# Newtons Laws of Motion

August 14, 2017 | Author: Vikalp Joshi | Category: Force, Friction, Newton's Laws Of Motion, Acceleration, Torque

For IIT JEE....

#### Description

Dynamics : Newton’s Laws of Motion Newton‟s laws of motion act as basis for studying dynamics. These laws, which are basically observations, relate dynamical and kinematical quantities. These three laws had been presented by Sir Isaac Newton in 1686, in his book "Principia Mathematica Philosophiae Naturalis." First Law : “Every object persists in its state of rest or of uniform motion in a straight line unless it is compelled to change that state by unbalanced forces impressed on it.” This is normally taken as the definition of inertia. The key point here is that if there is no net force acting on an object (if all the external forces cancel each other out) then the object will maintain a constant velocity. If that velocity is zero, then the object remains at rest. If an external force is applied, the velocity will change because of the force. Second Law : “Force is equal to the change in momentum (mv) per unit change in time.” (For a constant mass, force equals mass times acceleration)”. i.e. rate of change of momentum is force. The second law explains how the velocity of an object changes when it is subjected to an external force. The law defines a force to be equal to change in momentum (mass times velocity) per unit change in time. Newton also developed the calculus of mathematics, and the “changes” expressed in the second law are most accurately defined in differential forms. (Calculus can also be used to determine the velocity and location variations experienced by an object subjected to an external force.) For an object with a constant mass m, the second law states that the force F is the product of the object's mass m and its acceleration a: dp dv Fnet  F  m  ma dt dt Fnet  ma Fnet m Where Linear momentum = mass  velocity a

i.e., p  mv Third Law : “For every action there is an equal and opposite re-action”. The third law states that for every action (force) in nature there is an equal and opposite reaction. In other words, if object A exerts a force on object B, then object B also exerts an equal and opposite force on object A. Notice that the forces are exerted on different objects as described in the figure.

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Notes on Newton's Law of Motion (73) i.e.,

F12  F21

B (action force)

F12 (reaction force)

F21

Types of Forces :

A (a) Field Forces In such type of forces contact between two objects is not necessary. For e.g., gravitational force between two bodies and electrostatic force between two charges. (b) Contact Forces The contact force between two objects is made up of two forces. i) Normal reaction (N)  acts perpendicular to the common surface. ii) Force of friction (f). We will discuss these forces later in Friction. (c) When two bodies are attached In this category, there comes tension (T) in the string and spring force. Some Important Points regarding the tension  Acceleration of any number of masses connected through string is always same. (assuming the string to be inextensible). a a a m1

m1

m1 F

For a massless string, the tension is same everywhere. However, if a string has a mass, tension at different points will be different.

In case of frictionless pulley, tension will be same on both sides of the pulley. However, if there is friction between string and pulley, tension is different on two sides of the pulley. T

T

m1

m1

T2

m1

T4

T3

T

T1

T2

T

T1

T1

m2

(i) massless string (ii) no friction between pulley & string

T2

m2

(i) massless string (ii) there is friction between string & pulley

m2

(i) string is not massless (ii) there is friction between pulley & string.

0413/IIT.15/CR/P.2/Ch.3/Pg.73

(74) Vidyalankar : JEE Physics

Identification of the system and forces: Free body diagram (FBD) It is the diagrammatic representation of all the forces acting on a single body or a subsystem of bodies isolated from its surrounding. Illustration of FBD concept For the system shown in figure. Taking block A and writing all forces which acts on it. For block ‘A’,

N1

A Balancing forces

B

A N1 = WA = MAg

C MAg

For block ‘B’,

N1

N1

By force balance, N2 = N1 + W B = mA g + mB g = (mA + mB) g

B

N2

WB

where N1 : Normal reaction of block „A‟ on block „B‟ N2 : Normal reaction of block „C‟ on block „B‟. For Block ‘C’,

N2

By force balance, N3 = N2 + W C

C

= (mA + mB + mC) g N3

0413/IIT.15/CR/P.2/Ch.3/Pg.74

WC

N2 WB

Notes on Newton's Law of Motion (75) Equilibrium A body is said to be in equilibrium if net force ( Fnet ) acting on it, is zero and net moments of all the forces ( τnet ) acting on it about any axis is zero. Where moment of force (known as torque) is calculated as :   rF

where r is the perpendicular vector from axis of rotation to the point where force is acting. Here condition of τnet = 0 ; gives equilibrium with respect to rotational motion and Fnet = 0 gives translational equilibrium. So at Fnet = 0, according to Newton‟s law, it will remain in rest if it is at rest and will move with same velocity if it is in motion already. This is the state of translational equilibrium. This section mainly concerns with this. Rotational equilibrium will be discussed in detail in rotational mechanics. Some Examples of Newton’s Laws 1. Three blocks of masses 5kg, 3kg and 2kg are placed side by side on a smooth surface as shown in figure. A horizontal force of 15N is applied on 5kg block. Find the net force on 2kg block. F  15N

5kg

3kg

2kg

Solution : As all the blocks will move with same acceleration in the horizontal direction, so let us take all the blocks as a system. Net external force on the system is Fnet  15N using

Fnet = ma

i.e.,

Fx  max , we get

15  1.5m / s2 . 10 If „F‟ is the net force on 2 kg block in xdirection, then for 2 kg block only, we get

15 = (5 + 3 + 2) a 

F = (2) (2) = 4N.

a=

[Ans.]

2. If the blocks in the above problem are connected with light and in extensible strings and force is applied as shown in figure. Then (a) Find the acceleration of each block (b) Find the tension in each string. 5Kg

3Kg

2Kg

F = 15N

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(76) Vidyalankar : JEE Physics Solution : (a) Let T1 and T2 be the tensions in the two strings and „a‟ be the acceleration of each block. 5Kg

T2

T2 3Kg

T1

T1

2Kg

F = 15N

As the strings are light and inextensible, all the blocks will have same acceleration. Taking the three blocks and the two strings as the system, we have Fx = max 15 = (5 + 3 + 2)a F = 15N 5Kg 3Kg 2Kg 2  a = 1.5 m/s . [Ans.] (b) To find the tension in each string, we will make the free body diagrams of 5kg block and 2kg block as shown in figure. 2

2

a = 1.5m/s 5 Kg

a = 1.5m/s T2

T1

2 Kg

F = 15N

We have not shown the vertical forces as there is no movement in the vertical direction. Also, all the blocks have same acceleration.  for 5 kg block T2 = 5a 

T2 = (5 kg) (1.5m/s2) T 2 = 7.5N.

[Ans.]

For 2kg block, F  T1 = 2a  T1 = F  2a 2 = 15N  (2 Kg) (1.5 m/s ) = 15N  3N. T1 =

12N.

[Ans.]

Illustrations related to pulleys Constraint relations Two blocks of masses m 1 and m2 (m1 > m2) are attached at the ends of an inextensible string, which passes over a smooth massless pulley. If a1 be the downward acceleration of m 1 and a2 be the upward acceleration of m2, then by judgment alone we can say a1 = a2 = a.

0413/IIT.15/CR/P.2/Ch.3/Pg.76

a2 m2 m1

a1

Notes on Newton's Law of Motion (77) 1. In the arrangement of three blocks as shown in figure, the string is inextensible. If the directions of accelerations are as shown in the figure, then determine the constraint relation. a1

a2

m1

m2

m3 a3 Pseudo Force Let us first discuss briefly about frame of reference. Frames of references are of two types. (a) Inertial frame of reference An inertial frame is a nonaccelerating frame of reference or a frame which is either at rest or moving with uniform velocity. It is the frame in which Newton‟s first law of motion is obeyed. (b) Noninertial frame of reference An accelerated frame of reference is called a noninertial frame. Newton‟s first two laws are valid in inertial frame of reference, but we can‟t use them in noninertial frame of reference, without modification. In noninertial frame of reference, we use pseudo force which is given by Fp  ma , where a is the acceleration of the noninertial frame of reference. Friction The force of friction comes into play only when there is a relative motion or there is a possibility of relative motion between the two contact surfaces. The force of friction on each body is in a direction opposite to its motion relative to the other body. When two bodies slip over each other the force of friction is called kinetic friction, but when they do not slip and have a tendency to do so the force of friction is called static friction.

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(78) Vidyalankar : JEE Physics Important points regarding friction 1. If a body is at rest and no tendency of motion, then force of friction acting on it is zero. 2. Static Friction :If a body does not move even on application of a force, the friction which acts is equal in magnitude and opposite in direction to the applied force and is known as static friction. For this reason friction is known as selfadjusting force. 3. Limiting Friction : The maximum force of friction when a body just begins to slide over the surface of another body is called the limiting friction. This limiting force of friction ( f ) is proportional to normal reaction (N). Hence, or f = sN f N where s is a dimensionless constant and known as coefficient of static friction. Note : Static frictional force is given by fs  sN . 4. Once the body starts sliding on application of the force, the friction opposing the motion is called kinetic or sliding friction. It is given by fk = kN where K is coefficient of kinetic friction and k < s. Note : i) ii)

Kinetic friction is less than limiting friction. It is independent of the surface area of contact.

The graph shown in the adjacent figure shows the variation of friction with the applied force. f Rest

Relative motion

f  (fs )max fk

45

F Angle of Friction () Angle of friction () is defined as the angle between the normal reaction „N‟ and the resultant of the friction force „f ‟ and the normal reaction. Thus,

tan 

=

tan 

=

f N  N N .

=

tan1().

0413/IIT.15/CR/P.2/Ch.3/Pg.78

N

F

f = N

Notes on Newton's Law of Motion (79) Angle of Response () : Angle of response of an inclined plane, w.r.t. the surface of a body in contact with it, is the angle of inclination of the plane with the horizontal, when the body just starts sliding down the plane under its own weight. At this point, N = Mg cos  and Mg sin  = sN where sN is the limiting friction. i.e., Mg sin  = Mg cos   tan  =  

 = tan1().

Angle of response () is same as the angle of friction (). Dynamics of Circular Motion If an object moves on a circular path with a constant speed, then its motion is known as uniform circular motion in a plane. In uniform circular motion even if speed of the particle is constant, it has nonzero acceleration and a resultant nonzero force acts on it. This acceleration is due to the change in direction of the velocity vector and is towards the v2 centre. It‟s magnitude is given by where „v‟ is the speed of the particle and „r‟ the r radius of the circle. The resultant force „F‟ is given by, v2 r or F = mr 2 , where „‟ is the angular speed of the particle and given by  = v/r. The direction of the resultant force „F is towards centre and is known as centripetal force.

F = ma

or

F=m

Effect of Centripetal force on circular turning of Roads When vehicles go through turnings, they almost take the circular path. So, there has to be some force which will provide the required centripetal acceleration. This necessary centripetal force is being provided to the vehicles by following ways. (1) By friction only (2) By banking of roads only 1. By friction only If a truck of mass m is moving at a speed „v‟ in a horizontal circular arc of radius „r‟, then the necessary centripetal force to the truck will be provided by the force of friction „f ‟ which will act towards the centre. mv 2 Thus, f= r As we know, limiting value of frictional force is f  N = mg Therefore, for a safe turn x mv 2 mv 2 f  mg  r r v2  or   rg or rg 0413/IIT.15/CR/P.2/Ch.3/Pg.79

(80) Vidyalankar : JEE Physics 2. By banking of roads only Frictional forces are not sufficient at circular turns if high speeds and sharp turns are involved. To avoid dependence on friction, the roads are banked at the turn so that the outer part of the road is somewhat lifted compared to the inner part. Applying Newton‟s 2

nd

law.

In vertical direction N cos  = mg

……(1)

mv 2 In horizontal direction  Nsin  ……(2) r

N mv 2 r

mg

 From (1) and (2), we will get tan  =

v2 rg

or v =

r gtan 

Centrifugal Force As we have discussed before, the Newton‟s laws are valid only in inertial frames. In noninertial frames a pseudo force, m.a , has to be applied on a particle of mass „m‟ where a is the acceleration with which the frame translates w.r.t. an inertial frame. After applying the pseudo force one can apply Newton‟s laws in their usual form. Now suppose a body of mass m is placed on a frame of reference which is rotating with a uniform velocity  in a circle of radius „r‟. For an observer on the earth watching the body would find that the body is in uniform circular motion, accelerated radially inward with centripetal 2 2 acceleration a = v /r =  r. For another observer standing on the tables, there should be no force acting on the body as he will feel that the body is at rest, because its position is not changing w.r.t. him. mv 2 But actually the body is acted upon by an inward force . Hence according to r mv 2 observer standing on the rotating frame, a force of magnitude is also acting r outward so that net force on the body is zero. This apparent outward force is called centrifugal force. After applying this pseudo force we can now apply Newton‟s laws in their usual form in rotating frame also.

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