college - himpub.comnewton’s laws are valid only in an inertial frame of reference. any reference...

COLLEGE

PHYSICSAs per Credit Based Semester and Grading System F.Y. B.Sc. New Syllabus

University of Mumbai(w.e.f. 2016-2017)

Dr. Kailas R. Jagdeo(M.Sc., B.Ed., Ph.D., LL.B., GDC&A)

Department of Physics,DSPM’s K.V. Pendharkar College of Arts, Science and Commerce,

Dombivli (E) - 421203.

Dr. Sopan A. Bhamare(M.Sc., Ph.D.)

Head, Department of Physics,B.N.N. College of Arts, Science and Commerce,

Bhiwandi, Thane - 421302.

Dr. Suresh N. Kadam(M.Sc., B.Ed., M.Phil, Ph.D.)

Department of Physics,KET’s V.G. Vaze College of Arts, Science and

Commerce, Mulund (E), Mumbai - 400081.

Moharram Ali Khan(M.Sc., M.Phil.)

Department of Physics,Rizvi College of Arts, Science and Commerce,

Bandra (W), Mumbai - 400 050.

Vishwas V. Deshmukh(M.Sc., M.Phil., MBA (Edu. Mgt.))

Department of Physics,Rizvi College of Arts, Science and Commerce,

Bandra (W), Mumbai - 400 050.

ISO 9001:2008 CERTIFIED

© AuthorsNo part of this publication may be reproduced, stored in a retrieval system, or transmitted in any form or by anymeans, electronic, mechanical, photocopying, recording and/or otherwise without the prior written permission of thepublisher.

First Edition : 2016Second Revised Edition : 2017

Published by : Mrs. Meena Pandey for Himalaya Publishing House Pvt. Ltd.,“Ramdoot”, Dr. Bhalerao Marg, Girgaon, Mumbai - 400 004.Phone: 022-23860170, 23863863; Fax: 022-23877178E-mail: [email protected]; Website: www.himpub.com

Branch Offices :New Delhi : “Pooja Apartments”, 4-B, Murari Lal Street, Ansari Road, Darya Ganj, New

Delhi - 110 002. Phone: 011-23270392, 23278631; Fax: 011-23256286Nagpur : Kundanlal Chandak Industrial Estate, Ghat Road, Nagpur - 440 018.

Phone: 0712-2738731, 3296733; Telefax: 0712-2721216Bengaluru : Plot No. 91-33, 2nd Main Road, Seshadripuram, Behind Nataraja Theatre,

Bengaluru - 560 020. Phone: 080-41138821; Mobile: 09379847017,09379847005

Hyderabad : No. 3-4-184, Lingampally, Besides Raghavendra Swamy Matham,Kachiguda, Hyderabad - 500 027. Phone: 040-27560041, 27550139

Chennai : New No. 48/2, Old No. 28/2, Ground Floor, Sarangapani Street, T. Nagar,Chennai - 600 012. Mobile: 09380460419

Pune : First Floor, “Laksha” Apartment, No. 527, Mehunpura, Shaniwarpeth (NearPrabhat Theatre), Pune - 411 030. Phone: 020-24496323, 24496333;Mobile: 09370579333

Lucknow : House No. 731, Shekhupura Colony, Near B.D. Convent School, Aliganj,Lucknow - 226 022. Phone: 0522-4012353; Mobile: 09307501549

Ahmedabad : 114, “SHAIL”, 1st Floor, Opp. Madhu Sudan House, C.G. Road, NavrangPura, Ahmedabad - 380 009. Phone: 079-26560126; Mobile: 09377088847

Ernakulam : 39/176 (New No. 60/251), 1st Floor, Karikkamuri Road, Ernakulam,Kochi - 682011. Phone: 0484-2378012, 2378016; Mobile: 09387122121

Bhubaneswar : 5 Station Square, Bhubaneswar - 751 001 (Odisha).Phone: 0674-2532129; Mobile: 09338746007

Kolkata : 108/4, Beliaghata Main Road, Near ID Hospital, Opp. SBI Bank,Kolkata - 700 010. Phone: 033-32449649; Mobile: 07439040301

DTP by : PravinPrinted at : Rose Fine Art, Mumbai. On behalf of HPH.

PREFACE

It gives immense pleasure to present this “College Physics” book for the studentsof F.Y.B.Sc. Semester I, Paper-I and Paper-II, University of Mumbai for their syllabus.

The said syllabus has been implemented from the year 2016-2017. This bookcontains “Classical Physics” and “Modern Physics”, both papers of the Semester I.

We are confident that the students would like and useful to understand the Physics.

We are sure our teacher friends would receive this edition with open mind andwould suggest method for its improvement whenever necessary.

Mumbai AUTHORS

Syllabus

CLASSICAL PHYSICSCourse Code - USPH101

Unit:I 15 lectures

1. Newton’s Laws: Newton’s first, second and third laws of motion, interpretation andapplications, pseudo forces, Inertial and non-inertial frames of reference. Worked outexamples (with friction present).

2. Elasticity: Review of Elastic constants Y, K, η and σ; Equivalence of shear strain tocompression and extension strains. Relations between elastic constants, Couple for twist incylinder.

3. Fluid Dynamics: Equation of continuity, Bernoulli’s equation, applications of Bernoulli’sequation, streamline and turbulent flow, lines of flow in airfoil, Poiseuille’s equation.

Unit:II 15 lectures

1. Lens Maker’s Formula (Review), Newton’s lens equation, magnification-lateral, longitudinaland angular.

2. Equivalent focal length of two thin lenses, thick lens, cardinal points of thick lens, Ramsden’sand Huygen’s eyepiece.

3. Aberration: Spherical Aberration, Reduction of Spherical Aberration, Chromatic aberrationand condition for achromatic aberration.

4. Interference: Interference in thin films, Fringes in Wedge shaped films, Newton’s Rings(Reflective).

UNIT III 15 lectures

1. Behaviour of real gases and real gas equation, Van der Waals equation2. Thermodynamic Systems, Zeroth law of thermodynamics, Concept of Heat, The first law,

Non Adiabatic process and Heat as a path function, Internal energy, Heat Capacity andspecific heat, Applications of first law to simple processes, general relations from the first law,Indicator diagrams, Work done during isothermal and adiabatic processes, Worked examplesand Problems.

MODERN PHYSICSCourse Code -USPH102

Unit I 15 lectures

1. Structure of Nuclei: Basic properties of nuclei, Composition, Charge, Size, Rutherford’s expt.for estimation of nuclear size, density of nucleus, Mass defect and Binding energy, Packingfraction, BE/A vs. A plot, stability of nuclei (N vs Z plot) and Problems.

2. Radioactivity: Radioactive disintegration concept of natural and artificial radioactivity,Properties of α, β, γ-rays, laws of radioactive decay, half-life, mean life (derivation notrequired), units of radioactivity, successive disintegration and equilibriums, radioisotopes.Numerical Problems.

3. Carbon dating and other applications of radioactive isotopes (Agricultural, Medical, Industrial,Archaeological).

Unit II 15 lectures

1. Interaction between particles and matter, Ionization chamber, Proportional counter and GMcounter, Problems.

2. Nuclear Reactions: Types of Reactions and Conservation Laws. Concept of Compound andDirect Reaction, Q value equation and solution of the Q equation, problems. Fusion andfission definitions and qualitative discussion with examples.

Unit III 15 lectures

1. Origin of Quantum theory, Black body (definition), Black Body spectrum, Wien’sdisplacement law, Matter waves, wave particle duality, Heisenberg’s uncertainty Principle.Davisson-Germer experiment, G. P. Thompson experiment.

2. X-Rays: Production and properties. Continuous and characteristic X-Ray spectra, X-RayDiffraction, Bragg’s Law, Applications of X-Rays.

3. Compton Effect, Pair production, Photons and Gravity, Gravitational Red Shift.

CONTENTS

Paper I : Classical Physics

Chapter 1 Newton’s Laws of Motion 3 – 41

Chapter 2 Elasticity 42 – 54

Chapter 3 Fluid Dynamics 55 – 68

Chapter 4 Lenses 69 – 94

Chapter 5 Aberrations 95 – 106

Chapter 6 Interference 107 – 123

Chapter 7 Behaviour of Real Gases 124 – 140

Chapter 8 Thermodynamics 141 – 160

Paper II : Modern Physics

Chapter 1 Structure of Nuclei 163 – 177

Chapter 2 Radioactivity 178 – 196

Chapter 3 Nuclear Detectors 197 – 213

Chapter 4 Nuclear Reactions 214 – 227

Chapter 5 Origin of Quantum Theory 228 – 247

Chapter 6 X-Rays 248 – 261

Chapter 7 Compton Effect 262 – 272

Physical Constants 273 – 273

University Question Papers 275 – 298

Paper IClassical Physics

(Course Code - USPH101)

UNIT : I

1.1 INTRODUCTION

Newton’s laws of motion are three physical laws that build the foundation for classicalmechanics. It describes the relationship between a body and forces acting upon it, and the body’smotion in response to the said forces:

1. When viewed in an inertial reference frame, a body keeps moving with a uniform velocity orit remains at rest in the absence of a resultant force.

2. If a body acted upon by a force which moves the body in such a way that its time rate ofchange of momentum equals the force.

3. If two bodies exert forces on each other, these forces are equal in magnitude and opposite indirection.

The three laws of motion were first compiled by Sir Issac Newton in his Philosophiae NaturalisPricipia Mathematica (Mathematical Principles of Natural Philosophy), first published in 1687. Theselaws are valid only with respect to a certain set of frames of reference called Newtonian or inertialreference frames. Newton’s laws are valid only in an inertial frame of reference. Any reference framethat is in uniform motion with respect to an inertial frame is also an inertial frame, i.e., Galileaninvariance or principle of Newtonian relativity. A frame of reference in which Newton’s laws are notvalid is called non-inertial frame of reference.

In this chapter we discuss the Newton’s laws of motion and its application.

1.2 NEWTON’S FIRST LAW OF MOTION

When viewed in an inertial reference frame, Newton’s first law states that a body at rest willremain at rest and if it is in motion, it will continue its motion in the same direction until and unlessan external unbalanced force is applied on it.

If sum of all the forces acting on a body is F and its acceleration is a , then according to firstlaw of motion,

F = 0 iff a = 0 or a = 0 iff F = 0

CHAPTER

NEWTON’S LAWS OFMOTION1

4 College Physics

Thus, the first law states that a body which remains at rest or in uniform motion, i.e.,unaccelerated, rectilinear motion, has zero force. The first law also implies that all bodies resist beingaccelerated. The property of the bodies by virtue of which they always oppose the change in its ownstate is known as inertia, hence, the first law of motion is known as law of inertia. Body can be inposition of rest or motion, so the inertia of body are of two types, namely, inertia of rest and inertia ofmotion.

First law provides a qualitative definition of force and recognizes inertia as a fundamentalproperty of material bodies.

1.3 NEWTON’S SECOND LAW OF MOTION

Newton’s second law states that the rate of change of momentum of a body is proportional to theforce applied and the direction of change of momentum is the same in which the force is applied.

If v be the velocity of a body of mass m, then momentum of the body is,

P = vm ... (1.1)

Let F be the vector sum of all forces acting upon the body at any instant of time.

F dtpd

F =dtpdK ... (1.2)

Where, K is the proportionality constant

F = )m(dK vdt

For non-relativistic velocities (i.e., v << c , where c is speed of light), the mass m of body canbe taken as constant.

F =dtvdKm ... (1.3)

But,dtvd = a (acceleration of body)

F = aKm ... (1.4)

In S.I. System, if m = 1 kg, a = 1m/s2 and F = 1 NThen K = 1

Hence, F = am ... (1.5) Force = mass × acceleration

Second law provides a quantitative definition of force.

Thus, Newton’s first law of motion is concerned with the bodies at rest or in uniform motionwhen no forces are acting on it. The second law, however, assumes that a force is acting and the lawdescribes how the motion changes.

Newton’s Laws of Motion 5

1.4 NEWTON’S THIRD LAW OF MOTION

Newton’s third law states that, to every action, there is always an equal and opposite reaction.

Thus, action of one object is equal to the reaction of other, but in opposite direction. If the firstobject applies force 21F on the second object, then 12F will be the force of reaction of second objecton the first.

12F = – 21F or 21F = – 12F

These forces always exist in pair and do not cancel each other.

1.5 INERTIAL FRAME OF REFERENCE

A system of coordinate axes which represents the position of a body or an event in two or threedimension space is called a frame of reference. The laws of motion to have meaning, a referenceframe must be chosen with reference to which the motion of the bodies could be measured. Themotion of body is not absolute rather it is relative term. A body at rest in one reference frame may notnecessarily appear to be at rest in another. Thus, for example, a passenger sitting in a moving trainwould appear to be in a state of rest with reference to his fellow passenger but would be in a state ofmotion with reference to an observer standing on the railway platform.

A reference frame is called an inertial frame of reference if Newton’s laws are found valid in thatframe. It is found that Newton’s laws, which are valid in any frame of reference, are equally valid inany other frame of reference moving with a uniform motion (i.e., unaccelerated). This is valid due toGalilean Invariance or the principle of Newtonian Relativity.

The inertial frame of reference is non-rotating frame, because a body at rest in inertial frame mayappear to be moving in circle with respect to the rotating frame and hence has acceleration. The earthis not suitable example for inertial frame of reference. Since the earth has in addition to its spin motionabout its axis, a rotational motion round the sun. It gives centripetal acceleration 3.4 × 10–2m/s2 due tospin motion and 6 × 10–3m/s2 due to rotation motion. For our every purpose the acceleration can beneglected for being their smaller values and hence reference frames attached to the earth can beconsidered to be satisfactory inertial frame of reference. Also, interior of vehicle such as car, bus, trainetc. moving with uniform speed can be imagined equally satisfactory inertial frame of reference.“Fixed Stars” are conventionally considered to an inertial frame of reference to a good approximation.But, recent astronomical observation showed that fixed stars are slowly changing their pattern as theymove around in the sky. All stars are moving around the centre of our galaxy.

Therefore, the best approximation to an inertial frame of reference is the frame of reference in thegalactic space.1.6 NON-INERTIAL FRAME OF REFERENCE

The frame of reference in which Newton’s law of motion are not valid called as non-inertial oraccelerated frame of reference. All the accelerated and rotating frames of reference are non-inertialreference frames. Suppose a frame of reference S′ is moving with acceleration oa with respect to aninertial frame S. Let a particle ‘P’ be at rest in frame S.

6 College Physics

With respect to an observer in frame S, the particle ‘P’ appearsstationary. But, with respect to an observer in frame S', it will appearto move with acceleration 0a– .Therefore, observer in frame S' feel

that a force oF = oa– m is acting on a particle of mass m. But, in realityno force is acting on it. Thus, all those forces which do not really acton the particle, but appear to act on it due to acceleration or rotation offrame are called as the fictitious or pseudo forces.

Let 1F be a force acting on a particle of mass m moving withacceleration 1a in frame S, then 1F = .1am Therefore, the apparent force acting on the particle withrespect to an observer in non-inertial frame S' moving with acceleration oa with respect to frame S is,

F = 1F + oF

F = m 1a + )(– 0am

i.e., Apparent force = Real Force + Pseudo Force

Example: When a bus takes a turn on a curve, its motion becomes accelerated and hence theframe attached is non-inertial frame. The passenger sitting in a bus experience outward force. Thisforce is pseudo force called as centrifugal force. For an observer outside the bus, there is no forceacting on the passenger.

1.7 WORKING WITH NEWTON’S LAWS OF MOTION

The equations of motion describe the behaviour of a physical system as a set of mathematicalfunctions in terms of variables. There are two main descriptions of motion such as dynamic andkinematics. In dynamics; momenta, force and energy of the particles are taken into account. However,the kinematics concerns only spatial and time-related variables. The kinematic includes the quantitiesdisplacement, velocity, acceleration and time. While working with Newton’s Laws of motion asystematic algorithm should be written and applied. Deciding the system, identifying the forces,making a free body diagram, and choosing the axes and writing the equations are the steps ofsystematic algorithm.

1.7.1 Deciding the System

While writing the algorithm for equation from Newton’s laws, the first step is to decide thesystem on which Newton’s Laws of motion are applied. Newton’s laws are applied to a particle andthe forces acting on the particle with its acceleration and its mass. Therefore, before attempting towrite an equation we should understand which particle is considered. But, in practical situation, wedeal with extended bodies which are the collection of a large number of particles. The Newton’s lawsare used even for extended body, provided each part of this body has the same acceleration(in magnitude and direction). Thus, the system may be a single particle, a block, a combination of twoblocks i.e., one kept over the other or two blocks connected by a string etc. provided that all parts ofsystem should have same acceleration (in magnitude and direction).

oaS S'

P

Fig. 1.1


1.7.2 Identifying the Forces

After deciding the system, make a list of forces acting on the system due to all the bodies otherthan the system. Any force applied by the system should not be included in the list of the forces.

(a)

(Boy)

(c)

N’

(Box)

W’

N

(Boy)

(b)

N1 W

Fig. 1.2: Forces operating on the systemsConsider the boy stands on floor with a heavy box on his head. There are many forces acting in

this systems. The forces will be decided depending on which system we are interested. If we considera boy as a system, the forces acting is shown in Fig. 1.2(b) and in Table 1.1. If we consider box as asystem, the forces acting is shown in Fig. 1.2(c) and in Table 1.1.

Table 1.1System Force Exerted Magnitude of Force Direction of Force

I Boy Earth W DownwardFloor N UpwardBox N1 Downward

II Box Earth W' DownwardBoy N' Upward

1.7.3 Making a Free Body Diagram

Fig. 1.2(b) and 1.2(c) represents a free body diagram for above example. In free body diagramthe system is represented by a point and the Vectors representing the forces acting on the system withthis point as the common origin. The forces may be lying along a line or may be distributed in a plane(coplanar) or may be distributed in the space (non-planar).

1.7.4 Choosing the Axes and Writing the Equation

Any three mutually perpendiculars can be chosen as the X-Y-Z axes. Consider the example thatthe forces are coplanar. So we need two axes say X and Y. Now write the components of all the forcesalong the X-axis and equate their sum to the product of the mass of the system and its acceleration.This gives one equation. Now, with the components of the forces along the Y-axis equate the sum tozero. This gives second equation. If the forces are collinear, this second equation is not required.

8 College Physics

1.8 WORKED OUT EXAMPLES

(A) Newton’s Laws of Motion and Forces

Example 1.1: A 12 kg block rests on a smooth frictionless table. A string attached to the block passesover a frictionless pulley, and a 6 kg mass hangs from the string as shown in Fig. 1.3. Find theacceleration ‘a’ and the tension ‘T’ in the string.

Solution:

The forces acting on mass m2 are:(i) Its weigh m2g vertically downward.

(ii) The tension T acting upwards.(iii) Force due to acceleration of body.

m2a = m2g ...(1.6)

The forces acting on mass m1 are:(i) Its weight m1g vertically downward.

(ii) The tension T along accelerationdirection.

(iii) Vertically upward normal reactionR = m1g

T = m1a ... (1.7)m1a + m2a = m2g [from eqns. (1.6) and (1.7)]

a = gmm

m

21

2

a = 6126

× 9.8

a = 3.27 m/s2

From eqn. (1.7), we get,T = 12 (3.27)

T = 39.2 N

Example 1.2: In Atwood’s machine, a string passing over a frictionless, massless pulley has a 10 kg blocktied to one end and 12 kg block tied to the other. Compute the acceleration and the tension in the string.

Solution:

Forces acting on mass m1 are:(i) Its weight m1g vertically downwards.

(ii) The tension T acting upwards.(iii) Force due to acceleration of body.

T = m1g + m1a ... (1.8)

T = m1(g + a) ... (1.9)

m1 = 12 Kg

T a

T

am1g

m2g

m2 = 6 Kg

R

Fig. 1.3


Forces acting on mass m2 are shown if Fig. 1.4

T + m2a = m2g ... (1.10)

From eqns. (1.8) and (1.10), we get,

m1g + m1a + m2a = m2g

a (m1 + m2) = (m2 – m1)g

a = g

mmm–m

21

12

a =101210–12

(9.8)

a =222 (9.8) = 0.89 m/s2

From eqn. (1.9), we get,

T = 10 (9.8 + 0.89) = 106.9 N

Example 1.3: In Fig. 1.5, the weights of the object are 400 N and 600 N. The pulleys areessentially frictionless and massless. Pulley P1 has a stationary axle but pulley P2 is free to moveup and down. Assume that mass A falls down. Find the tension T1, T2 and the acceleration of eachbody.

Solution:

Fig. 1.5 Fig. 1.5(a) Fig. 1.5(b) Fig. 1.5(c)

P1

T2 T2

T2

T2 T1

B

mBg600 N

1/2 a

A

mAg400 N

a

1/2 a

T2 T2

P2

T1

P2

T1

B

600 N

400 N

a

A

The mass B will rise and mass A will fall. The forces acting on masses A, B and pulley P2 areshown in Fig. 1.5(a), (b) and (c) respectively.

Fig. 1.4

10 College Physics

The forces acting on pulley P2 are:(i) 2T2 in upwards.

(ii) T1 in downwards.

T1 = 2T2 ... (1.11)

The pulling upward force on B is double than as on A. If ‘a’ is downward acceleration of A then‘(1/2)a’ is upward acceleration of B.

The forces acting on mass B is shown in Fig. 1.5(b).

T1 =21 mBa + 600 ... (1.12)

The forces acting on mass A is shown in Fig. 1.5(a)

T2 + mAa = 400 ... (1.13)Since, mAg = 400

mA =g

400 =8.9

400 = 40.8 kg . .. (1.14)

Similarly, mBg = 600

mB =g

400 =8.9

600 = 61.2 kg ... (1.15)

2T2 =21 (61.2)a + 600 [From eqns (1.11), (1.12) and (1.15)]

T2 = 15.3a + 300 ... (1.16)

T2 = 400 – (40.8)a [From eqns. (1.13) and (1.14)] ... (1.17)


15.3a + 300 = 400 – (40.8)a56.1a = 100

a =1.56

100 = 1.78 m/s2

Acceleration on body A is 1.78 m/s2

Acceleration of body B is2a =

278.1 = 0.89 m/s2

From eqn. (1.16), we get,T2 = 15.3 × 1.78 + 300

= 237.2 NFrom eqn, (1.11), we get,

T1 = 2T2

T1 = 2(327.2) = 654.5 N


Example 1.4: Three blocks of masses m1, m2 and m3 are connected as shown in Fig. 1.6. All thesurfaces are frictionless and, the string and pulleys are of negligible mass. Show that the accelerationof m1 is,

a0 =

32

1

m1

m1

4m1

g

(Assume m2 < m3)

Solution:

Suppose the mass m1 moves to right withacceleration a0, as a result pulley B, moves down with thesame acceleration (i.e., a0).

Let us assume that m2 moves up with acceleration ‘a’.So, the upward acceleration of m2 with respect to B equalsthe downward acceleration of m3 with respect to B.

The acceleration of m2 with respect to ground is equal to a0 – a (downward) and theacceleration of m3 with respect to ground is equal to a0 + a (downward).

The forces acting on light pulley B [(as shown in Fig. 1.6(b)] are:(i) T1 tension vertically upwards.

(ii) 2T2 tension vertically downwards. T1 = 2T2 ... (1.18)

Fig. 1.6(d)

m2

T2

m2g

ao – a

Fig. 1.6

A

B

m3

m2

m1

m1T1

R

m1g

ao

Fig. 1.6(c)

T2T2

T1

Fig. 1.6(b)

Fig. 1.6(a)

A

m2

m3g

m2g

T2

T1

B

m3

T2

a0 – a a0 + a

a0

m1

T1

R

m1g

B

12 College Physics

The forces acting on mass m1 [as shown in Fig. 1.6(c) are:(i) T1 tension toward right by the horizontal string.

(ii) m1g vertically downwards.(iii) Normal reaction force by table, vertically upwards.(iv) Force m1a0

T1 = m1a0 ... (1.19)

For the motion of m2, acceleration a0 – a is in the downward direction.

The forces acting on mass m2 [as shown in Fig. 1.6(d) are;(i) m2g vertically downwards.

(ii) T2 tension vertically upwards. T2 + m2 (a0 – a) = m2g

m2 (a0 – a) = m2g – T2

a0 – a =2

22

mT–gm ... (1.20)

From equations (1.18), (1.19) and (1.20), we get,

a0 – a =2

012

m2/)am(–gm

a0 – a =2

01

m2am

g ... (1.21)

For the motion of m3, the acceleration is (a0 + a) vertically downwards. The forces acting on m3

(as shown in Fig. 1.6(e) are:(i) m3g vertically downwards.

(ii) T2 tension vertically upwards. T2 + m3 (a0 + a) = m3g m3 (a0 + a) = m3g – T2

a0 + a =3

23

mT–gm ... (1.22)

From equations (1.18), (1.19), and (1.22), we get,

a0 + a =3

013

m2/)am(–gm

a0 + a = g –3

01

m2am ... (1.23)

Adding eqns. (1.21) and (1.23), we get,

2a0 = 2g –

32

01

m1

m1

2am

T2

m3g

ao + a

Fig. 1.6(e)


a0 = g –

32

01

m1

m1

4am

a0 +

32

01

m1

m1

4am = g

a0

32

1

m1

m1

4m

1 = g

a0 =

32

1

m1

m1

4m1

g

Example 1.5: A block ‘A’ ofmass ‘m’ is kept on frictionlesshorizontal table. The block istied to rigid support at C bylight string, passing around amovable massless smoothpulley B (as shown in Fig. (1.7).If the applied force ‘F’ to

pulley displaces the pully by distance x, then show that the block ‘A’ displaced by distance 2x. Alsofind acceleration of block and pulley.

Solution: Suppose pulley is displaced to B' i.e., displaced by a distance x as a result mass m displacesform A to A'. The length of the string is CB +BA. The length of the string is also. CB + BB' +B'B + BA'

Hence, CB + BA' + AA' = CB + BB' + B'B+ BA'

AA' = 2BB'.

The displacement of A is, therefore, twice the displacement of B in any given time interval.

That is, if the pulley is displaced by a distance x, the blockwill be displaced by 2x.

We get, the acceleration of block is twice the accelerationof Pully.

Consider the pulley as the system, the forces acting on thepulley [as shown in Fig. 1.7(b)] are:

(i) Tension T on left side of pulley.(ii) Force F towards right, applied by the experimenter.

F = 2T ... (1.24)

Fig. 1.7(b)

T

F

T

Fig. 1.7

C

A m

B

F

Fig. 1.7(a)

A A’

B B’C

14 College Physics

Now, consider the block as system.

The forces acting on mass ‘m’ [as shown in Fig. 1.7(c)] are:(i) Tension T.

(ii) Force ma, due to acceleration a. ma = T ... (1.25)

a =mT ... (1.26)

a =m2F [From eqns. (1.24) and (1.26)]

The acceleration of block is,m2F

The acceleration of the pulley is2a =

4mF

Example 1.6: Two bodies of masses m1 and m2 areconnected by a light string going over a smooth lightpulley at the end of an incline. The mass m1 lies on theincline and m2 hangs vertically, as shown in Fig. 1.8.The system is at rest. Show that,

(i) θ = sin-1 (m2/m1)where θ is the angle of incline.

(ii) The force exerted by the incline on the body

of mass ‘m1’ is , R = m1g 212 )m/m(–1

Solution:Consider a body of mass m1, as the system. The forces

acting on this system is shown in Fig. 1.8(a).(i) m1g acting vertically downwards.

(ii) Normal reaction force R.(iii) Tension in the string T.

T = m1g sin θ ... (1.27)Consider the body of mass m2 as the system. The

forces acting on it [as shown in Fig. 1.8 (a)] are:(i) m2g vertically downwards.

(ii) Tension T in string vertically upwards.T = m2g ... (1.28)

From Fig. 1.8(a), we get,

R = m1g cos θ ... (1.29)

m

a

Fig. 1.7(c)

T

m1

m2

θ

Fig. 1.8

Fig. 1.8(a)

m2

θm1g

RT

m2g

m1g cos θ


R = m1g θsin–1 2 (sin2θ + cos2θ = 1) ... (1.30)


m1g sin θ = m2g

sin θ =1

2

mm ... (1.31)

θ = sin–1

1

2

mm ... (1.32)


R = m1g 212 m/m–1 ... (1.33)

Example 1.7: A body of mass m = 2 kg is suspended by two strings making angles α and β with thehorizontal. Find the tensions in the strings when α = β = 45.Solution:

Fig. 1.9 Fig. 1.9(a)Consider a body of mass m as the system. The forces acting on the system [as shown in Fig.

1.9(a)] are:(i) mg downwards.

(ii) T1 tension along the first string.(iii) T2 tension along the second string.Tensions are resolved along the horizontal and the vertical, as shown in Fig. 1.9(a). At equilibrium condition of body, we get,

T1 cos α = T2 cos β ... (1.34)

T1 sin α + T2 sin β = mg ... (1.35)


T1 sin α + T1β cosαcos sin β = mg

β

T2T1

β

m

β

T2T1

β

mg

T2 cos β T1 cos

16 College Physics

T1

βsin

β cosαcosαsin = mg

T1 =βsin

β cosαcosαsin

mg

T1 =βsinαcosαcosαsin

βcosmg

T1 = βαsinβ cosmg

=

4545sin45osc8.92 = 13.86 N ... (1.36)


T2 = βαsinαcosmg

=

4545sin

45cos8.92 =13.68 N ... (1.37)

Example 1.8: A light rope tied at one end to a wooden clamp on the ground and the other end passesover a tree branch and hangs. The rope makes an angle of 45 with the ground. The wooden clamp can

come out of the ground if an upward force greater than460 N is applied to it. Find the maximum accelerationin the upward direction with which the man, weighing60 kg, can climp safely. Neglect friction at the treebranch.

Solution:

Let ‘T’ be the tension in the rope. The upward force

on the clamp is, T sin 45= .2

T

2145sin

The maximum tension that will not detach theclamp from the ground is given by,

2T = 460

T = 650.54 N ... (1.38)

If the acceleration of man in upward direction is ‘a’,the equation of motion of the man is [From Fig. 1.10(a)].

T = ma + mg

ma = T – mg

a =m

mg–T ... (1.39)

Fig. 1.10

Fig. 1.10(a)

T sin 45

45

T

T

m

mg

a


a =60

9860–54.650

a = 1.04 m/s2

Example 1.9: Find T1 and T2 (Fig. 1.11), if the blocks are accelerated(a) upward at 6.2 m/s2 and(b) downward 6.8 m/s2.

Solution:

(a) Consider two masses connected by a string as a system, shown in Fig. 1.11(a).The total mass of a system is m = 200 + 800 = 1000g = 1 kg.

T1 = mg + ma T1 = m (g + a) T1 = 1 (9.8 + 6.2) T1 = 16 N.

Consider a system with mass m1 = 800g = 0.8 kg, as shownin Fig. 1.11(b).

T2 = m1g + m1a T2 = m1 (g + a) T2 = 0.8 (9.8 + 6.2) T2 = 12.8 N.

(b) Consider two masses connected by a string as a system,shown in 1.11(c).

T1 + ma = mg T1 = m(g – a) T1 = 1(9.8 – 6.8) T1 = 3 N.

Consider a system with mass m1 = 800g = 0.8 kg, asshown in Fig. 1.11(d).

T2 + m1a = m1g T2 = m1(g – a) T2 = 0.8 (9.8 – 6.8) T2 = 2.4 N.

Example 1.10: In Fig. 1.12, assume that there is negligiblefriction between the blocks and table. Compute the tension inthe string connected to m2 and acceleration of m2, if m1 =300g, m2 = 100g and F = 0.20 N.

200g

800g

T1

T2

Fig. 1.11

Fig. 1.11(a)

200g

800g

T1

a 800g

T2

m1g

a

mg

Fig. 1.11(b)

Fig. 1.11(c)

200g

800g

T1

a 800g

T2

m1g

a

Fig. 1.11(d)

Fig. 1.12

m2

m1F

18 College Physics

Solution:

Consider a system of pulley P get acted by tension 2T1 and T2.

T2 = 2T1 ... (1.40)

Let ‘a’ be the acceleration of m2. The acceleration of m1

is a/2.

Consider a system of a mass m2 with forces acted upon, asshown in Fig. 1.12(b).

T1 = m2a ... (1.41)

Consider a system of a mass m1, with forces acted upon, as shown in Fig, 1.12(c).

F = T2 +2

am1 ... (1.42)

From, eqns. (1.40), (1.41) and (1.42), we get,

F = 2m2a +2

am1

2F = (4m2 + m1) a

a =)mm4(

F2

12

a =)3.01.04(

20.02

a = 0.57 m/s2

Tension in the string connected to mass m2, T2 = m2a T2 = 0.1 × 0.57 = 0.057 N.

Example 1.11: The force on a small body of mass 10g is F = (20 i + 10 j ) N. If it starts from rest,what would be its position at time 2 seconds.

Solution:

F = Fx i + Fy j

Fx = 20 N and Fy = 10 N

ax =mFx =

01.020 ( m = 10g =

100010 = 0.01 kg)

= 2000 m/s2

Since the acceleration is constant and initial velocity of a body, ux = 0 in x-direction.

x = uxt +21 ax t2

T1

T1

P

T2

Fig. 1.12(a)

m2

T1

a

Fig. 1.12(b)

Fig. 1.12(c)

T2

m1

F

a/2


x = 0 +21 × 2000 × (2)2 = 4000 m

Similarly, ay =mFy =

01.010 =1000 m/s2

and y = uyt +21 ayt2

= 0 +21 × 1000 × (2)2 = 2000 m

Hence, the position of the particle at time 2 seconds is,r = x i + y j

r = 4000 i + 2000 j

(B) Newton’s Laws of Motion and Friction

If bodies slip over each other, the force of friction isgiven by,

f = μRwhere, R is the normal reaction force and μ is the coefficientof friction.Example 1.12: The coefficient of static friction between theblock of mass 4 kg and the table (shown in Fig. 1.13) isμs = 0.4. What should be the maximum value of ‘m’ so thatthe blocks do not move? The string and the pulley are lightand smooth.

Solution:

Consider a system of mass ‘m’ at equilibrium.The forces acting on mass ‘m’ [as shown in Fig.1.13(a)] are:

(i) Tension T in string in upwards.

(ii) mg in downwards.

‘T’ Tension and ‘mg’ forces balances each other.

T = mg ... (1.43)

Now, consider the mass ‘M’ of 4 kg, as a system.The forces acting on this block are:

(i) Tension ‘T’ towards right by the string.(ii) ‘f’ friction force towards left by the table.

(iii) ‘Mg’ force vertically downwards.

Fig. 1.13

M 4 kg

m

Fig. 1.13(a)

4 kg

mg

m

R

T

TMg

f M

20 College Physics

(iv) R normal reaction force vertically upwards.


R = Mg ... (1.44)

and we have f = μsR ... (1.45)


f = μsMg ... (1.46)


f = T ... (1.47)


T = μsMg ... (1.48)


mg = μsMg

m = μsM

m = 0.4 × 4 = 1.6 kg.

Example 1.13: The coefficient of static friction between a block of mass m = 1 kg and an incline isμs = 0.4.

(a) What can be the maximum angle θ of the incline with thehorizontal so that the block does not slip on the plane?

(b) If the incline makes an angle2θ with the horizontal, find the

frictional force on the block.

Solution:(a) Forces acting on the block are,

(i) Weight mg downwards.(ii) Normal reaction force R by the incline.

(iii) The friction force f = μsR, by the incline.

Let θ be the maximum angle to prevent slipping.

R = mg cos θ (Perpendicular components) ... (1.49)

and f = mg sin θ (Parallel components) ... (1.50)

μsR = mg sin θ ... (1.51)

μs =R

θsinmg =θcosmgθsinmg

μs = tan θ θ = tan–1(μs)

θ

R

F

θ

mg

Fig. 1.14

m


θ = tan–1 (0.4) θ = 21 48'

(b) If the angle of incline is reduced to2θ , the equilibrium is not limiting, and hence the force of

static friction f is less than μs R.

R = mg cos

2θ

and f = mg sin

2θ = 1 × 9.8 × sin

248'21 = 1.85 N

Thus, the force of friction is 1.85 N.

Example 1.14: A 12 kg block rests on a horizontal surface,having coefficient of kinetic friction 0.44. This block isconnected by a string, which passes over a pulley and otherend connected to a 6 kg mass, as shown inFig. 1.15.

(a) What is the acceleration ‘a’?

(b) What is the tension ‘T’ in the string?

Solution:Consider the system of mass m1, the forces acting on m1 are:

(i) Weight m1g vertically downwards.(ii) Frictional force ‘f’ on left.

(iii) Tension in string ‘T’, on right.(iv) Normal reaction force ‘R’ due to horizontal

surface.(v) Force m1a.

R = m1g ... (1.52)

T = f + m1a

T = μkR +m1a ... (1.53)

T = μkm1g + m1a ... (1.54)

T = m1 (μkg + a) ... (1.55)

Consider a system of mass m2, the forces acting on m2 are:(i) Weight ‘m2g’ vertically downwards.

(ii) Tension ‘T’ in string vertically upwards.(iii) Force m2a

Fig. 1.15

m1

m2

Fig. 1.15(a)

Rm1

Ta

T

m2

m1g

f = µkR

m2g

a

22 College Physics

T + m2a = m2g ... (1.56) T = m2g – m2a ... (1.57) T = (g – a) m2 ... (1.58)

From eqns. (1.54) and (1.57), we get,μkm1g + m1a = m2g – m2a

m1a + m2a = m2g – μkm1g

a = )mm(

gmμ– m

21

1k2

... (1.59)

a = )612(

8.912 0.44– 6

a = 0.39 m/s2

From eqn. (1.58), we get,T = (g – a) m2

T = (9.8 – 0.39) 6T = 56.46 N.

Example 1.15: A bigger block 'C', of mass M is kept on smoothhorizontal surface. Two blocks ‘A’ and ‘B’, having same mass mare connected by massless inextensible string, which passes overfrictionless pulley, while block ‘A’ is on block ‘C’ and block ‘B’is hanging, as shown in Fig. 1.16.

Show that the minimum force Fmin = (M + 2m) )μ1(

μ–1

g and

maximum force Fmax = (M + 2m) )μ–1(

μ1 g that can be applied in order to keep the smaller blocks at

rest with respect to bigger mass. Where μ is the coefficient of friction between the blocks.

Solution:

If there is no force applied to block C, the block A will slip on Ctowards right and the block B will move downward. Suppose the minimumforce needed to prevent slipping is F. Consider blocks, A + B + C masses asthe system, and only horizontal external force on the system is F. Hence, theacceleration of the system is,

a =m2M

F

... (1.60)

Consider the block A as the system. The forces on block A are,(i) Tension ‘T’ by the string towards right.

(ii) Friction ‘f1’ by the block C towards left.(iii) Weight mg downward.

Fig. 1.16

mM

C

B

A

m

Fig. 1.16(a)

R1

a

T

mg

f1

m

A


(iv) Normal reaction force R1 vertically upwards.(v) ma force.

R1 = mg ... (1.61)

and f1 = μR1 = μmg ... (1.62)

f1 + ma = T [From Fig. 1.16(a)]

T = μmg + ma ... (1.63)

Now take the block B as the system. The forces acting on block B are,(i) Tension ‘T’ upward.

(ii) Weight ‘mg’ vertically downwards.(iii) Normal force ‘R2’ towards right.(iv) Friction ‘f2’ vertically upwards.(v) ‘ma’ force due to horizontal acceleration ‘a’.

R2 = ma ... (1.64)

f 2 = μR2 = μma ... (1.65)

From vertical equilibrium

T + f2 = mg [from Fig. 1.16(b)]

T = mg – μma ... (1.66)

From eqns (1.63) and (1.66), we get,

μmg + ma = mg – μma

ma + μma = mg – μmg

m (1 + μ)a = m (1 – μ) g

a = gμ1μ–1

... (1.67)

Force, F is assumed to be minimum, this is due to minimum acceleration for the system.

amin gμ1μ–1

... (1.68)


Fmin = (M + 2m) amin = (M +2m) )μ1(

μ–1

g ... (1.69)

When force, F is gradually increased, at critical stage, the block A slides to the left and block B ispulled in the upward direction. Then the frictional force on A will be towards right and the frictionalforce on B will be down, i.e., the direction of friction are reversed.

Fig. 1.16(b)

R2

a

T

mg

f2m

B

24 College Physics

Fig. 1.16(d)

m

mg

T

f4

Fig. 1.16(c)

m

mg

T

a

R3

f3

A B

From Fig. 1.16(c), we get,

f3 = – μmg ... (1.70)

T = f3 + ma

T = – μmg + ma ... (1.71)

From Fig. (1.16)(d) we get,

f4 = μma

T = f4 + mg ... (1.72)

T = μma + mg ... (1.73)

– μmg + ma = μma + mg [From eqns. (1.71) and (1.73)]

ma – μma = mg + μmg

m (1– μ)a = m (1+μ)g

a =μ–1μ1 g

amax =μ–1μ1 g ... (1.74)

Fmax. = (M + 2m) amax

Fmax = (M + 2m) μ–1

μ1 g ... (1.75)

Example 1.16: In the given Fig. 1.17, show that the maximum value ofmM =

θcosμ– θsinμ , so that the

system remains at rest. Friction coefficient at both the contacts is μ. What happens, when tan θ < μ?


Fig. 1.17

Solution:

Consider the system with mass ‘m’, the forces acting on ‘m’[as shown in Fig, 1.17(a)] are:

(i) Weight ‘mg’.(ii) Normal reaction force ‘R1’.

(iii) Tension ‘T’ in string.(iv) Frictional force ‘f1’.

R1 = mg ... (1.76)

f1 = μR1 = μmg ... (1.77)

T = f1 = μmg ... (1.78)

Now, consider the system with mass ‘M’, at equilibrium, the forces acting on M [as shown in Fig.1.17(b)] are:

(i) Weight ‘Mg’.(ii) Normal reaction ‘R2’

(iii) Tension T in string.(iv) Friction force ‘f2’.

R2 = Mg cos θ ... (1.79)

f2 = μR2 = μMg cos θ ... (1.80)

T + f2 = Mg sin θ

T = Mg sin θ – f2

T = Mg sin θ – μMg cos θ ... (1.81)


Mg sin θ – μMg cos θ = μmg

M(sin θ – μ cos θ)g = μmg

θcosμ–θsin

μmM

... (1.82)

If tan θ < μ, (sin θ – μ cos θ) < 0 and the system will slide for any value of M/m.

Fig. 1.17(a)

Fig. 1.17(b)

Mg

T

θ

θ

m

mg

T

R1

f1

26 College Physics

Example 1.17: A block of mass 3 kg is kept on a rough horizontal surface. If a horizontal force is less15 N is applied to it, the block does not slide. A horizontal force of 15 N is applied to the block andgently pushed to start the motion, it slides the first 10 m in 5 seconds. Calculate the coefficient ofstatic and kinetic friction between block and the surface.

Solution:

When F = 15 N is applied to the block, the block remains in limiting equilibrium.

The forces acting on the block are shown in Fig. 1.18.

R g ... (1.83)

The force of friction = fs = μsR ... (1.84)

Where μs = coefficient of static friction

fs = μsMg ... (1.85)

From Fig. 1.18, F = fs

μs =MgF ... (1.86)

μs =)8.9)(3(

15 = 0.51.

When the block is gently pushed to start the motion, kinetic friction acts between the block andthe surface. Now, block covers 10 m distance in 5 second. Therefore acceleration ‘a’ is given by,

S =21 at2

(S = ut +21 at2 and initial velocity u = 0)

10 =21 × a × (5)2

a =2520 = 0.8 m/s2

Now, the frictional force is, fk = μkR = μkMg ... (1.87)


fk + Ma = F

μkMg + Ma = F

μkMg = F – Ma

μk =Mg

Ma–F

Fig. 1.18

Mg

F

R

fs = sRM

Fig. 1.18 (a)

Mg

F

R

fk

a


μk =)8.9)(3(

)8.0)(3(–15

μk = 0.42.

Example 1.18: In Fig. 1.19, the two boxes have identical masses, 80 kg and both the massesexperiences a sliding friction force with μ = 0.10. Find the acceleration of the boxes and the tension inthe string.

Fig. 1.19

Solution:

A

R1

mg

a

T

T

a

m

fA

Bm

fB

Fig. 1.19(a)Consider a system with box A, the forces acting on box A are,(i) ‘mg’ weight vertically downward.

(ii) Normal reaction force ‘R1’.(iii) Friction force ‘fA’.(iv) Tension ‘T’ in the string.(v) Force ‘ma’.

We have, R1 = mg

and fA = μR1 = μmg ... (1.88)

from Fig. 1.19(a), we get,

T = fA + ma

T = μmg + ma ... (1.89)

T = m (μg + a) ... (1.90)

28 College Physics

Consider a system with box B, the forces acting on box B are shown in Fig. 1.19(a)(i) ‘mg’ weight vertically downward.

(ii) Normal reaction force ‘R2’.(iii) Friction force ‘fB’.(iv) Tension ‘T’ in the string.(v) Force ‘ma’

We have, R2 = mg cos 30

and fB = μR2 = μmg cos 30


T + fB + ma = mg sin 30 ... (1.91)

T = mg sin 30 – μmg cos 30 – ma ... (1.92)


μmg + ma = mg sin 30 – μmg cos 30 – ma

2ma = mg sin 30 – μmg cos 30 – μmg

a =2m

μ)–30μcos–(sin30mg

a =2

μ)–30μcos–(sin30g

a =2

)10.0–8660.010.0–5.0(8.9

a = 1.53 m/s2


T = 80 (0.10 × 9.8 + 1.53)

T = 201.25N.

Example 1.19: Two blocks of mass M and m are kept on a table,as shown in Fig. 1.20. The coefficient of static friction betweenthe two blocks is μ and the table is smooth. What maximumhorizontal force F can be applied to the block of mass M so thatthe blocks move together?

Solution:

Let F be the maximum horizontal force applied so that boththe blocks move together towards right. The force of friction is the only force on the upper block ofmass ‘m’, towards right, due to the friction by the lower block of mass ‘M’. Therefore, the force offriction on M by m should be towards left by Newton’s third law.

Fig. 1.20

FM

m


Consider the motion of ‘m’. The forces on ‘m’ are shown is Fig. 1.20(a).(i) Weight mg vertically downwards.

(ii) Normal reaction force R1 vertically upwards.(iii) f = μR1 (friction) towards right by the block M.

R1 = mg ... (1.93)

In the horizontal direction, let the acceleration be ‘a’, then

f = ma

μR1 = ma ... (1.94)

μmg = ma

a = μg ... (1.95)

Consider the motion of M. The forces acting on M are shown in Fig. 1.20(b)(i) Weight ‘Mg’ vertically downward.

(ii) Normal reaction force ‘R2’ vertically upwards.(iii) ‘R1’ downward by ‘m’(iv) f = μR1 = μmg (friction force) towards left by m.(v) Applied force F.

From Fig. 1.20(b), we get,

F = f + Ma

F = μmg + Ma

F = μmg + Mμg (a = μg)

F = μg (m + M)

Example 1.20: Three blocks with masses 3 kg, 6 kg, 10 kg are connected as shown in Fig. 1.21. Thecoefficient of friction between the table and 10 kg block is 0.2. Pulleys are assumed to be frictionless.Find.

(a) Acceleration of the system.(b) Tension in the string on left and on right side of the mass 10 kg.

C

6 kg

10 kg

B

A

3 kg

Fig. 1.21

Fig. 1.20(b)

Ff = µR1

R1 R2

M

Mg

a

mf = µR1

mg

R1

Fig. 1.20(a)

30 College Physics

Solution:

Let the tension in the string on the left of the mass 10 kg be T1 and on right be T2.

Consider the system of a block B having mass 10 kg is accelerating on right with acceleration ‘a’.The forces acting on mass 10 kg are shown in Fig. 1.21(a).

C

T2 a

mCg

T2T1

f R a

B

mBgB

A

mAg

a

T1

Fig. 1.21(a)

The forces are,

(i) ‘T1’ tension in string towards left.

(ii) ‘T2’ tension in string towards right.

(iii) Normal reaction ‘R’ vertically upwards.

(iv) Weight ‘mBg’ vertically downwards.

(v) Friction force ‘f’.

(vi) Force, ‘mBa’.

R = mBg ... (1.96)

f = μKR = μKmBg ... (1.97)

T2 = T1 + mBa + f ... (1.98)

T2 – T1 – f = mBa ... (1.99)

Consider the system of a block C, moves downwards, having mass 6 kg. The forces acting onmass 6 kg are shown in Fig. 1.21(a).

The forces are,(i) ‘T2’ tension in string vertically upwards.

(ii) Weight ‘mCg’ vertically downwards.(iii) Force, ‘mCa’.

T2 + mCa = mCg ... (1.100)

mCg – T2 = mCa ... (1.101)

Consider the system of a block A, moves upwards, having mass 3 kg. The forces acting on mass3 kg are shown in Fig. 1.21(a).


The forces are,(i) ‘T1’ tension in string vertically upwards.

(ii) Weight ‘mAg’ vertically downwards.

(iii) Force, ‘mAa

T1 = mAa + mAg ... (1.102)

T1 – mAg = mAa ... (1.103)

Adding eqns. (1.99), (1.101) and (1.103), we get,

mCg – mAg – f = (mA + mB + mC)a

mCg – mAg – μKmBg = (mA + mB + mC) a

a =)mmm(

g )μm–m–m[

CBA

KBAC

a =)6103(

8.9)]2.0(10–3–6[

a = 0.51 m/s2.


T1 = (a + g) mA

T1 = (0.51 + 9.8)3

T1 = 30.9. N

From eqn. (1.101)

T2 = (g – a) mC

T2 = (9.8 – 0.51)6

T2 = 55.74 N.

Example 1.21: Two blocks of mass m and M = 2 kg areconnected by a light string passed over the light andfrictionless pulley, kept on the inclined planes, as shown inFig. 1.22. At each of the surfaces, the coefficient of static andkinetic friction are 0.30 and 0.25 respectively. Find theminimum and maximum value of ‘m’ for which the systemremain at rest. Also find the acceleration of either block ifminimum value of m is considered and gently pushed to startthe motion.

Fig. 1.22

32 College Physics

Solution:

(a) Consider ‘m’ has its minimum value so that the M = 2 kg block has tendency to slip down.The forces on block, with mass M = 2 kg as a system, is shown in Fig. 1.22(a). As the block is inequilibrium, the resultant force should be zero.

R1 = Mg cos 45

R1 =2

Mg ... (1.104)

From Fig. 1.22(a), we get,T + f1 = Mg sin 45

T + f1 =2

Mg

T =2

Mg – f1 ... (1.105)

As it is a case of limiting equilibrium,

f1 = μs R1 = μs2

Mg ... (1.106)


T =2

Mg – μs2

Mg

T =2

Mg (1– μs) ... (1.107)

Now consider the block with mass m, as the system. The forces acting on this block are shown inFig. 1.22(b).

R2 = mg cos 45 =2

mg ... (1.108)

T= f2 + mg sin 45°

T = f2 +2

mg ... (1.109)

As it is the case of limiting equilibrium

f2 = μR2 = μs2

mg ... (1.110)

From eqns,(1.109) and (1.110), we get,

T =2

mgsμ +2

mg

T =2

mg (1 + μs) ... (1.111)

Fig. 1.22(a)

Fig. 1.22(b)



2mg (1 + μs) =

2Mg (1 – μs)

m = M)μ1()μ–1(

s

s

.. (1.112)

m =)30.01()30.0–1(

× 2 = 1 kg.

When maximum possible value of ‘m’ is supplied, the directions of friction get reverse, as aresult ‘m’ slips down and M = 2 kg block slips up. Thus, the maximum value of ‘m’ can be obtainedfrom eqn. (1.112) by putting μs = – 0.30. Therefore, for maximum ‘m’, we get,

m =30.0–130.01 × 2

= 3.7 kg.

(b) If m = 1 kg and the system is gently pushed, kinetic friction will operate.

mg45°

R1

45°Mg

Fig. 1.22(c)

f1 = μKR1 = μKMg cos 45 =2

MgμK ... (1.113)

and f2 = μKR2 = μKmg cos 45 =2

mgμK ... (1.114)

Let ‘a’ is the acceleration. The forces acting on masses ‘m’ and ‘M’ are shown Fig. 1.22(c).

For the system with mass m = 1 kg.

T = f2 + mg sin 45 + ma

T =2

mgμK +2

mg + ma ... (1.115)

college - himpub.comnewton’s laws are valid only in an inertial frame of reference. any reference...

Documents