Monday, May 21, 2018

Unizor - Physics4Teens - Mechanics - Dynamics - Conservation of Momentum





Notes to a video lecture on http://www.unizor.com



Conservation of Momentum



In the differential form relationship between vectors of momentum of motion and impulse of force causing this motion is

d[v(t)] = F(t)·dt

The same can be expressed in terms of derivative of momentum by time:

d[v(t)]/dt = F(t)

or, using different notation for derivative,

[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+mBvAB

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

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+mBvAB =

= √[mA·vA]² + [mB·vB

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·mAvAB =

= √[mA·3·vB]² + [4·mA·vB =

= 5·mA·vB

Reducing both sides of the equation by 5·mA, we obtain the value of speed of object B:

vB = vAB

Now we can determine speed of object A:

vA = 3·vB = 3·vAB

Unizor - Physics4Teens - Mechanics - Dynamics - Impulse of Force





Notes to a video lecture on http://www.unizor.com



Impulse



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

Δv = m·a·Δt or

Δv = F·Δt

Considering m is constant, we can make the following manipulations with the left side of this equation:

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·
(vfin0) = 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 impulses F1·t1 + F2·t2 and equating their sum to final increase in momentum m·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.


Thursday, May 17, 2018

Unizor - Physics4Teens - Mechanics - Dynamics - Momentum of Motion





Notes to a video lecture on http://www.unizor.com



Momentum of Motion



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 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

Δv = m·a·Δt or

Δv = F·Δt or

v/Δt) = F



If the time interval Δt is infinitesimal, this can be written in terms of derivative of speed v by time t as

(dv/dt) = F

or, since mass m is constant,

d(m·v)/dt = F



So, the force F equals to a derivative by time of the value of a product of mass by speed m·v, called momentum of motion.

Since we have reduced the time interval Δt to
infinitesimal value, it's no longer important that the force and
acceleration are constant. In the next moment of time it can be
different, but at any moment of time the dependency between the force
and a derivative of the momentum above will take place.



Let's consider a more complicated case of motion in the three-dimensional space.

The Newton's Second Law states that the vector of force F(t), as a function of time t, equals to a product of constant mass m and the vector of acceleration a(t), which, in turn, also depends on time t:

F(t) = m·a(t)



An immediate consequence from this, considering the vector of acceleration is a derivative of vector of velocity v(t), is the following chain of equalities:

F(t) = m·a(t) = m·dv(t)/dt =

=
d(m·v(t))/dt


from which we conclude that the force F(t), acting on an object, is a derivative by time of the characteristic of motion called momentum, that is equal to v(t):

F(t) = [v(t)]I



The same equality can be expressed in term of infinitesimal increments:

F(t)·dt = d(m·v(t))

The left side of this equation is an increment in impulse of the force F(t) during the time from t to t+dt, while the right side represents an increment of the momentum of object of mass m moving with velocity v(t) during the same time interval.



Let's return to a simple case of a constant force acting in the direction of movement with an equation that connects the force with a derivative of momentum:

d(m·v)/dt = F

The first consequence from this equation is that in the absence of force (F=0)
the derivative of momentum is zero and, therefore, momentum is
constant. With constant mass it is equivalent to the Principle of
Inertia because it implies the constant velocity.

But, expressed as a constant momentum, it covers a more general case of complex systems of objects with no external forces.



Assume that a two-stage rocket of mass m flies in open space far from any gravity fields along a straight line with speed v
with no engine working. According to the Principle of Inertia (and the
Newton's First Law), it will move like this indefinitely.

Then at some moment the first stage of mass m1 is separated from the second stage of mass m2=m−m1. This separation is achieved by pushing the first stage back along the same straight line with a force F during time t.

The resulting speed of the first stage is

v1 = v − (F/m1)·t

The resulting momentum of the first stage is

m1·v1 = m1·v − F·t

Meanwhile, according to the Newton's Third Law, the same by magnitude
force should push the second stage forward during the same time. Its new
speed is

v2 = v + (F/m2)·t

The momentum of the second stage is

m2·v2 = m2·v + F·t

Let's add the momentums of two stages after the separation:

m1·v1 + m2·v2 =

= m1·v − F·t + m2·v + F·t =

= (m1 + m2)·v = m·v




As we see, even in case of a complex system of objects with no external forces, the momentum of an entire system is constant.

Our example with a rocket can be generalized into a system of any number
of objects and not only straight line motion. We will address this
issue further in the course as the Law of Conservation of Momentum.

Wednesday, May 16, 2018

Unizor - Physics4Teens - Mechanics - Dynamics - Superposition of Forces ...





Notes to a video lecture on http://www.unizor.com



Superposition of Forces -

Problems 2




Problem A



Consider a slope that makes an angle θ with the ground. On
the top of this slope there is a pulley, through which goes a
weightless non-stretchable thread that connects two point-objects:

one of mass M1, sliding up the slope, and

another of a bigger mass M2, hanging off the top pulley vertically down and pulling the thread down, thus forcing the smaller object to go up the slope.



Determine the direction and magnitude of the pressure vector P on the pulley's axis.



Answer:



Direction of the vector of pressure on the pulley's axis P is along the angle bisector of the angle at the top of a pulley.

The magnitude of this vector is

P = k·[1+sin(θ)]·cos(π/4−θ/2)

where

k = 2M1·M2·g / (M1+M2)





Problem B



A point-object is hanging on a thread of length L and is uniformly moving along a circular trajectory in a horizontal plane, while a thread always maintains a constant angle φ with a vertical.



How many loops N(t) this object does in time t?



Answer:



N(t) = (t/2π)·√g/[L·cos(φ)]

where g is a free fall acceleration.





Problem C



A wedge of mass m and angle slides down between two point-objects of mass M each, pushing them apart without any friction.



Find the acceleration a of each of the two objects that a wedge pushes apart.



Answer:



a = g·tan(φ)/

/[1+2(M/m)·tan²(φ)]






Problem D

During long jump competition a sportsman accelerates with constant acceleration a during time t and then jumps vertically up.

During his jump he goes up to height h, while continuing moving forward because of his initial horizontal speed he gained before jumping.

What is the length L of his jump?



Answer:

L = 2a·t·√2h/g

Tuesday, May 15, 2018

Unizor - Physics4Teens - Mechanics - Dynamics - Superposition of Forces ...





Notes to a video lecture on http://www.unizor.com



Superposition of Forces -

Problems 1




Problem A



A boat stands still in the middle of a river. It is tied to a river bank by a rope of the length L.

The river flow pushes it with the force F downstream, and the wind pushes it with the force 0.75·F perpendicularly to the river flow away from the bank.

(a) What is the force of tension P of the rope?

(b) What is the distance d from the river bank to the boat?



Answer:

(a) P = 1.25·F

(b) d = 0.6·L





Problem B



An object of mass m and, therefore, of weight at the ground W=m·g, where g
is a free fall acceleration, moves with constant linear speed on a
bridge, that has a shape of an arc with upward convexity of a radius R.

(a) Determine the speed v of an object as a function of pressure P it produces on the bridge at the top of the arc.

(b) At what speed v0 the object will become "weightless" at the top of the bridge's arc?



Answer:

(a) v = √R·(g − P/m)

(b) v0 = √R·g,

because "weightlessness" is a state when an object does not press on the support and, therefore, P=0





Problem C



A point object of mass m and, therefore, of weight W=m·g, where g is a free fall acceleration, is hanging on the end of a weightless thread of length R,
fixed at the other end at some point, and can freely move like a
pendulum within a vertical plane that we can take as the XZ-plane of
coordinates with origin at the point where a thread is fixed.

An object's initial position is with its thread making an angle φ from a vertical.

(a) What is the tension of a thread P in the initial position?

(b) What is the vector of force F acting on an object in its initial position in the direction of its motion along a circular trajectory?



Answer:

(a) Tension P = m·g·cos(φ)

(b) Force along the trajectory F = m·g·sin(φ)





Problem D



A projectile is launched from the wall of the castle horizontally towards the army that laid a siege on this castle.

The height of the castle wall is H, the horizontal speed of a projectile is v.

The free falling acceleration caused by gravity is g.

Ignore the air resistance.

(a) What is the time tLand the projectile will be in the air until landing?

(b) What is the distance d of a point the projectile lands from the bottom of the castle wall?

(c) What is the magnitude of a speed of the projectile vLand at the moment of landing?



Answer:

(a) tLand = 2H/g

(b) d = v·√2H/g

(c) vLand = √v² + 2Hg

Monday, May 14, 2018

Unizor - Physics4Teens - Mechanics - Dynamics - Resultant Force





Notes to a video lecture on http://www.unizor.com



Resultant Force



According to the Principle of Superposition of Forces, we can replace actions of many forces acting upon a point-object (each one being a vector) with a vector sum of these forces (called resultant force), getting exactly the same effect on the motion of a point-object.



An immediate consequence from the Principle of Superposition of Forces
is that any single force acting upon a point-object can be replaced by a
vector sum of a few forces, which produce exactly the same effect on
the motion of a point-object, as long as this vector sum of forces
exactly equals to the vector of original force.



We will demonstrate this Principle on a few examples.





Example 1



In the frame of reference associated with the ground we have a point-object sliding down a slope.

Given are: mass of an object m, force of gravitation of this object (its weight, directed vertically down) W and an angle φ between a slope of a slide and horizontal ground.

Our task is to determine the acceleration of this object on its
trajectory down a slope, considering there is no friction that slows
down the object's movement down a slope.



Solution



Trajectory of our object is a straight line (the slope), mass m
is known, so all we need to find the acceleration from the Newton's
Second Law is the force that pushes this object along its trajectory
down a slope.

Obviously, the slope's angle φ plays an important role here. Consider a slope with φ=0, that is the "slope" is not really a slope, but a horizontal plane. The object will not move then because its weight W, directed vertically down perpendicularly to a horizontal plane, is completely balanced by the reaction force of the plane R
directed vertically up and equal in magnitude to the weight, according
to the Newton's Third Law. The resultant force is zero-vector and there
is no acceleration.

On the other extreme, if our slide is vertical with φ=90°, an object does not press on it, there is no reaction force, all the object's weight W is directed along the vertical trajectory downward, and the acceleration can be determined from the Newton's Second Law as a=W/m.



To meaningfully include the slope's angle φ into our calculation, we will represent its weight W as a vector sum of two forces:

(a) the force WR, directed perpendicularly to a slope, causing slope's reaction force R to balance it, and

(b) the force WF, directed parallel to a slope, pushing the object forward along its trajectory and causing its acceleration.

If we succeed in such representation, we will say that the force WF is the one and only force that pushes the object of mass m down a slope and can be used to determine the object's acceleration from the Newton's Second Law as a=WF/m.



This representation of the weight of an object as a sum of two vectors W=WR+WF, knowing the angle φ, is easy.


The magnitude of vectors WR and WF, which we will denote as WR and WF correspondingly, equal to

WR = W·cos(φ)

WF = W·sin(φ)

where W is the magnitude of a vector of weight W.



The force WR
is completely balanced by the reaction force of the slope (they both
are perpendicular to a slope, equal in magnitude and oppositely
directed, according to the Newton's Third Law).

So, the only force affecting the motion of an object on its trajectory down a slope and acting along this trajectory is WF, the magnitude of which we have calculated above.

Therefore, an acceleration of an object down a slope will be a vector directed parallel to a slope with a magnitude

a=W·sin(φ)/m.





Example 2



A motorcycle stuntman decides to demonstrate his skills by going inside a vertical loop.

His mass is m.

His linear speed inside this loop is v.

The magnitude of the gravity force (his weight with a motorcycle) is W.

The radius of a loop is R.

Find the pressure on the loop at its top and at its bottom from a moving motorcycle.



Solution



Let's set our frame of reference on the ground with Z-axis directed upward and a loop being in the XZ-plane.



Since an object of mass m moves with a constant linear speed v along a circular trajectory of radius R,
there is a force that keeps it on this trajectory. This force is always
directed towards the center of a circle and is equal in magnitude to m·a, where a is a magnitude of its acceleration towards a center of a circle called centripetal acceleration.

From the Kinematics of rotation we know that a=R·ω², where ω is the angular speed of rotation and is related to linear speed of rotation v as v=R·ω.

From this we derive the centripetal acceleration a in terms of linear speed and radius: a=v²/R.



Knowing the force that keeps an object on a circular trajectory, let's analyze what is the source of this force.

On one hand, there is always a weight directed downward (towards negative direction of Z-axis) and equal in magnitude to W.

On the other hand, an object presses onto the loop with the force P(t)
perpendicularly to the loop, and the loop reacts back unto object with
the same by magnitude but oppositely directed reaction force. This
reaction force is directed upward, when the object is at the bottom of
the loop (we will denote its magnitude as Pb), or downward, when an object is at the top (we will denote its magnitude as Pt).



The combination of weight and the reaction force of the loop is the
resultant force that keeps the object on its circular trajectory.



Therefore, at the bottom of the loop, when weight is directed downward,
reaction is directed upward and the centripetal force is directed
upward,

−W+Pb = m·R·ω²



At the top of the loop, when weight is directed downward, reaction is
directed downward and the centripetal force is directed downward,

−W−Pt = −m·R·ω²



From the first equation we can derive the reaction force at the bottom
that is, by magnitude, equals to the pressure on a loop at this point:

Pb = m·R·ω² + W

From the second equation we can derive the reaction force at the top
that is, by magnitude, equals to the pressure on a loop at this point:

Pt = m·R·ω² − W



Theory of gravitation, that will be studied later in this course, states
that weight is a force of gravity proportional to a product of masses
of a planet and an object on its surface and inversely proportional to a
square of a distance between centers of mass. The latter is
approximately equal to a radius of a planet.

For Earth it is expressed as W = m·g, where constant g is an acceleration of the free falling on Earth, which is equal to 9.8 m/sec².

On Jupiter, by the way, this constant - free fall acceleration - is equal to 25 m/sec².

Using this, we can write our formulas for pressure as follows:

Pb = m·R·ω² + m·g = m·(R·ω² + g)

Pt = m·R·ω² − m·g = m·(R·ω² − g)



It should be noted that, while Pb is always positive, Pt
might be negative, if the speed of an object is not sufficient. If
negative, the movement will not be circular and object will fall down
from its trajectory.



If R·ω²=g or ω=√g/R, or v=√g·R,
the object will experience weightlessness at the top position, since it
will not press onto loop and, therefore, loop will not press on it. For
loop of, say, 10 meters in diameter (R=5 m) linear speed of approximately 7 m/sec
will produce an effect of weightlessness at the top of a loop. This
linear speed with a given radius of a loop is equivalent to the angular
speed of 1.4 rad/sec or about one complete loop in 4.5 second.

Slower speed will result in falling from the circular trajectory.  

Friday, May 11, 2018

Unizor - Physics4Teens - Mechanics - Dynamics - Superposition of Forces





Notes to a video lecture on http://www.unizor.com



Sum of Forces



Previously, in most cases we were analyzing the motion of objects with only one force participating in the process.

With few exceptions our objects moved either along a straight line with a
force acting along this line or rotated around some center with only
one force keeping them on their circular trajectory.
A couple of problems related to launching of projectile at an angle to
horizon did deal with a more complex motion, but even in this case there
was only one force - gravity - that acted on an object.



This lecture is dedicated to general approach to dealing with multiple
forces of different types acting simultaneously on the same object.

Let's review what kind of forces occur in real life and, therefore, might affect the motion of an object.



Applied Forces

These are forces caused by one object contacting another (pull, push, roll etc.).

These are the first kind of forces people dealt with and understood
relatively well in a sense that they always saw the source of them.

A car moves because its engine pushes the wheels forward. A person walks
because his muscles make his legs move in a special way to push back
the road.



Friction

The effect of friction is slowing down the movement, unless additional
forces are applied. The source of friction is non-ideal contact between a
moving object and its surrounding, that causes tiny bumps between them
catch each other, thus preventing the uniform motion.

To the same category we can put resistance of air, when a plane flies in
a sky or resistance of water, when a submarine moves in the deep sea.



Tension

If you pull a cart by a rope, you pull the rope, not a cart. That force
somehow is transferred to a cart. The mechanism of transferring a force
from one end of a rope to another is tension. Your force of
pulling attempts to stretch a rope, but internal links within a rope do
not allow it to stretch, one link is attached to another and transfers
the force down to all tiny links inside the rope. That causes the rope
to become tense, and the force you applied to one end of the rope is the
cause of the rope's tension, which, in turn, pulls the cart.



Elasticity

Elasticity is a property of an object, like spring or rubber, to change
its form under some external force and then restore the original form,
when the force is no longer applied. If an object stretches with its
weight a vertically positioned spring with fixed top end, reaching some
fixed balanced position, it is the elasticity force of a spring that
counteracts the force of the object's weight, thus holding an object in a
fixed position.

The cause of elasticity is in the inner structure of a material. Steel
of spring is such an elastic material, as well as rubber and others.



Gravitation

Any two masses attract ("gravitate") each other. The attraction is
mutual, the force of gravitation of object A towards object B is
directed from A to B and is equal in magnitude, but opposite in
direction, to the force of gravitation of object B towards object A,
directed from B to A.

Earth attracts everything on it, like people, houses, airplanes in the
sky, ships in the sea, while each of them attract Earth with the same in
magnitude, but opposite in direction force.

Quantitatively, the force of gravitation between two masses proportional
to each mass and inversely proportional to square of distance between
them. This is an experimental fact, we'll discuss it later in this
course.



Electricity

This force is in some way similar to gravitation, but in case of
gravitation we have only one type of force - attraction, while in case
of electricity it might be attraction or repelling, depending on what
kind of electrical charges we deal with.

Two electrically charged objects with opposite charges (which we call
"positive" and "negative") attract each other similarly to gravitation.

Quantitatively, the force of electrical attraction between two
electrically charged objects is proportional to each charge and
inversely proportional to square of distance between them. This is an
experimental fact, we'll discuss it later in this course.

If two positively or two negatively charged objects are involved, the
forces will act in opposite direction, the objects will be repelled from
each other with the force of repelling exactly the same in magnitude as
the force of attraction in case of two opposite charges.



Magnetism

This force is in some way similar to electricity, but in case of
electricity we have two types of objects - positively and negatively
charged, while in case of magnetism each magnetized object has two ends
- two "poles" with opposite "polarity". There are two kinds of
polarity, that we conditionally call "North" (N) and "South" (S), and
opposite ones (N - S) attract each other, while similar ones (N - N or S
- S) repel.

Two magnetized objects attract or repel each other, depending on which side of one object is close to which side of another.

Quantitatively, the force of magnetic attraction or repelling between
two magnetized objects is proportional to degree of their magnetizing
and inversely proportional to square of distance between them. This is
an experimental fact, we'll discuss it later in this course.



So, there is a multitude of forces that can act on our object
simultaneously and we have to be able to analyze their combined effect
on the motion.

The main idea of a motion, when multiple forces act on an object, is a principle of superposition of forces.



It is experimentally confirmed and accepted in classical Physics as an
axiom that, if two or more forces act on the same object at the same
time, the object moves identically to the movement caused by only one
force, that is a vector sum of all individual forces acting on an
object.



This axiom is called Principle of Superposition of Forces.



Here we are talking about general principle, that we accept for any kind
of forces in three-dimensional world, each being a vector: acting together, these forces produce the same effect, as if they are all replaced by one force - their vector sum.



Let's illustrate this with a couple of examples.



1. When a rocket is launched vertically up, there are two major forces
acting upon it: the force of gravity pulls it down and its engine pushes
it up (stronger than gravity, of course). These two forces are directed
along the same vertical line, but opposite in direction. Their vector
sum is directed upwards and quantitatively is determined by a vector sum
of these major components.



2. A car moves up the hill and is acted upon by three forces:

gravity pulls it downward,

its engine pulls uphill and

the reaction of the road pushes perpendicularly to the road.

The vector sum of these three forces must point up the hill.



3. An object is hanging on a thread fixed to a point on a ceiling and is moving in a vertical plane like a pendulum.

There are two forces acting upon it at each moment:

gravity pulls it downward,

tension of a thread pulls it towards the point on a ceiling, where a thread is fixed.

The vector sum of these two forces must be tangential to an arc of
trajectory of this object and directed towards this arc's midpoint.



All the above examples are examples of the Principle of Superposition of Forces
- a universal principle of Physics, which we accept as an axiom,
regardless of the physical nature of the forces involved, and which is
confirmed by experiments.