 m/sec².
The coefficient of kinetic friction between an object and a surface of a platform is μ=0.3, while coefficient of static friction is μs=0.4.
The free fall acceleration is g=9.8m/sec², so the weight of an object of mass m is m·g, but mass m is unknown.
(a) How the object will behave?
(b) Why was coefficient of static friction μs given?
(c) What is the acceleration of the object relative to the ground and relative to the platform?
Answer:
(a) The object will slide back along the platform's surface, but forward relative to the ground.
(b) If the coefficient of static friction is too high, the object will
not change its position relatively to a platform, and the next question
would make no sense.
(c) Relative to the ground acceleration is a0=g·μ=2.94m/sec².
Relative to the platform acceleration is a1=a0−a=−7.06m/sec²
Consider a slope fixed on the ground that makes an angle φ
with horizon. Let's analyze the motion of an object, as it slides down a
slope under its own weight. Our task is to determine its acceleration
along the slope.
First of all, we have to choose a reference frame - a system of coordinates suitable for this problem.
The main force acting on this object is its weightW - a vector of gravity
directed vertically down to the ground. That prompts us to choose a
frame of reference with horizontal X-axis and vertical Y-axis. So, the
vector of weight has non-zero Y-component and zero X-component: W = {0,W}, where W is a magnitude of vector of weight W. However, this obvious choice is not the best in a case like this.
There are usually numerous forces involved in an experiment and only few
objects. In this case there is only one object. So, instead of catering
to one particular force, like weight, it's better to simplify the motion of a single object.
Much more convenient frame of reference would be the one, where our object in motion has only one non-zero coordinate.
Recall that there must be another force acting on an object - reaction force of the slope, that prevents an object to go vertically down to the ground through a slope and forces it, in cooperation with gravity force, to slide along a slope. This reaction force R
is always perpendicular to the surface, where an object is on (in this
case, perpendicular to a slope), and its magnitude is such that the resultant of the weight W and reaction R is directed along a slope.
Consider a frame of reference with X-axis going along the slope, where the object slides down and Y-axis perpendicular to it.
Granted, the weight now has both coordinates non-zero:
(a) perpendicular to a surface of a slope and (on the picture above) directed along negative Y-coordinates vector WR with magnitude WR=W·cos(φ), that causes reaction force R, that is equal in magnitude and opposite in direction to WR, and
(b) parallel to a slope, directed (on the picture) towards negative X-coordinate, vector WF with magnitude WF=W·sin(φ), that is the cause of motion of our object down a slope.
As it is pictured, both components of weight are negative since they are
directed towards negative direction of the X- and Y-axes:
W = {−W·sin(φ),−W·cos(φ)}
Force R , as opposite and equal in magnitude to WR, is
R = {0,W·cos(φ)}.
The resultant of three vectors WR, WF and R is WF, directed down a slope, equaled in magnitude to W·sin(φ).
Therefore, an object of mass M will slide down a slope with acceleration equaled in magnitude to
a = WF/M = W·cos(φ)/M
Since weight and mass of an object are related as W=M·g, where g is a known acceleration of free falling (9.8 m/sec² on the Earth ground), the resulting acceleration equals in magnitude to
a = M·g·sin(φ)/M = g·sin(φ)
In vector form in the chosen reference frame:
a = {−g·sin(φ), 0}
Problem B
Consider an object A of mass m, sliding without friction
on a slope of slide B, which itself lies on a horizontal surface and can
slide on it without friction. An angle of a slope of slide B is φ with horizon, its mass is M.
Our task is to determine acceleration a of slide B.
Let's analyze the motion of object A on a slope and the motion of a
slide B, as it moves horizontally, as a result of the weight of object A
on it.
As object A presses down with its weight, it slides downhill along a
slope of a slide B. At the same time slide B moves to the right (on the
picture above).
The horizontal component of the pressure WR of
an object A on slide B perpendicularly to its slope is the cause of the
motion of a slide B. However, this pressure is not the same as in the
problem A above. It will be less. Its opposite reaction force R, that is equal in magnitude to WR, but acting on the object A, will also be less than in the problem A above.
The resultant of weight W and reaction force R is force WF that is not parallel to a slope, but tilted downwards.
So, the combination of object A sliding downhill on a slope and slide B
movement to the right produces the resultant move of object A that is
not parallel to a slope, neither it is directed vertically down, but
will be somewhere in-between.
Let's consider the same reference frame as in the problem A above. Now
both object A and slide B, as they move, have both X- and Y-components
not equal to zero.
Consider only Y-coordinate of the A object now and Y-components of forces acting on it.
In the direction of Y-axis the force WF = W + R, acting against object A, has value
WFy = Wy + Ry = R − W·cos(φ)
This force R − W·cos(φ), according to the Newton's Second Law, should be equal to object A's mass m, multiplied by a Y-component of its acceleration, which so far is unknown.
Let a be an acceleration of slide B in the direction of
its horizontal movement. Since displacement of object A in the direction
of the Y-axis (perpendicularly to a slope) equals to horizontal
displacement of a slide B multiplied by sin(φ), the
acceleration of slide B in the horizontal direction and acceleration of
the object A in a direction perpendicular to a slope, maintain the same
factor.
Therefore, the acceleration of object A perpendicularly to a slope of slide B equals to a·sin(φ), where a is the acceleration of the slide B that we have to determine in this problem.
The Newton's Second Law for object A in the direction of the Y-axis (perpendicular to a slope of slide B) is
R − W·cos(φ) = −m·a·sin(φ)
(minus on the right because the acceleration of object A relative to Y-axis is negative).
This equation is the first in a system of two equations that include two unknowns R and a.
On the other hand, a horizontal component of vector −R is the cause of horizontal acceleration a of slide B that has mass M.
Therefore,
R·sin(φ) = M·a
This is the second equation in a system of two equations that include two unknowns R and a.
Solving this system as follows.
From the second equation:
R = M·a/sin(φ)
Substitute it in the first equation:
M·a/sin(φ) − W·cos(φ) =
= −a·m·sin(φ)
The solution for horizontal acceleration of slide B is
a = W·cos(φ) /
/[m·sin(φ) + M/sin(φ)] =
= W·sin(φ)·cos(φ) /
/[m·sin²(φ)+M]
So, our final result for an acceleration a of slide B, as it moves horizontally, is W·sin(φ)·cos(φ)/[m·sin²(φ)+M]
Friction prevents us to easily move furniture around. We can buy special
devices to make it easier, but friction is definitely prevents to move
furniture easily.
Friction is the reason why parts in many machines wear out and need replacement.
Friction between the air and a spaceship, when it enters the atmosphere,
causes tremendous heating of the spaceship's body and requires special
layer of heat-resistant material to cover the spaceship.
Friction prevents a rollerblader to roll indefinitely along the smooth straight road.
In a word, friction is bad.
But is it really?..
Friction is needed to walk on a road, to drive a car, to wear clothes,
to have furniture in place where we want it, to have plates on a table
in a steady position without slipping over the edge, to fork a piece of
meat, to hold a glass of water in a hand.
Enough! As we see, we cannot live without friction, but sometimes would like to reduce it.
So, what is friction?
Here is what causes friction:
On a micro level the surfaces of two objects touching each other are not
perfect. Each little bump of one is catching a little bump of another,
thus preventing smooth gliding of one surface over another.
As was mentioned above, there are good and bad sides of this story, but
it is what it is. We have to use this friction, when needed, and reduce,
when desired.
Static Friction
Static friction is the friction between an object at rest and surface it is positioned on.
A chair stands at a fixed place on a floor, even if the floor is slightly tilted, because static friction holds it in place.
Micro-bumps on the floor are caught in-between micro-bumps on the chair's legs and fix the chair's position on a floor.
Friction is the source of the force preventing the motion. If there is
no force that attempts to move an object from the position of rest,
friction results in no force preventing this move, but, if there is a
force that attempts to move an object, friction exhorts a force against
it, thus preventing the move.
If the force that attempts to move an object is not too strong, bumps on
the surfaces of object and its support are still caught between each
other, an object stays in place, which means that the force of friction,
directed against the force that attempts to move an object, is equal to
a force of moving by magnitude and opposite in direction.
Obviously, the force of friction has its limits, so, if a very strong
force attempts to move an object, the bumps on the surface will no
longer be able to hold an object, and an object moves.
So, as the force attempting to move an object from the state of rest
grows in magnitude, so does a force of static friction that prevents
this move, but only up to a maximum value, after which the force of
friction cannot grow any more, and the object moves. This maximum value of the friction force is the one when it is specified explicitly.
To move an object from the position of rest we have to apply force. The
heavier an object - the deeper bumps on its surface are caught
in-between bumps of the surface it stands on. Therefore, the force
needed to move it from the position of rest depends on an object's
weight. At the same time, surfaces can be more or less smooth, that is
they can have deeper or shallower bumps. Obviously, the degree of
smoothness must be a factor in the amount of force needed to move an
object from the position of rest.
So, two major factors contribute to the amount of force needed to shift
an object from the position of rest: the pressure between an object and a
surface it stands on (for example, the component of its weight) and the
quality of surfaces of an object and its support.
According to numerous experiments, for any two surfaces (surface of an object and surface of a support it rests on) there is a coefficient of static frictionμ, such that the force needed to shift an object from the position of rest F depends on this coefficient and the normal pressure N between an object and its supporting surface as
F = μ·N
If the experiment is conducted on Earth and the supporting surface is horizontal, normal pressure of an object is its weight.
If the surface is an incline, the vector of weight should be represented
as a combination of a normal to a supporting surface and tangential to
it components. The normal component contributes to friction, while the
tangential component represents the force that attempts to move object
from the position of rest.
Consider an incline of angle φ and an object of weight W on it. Will it slide down?
The following forces are acting upon this object:
(a) vector of gravity of Earth W directed vertically down;
(b) reaction of the surface R directed perpendicularly to a surface of an inclined;
(c) force of static friction Ff directed against potential movement down the slope.
The right approach to analysis of this experiment is to represent a vector of weight W as a combination of a vector perpendicular to an inclined WR and the vector of force attempting to move an object down a slope WF.
Now the only force that attempts to move an object is a component of its weight tangential to an inclined WF.
The reaction force R equals to a component WR of weight.
Obviously, the magnitudes of these vectors are:
WF = W·sin(φ)
WR = W·cos(φ)
Assuming, we know that the coefficient of static friction equals μ. Then the maximum force of static friction equals to
Ff = W·cos(φ)·μ
Therefore, if WF is greater than Ff, the object will slide down a slope, otherwise the static friction will be able to hold it in place.
The minimum angle when the movement occurs is a solution of the following equation for angle φ:
W·sin(φ) = W·cos(φ)·μ or
tan(φ) = μ or
φ = arctan(μ)
This is an angle, when the sliding of an object under its own weight begins.
Kinetic Friction
Kinetic friction is the friction between a moving object and
surface it moves on or between surfaces of two touching each other
objects moving relative to each other.
Generally speaking, we all know that it's easier to move a sofa on the
floor after the movement has started than to initiate a movement from
the state of rest.
The theory behind the friction, based on tiny bumps on the surfaces of
touching each other objects, explains this as a result of deeper
penetration of bumps of one surface between the bumps of another, when
objects are at rest, than when objects are in motion and bumps have
little time to deeply penetrate each other, they slide across each
other, causing kinetic friction, which is less than static.
These bumps are so small that relative speed of one object against another plays little role in the strength of kinetic friction, but the pressure between the objects and material their surfaces are made of are very important.
Quantitatively, kinetic friction is very much like static one with the only difference in the coefficient of friction.
According to numerous experiments, for any two surfaces there is a coefficient of kinetic frictionμ, such that any moving object experiences the force F, acting against its movement, which depends on this coefficient and the normal pressure N between an object and its supporting surface as
F = μ·N
This is an experimental law. Coefficients of friction are measured for
many pairs of surfaces, which allows to solve such problems as to find
the distance of breaking from some speed to full stop, determine forces
needed to overcome the friction to move an object with constant speed
etc.
These are the problems we will solve in the next lectures.
In the differential form relationship between vectors of momentum of motion and impulse of force causing this motion is
d[m·v(t)] = F(t)·dt
The same can be expressed in terms of derivative of momentum by time:
d[m·v(t)]/dt = F(t)
or, using different notation for derivative,
[m·v(t)]' = F(t)
Any of the above equations are equivalent to the Newton's Second Law.
A trivial consequence of these equations is that in case of absence of force (F(t)=0) the momentum's derivative is zero and, therefore, momentum remains constant.
If we have a system of objects with no external forces, any force,
acting from some object against another has equal in magnitude and
opposite in direction reaction force acting from the latter to the
former. Since the time of interaction is the same for action and
reaction, while forces are equal in magnitude and opposite in direction,
their vectors of impulses, if added, produce zero total impulse. Each
impulse produces a corresponding change in momentum of an object it's
applied against. These momentums will also be of the same magnitude and
opposite in direction. Consequently, the total momentum of the system
will not change.
We can conclude that in the absence of external forces the momentum of motion of a system of objects is constant.
This is the Law of Conservation of Momentum.
Example 1
Consider a frame of reference fixed on a billiard table.
A ball #1 is positioned near the origin of coordinates in the middle of a table.
Another ball #2 of the same mass is positioned at some point on the X-axis on its negative side.
A player hits ball #2 towards the point where ball #1 is located, giving it a speed v.
After the contact the balls went each in its own direction. Ball #1 went at an acute positive angle α from the positive direction of the X-axis (that is, counterclockwise), while ball #2 went at an acute but negative angle −β from the positive direction of the X-axis (that is, clockwise).
What are the magnitudes of the speeds of the balls after contact (ignore the friction)?
Solution
Let's apply the Law of Conservation of Momentum.
Initially, the total momentum of the system was equal to the momentum of
ball #2 (as the only one having non-zero velocity), which is P = m (kg)·v (m/sec).
As a vector, it has only X-component not equal to zero, since velocity of ball #2 is along the X-axis.
After the contact the balls have velocities v1 (for ball #1) and v2 (for ball #2) with unknown magnitudes, correspondingly, v1 and v2.
These velocities, as vectors, have both X- and Y-components:
v1x = v1·cos(α)
v1y = v1·sin(α)
v2x = v2·cos(−β) = v2·cos(β)
v2y = v2·sin(−β) = −v2·sin(β)
The sum of X-components of momentums of both balls after the contact
must be equal to X-component of the momentum of ball #2 before the
contact and the sum of Y-components of momentums of both balls after the
contact must be equal to Y-component of the momentum of ball #2 before
the contact. that gives us two equations with unknown:
m·v1·cos(α)+m·v2·cos(β) = m·v
m·v1·sin(α)−m·v2·sin(β) = 0
Reducing by m both equations, we have a simple system of two linear equations with two unknowns:
v1·cos(α) + v2·cos(β) = v
v1·sin(α) − v2·sin(β) = 0
Solutions for this system are:
v1 = v·sin(β)/sin(α+β)
v2 = v·sin(α)/sin(α+β)
Interesting particular case appears if α=0, that is when
the ball #2 hits the ball #1 heads-on. Then the speed of the ball #2
after the contact is zero, it stops at the place of contact and the ball
#1 takes on the full speed v, that the ball #2 used to have, and continues the motion in the same direction along the X-axis.
Example 2
Consider an inertial system of reference on a plane.
Object A moves along X-axis from the negative side of the X-axis in the positive direction towards the origin of coordinates.
Object B moves along Y-axis from the negative side of the Y-axis in the positive direction towards the origin of coordinates.
The mass of object B is 4 times greater than mass of object A, but the
speed of object A along X-axis is 3 times greater than speed of object B
along Y-axis.
At the origin of coordinates these two objects meet and attach to each
other, forming a new combined object AB. The magnitude of the speed of
the combined object AB is vAB.
What are the speeds of objects before their contact?
Solution
Let the unknown mass of object A be mA and its speed along the X-axis be vA.
Let the unknown mass of object B be mB and its speed along the Y-axis be vB.
We will deal with scalar speeds because the direction of the velocities
is along the coordinate axes and, therefore, Y-component of the velocity
of object A is zero, as well as X-component of the velocity of object
B.
The vector of momentum of object A before contact is directed along the positive direction of the X-axis and its magnitude is
PAx = mA·vA
The Y-component of the momentum of object A is zero:
PAy = 0
For object B the momentum before the contact is directed along the positive direction of the Y-axis and its magnitude is
PBy = mB·vB
The X-component of the momentum of object B is zero:
PBx = 0
After the contact and attachment the momentum of the combined object AB is equal by magnitude (with unknown direction) to
PAB = (mA+mB)·vAB
While we do not know the direction of the velocity of the combined
object AB, we can evaluate the X- and Y-components of its momentum as
sums of corresponding X- and Y-components of momentums of objects A and
B.
From the Law of Conservation of Momentum follows
PAx+PBx = PABx
PAy+PBy = PABy
Therefore,
PABx = mA·vA
PABy = mB·vB
From these two orthogonal components we can derive the magnitude of the momentum vector of combined object AB:
|PAB| = √[mA·vA]² + [mB·vB]²
On the other hand, the mass of combined object AB is a sum of masses of
objects A and B. Also, the magnitude of a speed of combined object vAB is given.
Therefore,
(mA+mB)·vAB =
= √[mA·vA]² + [mB·vB]²
We know that
mB = 4·mA
vA = 3·vB
Substituting these proportions to the equation above for the momentum of combined object AB, we obtain
(mA+4·mA)·vAB =
= √[mA·3·vB]² + [4·mA·vB]² =
= 5·mA·vB
Reducing both sides of the equation by 5·mA, we obtain the value of speed of object B:
In a simple case of a single constant force F acting in the direction of motion on an object of mass m that moves with constant acceleration a the Newton's Second Law states that
F = m·a
Let's assume that at moment of time t the speed of this object was v and at moment of time t+Δt the speed of this object was v+Δv.
Since an objects moved with constant acceleration a, the average increment of speed during this time should be equal to the acceleration multiplied by increment of time
(v+Δv) − v = a·[(t+Δt) − t] or
Δv = a·Δt
Multiplying both sides of the above equation by mass m, we obtain
m·Δv = m·a·Δt or
m·Δv = F·Δt
Considering m is constant, we can make the following manipulations with the left side of this equation:
m·Δv = m·[v(t+Δt)−v(t)] =
= m·v(t+Δt)−m·v(t) = Δ(m·v)
Therefore,
Δ(m·v) = F·Δt
The expression on the left side represents an increment of the momentum of motion. The expression on the right side is called impulse of the force F acting on an object during interval of time Δt.
If the time interval Δt is infinitesimal, this can be applied to variable force F(t) and written in terms of differentials
d(m·v) = F(t)·dt
Integrating this equality by time t, we obtain on the left side the total momentum change during time from moment of time t=t1 to moment t=t2:
P = ∫t1t2d(m·v) =
= m·v(t2) − m·v(t1)
The total impulse exhorted by a variable force F(t) during this time from t1 to t2 is
J = ∫t1t2F(t)dt
Hence, our qualitative observation, that the force acting on an object
changes its velocity, can be quantitatively characterized as impulse exhorted by a force during certain time period equals to change of momentum during this period.
Obviously, for a simple case of constant force F this is equivalent to
m·v(t2) − m·v(t1) = F·(t2−t1)
which fully corresponds to the Newton's Second Law for constant force F and, therefore, constant acceleration a since
[v(t2)−v(t1)]/(t2−t1) = a = F/m
Generally speaking, force is a vector. So is velocity and, therefore, momentum of motion. All the definitions discussed above, considering forces and motions in our three-dimensional space, are for vectors.
Hence, for constant force F in three-dimensional space the definition of impulse of this force acting during time t is: J = F·t.
For variable force F(t) acting on object from moment of time t=t1 to moment of time t=t2 we define the impulse in the integral form
J = ∫t1t2F(t)dt
where integral of vector is a vector of integrals of its X-, Y- and Z-components.
Using the approach based on the concept of impulse of the force and momentum of motion, we can simplify solutions to some problems.
Consider a case when an object of mass m, staying at rest in some inertial reference frame, is pushed forward with constant force F1 during time t1, then with force F2 in the same direction during time t2.
What would be its final speed vfin?
The impulse given by the first force is F1·t1. It caused an increase in speed from 0 to v1.
Therefore,
F1·t1 = m·(v1−0)
Then the impulse given by the second force is F2·t2. It caused an increase in speed from v1 to vfin (the final speed).
Therefore,
F2·t2 = m·(vfin−v1)
Adding them together, we obtain
F1·t1 + F2·t2 =
= m·v1 + m·(vfin−v1) =
= m·(vfin−0) = m·vfin
which gives the final speed
vfin = [F1·t1 + F2·t2]/m
But the easiest way to solve this is to use combined impulse given to an
object as the cause of increased speed from zero to its final value vfin by adding impulsesF1·t1 + F2·t2 and equating their sum to final increase in momentumm·vfin with the same result for final speed.
SUMMARY
Any action of force on an object during certain time interval exhorts an impulse that causes to change the object's momentum.
Each consecutively or simultaneously applied impulse contributes to this
change of momentum, so the final momentum of an object is the combined
effect of all impulses acting on it in an integrated fashion.
Example
Liquid fuel is pumped into combustion chambers of airplane engines at the rate μ (kg/sec).
Burned gases are exhausted with speed vout. An airplane is in uniform motion along a straight line with constant velocity.
What is the force of air resistance acting against its motion?
Assume that the mass of an airplane is significantly greater than the
mass of exhausted gases, so we can ignore the loss of mass during
engine's work.
Solution
Airplane is in uniform motion, which allows us to use a reference frame
associated with it as the inertial frame, where the Newton's Laws are
held and all calculations can be done.
Since an airplane's motion is uniform, the sum of the vector of air
resistance force, directed against its movement, and the vector of the
reaction force from the gases, exhausted by its engines and directed
towards its movement, should balance each other and be equal in
magnitude since their directions are opposite.
During time interval Δt the engines exhaust burned gases of mass Δm=μ·Δt, accelerating them from speed zero to vout and correspondingly increasing their momentum.
Therefore, from the equality between the increment of momentum of motion of burned gases and the impulse of force applied to them by the engines, we have the following equality:
Δm·vout = F··Δt
Using the expression for Δm above, we derive from this
μ·Δt·vout = F··Δt
And the value for the force produced by an engine applied to burned gases pushing them backward
F = μ·vout
According to Newton's Third Law, this is the same force, with which burned gases push an airplane forward.
Since the airplane movement is uniform, the air resistance is also equal in magnitude to this value.
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