111111 PAGE # 1
FORCE AND NEWTON’S LAWS OF MOTION
INTRODUCTION
Force is a push or pull which tries to change or successfully changes the state of rest or of uniform motion of a body, i.e., force is the cause of translatory
motion.
It arises due to interaction of the bodies either due to contact (e.g., normal reaction, friction, tension, spring force etc.) or from a distance (e.g., gravitational or electric force).
FUNDAMENTAL FORCES
All forces observed in nature such as muscular force, tension, reaction, friction, weight, electric, magnetic, nuclear, etc., can be explained in terms of only following four basic interactions.
(a ) Gra vit ational Force :
The force of interaction which exists between two particles of masses m1 and m2, due to their masses is called gravitational force. The gravitational force acts over long distances and does not need, any intervening medium. Gravitational force is the weakest force of nature.
( b) Elec t roma gne tic Forc e :
Force exerted by one particle on the other because of the electric charge on the particles is called electromagnetic force. Following are the main characteristics of electromagnetic force
(i) These can be attractive or repulsive.
(ii) These are long range forces.
(iii) These depend on the nature of medium between
the charged particles.
(iv) All macroscopic forces (except gravitational) which
we experience as push or pull or by contact are electromagnetic, i.e., tension in a rope, the force of friction, normal reaction, muscular force, and force experienced by a deformed spring are electromagnetic forces. These are manifestations of the electromagnetic attractions and repulsions between atoms/molecules.
( c ) Nuc lear Forc e :
It is the strongest force. It keeps nucleons (neutrons and protons) together inside the nucleus inspite of large electric repulsion between protons. Radioactivity, fission, and fusion, etc. results because of unbalancing
of nuclear forces. It acts within the nucleus that too upto a very small distance. It does not depends on charge and acts equally between a proton and proton, a neutron and neutron, and proton and neutron, electrons does not experience this force. It acts for very short distance order of 10–15 m.
(d) Weak Force :
It acts between any two elementary particles. Under its action a neutron can change into a proton emitting an electron and a particle called antineutrino. The range of weak force is very small, in fact much smaller than size of a proton or a neutron.
It has been found that for two protons at a distance of 1 fermi :
FN:FEM:FW:FG::1:10–2:10–7:10–38
On the basis of contact forces are classified into two categories
(i) Contact forces
(ii) Non contact or field forces
(a) Contact force :
Forces which are transmitted between bodies by short range atomic molecular interactions are called contact forces. When two objects come in contact they exert contact forces on each other. e.g. Normal, Tension etc. ( b ) Field forc e :
Force which acts on an object at a distance by the interaction of the object with the field produced by other object is called field force. e.g. Gravitational force, Electro magnetic force etc.
DETAILED ANALYSIS OF CONTACT FORCE
( a ) Normal force (N) :
It is the component of contact force perpendicular to the surface. It measures how strongly the surfaces in contact are pressed against each other. It is the electromagnetic force.
e.g.1 A table is placed on Earth as shown in figure id23534765 pdfMachine by Broadgun Software - a great PDF writer! - a great PDF creator! - http://www.pdfmachine.com http://www.broadgun.com
222222 PAGE # 2
Here table presses the earth so normal force exerted by four legs of table on earth are as shown in figure.
e.g.2 A boy pushes a block kept on a frictionless
surface.
Here, force exerted by boy on block is electromagnetic interaction which arises due to similar charges appearing on finger and contact surface of block, it is normal force.
A block is kept on inclined surface. Component of its weight presses the surface perpendicularly due to which contact force acts between surface and block.
Normal force exerted by block on the surface of inclined plane is shown in figure. Here normal force is a component of weight of the body perpendicular to the inclined surface i.e. N = mgsin
Force acts perpendicular to the surface
1. Two blocks are kept in contact on a smooth surface as
shown in figure. Draw normal force exerted by A on B.
Sol. In above problem, block A does not push block B, so
there is no molecular interaction between A and B. Hence normal force exerted by A on B is zero.
Note :
• Normal is a dependent force it comes in role when
one surface presses the other.
(b) Tension :
Tension is the magnitude of pulling force exerted by a string, cable, chain, rope etc. W hen a string is connected to a body and pulled out, the string said to be under tension. It pulls the body with a force T, whose direction is away from the body and along the length of the string. Usually strings are regarded to be massless and unstretchable, known as ideal string.
Note : (i) Tension in a string is an electromagnetic
force and it arises only when string is pulled. If a massless string is not pulled, tension in it is zero.
(ii) String can not push a body in direct contact.
(c ) Forc e Exert ed by spring :
A spring is made of a coiled metallic wire having a definite length. When it is neither pushed nor pulled then its length is called natural length.
At natural length the spring does not exert any force on
the objects attached to its ends.
f the spring is pulledat the ends, its length becomes larger than its natural length, it is known as stretched or extended spring. Extended spring pulls objects attached to its ends.
A B A B A B Normal spring Stretched spring Compressed spring
Spring force on A Spring force on B
Spring force on A Spring force on B
If the spring is pushed at the ends, its length becomes less than natural length. It is known as compressed spring. A compressed spring pushes the objects attached to its ends.
F x F F x F F = 0 spring in natural length does not exerts any force on its ends
F = – kx x = compression in spring F = – kx ;k = spring constant or stiffness constant (unit = N/m) x = extension in spring Fext Fext
333333 PAGE # 3 Note : Spring force is also electromagnetic in nature :
(d) Friction force :
When a body is moving on a rough surface resistance to the motion occurs because of the interaction between the body and its surroundings. We call such resistance as force of friction. Friction is also considered as component of contact force which acts parallel to the surfaces in contact.
(i) Origin of friction : The frictional force arises due to
molecular interactions between the surfaces at the points of actual contact. When two bodies are placed one over other, the actual area of contact is much smaller then the total surface areas of bodies. The molecular forces starts operating at the actual points of contact of the surfaces. Molecular bonds are formed at these contact points. When one body is pulled over the other, these bonds are broken, and the material get deformed and new bonds are formed. The local deformation sends vibrations into the bodies. These Vibrations ultimately dumps out and energy of vibrations appears as heat. Hence to start or carry on the motion, there is a need of force.
Body 1
Body 2
Actual area of contact
(ii) Statics and Kinetic Frictions : • Experiment :
(A) Consider a block placed on a table, and a small
force F1 is acted on it. The block does not move. It indicates that the frictional force fs starts acting in opposite direction of applied force and its magnitude is equal of F1(figure b). That is for the equilibrium of the block, we have
F1– fs = 0 or F1 = fs
The force of friction when body is in state of rest over the surface is called static friction (fs).
(B) As the applied force increases the frictional force
also increases. When the applied force is increased up to a certain limit (F2) such that the block is on the verge of motion. The value of frictional force at this stage is called limiting friction flim (figure c).
(C) Once the motion started, the smaller force is now
necessary to continue the motion (F3) and thus frictional force decreases. The force of friction when body is in state of motion over the surface is called kinetic or dynamic friction fk (figure d).
(iii) More about frictional force : (A) About static friction
1. The limiting friction depends on the materials of the surfaces in contact and their state of polish.
2. The magnitude of static friction is
independent of the apparent area of contact so long as the normal reaction remains the same.
3. The limiting friction is directly proportional to the magnitude of the normal reaction
between the two surfaces i.e. f
lim= SN. Here s is coefficient of static friction. We can write, s =
N flim
(B) About kinetic friction :
1. The kinetic friction depends on the materials
of the surface in contact.
2. It is also independent of apparent area of contact as long as the magnitude of normal reaction remains the same.
3. Kinetic friction is almost independent of the velocity, provided the velocity is not too large not too small.
444444 PAGE # 4 4. The kinetic friction is directly proportional to
the magnitude of the normal reaction between the surfaces.
fk = k N. Here k is coefficient of kinetic friction.
We can write, k =
N fk
• There are two types of kinetic frictions:
(i) Sliding friction : The force of friction when one
body slides over the surface of the another body is called sliding friction.
(ii) Rolling friction : When a wheel rolls without
slipping over a horizontal surface, there is no relative motion of the point of contact of the wheel with respect to the plane. Theoretically for a rolling wheel the frictional force is zero. This can only possible when bodies in contact are perfectly rigid and contact of wheel with the surface is made only at a point. But in practice no material body is perfectly rigid and therefore bodies get deformed when they pressed each other. The actual area of their contact no longer remains a point, and thus a small amount of friction starts acting between the body and the surface. Here frictional force is called rolling friction. It is clear from above discussion that rolling friction is very much smaller than sliding friction.
flim > fkinetic > frolling. Note :
s and k are dimensionless quantities and
independent of shape and area of contact . It is a property of two contact surfaces. s will always be
greater than k .Theoretical value of can be o to but
practical value is 0 <
1.6( a ) C ons e rv a t i v e Forc e :
A force is said to be conservative if the amount of work done in moving an object against that force is independent on the path. One important example of conservative force is the gravitational force. It means that amount of work done in moving a body against gravity from location A to location B is the same whichever path we may follow in going from A to B. This is illustrated in figure.
A force is conservative if the total work done by the force on an object in one complete round is zero,
i.e. when the object moves around any closed path
(returning to its initial position).
A force is conservative if there is no change in kinetic energy in one complete round. KE = 0
This definition illuminates an important aspect of a conservative force viz. Work done by a conservative force is recoverable. Thus in figure, we shall have to do mgh amount of work in taking the body from A to B. However, when body is released from B, we recover
mgh of work.
Other examples of conservative forces are spring force, electrostatic force etc.
( b) Non-C ons e rv a t iv e Forc e :
A force is non-conservative if the work done by that force on a particle moving between two points depends on the path taken between the points.
The force of friction is an example of non-conservative force. Let us illustrate this with an instructive example. Suppose we were to displace a book between two points on a rough horizontal surface (such as a table). If the book is displaced in a straight line between the two points, the work done by friction is simply FS where :
F = force of friction ;
S = distance between the points.
However, if the book is moved along any other path between the two points (such as a semicircular path), the work done by friction would be greater than FS. Finally, if the book is moved through any closed path, the work done by friction is never zero, it is always negative. Thus the work done by a non-conservative force is not recoverable, as it is for a conservative force.
SYSTEM
Two or more than two objects which interact with each other form a system.
Classification of forces on the basis of boundary of system :
(a) Internal Forces : Forces acting with in a system
among its constituents.
(b) External Forces : Forces exerted on the
constituents of a system by the outside surroundings are called as external forces.
555555 PAGE # 5 FREE BODY DIAGRAM
A free body diagram consists of a diagrammatic representations of single body or a subsystem of bodies isolated from surroundings showing all the forces acting on it.
Steps for F.B.D.
Step 1 : Identify the object or system and isolate it from
other objects, clearly specify its boundary.
Step 2 : First draw non-contact external force in the
diagram, generally it is weight.
Step 3 : Draw contact forces which acts at the boundary
of the object of system. Contact forces are normal , friction, tension and applied force. In F.B.D, internal forces are not drawn only external are drawn.
2. A block of mass ‘m’ is kept on the ground as shown in
figure.
(i) Draw F.B.D. of block.
(ii) Are forces acting on block forms action- reaction
pair.
(iii) If answer is no, draw action reaction pair. Sol.(i) F.B.D. of block
(ii) ‘N’ and mg are not action -reaction pair. Since pair
act on different bodies, and they are of same nature.
(iii) Pair of ‘mg’ of block acts on earth in opposite
direction.
mg earth
and pair of ‘N’ acts on surface as shown in figure.
N
3. Two sphere A and B are placed between two vertical
walls as shown in figure. Draw the free body diagrams of both the spheres.
A B
Sol.F.B.D. of sphere ‘A’ :
F.B.D. of sphere ‘B’ :
(exerted by A)
Note : Here N
AB and NBA are the action - reaction pair
(Newton’s third law).
4. Draw F.B.D. for systems shown in figure below.
Sol.
TRANSLATORY EQUILIBRIUM
When several forces acts on a body simultaneously in such a way that resultant force on the body is zero, i.e.,
F= 0 with F=
i
Fthe body is said to be in translatory equilibrium. Here it is worthy to note that :
(i) As if a vector is zero all its components must vanish i.e. in equilibrium as
-666666 PAGE # 6 F= 0 with F =
i F= 0 xF
= 0 ;
Fy= 0 ; zF
= 0So in equilibrium forces along x axes must balance each other and the same is true for other directions. If a body is in translatory equilibrium it will be either at rest or in uniform motion. If it is at rest, equilibrium is called static, otherwise dynamic.
Static equilibrium can be divided into following three types :
( a ) St a ble e quilibrium :
If on slight displacement from equilibrium position a body has a tendency to regain its original position it is said to be in stable equilibrium. In case of stable equilibrium potential energy is minimum and so center of gravity is lowest.
O
(b) Unstable equilibrium : If on slight displacement
from equilibrium position a body moves in the direction of displacement, the equilibrium is said to be unstable. In this situation potential energy of body is maximum and so center of gravity is highest.
O
(c) Neutral equilibrium : If on slight displacement from
equilibrium position a body has no tendency to come back to its original position or to move in the direction of displacement, it is said to be in neutral equilibrium. In this situation potential energy of body is constant and so center of gravity remains at constant height.
. ( a ) Ne wtons 2n d
law of motion :
The rate of change of linear momentum of a body is directly proportional to the applied force and the change takes place in the direction of the applied force. In relation
F
= ma the force
F
stands for the net external force. Any internal force in the system is not to be included in F
.In S.I. the absolute unit of force is newton (N) and gravitational unit of force is kilogram weight or kilogram force (kgf.)
Note : The absolute unit of force remains the same
everywhere, but the gravitational unit of force varies from place to place because it depends on the value of g. ( b ) App lic a t ions o f Ne w t on’s 2n d
L a w
(i) When objects are in equilibrium :
Steps to solve problem involving objects in equilibrium :
Step 1 : Make a sketch of the problem.
Step 2 : Isolate a single object and then draw the
free-body diagram for the object. Label all external forces acting on it.
Step 3 : Choose a convenient coordinate system and
resolve all forces into rectangular components along x and Y direction.
Step 4 : Apply the equations
Fx 0and
Fy 0. Step 5 : Step 4 will give you two equations with severalunknown quantities. If you have only two unknown quantities at this point, you can solve the two equations for those unknown quantities.
Step 6 : If step 5 produces two equations with more
than two unknowns, go back to step 2 and select another object and repeat these steps. Eventually at step 5 you will have enough equations to solve for all unknown quantities.
5. A ‘block’ of mass 10 kg is
suspended with string as
shown in figure.
Find tension in the string. (g = 10 m/s2).
Sol.F.B.D. of block
For equilibrium of block along Y axis
Fy 0T – 10 g = 0
T = 100 N
6. The system shown in figure is in equilibrium. Find the
magnitude of tension in each string ; T1 , T2, T3 and T4. (g = 10 m/s2).
Sol.F.B.D. of 10 kg block
For equilibrium of block along Y axis.
0 Fy
T0 10g T0 = 10 g T0 = 100 N777777 PAGE # 7 F.B.D. of point ‘A’ 0 Fy
30º T2 T1 x T0 A y T2 cos 30º = T0 = 100 N T2 = 3 200 N
Fx 0 T1 = T2 . sin 30º = 3 200 . 2 1 = 3 100 N. F.B.D. of point of ‘B’ 60º T4 T3 B x y 30º T2
Fy= 0 T4 cos 60º = T2 cos 30º T4 = 200 Nand
Fx= 0 T3 + T2 sin30º = T4 sin 60ºT3 = 3
200
N
7. Two blocks are kept in contact as shown in figure. Find
:-(a) forces exerted by surfaces (floor and wall) on
blocks.
(b) contact force between two blocks.
SolA : F.B.D. of 10 kg block N1 = 10 g = 100 N ...(1) N2 = 100 N ...(2) F.B.D. of 20 kg block N2 = 50 sin 30º + N3 N3 = 100 – 25 = 75 N & N4 = 50 cos 30º + 20 g N4 = 243.30 N
8. Find magnitude of force exerted by string on pulley.
Sol B. F.B.D. of 10 kg block :
T = 10 g = 100 N F.B.D. of pulley :
Since string is massless, so tension in both sides of string is same.
So magnitude of force exerted by string on pulley
=
100
2
100
2 = 100 2N Note : Since pulley is in equilibrium position, so net
forces on it is zero.
( i i ) Ac c e le r a t i ng Ob je c t s :
Steps to solve problems involving objects that are in accelerated motion :
Step 1 : Make a sketch of the problem.
Step 2 : Isolate a single object and then draw the free
- body diagram for that object. Label all external forces acting on it. Be sure to include all the forces acting on the chosen body, but be equally careful not to include any force exerted by the body on some other body. Some of the forces may be unknown , label them with algebraic symbols.
Step 3 : Choose a convenient coordinate system, show
location of coordinate axis explicitly in the free - body diagram, and then determine components of forces with reference to these axis and resolve all forces into x and y components.
Step 4 : Apply the equations
Fx= max &
Fy= may. Step 5 : Step 4 will give two equations with severalunknown quantities. If you have only two unknown quantities at this point, you can solve the two equations for those unknown quantities.
888888 PAGE # 8 Step 6 : If step 5 produces two equations with more
than two unknowns, go back to step 2 and select another object and repeat these steps. Eventually at step 5 you will have enough equations to solve for all unknown quantities.
9. A force F is applied horizontally on mass m1 as shown in figure. Find the contact force between m1 and m2.
Sol.Considering both blocks as a system to find the
common acceleration. Common acceleration a =
m1 m2
F ...(1) m1 m2 F aTo find the contact force between ‘A’ and ‘B’ we draw
F.B.D. of mass m2. F.B.D. of mass m2
Fx= max N = m2 . a N =
1 2
2 m m F m 2 1 m m F a ce sin10. A 5 kg block has a rope of mass 2 kg attached to its
underside and a 3 kg block is suspended from the other end of the rope. The whole system is accelerated upward at 2 m/s2 by an external force F
0. (a) What is F0 ?
(b) What is the net force on rope ?
(c) What is the tension at middle point of the rope ?
(g = 10 m/s2)
Sol.For calculating the value of F0. F.B.D of whole system (a) 2m/s2 F0 10 g = 100 N F0 –100 = 10 × 2 F0 = 120 N ...(1)
(b) According to Newton’s second law, net force on
rope.
F = ma = 2 × 2
= 4N ...(2)
(c) For calculating tension at the middle point we draw
F.B.D. of 3 kg block with half of the rope (mass 1 kg) as shown.
T – 4 g = 4 . 2
T = 48 N
11. A block of mass 50 kg is kept on another block of mass
1 kg as shown in figure. A horizontal force of 10 N is applied on the 1Kg block. (All surface are smooth). Find : (g = 10 m/s2)
(a) Acceleration of blocks A and B. (b) Force exerted by B on A. 50 kg 1 kg A B Sol.(a) F.B.D. of 50 kg N2 = 50 g = 500 N
along horizontal direction, there is no force aB = 0
(b) F.B.D. of 1 kg block :
N1 N2
10 N
1g
along horizontal direction 10 = 1 aA.
aA = 10 m/s2
along vertical direction
N1 = N2 + 1g
999999 PAGE # 9 12. One end of string which passes through pulley and
connected to 10 kg mass at other end is pulled by 100 N force. Find out the acceleration of 10 kg mass. (g =9.8 m/s2)
Sol.Since string is pulled by 100 N force. So tension in the
string is 100 N F.B.D. of 10 kg block 100 – 10 g = 10 a 100 – 10 × 9.8 = 10 a a = 0.2 m/s2. WEIGHING MACHINE
A weighing machine does not measure the weight but measures the force exerted by object on its upper surface.
13. A man of mass 60 Kg is
standing on a weighing machine placed on ground. Calculate the reading of machine (g = 10 m/s2).
weighing machine
Sol.For calculating the reading of weighing machine, we
draw F.B.D. of man and machine separately. F.B.D of man N Mg N = Mg F.B.D of man taking mass of man as M weighing machine N N1 Mg F.B.D. of weighing machine
Here force exerted by object on upper surface is N Reading of weighing machine
N = Mg = 60 × 10
N = 600 N.
SPRING BALANCE
It does not measure the weight. It measures the force exerted by the object at the hook. Symbolically, it is represented as shown in figure.
A block of mass ‘m’ is suspended at hook. When spring
balance is in equilibrium, we draw the F.B.D. of mass m for calculating the reading of balance.
m spring balance hook F.B.D. of ‘m’. mg – T = 0 T = mg
Magnitude of T gives the reading of spring balance.
14. A block of mass 20 kg is suspended through two light
spring balances as shown in figure . Calculate the :
(1) reading of spring balance (1). (2) reading of spring balance (2).
Sol.For calculating the reading, first we draw F.B.D.of 20 kg
block. F.B.D. 20 kg T 20 g mg – T = 0 T = 20 g = 200 N
101010101010 PAGE # 10
Since both the balances are light so, both the scales will read 200 N.
15. (i) A 10 kg block is supported by a cord that runs to a spring scale, which is supported by another cord from the ceiling figure (a). What is the reading on the scale ?
(ii) In figure (b) the block is supported by a cord that
runs around a pulley and to a scale. The opposite end of the scale is attached by cord to a wall. What is the reading of the scale.
(iii) In figure (c) the wall has been replaced with a
second 10 kg block on the left, and the assembly is stationary. What is the reading on the scale now ?
spring balance hook 10 kg T T (a) T T T
(b)
10kg T T T T 10kg 10kg (c)Sol. In all the three cases the spring balance reads 10 kg.
To understand this let us cut a section inside the spring as shown;
As each part of the spring is at rest, so F= T. As the block is stationary, so T= 10g = 100N.
16. Pull is easier than push
Push : Consider a block of mass m placed on
rough horizontal surface. The coefficient of static friction between the block and surface is . Let a push force F is applied at an angle with the
horizontal.
As the block is in equilibrium along y-axis, so we have
Fy 0;or N = mg + F sin
To just move the block along x-axis, we have F cos = N = (mg + F sin ) or F = sin – cos mg ...(i)
Pull : Along y-axis we have ;
Fy 0; N = mg – F sin
To just move the block along x-axis, we have F cos = N = (mg – F sin ) or F = sin cos mg . ...(ii)
It is clear from above discussion that pull force is smaller than push force.
17. Discuss the direction of friction in the following cases :
(i) A man walks slowly, without change in speed.
(ii) A man is going with increasing speed. (iii) When cycle is gaining speed.
(iv) When cycle is slowing down .
Sol. (i) Consider a man walks slowly without acceleration, and both the legs are touching the ground as shown in figure (a). The frictional force on rear leg is in forward direction and on front leg will be on backward direction of motion.
As a = 0, Fnet = 0 or f1 – f2 = 0 f1 = f2 & N1 = N2. N1 N2 (b) Ground f1 f1 f2 f2 N1 N2
111111111111 PAGE # 11 (ii) When man is gaining the speed : The frictional
force on rear leg f1 will be greater than frictional force on front leg f2 (fig. b).
acceleration of the man, a =
m f f1 2
.
(iii) When cycle is gaining speed : In this case torque
is applied on the rear wheel of the cycle by the chain-gear system. Because of this the slipping tendency of the point of contact of the rear wheel is backward and so friction acts in forward direction. The slipping tendency of point of contact of front wheel is forward and so friction acts in backward direction. If f1 and f2 are the frictional forces on rear and front wheel, then acceleration of the cycle a =
M f
–
f1 2
, where M is the mass of the cycle together with rider (fig. a).
N1 N2 f1 f2 (a) N1 N2 f1 f2 (b)
(iv) When cycle is slowing down : When torque is not
applied (cycle stops pedaling), the slipping tendency of points of contact of both the wheels are forward, and so friction acts in backward direction (fig. b). If f1 and f2 are the frictional forces on rear and front wheel, then retardation
a =
M f f1 2
18. A block of mass 25 kg is raised by a 50 kg man in two
different ways as shown in fig.. What is the action on the floor by the man in the two cases ? If the floor yields to a normal force of 700 N, which mode should the man adopt to lift the block without the floor yielding.
50g
50g
Sol. The FBD for the two cases are shown in figure.
In I
st case, let the force exerted by the man on the floor isN1. Consider the forces inside the dotted box, we have N1 = T + 50 g.
Block is to be raised without acceleration, so T = 25 g.
N1 = 25 g + 50 g
= 75 g = 75 × 9.8 = 735 N
In IInd case, let the force exerted by the man on the floor in N2 . Consider the forces inside the dotted box, we have
N2 = 50 g – T and T = 25 g
N2 = 50 g – 25 g
= 25 g = 25 × 9.8 = 245 N.
As the floor yields to a downward force of 700 N, so the man should adopt mode .
19. Figure shows a weighing machine kept in a lift is moving upwards with acceleration of 5 m/s2. A block is kept on the weighing machine. Upper surface of block is attached with a spring balance. Reading shown by weighing machine and spring balance is 15 kg and 45 kg respectively.
Answer the following questions. Assume that the weighing machine can measure weight by having negligible deformation due to block, while the spring balance requires larger expansion. (take g = 10 m/s2)
(i) Find the mass of the object in kg and the normal
force acting on the block due to weighing machine?
(ii) Find the acceleration of the lift such that weighing
machine shows its true weight ?
Sol. (i)
T + N – Mg = Ma
45 g + 15 g = M(g + a) 450 + 150 = M(10 + 5) M = 40 kg
Normal force is the reaction applied by weighing machine i.e. 15 × 10 = 150 N.
121212121212 PAGE # 12 (ii) T + N – Mg = Ma 45 g + 40 g = 40(g + a) 450 + 400 = 400 +40 a a = 40 450 = 4 45 m/s2
E X E R C I S E
Normal Force :1. Two blocks are in contact on a frictionless table. One
has mass m and the other 2m.A force F is applied on 2m as shown in the figure. Now the same force F is applied from the right on m. In the two cases respectively, the ratio of force of contact between the two blocks will be :
(A) Same (B) 1 : 2
(C) 2 : 1 (D) 1 : 3
2. Two forces of 6N and 3N are acting on the two blocks of 2kg and 1kg kept on frictionless floor. What is the force exerted on 2kg block by 1kg block ?:
2kg 1kg
6N
3N
(A)1N (B) 2N
(C) 4N (D) 5N
3. There are two forces on the 2.0 kg box in the overhead
view of figure but only one is shown. The second force is nearly : y 30º F = 20 N1 a = 12 m/s2 x (A) –20ˆjN (B) – 20 ˆi + 20ˆj N (C) –32iˆ – 12
3
ˆjN (D) –21iˆ – 16ˆjN4. A dish of mass 10 g is kept horizontally in air by firing bullets of mass 5 g each at the rate of 100 per second. If the bullets rebound with the same speed, what is the velocity with which the bullets are fired :
(A) 0.49 m/s (B) 0.098 m/s
(C) 1.47 m/s (D) 1.96 m/s
5. A block of metal weighing 2 kg is resting on a frictionless plank. If struck by a jet releasing water at a rate of 1 kg/s and at a speed of 5 m/s. The initial acceleration of the block will be :
(A) 2.5 m/s2 (B) 5.0 m/s2
(C) 10 m /s2 (D) none of the above
6. A constant force F is applied in horizontal direction as shown. Contact force between M and m is N and
between m and M’ is N’ then
(A) N= N’ (B) N > N’
(C) N’> N
(D) cannot be determined
• ASSERTION / REASON
7. STATEMENT-1 : Block A is moving on horizontal surface
towards right under action of force. All surface are smooth. At the instant shown the force exerted by block A on block B is equal to net force on block B.
STATEMENT-2 : From Newtons’s third law, the force
exerted by block A on B is equal in magnitude to force exerted block B on A
(A) statement-1 is true, Statement 2 is true, statement-2 is correct explanation for statement-1.
(B) statement-1 is true, Statement 2 is true, statement-2 is NOT a correct explanation for statement-1. (C) statement-1 is true, Statement 2 is false (D) statement-1 is False, Statement 2 is True
8. A certain force applied to a body A gives it an acceleration of 10 ms–2 . The same force applied to body B gives it
an acceleration of 15 ms–2 . If the two bodies are joined
together and same force is applied to the combination, the acceleration will be :
(IJSO/Stage-I/2011)
(A) 6 ms–2 (B) 25 ms–2
131313131313 PAGE # 13 9. Four blocks are kept in a row on a smooth horizontal
table with their centres of mass collinear as shown in the figure. An external force of 60 N is applied from left on the 7 kg block to push all of them along the table. The forces exerted by them are :(IAO/Sr./Stage-I/2008)
7 kg 5 kg 2 kg 1 kg 60N P Q R S (A) 32 N by P on Q (B) 28 N by Q on P (C) 12 N by Q on R (D) 4 N by S on R Tension :
10. A mass M is suspended by a rope from a rigid support
at A as shown in figure. Another rope is tied at the end B, and it is pulled horizontally with a force F. If the rope
AB makes an angle with the vertical in
equilibrium,then the tension in the string AB is :
(A) F sin (B) F /sin
(C) F cos (D) F / cos
11. In the system shown in the figure, the acceleration of
the 1kg mass and the tension in the string connecting between A and B is : (A) g 4 downward, 8g 7 (B) g 4 upward, g 7 (C) g 7 downward, 6 7 g (D) g 2 upward, g
12. A body of mass 8 kg is hanging from another body of
mass 12 kg. The combination is being pulled by a string with an acceleration of 2.2 m s–2. The tension T
1
and T2 will be respectively :(Use g =9.8 m/s2)
(A) 200 N, 80 N (B) 220 N, 90 N
(C) 240 N, 96 N (D) 260 N, 96 N
13. Two masses M1 and M2 are attached to the ends of a light string which passes over a massless pulley attached to the top of a double inclined smooth plane of angles of inclination and . If M2 > M1 then the
acceleration of block M2 down the inclined will be :
(A) 2 1 2 M (sin ) M M g (B) 1 1 2 M g(sin ) M M (C) 2 1 1 2 M M sin M sin M g (D) Zero
14. Three masses of 1 kg, 6 kg and 3 kg are connected to
each other by threads and are placed on table as shown in figure. What is the acceleration with which the system is moving ? Take g = 10 m s–2:
(A) Zero (B) 1 ms–2
(C) 2 m s–2 (D) 3 m s–2
15. The pulley arrangements shown in figure are identical
the mass of the rope being negligible. In case I, the mass m is lifted by attaching a mass 2m to the other end of the rope. In case II, the mass m is lifted by pulling the other end of the rope with a constant downward force F= 2 mg, where g is acceleration due to gravity. The acceleration of mass in case I is :
(A) Zero
(B) More than that in case II (C) Less than that in case II (D) Equal to that in case II
141414141414 PAGE # 14 16. A 50 kg person stands on a 25 kg platform. He pulls
massless rope which is attached to the platform via the frictionless, massless pulleys as shown in the figure. The platform moves upwards at a steady velocity if the force with which the person pulls the rope is :
(A) 500 N (B) 250 N
(C) 25 N (D) 50 N
17. Figure shows four blocks that are being pulled along a
smooth horizontal surface. The mssses of the blocks and tension in one cord are given. The pulling force F is :
4kg 3kg 2kg 1kg 60º F 30N (A) 50 N (B) 100 N (C) 125 N (D) 200 N
18. A10 kg monkey climbs up a massless rope that runs
over a frictionless tree limb and back down to a 15 kg package on the ground. The magnitude of the least acceleration the monkey must have if it is to lift the package off the ground is :
(A) 4.9 m/s2 (B) 5.5 m/s2
(C) 9.8 m/s2 (D) none of these
19. Two blocks, each of mass M, are connected by a
massless string, which passes over a smooth massless pulley. Forces F act on the blocks as shown. The tension in the string is :
(A) Mg (B) 2 Mg
(C) Mg + F (D) none of these
20. Two blocks of mass m each is connected with the
string which passes over fixed pulley, as shown in figure. The force exerted by the string on the pulley P is :
(A) mg (B) 2 mg
(C) 2mg (D) 4 mg
21. One end of a massless rope, which passes over a
massless and frictionless pulley P is tied to a hook C while the other end is free. Maximum tension that rope can bear is 360 N, with what minimum safe acceleration (in m/s2) can a monkey of 60 kg move
down on the rope :
P
C
(A) 16 (B) 6
(C) 4 (D) 8
22. Which figure represents the correct F.B.D. of rod of
mass m as shown in figure :
(A) (B)
(C) (D) None of these
23. Two persons are holding a rope of negligible weight
tightly at its ends so that it is horizontal. A 15 kg weight is attached to the rope at the mid point which now no longer remains horizontal. The minimum tension required to completely straighten the rope is :
(A) 15 kg
(B) kg
2 15
(C) 5 kg
151515151515 PAGE # 15 24. In the figure, the blocks A, B and C of mass each have
acceleration a1 ,a2 and a3 respectively . F1 and F2 are external forces of magnitudes 2 mg and mg respectively then which of the following relations is correct :
(A) a1 = a2 = a3 (B) a1 > a2 > a3 (C) a1 = a2 , a2 > a3 (D) a1 > a2 , a2= a3
25. A weight is supported by two strings 1.3 and 2.0 m
long fastened to two points on a horizontal beam 2.0 m apart. The depth of this weight below the beam is :
(IAO/Jr./Stage-I/2007)
(A) 1.0 m (B) 1.23 m
(C) 0.77 m (D) 0.89 m
26. A fully loaded elevator has a mass of 6000 kg. The
tension in the cable as the elevator is accelerated downward with an acceleration of 2ms–2 is (Take g = I0
ms–2) (KVPY/2007)
(A) 7·2 × 104 N (B) 4.8 × 104 N
(C) 6 × 104 N (D) 1.2 × 104 N
27. A light string goes over a frictionless pulley. At its one
end hangs a mass of 2 kg and at the other end hangs a mass of 6 kg. Both the masses are supported by hands to keep them at rest. When the masses are released, they being to move and the string gets taut. (Take g = 10 ms–2) The tension in the string during the
motion of the masses is : (KVPY/2008)
(A) 60 N (B) 30 N
(C) 20 N (D) 40 N
Force Exerted by Spring :
28. In the given figure. What is the reading of the spring
balance:
(A) 10 N (B) 20 N
(C) 5 N (D) Zero
29. Two bodies of masses M1 and M2 are connected to each other through a light spring as shown in figure. If
we push mass M1 with force F and cause acceleration
a1 in mass M1 what will be the acceleration in M2 ?
(A) F/M2 (B) F/(M1 + M2)
(C) a1 (D) (F–M1a1)/M2
30. A spring balance is attached to 2 kg trolley and is used
to pull the trolly along a flat surface as shown in the fig. The reading on the spring balance remains at 10 kg during the motion. The acceleration of the trolly is (Use g= 9.8 m–2) :
(A) 4.9 ms–2 (B) 9.8 ms–2
(C) 49 ms–2 (D) 98 ms–2
31. A body of mass 32 kg is suspended by a spring balance
from the roof of a vertically operating lift and going downward from rest. At the instants the lift has covered 20 m and 50 m, the spring balance showed 30 kg & 36 kg respectively. The velocity of the lift is :
(A) Decreasing at 20 m & increasing at 50 m (B) Increasing at 20 m & decreasing at 50 m
(C) Continuously decreasing at a constant rate throughout the journey
(D) Continuously increasing at constant rate throughout the journey
Friction Force :
32. A ship of mass 3 × 107 kg initially at rest is pulled by a
force of 5 × 104 N through a distance of 3m. Assume
that the resistance due to water is negligible, the speed of the ship is :
(A) 1.5 m/s (B) 60 m/s
(C) 0.1 m/s (D) 5 m/s
33. When a horse pulls a cart, the force needed to move
the horse in forward direction is the force exerted by : (A) The cart on the horse
(B) The ground on the horse (C) The ground on the cart (D) The horse on the ground
161616161616 PAGE # 16 34. A 2.5 kg block is initially at rest on a horizontal surface.
A 6.0 N horizontal force and a vertical force Pare applied to the block as shown in figure. The coefficient of static friction for the block and surface is 0.4. The magnitude of friction force when P = 9N : (g = 10 m/s2)
(A) 6.0 N (B) 6.4 N
(C) 9.0 N (D) zero
35. The upper half of an inclined plane with inclination is
perfectly smooth while the lower half is rough. A body starting from rest at the top will again come to rest at the bottom, if the coefficient of friction for the lower half is :
(A) 2 tan (B) tan
(C) 2 sin (D) 2 cos
36. Minimum force required to pull the lower block is (take
g = 10 m/s2) :
(A) 1 N (B) 5 N
(C) 7 N (D) 10 N
37. N bullets each of mass m are fired with a velocity v m/
s at the rate of n bullets per sec., upon a wall. If the bullets are completely stopped by the wall, the reaction offered by the wall to the bullets is :
(A) N m v / n (B) n m v
(C) n N v / m (D) n v m / N
38. A vehicle of mass m is moving on a rough horizontal
road with momentum P. If the coefficient of friction between the tyres and the road be , then the stopping
distance is : (A)
mg
2
P
(B) 2 mg P2 (C) g m 2 P 2 (D)2
m
g
P
2 2
39. What is the maximum value of the force F such that the
block shown in the arrangement, does not move :
F 60º 1 2 3 m = 3kg (A) 20 N (B) 10 N (C) 12N (D) 15 N
40. A bock of mass 5 kg is held against wall by applying a
horizontal force of 100N. If the coefficient of friction between the block and the wall is 0.5, the frictional force acting on the block is : (g =9.8 m/s2)
5kg 100N
(A)100 N (B) 50 N
(C) 49 N (D) 24.9 N
41. A heavy roller is being pulled along a rough road as
shown in the figure. The frictional force at the point of
contact is : (IAO/Jr./Stage-I/2007)
F
(A) parallel to F (B) opposite to F
(C) perpendicular to F (D) zero
42. When a motor car of mass 1500 kg is pushed on a
road by two persons, it moves with a small uniform velocity. On the other hand if this car is pushed on the same road by three persons, it moves with an acceleration of 0.2 m/s2. Assume that each person is
producing the same muscular force. Then, the force of friction between the tyres of the car and the surface of
the road is : (IAO/Jr./Stage-I/2009)
(A) 300 N (B) 600 N
(C) 900 N (D) 100 N
43. A block of mass M is at rest on a plane surface inclined
at an angle to the horizontal The magnitude of force
exerted by the plane on the block is : (KVPY/2009)
(A) Mg cos (B) Mg sin
(C) Mg tan (D) Mg
44. A block of mass M rests on a rough horizontal table. A
steadily increasing horizontal force is applied such that the block starts to slide on the table without toppling. The force is continued even after sliding has started. Assume the coefficients of static and kinetic friction between the table and the block to be equal. The cor-rect representation of the variation of the frictional forces, ƒ, exerted by the table on the block with time t is
given by : (KVPY/2010)
(A) (B)
171717171717 PAGE # 17 45. A small child tries to move a large rubber toy placed on
the ground. The toy does not move but gets deformed under her pushing force (F) which is obliquely upward
as shown . Then (KVPY/2011)
(A) The resultant of the pushing force (F), weight of
the toy, normal force by the ground on the toy and the frictional force is zero.
(B) The normal force by the ground is equal and oppo-site to the weight of the toy.
(C) The pushing force (F) of the child is balanced by
the equal and opposite frictional force
(D) The pushing force (F) of the child is balanced by
the total internal force in the toy generated due to deformation
46. On a horizontal frictional frozen lake, a girl (36 kg) and
a box (9kg) are connected to each other by means of a rope. Initially they are 20 m apart. The girl exerts a horizontal force on the box, pulling it towards her. How far has the girl travelled when she meets the box ?
(KVPY/2011)
(A) 10 m
(B) Since there is no friction, the girl will not move (C) 16 m
(D) 4m
47. Which of the following does NOT involve friction ? (IJSO/Stage-I/2011)
(A) Writing on a paper using a pencil
(B) Turning a car to the left on a horizontal road. (C) A car at rest parked on a sloping ground (D) Motion of a satellite around the earth.
48. In the two cases shown below, the coefficient of kinetic
friction between the block and the surface is the same, and both the blocks are moving with the same uniform
speed. Then, (IAO/Sr./Stage-I/2008)
F1 F2 (A) F1 = F2 (B) F1 < F2 (C) F1 > F2 (D) F1 = 2F2 if sin = Mg/4F2 Weighing Machine :
49. The ratio of the weight of a man in a stationary lift and
when it is moving downward with uniform acceleration
‘a’ 3:2. The value of ‘a’ is : (g = acceleration, due to
gravity)
(A) (3/2)g (B) g
(C) (2/3) g (D) g/3
50. A person standing on the floor of an elevator drops a
coin. The coin reaches the floor of the elevator in time t1 when elevator is stationary and in time t2 if it is moving uniformly. Then (A) t1 = t2 (B) t1 > t2 (C) t1 < t2 (D) t1 < t2 or t1 > t2 depending • ASSERTION / REASON
51. STATEMENT-1 : A man standing in a lift which is moving
upward, will feel his weight to be greater than when the lift was at rest.
STATEMENT-2 : If the acceleration of the lift is ‘a’ upward
then the man of mass m shall feel his weight to be equal to normal reaction (N) exerted by the lift given N = m(g+a) (where g is acceleration due to gravity (A) 1 is true, Statement 2 is true, statement-2 is correct explanation for statement 1.
(B) 1 is true, Statement 2 is true, statement-2 is NOT a correct explanation for statement-1. (C) statement-1 is true, Statement 2 is false (D) statement-1 is False, Statement 2 is True
52. A beaker containing water is placed on the platform of
a digital weighing machine. It reads 900 g. A wooden block of mass 300 g is now made to float in water in the beaker (without touching walls of the beaker). Half the wooden block is submerged inside water. Now, the reading of weighing machine will be :
(IAO/Jr./Stage-I/2009)
(A) 750 g (B) 900 g
(C) 1050 g (D) 1200 g
Miscellaneous :
53. An object will continue accelerating until :
(A) Resultant force on it begins to decreases (B) Its velocity changes direction
(C) The resultant force on it is zero
(D) The resultant force is at right angles to its direction of motion
181818181818 PAGE # 18 54. In which of the following cases the net force is not zero ?
(A) A kite skillfully held stationary in the sky (B) A ball freely falling from a height
(C) An aeroplane rising upward at an angle of 45° with
the horizontal with a constant speed (D) A cork floating on the surface of water.
55. Figure shows the displacement of a particle going
along the X-axis as a function of time. The force acting on the particle is zero in the region.
(A) AB (B) BC
(C) CD (D) DE
56. A 2 kg toy car can move along x axis. Graph shows force
Fx, acting on the car which begins to rest at time t = 0. The velocity of the car at t = 10 s is :
(A) – ˆi m/s (B) – 1.5 ˆi m/s
(C) 6.5 ˆi m/s (D) 13 ˆi m/s
57. Figure shows the displacement of a particle going
along the x-axis as a function of time :
(A) The force acting on the particle is zero in the region AB (B) The force acting on the particle is zero in the region BC (C) The force acting on the particle is zero in the region CD (D) The force is zero no where
58. A force of magnitude F1 acts on a particle so as to accelerate if from rest to velocity v. The force F1 is then replaced by another force of magnitude F2 which decelerates it to rest.
(A) F1 must be the equal to F2 (B) F1 may be equal to F2 (C) F1 must be unequal to F2 (D) None of these
59. In a imaginary atmosphere, the air exerts a small force
F on any particle in the direction of the particle’s motion.
A particle of mass m projected upward takes a time t1 in reaching the maximum height and t2 in the return journey to the original point. Then
(A) t1 < t2 (B) t1 > t2 (C) t1 = t2
(D) The relation between t1 and t2 depends on the mass of the particle
60. A single force F of constant magnitude begins to act on
a stone that is moving along x axis. The stone continues to move along that axis. W hich of the following represents the stone’s position ?
(A) x = 5t – 3 (B) x = 5t2 + 8t – 3
(C) x = –5t2 + 5t – 3 (D) x = 5t3 + 4t2 – 3 61. Three forces act on a particle that moves with
unchanging velocity v= (3
i
ˆ – 4ˆj) m/s. Two of the
forces are F1= (3iˆ + 2ˆj – 4kˆ) N and F2
= (–5ˆi + 8ˆj
+ 3kˆ) N. The third force is : (A) (–2ˆi + 10ˆj – 7kˆ) N
(B) (2iˆ– 10ˆj + kˆ) N
(C) (7ˆi – 2kˆ + 10ˆj) N
(D) none of these
62. An 80 kg person is parachuting and experiencing a
downward acceleration of 2.5 m/s2 . The mass of the
parachute is 5.0 kg. The upward force on the open parachute from the air is :
(A) 620 N (B) 740 N
(C) 800 N (D) 920 N
63. A block of mass m is pulled on the smooth horizontal
surface with the help of two ropes, each of mass m, connected to the opposite faces of the block. The forces on the ropes are F and 2F. The pulling force on the block is :
(A) F (B) 2F
191919191919 PAGE # 19 64. A body of mass 5 kg starts from the origin with an initial
velocity
u
= 30i
ˆ+ 40ˆj ms–1 . If a constant force
F= –(ˆi+ 5ˆj) N acts on the body, the time in which the
y-component of the velocity becomes zero is :
(A) 5 s (B) 20 s
(C) 40 s (D) 80 s
65. STATEMENT-1 :According to the newton’s third law of
motion, the magnitude of the action and reaction force is an action reaction pair is same only in an inertial frame of reference.
STATEMENT-2 : Newton’s laws o f m otio n a re
applicable in every inertial reference frame.
(A) 1 is true, Statement 2 is true, statement-2 is correct explanation for statement 1.
(B) 1 is true, Statement 2 is true, statement-2 is NOT a correct explanation for statement-1. (C) statement-1 is true, Statement 2 is false (D) statement-1 is False, Statement 2 is True
66. A body of mass 10 g moves with constant speed 2 m/
s along a regular hexagon. The magnitude of change in momentum when the body crosses a corner is :
(IAO/Sr./Stage-I/2007)
(A) 0.04 kg-m/s (B) zero
(C) 0.02 kg-m / s (D) 0.4 kg-m/s
67. An object with uniform density is attached to a spring
that is known to stretch linearly with applied force as shown below
When the spring object system is immersed in a liquid of density 1 as shown in the figure, the spring
stretches by an amount x1 ( > 1). When the experiment
is repeated in a liquid of density 2 < 1 . the spring is
stretched by an amount x2. Neglecting any buoyant force on the spring, the density of the object is:
(KVPY/2011) (A) 2 1 2 2 1 1 x x x x (B) 1 2 1 2 2 1 x x x x (C) 2 1 1 2 2 1 x x x x (D) 2 1 2 2 1 1 x x x x
68. A body of 0.5 kg moves along the positive x - axis under
the influence of a varying force F (in Newtons) as shown
below : (KVPY/2011) 3 3 1 F (N ) 0,0 2 4 6 8 10 x(m)
If the speed of the object at x = 4m is 3.16 ms–1 then its
speed at x = 8 m is :
(A) 3.16 ms–1 (B) 9.3 ms–1
(C) 8 ms–1 (D) 6.8 ms–1
69. A soldier with a machine gun, falling from an airplane
gets detached from his parachute. He is able to resist the downward acceleration if he shoots 40 bullets a second at the speed of 500 m/s. If the weight of a bullet is 49 gm, what is the weight of the man with the gun ? Ignore resistance due to air and assume the acceleration due to gravity g = 9.8 m/s2 . (KVPY/2010)
(A) 50 kg (B) 75 kg
20 20 20 20 PAGE # 20
CARBON
INTRODUCTIONThe compounds like urea, sugars, fats, oils, dyes, proteins, vitamins etc., which are isolated directly or indirectly from living organisms such as animals and plants are called organic compounds.The branch of chemistry which deals with the study of these compounds is called ORGANIC CHEMISTRY.
VITAL FORCE THEORY OR BERZELIUS HYPOTHESIS
Organic compounds cannot be synthesized in the laboratory because they require the presence of a mysterious force (called vital force) which exists only in living organisms.
WOHLER’S SYNTHESIS
In 1828, Friedrich Wohler synthesized urea (a well known organic compound) in the laboratory by heating ammonium cyanate.
(NH ) SO + 2 KCNO 2NH CNO + K SO4 2 4 4 2 4 Ammonium Potassium Ammonium Potassium
sulphate cyanate cyanate sulphate
Note :
Urea is the first organic compound synthesized in the laboratory.
MODERN DEFINITION OF ORGANIC CHEMISTRY
Organic compounds may be defined as hydrocarbons and their derivatives and the branch of chemistry which deals with the study of hydrocarbons and their derivatives is called ORGANIC CHEMISTRY.
Organic chemistry is treated as a separate branch because of following
reasons-(i) Organic compounds are large in number. (ii) Organic compounds generally contain covalent
bond.
(iii) Organic compounds are soluble in non polar solvents. (iv) Organic compounds have low melting and boiling
points.
(v) Organic compounds show isomerism . (vi) Organic compounds exhibit homology.
VERSATILE NATURE OF CARBON
About 3 million organic compounds are known today. The main reasons for this huge number of organic compounds are
-(i) Catenation : The property of self linking of carbon
atoms through covalent bonds to form long straight or branched chains and rings of different sizes is called catenation.Carbon shows maximum catenation in the periodic table due to its small size, electronic configuration and unique strength of carbon-carbon bonds.
(ii) Electronegativity and strength of bonds : The
electronegativity of carbon (2.5) is close to a number of other elements like H (2.1) , N(3.0) , P (2.1), Cl (3.0) and O (3.5). So carbon forms strong covalent bonds with these elements.
(iii) Tendency to form multiple bonds : Due to small
size of carbon it has a strong tendency to form multiple bonds (double & triple bonds).
(iv) Isomerism : It is a phenomenon by the virtue of
which two compounds have same molecular formula but different physical and chemical properties.
CLASSIFICATION OF ORGANIC COMPOUNDS
The organic compounds are very large in number on account of the self -linking property of carbon called catenation. These compounds have been further classified as open chain and cyclic compounds.
Open chain
compounds Closed chain compounds Organic compounds Alicyclic compounds Aromatic compounds ( a ) Open C ha in C ompounds :
These compounds contain an open chain of carbon atoms which may be either straight chain or branched chain in nature. Apart from that, they may also be saturated or unsaturated based upon the nature of bonding in the carbon atoms. For example.
, ,
21 21 21 21 PAGE # 21 ,
n-Butane is a straight chain alkane while 2-Methylpropane is branched alkane.
( b) C los e d C ha in or Cy c lic Compounds :
Apart from the open chains, the organic compounds can have cyclic or ring structures. A minimum of three atoms are needed to form a ring. These compounds have been further classified into following types.
(i) Alicyclic compounds : Those carbocyclic
compounds which resemble to aliphatic compounds in their properties are called alicyclic compounds .
e.g. or Cyclopropane
or Cyclobutane
or Cyclopentane
or Cyclohexane
(ii) Aromatic compounds : Organic compounds which
contain one or more fused or isolated benzene rings are called aromatic compounds.
e.g.
Benzene Toluene Ethyl benzene
Phenol Aniline
Note :
Benzene is the parent compound of majority of aromatic organic compounds.
HYDROCARBONS
The organic compounds containing only carbon and hydrogen are called hydrocarbons. These are the simplest organic compounds and are regarded as parent organic compounds. All other compounds are considered to be derived from them by the replacement of one or more hydrogen atoms by other atoms or groups of atoms. The major source of hydrocarbons is petroleum.
Types of Hydrocarbons :
The hydrocarbons can be classified as :
(i) Saturated hydrocarbons :
(A) Alkanes : Alkanes are saturated hydrocarbons
containing only carbon - carbon and carbon - hydrogen single covalent bonds.
General formula- CnH2n + 2(n is the number of carbon atoms)
e.g. CH4 ( Methane)
C2H6 (Ethane)
(ii) Unsaturated hydrocarbons :
(A) Alkenes : These are unsaturated hydrocarbons
which contain carbon - carbon double bond. They contain two hydrogen less than the corresponding alkanes.
General formula - CnH2n
e.g. C2H4 (Ethene)
C3H6 (Propene)
(B) Alkynes : They are also unsaturated hydrocarbons
which contain carbon-carbon triple bond. They contain four hydrogen atoms less than the corresponding alkanes.
General formula - CnH2n
–2
e.g. C2H2 (Ethyne)
22 22 22 22 PAGE # 22 NOMENCLATURE OF ORGANIC COMPOUNDS
Nomenclature means the assignment of names to organic compounds . There are two main systems of nomenclature of organic compounds
-(1) Trivial system
(2) IUPAC system (International Union of Pure and Applied Chemistry)
( a ) Ba s i c ru le s o f I UPAC n om e ncl at ure o f o rg a n ic c o mp ou nd s :
For naming simple aliphatic compounds, the normal saturated hydrocarbons have been considered as the parent compounds and the other compounds as their derivatives obtained by the replacement of one or more hydrogen atoms with various functional groups.
(i) Each systematic name has two or three of the following
parts-(A) Word root : The basic unit of a series is word root
which indicate linear or continuous number of carbon atoms.
(B) Primary suffix : Primary suffixes are added to the
word root to show saturation or unsaturation in a carbon chain.
(C) Secondary suffix : Suffixes added after the
primary suffix to indicate the presence of a particular functional group in the carbon chain are known as secondary suffixes.
(ii) Names of straight chain hydrocarbons : The
name of straight chain hydrocarbon may be divided into two
parts-(A) Word root (B) Primary suffix
(A) Word roots for carbon chain lengths :
Chain
length root length root C Meth- C Hex-C Eth - Hex-C Hept-C Prop - Hept-C Oct-C But - Oct-C Non-C Pent- Non-C Dec-1 6 2 7 3 8 4 9 5 10
Word Chain Word
(B) Primary suffix : Examples : Eth -E B P E M Propene a Note :
The name of the compound, in general , is written in the following
sequence(Position of substituents )(prefixes ) (word root)(p -suffix).
(iii) Names of branched chain hydrocarbon : The
carbon atoms in branched chain hydrocarbons are present as side chain . These side chain carbon atoms constitute the alkyl group or alkyl radicals. An alkyl group is obtained from an alkane by removal of a hydrogen. General formula of alkyl group = CnH2n+1
An alkyl group is represented by R. e.g. (A) H H H C Methyl –H (B) H H H C C Ethyl –H H H (C)
23 23 23 23 PAGE # 23 A branched chain hydrocarbon is named using the
following general IUPAC rules :
Rule1: Longest chain rule : Select the longest possible
continuous chain of carbon atoms. If some multiple bond is present , the chain selected must contain the multiple bond.
(i) The number of carbon atoms in the selected chain determines the word root .
(ii) Saturation or unsaturation determines the primary suffix (P. suffix).
(iii) Alkyl substituents are indicated by prefixes.
e.g. CH 3– CH – CH – CH – CH2 2 3
CH3
Prefix : Methyl Word root :
pent-P. Suffix: - ane
e.g. CH 3– CH – CH – CH – CH2 3
CH3
Prefix : Methyl Word root :
Hept-P. Suffix : -ane CH 2– CH – CH2 3
e.g. CH 3– CH – C – CH2 3
CH2
Prefix : Methyl Word root :
But-P. Suffix : –ene
e.g. CH 3– CH – CH – CH – CH2 2 3
CH3
Prefixes : Ethyl, Methyl Word root :
Pent-P. Suffix : -ane CH – CH3
Rule 2 : Lowest number rule: The chain selected is
numbered in terms of arabic numerals and the position of the alkyl groups are indicated by the number of the carbon atom to which alkyl group is attached .
(i) The numbering is done in such a way that the
substituent carbon atom has the lowest possible number.
(ii) If some multiple bond is present in the chain, the
carbon atoms involved in the multiple bond should get lowest possible numbers.
e.g. 2–Methylbutane 3–Methylbutane (Correct) (Wrong) e.g. 2–Methylpentane 4–Methylpentane (Correct) (Wrong) e.g.
3–Methylbut–1– ene 2–Methylbut – 3 – ene
(Correct) (Wrong) e.g. CH3 | CH| 3 1 1 3 3 4 4 2 2 3-Methylbut-1-yne (Correct) 2-Methylbut-3-yne (Wrong)
Rule 3 : Use of prefixes di, tri etc. : If the compound
contains more than one similar alkyl groups,their positions are indicated separately and an appropriate numerical prefix di, tri etc. , is attached to the name of the substituents. The positions of the substituents are separated by commas.
e.g. CH 3– CH – C – CH – CH2 3 CH3 CH3CH3 3 4 5 2 1 2,3 - Dimethylpentane 2,3,3 - Trimethylpentane e.g. 2,3,5 -Trimethylhexane 2,2,4 - Trimethylpentane
Rule 4 : Alphabetical arrangement of prefixes: If there
are different alkyl substituents present in the compound their names are written in the alphabetical order. However, the numerical prefixes such as di, tri etc. , are not considered for the alphabetical order.
e.g.
3-Ethyl - 2,3-dimethylpentane
Rule 5 : Naming of different alkyl substituents at the equivalent positions :
Numbering of the chain is done in such a way that the alkyl group which comes first in alphabetical order gets the lower position.
e.g.
