KINEMATICS ANALYSIS : MOVING BODIES......

KINEMATICS ANALYSIS: MOVING BODIES......
Mechanics, physics, mechanical engineering

What makes a body moving?

OBSERVATION ONE:

"From our perception, we can say, a body moves if we apply either a push or a pull on the body. There are several  instances, when after applying push or pull, the body still doesn't move. What is the exact reason behind this?"

OBSERVATION TWO:

"Almost to apply a push or pull on a body, a physical contact is needed, with out any physical contact it is not possible to exert push/pull on a body, although there are exceptions too.

(i) We know on every material body existing on earth experience a downward pull towards the center of Earth and to exert this pull, the object and the earth don't need any physical contact. This phenomena is aptly named as the 'Force of Gravity.'

(ii) The second option is Magnet. A magnet can pull as well as push another magnet from a distance with out any physical contact.

(iii) When we place a charged particle near another charged particle, we watch the particles can exert push as well as pull without any physical contact. Like a proton repels another proton, ie they exert push on each other. But a proton attracts an electron by pulling each other to come close.

So there are two types of motion. Type one is the example of a Cricket ball going to boundary after being hit by the bat. So, here impact is the driver of the motion.

But an apple falling from a tree is also in motion, but for this no impact is there. In fact the attraction between the apple and the earth is responsible for the motion. By attraction, it means the tendency of going closer in two objects. Here motion of apple occurs without being hit.


LINEAR MOTION:

We know that Force is a kind of physical quantity having both magnitude as well as direction. So, force is a vector quantity. We also know that applying triangle's law or parallelogram law of forces addition, we can add two forces to get an equivalent force which is known Resultant force.

So, application of force on an object brings a change in position of the object. This change is position is called displacement. An object in a coordinate system has a position vector to define the position vector. Any change of position will bring change in its position vector.

Suppose in a coordinate system we have an object at a position vector (r). Let after a time interval of (dt), the object changes its position by an amount (dr). Hence, the rate of change of position will be (dr)/(dt). Rate of change of position

means the change of position in unit time. Rate of change of position is called velocity of the object. It is denoted by (v).

Hence, (v) = (dr)/(dt).
The unit of velocity is m/s in SI units and cm/s in cgs system. The most popular unit is km/hr.

A velocity may change, it may change in direction or it may change in magnitude. Suppose, we have an object moving with a velocity (v) at any instant. Suppose after an interval of time (dt), its new velocity becomes (v + dv), where (dv) is the change in velocity. Hence (dv)/(dt) will be the rate of change of velocity and it indicates the change of velocity per unit time. This physical quantity is called acceleration. Negative acceleration which is rate of decrease in velocity is called retardation or deacceleration too.

Acceleration may be changed; the rate of change of acceleration is called impulse. Like when a bat touches a moving ball, it has an impact and this changes its acceleration due to this magnitude and direction of the ball changes. If a large magnitude of force act on a body for a very short period of time, it is called impulse.

Free falling under the forces of gravity is a case of constant acceleration and that is denoted by (g) and it is equal to g = 9.81  m/s ² .

THREE EQUATIONS OF MOTION

These three equations are valid only in the case of constant acceleration. Every particle falling under the gravity will satisfy these equations.

We know (dv)/(dt) = a,
hence, (dv) = a(dt) or by integrating both side from initial state t = 0; v = u to final t = t; v = v,
we get,
v - u = a(t - 0) or v = u + at

Again, a = (dv)/(dt)
a = {(dv)/(dx)}.{(dx)/(dt)}
a = v.(dv)/(dx)
hence, v.dv = a.dx
Integrating both sides from initial v= u, x = 0 to final v = v, x = s
we get,
v² - u² = 2 a (s - 0)
v² - u² = 2as

From the defination of velocity, we get
v = (dx)/(dt)
but, v = u + at
hence,
u + at = (dx)/(dt)
dx = (u +at). dt
Integrating both sides of the equation from initial condition t = 0, x = 0 and final condition t = t, x = s, we get

(s - 0) = u(t - 0) + (a.t²)/2
s = ut + (at²)/2

Average Velocity = (u + v)/2
= (u + u + at)/2 = u + at/2

Total distance, s = Average Velocity x time

s = (u + at/2) x t
s = ut + (at²)/2

Uses of these equations:

1) Suppose we have a particle travelling with 5 m/s and an acceleration of 1 m/s^2 is applied on the body. What will be the

velocity after a time of 10 s?

Ans: Here, u = 5 m/s, a = 1 m/s ²  and t = 10 s, then

velocity after 10 s,
v = u + at
v = 5 + 1 x 10 = 15 m/s

2) A car moving with a velocity 60 km/hr suddenly applies the brake. As a result, the car comes to a halt after running 50 m

after applying brake. What will the value of retardation? Find the time it needs to come to rest after the application of brake.

Ans: Here, initial velocity u = 60 km/hr = (60 x 1000)/(60 x 60) m/s = 16.67 m/s
final veloity, v = 0 m/s
Total distance travelled s = 50 m
a = (v² - u²)/2s
a = (0 - 16.67²)/(2 x 50)
a = - 2.78 m/s²

again time, t = (v - u)/a
t = (0 - 16.67)/(-2.78)
t = 5.99 s

The Concept of Relative Velocity:

Suppose in a road two car is moving. The faster car at 40 km/hr and the slower car at 30 km/hr in the same direction. Now, if anyone watches the faster car from the slower car, he won't be see it running at 40 km/hr, in stead he will see the velocity at 15 km/hr. When we watch from a moving body or better we say moving reference frame, the velocity of other bodies seem be reduced. Again if we take the same two cars running in opposite directions to each other, the velocity of the each car will be at 55 km/hr as seen from the other. This is due to relative velocity. The relative velocity of a body is the velocity of the body relative to an observer.

Suppose, a car is moving with a velocity V_c and a train is moving with a velocity V_t. Then velocity of car with respect to a person sitting in the train will be V_ct = V_c – V_t and velocity of the train to a person in the car will be  V_tc = V_t – V_c  .

3) A train is moving towards east with a velocity 120 km/hr and wind is blowing towards west with a velocity 20 km/hr. What will the velocity of the wind to an observer in the train?

Ans: We shall take towards east direction as positive and towards west direction as negative.
Let the velocity of train is (V_t) and velocity of wind is (V_w)

So,  V_t  = 120 km/hr and
V_w = - 20 km/hr

V_wt =  V_w  - V_t
V_wt = - 20 - 120 = - 140 m/s and it means wind is flowing from west.

4) A car is moving on a horizontal road at 40 km/hr. Suddenly rain started to pour down at a velocity 30 km/hr. Find the

velocity of the rain drops with rest to an observer in the car. Also find the angle with which rain drops would appear to strike the car.

Ans: Let the velocity of the car be V_c = 40 km/hr and rain drop velocity is  V_r = - 30 km/hr. The angle between them is θ = 90°.
(V_rc) = (V_r) - (V_c)
(V_rc)² = (V_r)² + (V_c)² - 2(V_r)(V_c) cos θ
= 30² + 40² + 0
= 2500
(V_rc) = 50 m/s

ANGULAR MOTION:

Suppose a line AB displaces side wise such that A remains at same point, but the other end B comes to new position C. This is an angular displacement. Angles are measured in radian. 1 radian is the angle formed by an arc equals to magnitude of radius (r). A full circle produces 2π.

Suppose, a line of length (r) makes an angular displacement of (dθ) in time (dt). Then the rate of change of angular displacement is given by (dθ)/(dt) and it is called angular velocity, (w). Hence, (w) = (dθ)/(dt).

If we apply torque or moment in the line, the angular velocity will be changed. Let during the time interval, (dt), the change in angular velocity be (dω). The rate of change of angular velocity will (dω)/(dt) and it is called as angular acceleration and denoted as α.

α = (dω)/(dt)
(dω) = α.(dt)
Taking initial value ω = ωₒ, t = 0, to final value ω = ω, t = t and integrating both side, we get,
ω - ωₒ = α.(t - 0)
ω = ωₒ + αt

angular velocity ω = (dθ)/(dt)
ωₒ + αt = (dθ)/(dt)
(ωₒ + αt)(dt) = dθ
Taking initial value θ = 0, t = 0, to final value, θ = θ, t = t, and integrating both side we get,

ωₒ.(t - 0) + (α/2)(t² - 0) = θ - 0

θ = ωₒt + (αt²/2)

α = (dω)/(dt)
= {(dω)/(dθ)}.{(dθ)/(dt)}
α = ω.(dω)/(dθ)
ω.(dω) = α.(dθ)
taking the initial value ω = ωₒ, θ = 0 and final value ω =ω, θ = θ and integrating both sides,
(ω²)/2 - (ωₒ²)/2 = α.(θ - 0)
(ω²) = (ωₒ²) + 2.α.θ



MOTION OF A RIGID BODY

Plane Motion:

A rigid body is said to be in plane motion when all parts of the body move in parallel planes. The plane motion of a rigid body may be classified into several categories like :

1) Translation
2) Rotation
3) General plane motion.

Translation:

ENGINEERING MECHANICS CLASS TEST: ONE

CLASS TEST: ONE

Time: 1 hr 30 min                                                                                        Max. Marks: 30

                                                                                                                         

1) Answer the following question in brief                                                              2 x 6 = 12   

                                                                 

a)      Distinguish clearly between composition of forces and resolution of forces.

b)      Show that the algebraic sum of the resolved part of a number of forces in a given direction is equal to the resolved part of their resultant in the same direction.

c)      Differentiate between coplanar forces and concurrent forces clearly.

d)     State and explain the laws of transmissibility of forces.

e)      Explain Newton’s third law of motion.

f)       What is a couple? Explain its characteristics.

2) Answer any two of the following questions                                                     6 x 3 = 18

a)      A smooth circular cylinder of radius 1500 mm is lying in a triangular groove, the right side of which makes 20° angle and left side makes 40° angle with horizontal. Find the reaction at each contact if there is no friction and the cylinder weight is 400 N.

                                                                                             

b)      A right circular cylinder of weight 5 kN rests on a smooth inclined plane and is held in position by a cord AC as shown in the figure. Find the tension in the cord if there is a horizontal force of magnitude 1 kN acting at C.

c)      Four forces of magnitude 10 kN, 18 kN, 15 kN and 12 kN are acting along the diagonals and sides of a regular pentagon as shown in the figure. Find the resultant force of the given force system.

d)     Prove that if a body is at equilibrium under three forces, then the forces are concurrent forces.

SMALL ENGINEERING COLLEGES OF GHAZIABAD: A BLEAK FUTURE

Economics says "When Supply is more than the Demand of a product, the Price falls." This is particularly true in the case of Technical Education in U.P. and perhaps upto some extent in the country itself.

Over the past few years the supply is outstripping the demand for Engineering and Management seats in the country. Just take the example of Uttar Pradesh, the most populous state of India is home to about 333 Engineering colleges which cumulatively offer a total seats of 1,15,379 in Engineering Education where as according to University datas the total number of students took admission in various engineering colleges after qualifying SEE amounts to mere 25,903. So, what happens to the vacant seats? And this year the figures are not going to be improved it seems.

This year a total 1,60,561 candidates had registered for the State Entrance Examination, out of whom, 1,29,924 have qualified. But, there are approximately 1.33 lakh B.Tech seats in the Engineering colleges affiliated to GBTU and MTU.

There are clearly a huge gap between the supply and demand of Engineering seats. In this situation, it has been believed that many small colleges will be bankrupt due to the lack of students. Many colleges have defered the salaries of the teachers and other employees due to the revenue crunch. It seems a grim scenario ahead for those colleges.

Due to the revenue crunch, promoters of small colleges are taking the refuge of cost cutting, and as a part of that they are trying to trim their faculty strength. Surely, this will affect the quality of the education as the each teacher will be over burdened and perhaps have to take five classes per day, where as the AICTE limits the load at best 18 classes per week. Also, most of the colleges don't follow the exact teacher students ratio of 1:20 prescribed by the apex body.

Many promoters are planning to opt out by selling their stakes in the colleges. The causes of their exits are the facts that running colleges in western U.P. is no longer a profitable business. They have cited that due to lack of students in taking admission, the colleges are no longer the chickens that lay gold eggs, which were in fact so just three years ago. So, why these colleges suddenly loss their values? What are the reasons behind these failures?

There are several reasons for the fall in numbers of students opting B.Tech courses. The most vital reason is the very high tution fees in colleges under MTU and GBTU compared to colleges in other states like Karnataka, Punjab and Rajasthan. Most of the colleges here charge more than 90,000.00 in the first year B.Tech where as colleges in other states charges below 60,000.00, even colleges in Punjab and West Bengal charge below 50 thousand and this is going to be a major factor.

The 2nd factor is the placement after the completion of the degree. Although many colleges claim tall, citing a long list of companies taking interest to place their students in very good packages, but reality always bites hard. The negative publicity by the ex students are also eating the pie here and there is no solution other than boosting the placement record by making a good relation with the HR of these companies by the respective college authorities.

The third most important factor is the sagging quality of the available faculty members. Many teachers although possess M.Tech degrees are not competent to impart quality education due to lack of depth of required knowledge as well as the essential communicating power required to be a good teacher. In some cases, due to over burdened schedule, a good teacher becomes unable to teach in the class. Just imagine the mental fatigue a teacher experienced while taking 5th or 6th class in a day when each class is of 55 min. duration.

A college has to show money to run the next three years during the visits from AICTE. So all the colleges have to show enough balance to pay the salaries of the employees for atleast three years, otherwise they won't get the permission to run the colleges, still some of them couldn't pay the salaries of the teachers and staffs. Why? Becouse they must have showed enough balances to acquire the clearences during the AICTE visits. Where does the money go? Vanished! Or siphoned off? There are several "skips" of the rules and regulations these college authorities used to practise.

QUESTIONS BANK 8 : FORCE AND FORCE SYSTEM

CONCEPT OF FRICTION

CONCEPT OF FRICTION

FRICTION IS ALL PERVADING, FRICTION IS OMNIPRESENT
FRICTION IS AN UNIVERSAL PHENOMENA

FRICTION: Friction may be defined as the resistive force acting in opposite direction in which the body tends to move or it moves. Frictional force always acts tangentially at points of contact.


Friction may be classified into two categories.


• Static Friction and
• Kinetic Friction


Static Friction is the friction experienced by a body when it is at rest under the action of external friction.


Kinetic Friction is the friction experienced by a body when it moves. Kinetic friction may be classified as Sliding Friction and Rolling Friction.


As a body can move in two ways, one is sliding and the other is rolling, there are two types of kinetic friction


• Sliding Friction and
• Rolling Friction


Sliding Friction: When a body slides over a surface without having any rotational tendency about a horizontal axis, it experiences a sliding friction. Sliding frictional force always try to dampen the movement as quickly as possible.


Rolling Friction: It is the friction experienced by a body when it rolls over the surface with an angular velocity as well as linear velocity. Rolling is a combination of translation as well as rotation about a horizontal axis passing through the centre of gravity.

Suppose we have a block of weight (W) lying on the ground as shown in the figure. The block will be at equilibrium as the weight will be neutralised due to the normal reaction provided by the ground.


Now let a horizontal (P) force is gradually applied to the block. Initially, when the force P is small, the block will not move, but if we increase the magnitude of P, then one moment will come when the block will start to move along the direction of applied force.


But from Newton's 2nd law, we know that whether the magnitude of force P is small or large, in all the cases the block should start to move with an acceleration F/m. But, practically the block starts to move when the magnitude of the applied forces reaches a definite value. Why does the block behave so? What does it mean?


Well, it means that there must exist an opposite resistance force that acts opposite to the applied force, having the same magnitude as that of applied force. But, this resistance force has a limit. When the resistance force reaches a maximum value, then the further increase in applied force can not be neutralized and as a result the body starts to move. When the value of the resistance force becomes maximum, any further increase of applied force set the movement in the body. The condition just before a body starts to move is called as "Limiting Conditions." The resistance force is called Frictional force and it becomes maximum, when the body attains the limiting condition. At this position the maximum frictional force is called "Limiting Friction."


Now just look at the experiment again, if we applied an infinitesimal force, the block doesn't move, which means the frictional force thus generated must be equal to the infinitesimal applied, else the body will experience a net force and the body will start to move. Now, if we increase force by some amount, the body will still be static, until the applied force reached the value of limiting friction. So, what does this indicate?

It indicates that frictional forces are variable. As we increase applied load, the frictional force also increases from zero to the highest value of limiting friction.


When the horizontal applied force is zero, the frictional force must be zero else the body will experience a net force. It means that when there is no applied force, frictional force just vanishes. Hence, it is known as pseudo force as its existence is dependent upon the applied force. (Remember there is another force centrifugal force which depends upon the centripetal force in rotational motion, hence, it is also a pseudo force.)


So, as the applied force starts to increase, the frictional force also increases thus maintaining the equilibrium conditions. But, the frictional force can not neutralize the applied force P, when its magnitude crosses the maximum value frictional force that can be generated due to friction on the contact surfaces.

Therefore when P > (fs)max, where (fs)max is the maximum magnitude of frictional force, the body will be in motion. When P = (fs)max, we call it as the case of impending motion or limiting condition. 

So, on what factor does the maximum frictional force on a surface depend upon? Certainly, it doesn't depend upon the applied force, although frictional force depends upon the applied force, but how much will be the maximum value of friction entirely depends upon the surface and the geometry of that surface and the Normal reaction the surface produces to counter balance the weight of the body. Here, one should remember normal reaction depends upon the mass of the body and the inclination of the surface with horizontal. 

COULOMB'S LAW OF DRY FRICTION: 

So, Coulomb's Law of dry friction states that, when there is a body resting on a surface is subjected to an applied force P, the maximum frictional force that would be generated directly depends upon the value of Normal reaction experienced by the body. 

(f_s)_max ∝ N 

N = mg cos θ, where mg is the weight of the body and θ is the inclination angle of the plane with horizonal. For a flat plane N = mg.

(f_s)_max = µN 

where µ is the proportionality constant and N is the normal reaction the body experiences from the surface. 

COEFFICIENT OF FRICTION: 

Here the constant (µ) plays a vital role. On what factor does the constant mu depend upon? It has been observed that the value of (µ) is greatly affected by the roughness of the surface upon which the body rests. It's value is a combined property of the contact surface as well as the surface roughness of the body itself. If we replace the body with another body of same mass but different material the value of (µ) changes. Also, if we place the body upon a different surface then also the value of mu changes. So, the value of mu is such a property that defines the characteristics of friction between the body and the contact surface. Hence, it is aptly named as the coefficient of friction. 

There are basically two types of Co-efficient of Friction. 

• Co-efficient of Static Friction 
• Co-efficient of Kinetic Friction 

ANGLE OF FRICTION (φ) 

When a block of mass is at rest on a surface and a horizontal force P is applied on the body to move it, a frictional force will be there to oppose any movement of the body. This force will act on the contact surface. Normal reaction is also acting upward on the contact surface. So total force on the contact surface will be resultant of normal reaction and frictional force. The angle made by this resultant force with normal reaction is called the angle of friction. 

FRICTION ON AN INCLINED PLANE: 

The direction of a frictional force depends upon the tendency of movement. 

Suppose we get two identical block of weight W in identical planes at angle α with horizontal as shown in the figure. 

Due to the force component W.sin α acting downwards along the plane, the body will have a tendency to move downwards along the plane. 

As the body would try to move downwards, a frictional force will be generated at the contact surface which would try to oppose the tendency to move downwards of the body, i.e., it would try to resist the downwards movement of the body. So, it will act upwards along the plane. 

Normal reaction produced by the inclined surface at the contact point or area. The normal reaction will be equal and opposite the force component of the weight of the body at a perpendicular direction to the inclined plane hence, N = W cos α, where N is the normal reaction. 

Now suppose we plane adjust the inclination of the plane, it means we can either increase or decrease the inclination of the plane. When the inclination is very small, the downward force component W sin α will be small and an equal magnitude frictional force will be produced and neutralize the downward force. Hence, the body will be at rest. 

Now if we increase the inclination of the plane, the downward force component W sin α will increases too, and frictional force will also be increased. Gradually, a condition will arrive when the downward weight component becomes equal to the maximum frictional force generated on the contact surface. This is limiting condition and also known as impending motion. If we increase the inclination angle α by a small amount, the body will start to move downwards. The angle of the plane when the body is at limiting condition is known as angle of repose. 

ANGLE OF REPOSE (α) 

We can define angle of repose as the angle of the inclination of a plane when a body on the plane is at limiting condition of impending motion due to its self weight component along the inclined plane. 

It is numerically equal to the angle of friction. It is denoted by (α). 

CONE OF FRICTION: 

It is an imaginary cone generated by revolving resultant reaction R about the normal reaction N. R is the resultant of the frictional force and normal reaction. 

Properties of Cone of Friction: 

• The radius of this cone represents the frictional force (fs)max. 
• The semi apex angle of the cone represents the angle of friction. 
• For co-planar forces, in order for motion not to occur the reaction R must be within the cone of friction. 

QUESTIONS BANK 7 : FORCE AND FORCE SYSTEM

QUESTIONS BANK 6 : FORCE AND FORCE SYSTEM

(I am going to publish a question bank for EME-102/EME-202 of 1st yr. MTU; Greater Noida. Some pages from the book .......Subhankar Karmakar)