Laws of Motion1

8/21/2019 Laws of Motion1

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Newton’s Laws of Motion 1genius PHYSICS by Pradeep Kshetrapal

4.1 Point Mass.

(! "n ob#e$t $an be $onsidered as a point ob#e$t if during %otion in a gi&en ti%e' it $o&ersdistan$e %u$h greater than its own sie)

(*! +b#e$t with ero di%ension $onsidered as a point %ass)

(,! Point %ass is a %athe%ati$al $on$ept to si%plify the proble%s)

4.2 Inertia.

(! Inherent property of all the bodies by &irtue of whi$h they $annot $hange their state of

rest or unifor% %otion along a straight line by their own is $alled inertia)

(*! Inertia is not a physi$al -uantity' it is only a property of the body whi$h depends on %assof the body)

(,! Inertia has no units and no di%ensions

(.! /wo bodies of e-ual %ass' one in %otion and another is at rest' possess sa%e inertiabe$ause it is a fa$tor of %ass only and does not depend upon the &elo$ity)

4.3 Linear Momentum.

(! Linear %o%entu% of a body is the -uantity of %otion $ontained in the body)

(*! It is %easured in ter%s of the for$e re-uired to stop the body in unit ti%e)(,! It is %easured as the produ$t of the %ass of the body and its &elo$ity i.e.' Mo%entu% 0

%ass 1 &elo$ity)

If a body of %ass m is %o&ing with &elo$ity v then its linear %o%entu% p is gi&en byv m p =

(.! It is a &e$tor -uantity and it’s dire$tion is the sa%e as the dire$tion of &elo$ity of thebody)

(2! 3nits 4 kg5m/sec 6S)I)7' g-cm/sec 6C)8)S)7

(9! :i%ension 4 76 −MLT

(;! If two ob#e$ts of di


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2 Newton’s Laws of Motiongenius PHYSICS by Pradeep Kshetrapal

4.4 Newton’s First Law." body $ontinue to be in its state of rest or of unifor% %otion along a straight line' unless it

is a$ted upon by so%e e@ternal for$e to $hange the state)

(! If no net for$e a$ts on a body' then the &elo$ity of the body $annot $hange i.e. the body

$annot a$$elerate)

(*! Newton’s Arst law deAnes inertia and is rightly $alled the law of inertia) Inertia are of

three types 4

Inertia of rest' Inertia of %otion' Inertia of dire$tion

(,!Inertia of rest :

It is the inability of a body to $hange by itself' its state of rest) /his

%eans a body at rest re%ains at rest and $annot start %o&ing by its own)

Example 4 (i! " person who is standing freely in bus' thrown ba$Bward' when bus startssuddenly)

hen a bus suddenly starts' the for$e responsible for bringing bus in %otion is alsotrans%itted to lower part of body' so this part of the body $o%es in %otion along with the bus)hile the upper half of body (say abo&e the waist! re$ei&es no for$e to o&er$o%e inertia of restand so it stays in its original position) /hus there is a relati&e displa$e%ent between the twoparts of the body and it appears as if the upper part of the body has been thrown ba$Bward)

Note 4 If the %otion of the bus is slow' the inertia of %otion will be trans%itted tothe body of the person unifor%ly and so the entire body of the person will $o%e in%otion with the bus and the person will not e@perien$e any #erB)

(ii! hen a horse starts suddenly' the rider tends to fall ba$Bward on a$$ount of inertia of rest of upper part of the body as e@plained abo&e)

(iii! " bullet Ared on a window pane %aBes a $lean hole through it while a stone breaBs thewhole window be$ause the bullet has a speed %u$hgreater than the stone) So its ti%e of $onta$t with glassis s%all) So in $ase of bullet the %otion is trans%ittedonly to a s%all portion of the glass in that s%all ti%e)Hen$e a $lear hole is $reated in the glass window' whilein $ase of ball' the ti%e and the area of $onta$t is large)

:uring this ti%e the %otion is trans%itted to the entirewindow' thus $reating the $ra$Bs in the entire window)

(i&! In the arrange%ent shown in the Agure 4

(a! If the string B is pulled with a sudden #erB then it will e@perien$e tension while due toinertia of rest of %ass M this for$e will not be trans%itted to the string A and sothe string B will breaB)

(b! If the string B is pulled steadily the for$e applied to it will betrans%itted fro% string B to A through the %ass M and as tension in A will begreater than in B by Mg (weight of %ass M! the string A will breaB)

(&! If we pla$e a $oin on s%ooth pie$e of $ard board $o&ering a glass and

striBe the $ard board pie$e suddenly with a Anger) /he $ardboard slips awayand the $oin falls into the glass due to inertia of rest)

M

A

B

Cra$Bs by theball

Hole by thebullet


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(&i! /he dust parti$les in a durree falls o< when it is beaten with a sti$B) /his is be$ause thebeating sets the durree in %otion whereas the dust parti$les tend to re%ain at rest and hen$eseparate)

(.! Inertia of motion : It is the inability of a body to $hange itself its state of unifor%

%otion i.e., a body in unifor% %otion $an neither a$$elerate nor retard by its own)Example 4 (i! hen a bus or train stops suddenly' a passenger sitting inside tends to fall

forward) /his is be$ause the lower part of his body $o%es to rest with the bus or train but theupper part tends to $ontinue its %otion due to inertia of %otion)

(ii! " person #u%ping out of a %o&ing train %ay fall forward)

(iii! "n athlete runs a $ertain distan$e before taBing a long #u%p) /his is be$ause &elo$itya$-uired by running is added to &elo$ity of the athlete at the ti%e of #u%p) Hen$e he $an #u%po&er a longer distan$e)

(2! Inertia of direction : It is the inability of a body to $hange by itself dire$tion of %otion)

Example 4 (i! hen a stone tied to one end of a string is whirled and the string breaBs

suddenly' the stone Dies o< along the tangent to the $ir$le) /his is be$ause the pull in the stringwas for$ing the stone to %o&e in a $ir$le) "s soon as the string breaBs' the pull &anishes) /hestone in a bid to %o&e along the straight line Dies o< tangentially)

(ii! /he rotating wheel of any &ehi$le throw out %ud' if any' tangentially' due to dire$tionalinertia)

(iii! hen a $ar goes round a $ur&e suddenly' the person sitting inside is thrown outwards)

Sample problem based on Newton’s frst law

P roblem 1. hen a bus suddenly taBes a turn' the passengers are thrown outwards be$ause of[AFMC 1! CPM" 2###$ 2##1%

(a! Inertia of %otion (b! "$$eleration of %otion

($!Speed of %otion (d! Eoth (b! and ($!

Solution 4 (a!

P roblem 2. " person sitting in an open $ar %o&ing at $onstant &elo$ity throws a ball &erti$ally up intoair) /he ball fall

[&AMC&" 'Med.( 1)%

(a! +utside the $ar (b! In the $ar ahead of the person

($!In the $ar to the side of the person (d! F@a$tly in the hand whi$h threw it up

Solution 4 (d! Ee$ause the horiontal $o%ponent of &elo$ity are sa%e for both $ar and ball so they $o&ere-ual horiontal distan$es in gi&en ti%e inter&al)

4.) Newton’s *econd Law.

(! /he rate of $hange of linear %o%entu% of a body is dire$tly proportional to the e@ternalfor$e applied on the body and this $hange taBes pla$e always in the dire$tion of the appliedfor$e)

(*! If a body of %ass m' %o&es with &elo$ity v then its linear %o%entu% $an be gi&en by

v m p = and if for$e→

F is applied on a body' then

dt

pd F

dt

pdF

=⇒∝

ordt

pd F = ( 0 in C)8)S) and S)I) units!


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or amdt

v dmv m

dt

dF

=== !( ("s ==

dt

v d a a$$eleration produ$ed in the body!

amF =

>or$e 0 %ass × a$$eleration

Sample problem based on Newton’s second law

P roblem 3. " train is %o&ing with &elo$ity *G msec) on this' dust is falling at the rate of 2G kgmin) /hee@tra for$e re-uired to %o&e this train with $onstant &elo$ity will be

[+P&" 1%

(a! 9)99 ! (b! GGG ! ($! 99)9 ! (d!

*GG !

Solution 4 (a! >or$edt

dmv F = !99)9

9G

2G*G =×=

P roblem 4. " for$e of G Newton a$ts on a body of %ass *G kg for G se$onds) Change in its %o%entu%is [MP P&" 2##2%

(a! 2 kg ms (b! GG kg ms ($! *GG kg ms (d! GGG kg ms

Solution 4 (b! Change in %o%entu% se$HGGGGti%efor$e mkg=×=×=

P roblem ). " &ehi$le of GG kg is %o&ing with a &elo$ity of 2 msec) /o stop it inG

sec' the re-uired

for$e in opposite dire$tion is[MP P&" 1)%

(a! 2GGG ! (b! 2GG ! ($! 2G ! (d!GGG !

Solution 4 (a! kgm GG= 'H2 smu = G=v t 0 G) sec

t

uv m

dt

mdv Fo"ce

!( −==

)G

!2G(GG −=

!F 2GGG−= 4., Force.

(! >or$e is an e@ternal e


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Newton’s Laws of Motion )genius PHYSICS by Pradeep Kshetrapal

(*! :i%ension 4 >or$e 0 %ass × a$$eleration

76776676 ** −− == MLT LT MF

(,! 3nits 4 "bsolute units 4 (i! !e#ton (S)I)! (ii! $%ne (C)8)S!

8ra&itational units 4 (i! Kilogra%5for$e (M)K)S)! (ii! 8ra%5for$e (C)8)S!

!e#ton 4 +ne Newton is that for$e whi$h produ$es an a$$eleration of *H sm in a

body of %ass ilog"am) ∴ !e#ton *H smkg=

$%ne 4 +ne dyne is that for$e whi$h produ$es an a$$eleration of *H scm in a

body of %ass g"am) ∴ $%ne *se$H cmgm=

elation between absolute units of for$e !e#ton

2

G= $%ne ilog"am-&o"ce 4 It i s that for$e whi$h produ$es an a$$eleration of *H=)? sm in a

body of %ass kg) ∴ kg-& 0 ?)= !e#ton

'"am-&o"ce 4 It is that for$e whi$h produ$es an a$$eleration of *H?=G scm in a body of

%ass gm) ∴ gm-& 0 ?=G $%ne

elation between gra&itational units of for$e 4 kg-& 0 ;G gm-&

(.! amF = for%ula is &alid only if for$e is $hanging the state of rest or %otion and the %assof the body is $onstant and Anite)

(2! If m is not $onstant dt

dm

v dt

v d

mv mdt

d

F

+== !((9! If for$e and a$$eleration ha&e three $o%ponent along x ' % and ( a@is' then

Eody re%ains at rest) Here for$e is trying to $hange the state of

rest)

Eody starts %o&ing) Here for$e $hanges the state of rest)

In a s%all inter&al of ti%e' for$e in$reases the %agnitude of

speed and dire$tion of %otion re%ains sa%e)

In a s%all inter&al of ti%e' for$e de$reases the %agnitude of

speed and dire$tion of %otion re%ains sa%e)

In unifor% $ir$ular %otion only dire$tion of &elo$ity $hanges'

speed re%ains $onstant) >or$e is always perpendi$ular to

&elo$ity)

In non5unifor% $ir$ular %otion' ellipti$al' paraboli$ or hyperboli$

%otion for$e a$ts at an angle to the dire$tion of %otion) In all

these %otions) Eoth %agnitude and dire$tion of &elo$ity

$hanges)

F

u 0 G v 0 G

F u

v J u

F

u 0 G v G

F

v uu ≠ G

v F

v

F

v

F 0 mg


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, Newton’s Laws of Motiongenius PHYSICS by Pradeep Kshetrapal

k F )F iF F ( % x LLL ++= and k a )aiaa ( % x

LLL ++=

>ro% abo&e it is $lear that ( ( % % x x maF maF maF === ''

(;! No for$e is re-uired to %o&e a body unifor%ly along a straight line)

maF = G=∴ F ("s G=a !(=! hen for$e is written without dire$tion then positi&e for$e %eans repulsi&e while

negati&e for$e %eans attra$ti&e)

Example 4 *ositive &o"ce >or$e between two si%ilar $harges

!egative &o"ce >or$e between two opposite $harges

(?! +ut of so %any natural for$es' for distan$e 2G− met"e' nu$lear for$e is strongest while

gra&itational for$e weaBest) nalgra&itationeti$ele$tro%agnu$lear F F F >>

(G! atio of ele$tri$ for$e and gra&itational for$e between two ele$tron .,GH =ge F F

ge F F >>∴

(! Constant for$e 4 If the dire$tion and %agnitude of a for$e is $onstant) It is said to be a$onstant for$e)

(*! ariable or dependent for$e 4

(i! Time dependent &o"ce 4 In $ase of i%pulse or %otion of a $harged parti$le in analternating ele$tri$ Aeld for$e is ti%e dependent)

(ii! *osition dependent &o"ce 4 8ra&itational for$e between two bodies*

*

"

m'm

or >or$e between two $harged parti$les *G

*

. "

++

πε = )

(iii! elocit% dependent &o"ce 4 is$ous for$e !9( "v πη

>or$e on $harged parti$le in a %agneti$ Aeld !sin( θ +vB

(,! Central for$e 4 If a position dependent for$e is always dire$ted towards or away fro% aA@ed point it is said to be $entral otherwise non5$entral)

Example 4 Motion of earth around the sun) Motion of ele$tron in an ato%) S$attering of α 5parti$les fro% a nu$leus)

(.! Conser&ati&e or non $onser&ati&e for$e 4 If under the a$tion of a for$e the worB done ina round trip is ero or the worB is path independent' the for$e is said to be $onser&ati&eotherwise non $onser&ati&e)

Example 4 Conser&ati&e for$e 4 8ra&itational for$e' ele$tri$ for$e' elasti$ for$e)

Non $onser&ati&e for$e 4 >ri$tional for$e' &is$ous for$e)

(2! Co%%on for$es in %e$hani$s 4

α -parti$le

F

Nu$leus

F Fle$tro

n

Nu$leus

OF Sun

Farth


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Newton’s Laws of Motion -genius PHYSICS by Pradeep Kshetrapal

(i! eigt 4 eight of an ob#e$t is the for$e with whi$h earth attra$ts it) It is also $alled thefor$e of gra&ity or the gra&itational for$e)

(ii! Reaction o" !o"mal &o"ce 0 hen a body is pla$ed on a rigid surfa$e' the bodye@perien$es a for$e whi$h is perpendi$ular to the surfa$es in $onta$t) /hen for$e is $alled

Nor%al for$e’ or ea$tion’)

(iii! Tension 4 /he for$e e@erted by the end of taut string' rope or $hain against pulling(applied! for$e is $alled the tension) /he dire$tion of tension is so as to pull the body)

(i&! Sp"ing &o"ce 0 F&ery spring resists any atte%pt to $hange its length) /his resisti&e for$e

in$reases with $hange in length) Spring for$e is gi&en by x F −= Q where x is the $hange inlength and is the spring $onstant (unit !m!)

4.- &ui/i0rium of Concurrent Force.

(! If all the for$es worBing on a body are a$ting on the sa%e point' then they are said to be$on$urrent)

(*! " body' under the a$tion of $on$urrent for$es' is said to be in e-uilibriu%' when there is

no $hange in the state of rest or of unifor% %otion along a straight line)(,! /he ne$essary $ondition for the e-uilibriu% of a body under the a$tion of $on$urrent

for$es is that the &e$tor su% of all the for$es a$ting on the body %ust be ero)

(.! Mathe%ati$ally for e-uilibriu% ∑ = GnetF or ∑ = G x F Q ∑ = G % F Q ' ∑ = G ( F (2! /hree $on$urrent for$es will be in e-uilibriu%' if they $an be represented $o%pletely by

three sides of a triangle taBen in order)

T 0 F

A

B1*F

F ,F

x

F 0 x

mg

R

mg $osθ

R

θ mgθ


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Newton’s Laws of Motiongenius PHYSICS by Pradeep Kshetrapal

(9! La%i’s /heore% 4 >or $on$urrent for$esγ β α sinsinsin,* F F F ==

Sample problem based on orce and equilibrium

P roblem ,. /hree for$es starts a$ting si%ultaneously on a parti$le %o&ing with &elo$ity .v /hese for$esare represented in %agnitude and dire$tion by the three sides of a triangle AB1 (as shown!) /he parti$le will now %o&e with &elo$ity

[AI&&& 2##3%

(a! v re%aining un$hanged

(b! Less than v

($! 8reater than v

(d! v in the dire$tion of the largest for$e B1

Solution 4 (a! 8i&en three for$es are in e-uilibriu% i.e. net for$e will be ero) It %eans the parti$le will%o&e with sa%e &elo$ity)

P roblem -. /wo for$es are su$h that the su% of their %agnitudes is = ! and their resultant isperpendi$ular to the s%aller for$e and %agnitude of resultant is *) /hen the %agnitudes of the for$es are [AI&&& 2##2%

(a! * !' 9 ! (b! , !, 2! ($! G !' = ! (d!9 !' * !

Solution 4 (b! Let two for$es are F and !( ** F F F < )

"$$ording to proble%4 =* =+ F F R))(i!

"ngle between F and resultant (R! is ?G

∴ ∞=+= θ θ

$os

sin?Gtan

*

*

F F

F

G$os* =+⇒ θ F F

⇒

*

$os

F

F −=θ R))(ii!

and θ $os* ***

*

* F F F F R ++=

θ $os*.. ***

* F F F F ++= R))(iii!

by sol&ing (i!' (ii! and (iii! we get !F 2 = and !F ,* =

P roblem . /he resultant of two for$es' one double the other in %agnitude' is perpendi$ular to thes%aller of the two for$es) /he angle between the two for$es is

[C&" '&n.Med.( 2##2%

(a! o9G (b! o*G ($! o2G (d! o?G

Solution4 (b! Let for$es are F and *F and angle between the% is θ and resultant %aBes an angle α withthe for$e F.

α β

γ *F F

,F

C

A B

F

F *

R 0*

θ


9/65

Newton’s Laws of Motion genius PHYSICS by Pradeep Kshetrapal

?Gtan$os*

sin*tan =

+=

θ

θ α

F F

F ∞=

⇒ G$os* =+ θ F F ∴ *H$os −=θ or °= *Gθ

P roblem . " weightless ladder' *G &t long rests against a fri$tionless wall at an angle of 9Go with thehoriontal) " 2G pound %an is . &t fro% the top of the ladder) " horiontal for$e is neededto pre&ent it fro% slipping) Choose the $orre$t %agnitude fro% the following

[C5*& PM" 1%

(a! ;2 l2 (b! GG l2 ($! ;G l2 (d! 2G l2

Solution4 ($! Sin$e the syste% is in e-uilibriu% therefore ∑ = G x F and ∑ = G % F ∴ *RF = andR. =

Now by taBing the %o%ent of for$es about point B.

=+ !)(!)( E1. B1F !( A1R 6fro% the Agure E10 . $os 9G7

!9G$os*G(!9G$os.(!9Gsin*G)(

R. F =+

G*,G R. F =+ [ ]. R ="s

∴ l2.

F ;G,G

2G=

,G

==×==

P roblem 1#. " %ass M is suspended by a rope fro% a rigid support at * as shown in the Agure)"nother rope is tied at the end 3' and it is pulled horiontally with a for$e F ) If the rope *3

%aBes angle θ with the &erti$al then the tension in the string *3 is

(a! θ sinF

(b! θ sinHF

($! θ $osF

(d! θ $osHF

Solution4 (b! >ro% the Agure

>or horiontal e-uilibriu%

F T =θ sin

∴ θ sin

F T =

P roblem 11. " spring balan$e " shows a reading of * kg' when an alu%iniu% blo$B is suspended fro%it) "nother balan$e B shows a reading of 2 kg' when a beaBer full of li-uid is pla$ed in its

pan) /he two balan$es are arranged su$h that the Al blo$B is $o%pletely i%%ersed insidethe li-uid as shown in the Agure) /hen [II"67&& 1)%

(a! /he reading of the balan$e A will be %ore than * kg

(b! /he reading of the balan$e B will be less than 2 kg

($!/he reading of the balan$e A will be less than * kg) and that of B will be %ore than 2 kg

(d! /he reading of balan$e " will be * kg) and that of B will be 2kg)

Solution4 ($! :ue to buoyant for$e on the alu%iniu% blo$B the reading of spring balan$e A will be less than * Bg but it in$rease the reading of balan$e B)

P roblem 12. In the following diagra%' pulley * is %o&able and pulley ** is A@ed) /he &alue of angle θ will be

M

F

*

θ

3

v

A

m

B

9 &t

. &t

R

AF

E 1

B

all

R*

9Go

$

F

θ

T T

$osθ

mg

T sinθ


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1# Newton’s Laws of Motiongenius PHYSICS by Pradeep Kshetrapal

(a! 9Go

(b! ,Go

($! .2o

(d! 2o

Solution4 (b! >ree body diagra% of pulley * is shown in the Agure

>or horiontal e-uilibriu% θ θ $os$os * T T = ∴ * T T =

and . T T == *>or &erti$al e-uilibriu%

. T T =+ θ θ sinsin * ⇒ . . . =+ θ θ sinsin

∴ *

sin =θ or °= ,Gθ

P roblem 13. In the following Agure' the pulley is %assless and fri$tionless) /he relation between T '

*T and ,T will be

(a! ,* T T T ≠=

(b! ,* T T T =≠

($! ,* T T T ≠≠

(d! ,* T T T ==

Solution 4 (d! Sin$e through a single string whole syste% is atta$hed so *,* T T T . ===

P roblem 14. In the abo&e proble% (,!' the relation between .

and *.

will be

(a!θ $os*

*

. . = (b! θ $os* . ($! * . . = (d!

*

$os*

. .

θ =

Solution 4 (a! >or &erti$al e-uilibriu%

* $os$os . T T =+ θ θ [ ]**"s . T T ==

*$os* . . =θ

)$os*

*θ

. . =∴

P roblem 1). In the following Agure the %asses of the blo$Bs A and B are sa%e and ea$h e-ual to m) /he tensions in the strings 4A and AB are *T and T respe$ti&ely) /he syste% is in

e-uilibriu% with a $onstant horiontal for$e %g on B) /he T is

(a! mg

(b! mg*

($! mg,

(d! mg2

Solution 4 (b! >ro% the free body diagra% of blo$B B

mgT =$osθ RR(i!mgT −= sinθ R))(ii!

* *

*θ

*

**θ θ

T *

T

T ,

θ *

θ

A

mg

T

m

m

B

4

T *

T T *

T $osθ T

* $osθ

T sinθ

T * sinθ

θ θ

T T *

T $os

θ T * $os

θ θ θ

T

mg

T $os

θ

θ

B mgT

sin

θ


11/65


by s-uaring and adding ( ) ( ) **** *$ossin mgT =+ θ θ

∴ mgT * =

P roblem 1,. In the abo&e proble% (2!' the angle θ is

(a! ,Go (b! .2o ($! 9Go (d!

−*

tan

Solution 4 (b! >ro% the solution (2! by di&iding e-uation(ii! by e-uation (i!

mg

mg

T

T =

$os

sin

θ θ

∴ tan =1θ or °= .2θ P roblem 1-. In the abo&e proble% (2! the tension *T will be

(a! mg (b! mg* ($! mg, (d! mg2

Solution 4 (d! >ro% the free body diagra% of blo$B A

>or &erti$al e-uilibriu% ** $os$os θ θ T mgT +=

°+= .2$os*$os ** mgmgT θ

mgT *$os ** =θ R))(i!

>or horiontal e-uilibriu% ** sinsin θ θ T T = °= .2sin*mg

mgT =** sinθ R))(ii!

by s-uaring and adding (i! and (ii! e-uilibriu%

*** !(2mgT = or mgT 2* =

P roblem 1$ In the abo&e proble% (2! the angle *θ will be

(a! ,Go (b! .2o ($! 9Go (d!

−*

tan

Solution 4 (d! >ro% the solution (;! by di&iding e-uation(ii! by e-uation (i!

mg

mg

*$os

sin

*

* =θ θ

⇒ *

tan * =θ ∴

= −*

tan *θ

P roblem 1. " %an of %ass m stands on a $rate of %ass M) He pulls on a light rope passing o&er a

s%ooth light pulley) /he other end of the rope is atta$hed to the $rate) >or the syste% to bein e-uilibriu%' the for$e e@erted by the %en on the rope will be

(a! (M O m!g

(b! gmM !(*

+

($! Mg

(d! mg

Solution 4 (b! >ro% the free body diagra% of %an and $rate syste%4

>or &erti$al e-uilibriu%

gmMT !(* +=

*

!( gmMT +=∴

M

m

T *

mg

T * $os

θ *

θ *

A

T

T sin

θ

T * sin

θ *

θ

T $os

θ

(M Om!g

T T


12/65


P roblem 2#. /wo for$es' with e-ual %agnitude F ' a$t on a body and the %agnitude of the resultant for$e

is,

F ) /he angle between the two for$es is

(a!

−−

=

;$os (b!

−−

,

$os ($!

−,

*$os (d!

−?

=$os

Solution 4 (a! esultant of two &e$tors " and E' whi$h are worBing at an angle θ ' $an be gi&en by

θ $os*** ABB AR ++= 6"s F B A == and,

F R = 7

θ $os*,

****

F F F F

++=

θ $os**?

***

F F F

+= ⇒ θ $os*

?

; ** F F =−

⇒

−=

=

;$osθ or

−= −

=

;$os θ

P roblem 21. " $ri$Bet ball of %ass 2G gm is %o&ing with a &elo$ity of * ms and is hit by a bat so thatthe ball is turned ba$B with a &elo$ity of *G ms) /he for$e of blow a$ts for G)Gs on the ball) /he a&erage for$e e@erted by the bat on the ball is

(a! .=G ! (b! 9GG ! ($! 2GG ! (d! .GG !

Solution 4 (a! smv H* −= and smv H*G* += 6be$ause dire$tion is re&ersed7

kggmm 2)G2G == ' se$G)G=t

>or$e e@erted by the bat on the ballG)G

!7*(*G62)G76 * −−=−

=t

v v mF 0 .=G !e#ton

4. Newton’s "8ird Law. /o e&ery a$tion' there is always an e-ual (in %agnitude! and opposite (in dire$tion! rea$tion)

(! hen a body e@erts a for$e on any other body' the se$ond body also e@erts an e-ual andopposite for$e on the Arst)

(*! >or$es in nature always o$$urs in pairs) " single isolated for$e is not possible)

(,! "ny agent' applying a for$e also e@perien$es a for$e of e-ual %agnitude but in oppositedire$tion) /he for$e applied by the agent is $alled Action’ and the $ounter for$e e@perien$ed byit is $alled Reaction’)

(.! "$tion and rea$tion ne&er a$t on the sa%e body) If it were so the total for$e on a bodywould ha&e always been ero i.e. the body will always re%ain in e-uilibriu%)

(2! If ABF 0 for$e e@erted on body A by body B ("$tion! and BAF 0 for$e e@erted on body B bybody A (ea$tion!

/hen a$$ording to Newton’s third law of %otion BA AB F F −=

(9! F@a%ple 4 (i! " booB lying on a table e@erts a for$e on the table whi$h is e-ual to theweight of the booB) /his is the for$e of a$tion)

/he table supports the booB' by e@erting an e-ual for$e on the booB) /his is the for$e of rea$tion)

"s the syste% is at rest' net for$e on it is ero) /herefore for$e of a$tion and rea$tion %ust be e-ual and opposite)

(ii! Swi%%ing is possible due to third law of %otion)

(iii! hen a gun is Ared' the bullet %o&es forward (a$tion!) /he gun re$oils ba$Bward(rea$tion!

R

mg


13/65


(i&! ebounding of rubber ball taBes pla$e due to third law of %otion)

(&! hile walBing a person presses the ground in the ba$Bward dire$tion(a$tion! by his feet) /he ground pushes the person in forward dire$tion withan e-ual for$e (rea$tion!) /he $o%ponent of rea$tion in horiontal dire$tion

%aBes the person %o&e forward)(&i! It is diT$ult to walB on sand or i$e)

(&ii! :ri&ing a nail into a wooden blo$B without holding the blo$B isdiT$ult)

Sample problem based on Newton’s third law

P roblem 22. You are on a fri$tionless horiontal plane) How $an you get o< if no horiontal for$e ise@erted by pushing against the surfa$e

(a! Ey #u%ping (b! Ey splitting or sneeing

($! Ey rolling your body on the surfa$e (d! Ey running on the plane

Solution 4 (b! Ey doing so we $an get push in ba$Bward dire$tion in a$$ordan$e with !e#ton5s third law of %otion)

4. Frame of +eference.

(! " fra%e in whi$h an obser&er is situated and %aBes his obser&ations is Bnown as his>ra%e of referen$e’)

(*! /he referen$e fra%e is asso$iated with a $o5ordinate syste% and a $lo$B to %easure theposition and ti%e of e&ents happening in spa$e) e $an des$ribe all the physi$al -uantities liBeposition' &elo$ity' a$$eleration et$) of an ob#e$t in this $oordinate syste%)

(,! >ra%e of referen$e are of two types 4 (i! Inertial fra%e of referen$e (ii! Non5inertialfra%e of referen$e)

(i! Inertia/ frame of reference :

(a! " fra%e of referen$e whi$h is at rest or whi$h is %o&ing with a unifor% &elo$ity along astraight line is $alled an inertial fra%e of referen$e)

(b! In inertial fra%e of referen$e Newton’s laws of %otion holds good)

($! Inertial fra%e of referen$e are also $alled una$$elerated fra%e of referen$e or Newtonianor 8alilean fra%e of referen$e)

(d! Ideally no inertial fra%e e@ist in uni&erse) >or pra$ti$al purpose a fra%e of referen$e %aybe $onsidered as inertial if it’s a$$eleration is negligible with respe$t to the a$$eleration of theob#e$t to be obser&ed)

(e! /o %easure the a$$eleration of a falling apple' earth $an be $onsidered as an inertialfra%e)

(f! /o obser&e the %otion of planets' earth $an not be $onsidered as an inertial fra%e but forthis purpose the sun %ay be assu%ed to be an inertial fra%e)

Example 4 /he lift at rest' lift %o&ing (up or down! with $onstant &elo$ity' $ar %o&ing with$onstant &elo$ity on a straight road)

(ii! Non inertia/ frame of reference :

(a! "$$elerated fra%e of referen$es are $alled non5inertial fra%e of referen$e)

(b! Newton’s laws of %otion are not appli$able in non5inertial fra%e of referen$e)Example 4 Car %o&ing in unifor% $ir$ular %otion' lift whi$h is %o&ing upward or downward

with so%e a$$eleration' plane whi$h is taBing o


14/65


4.1# Im9u/se.

(! hen a large for$e worBs on a body for &ery s%all ti%e inter&al' it is $alled i%pulsi&efor$e)

"n i%pulsi&e for$e does not re%ain $onstant' but $hanges Arst fro% ero to %a@i%u% and

then fro% %a@i%u% to ero) In su$h $ase we %easure the total e o r $ e

/i%e

t

F


15/65

Newton’s Laws of Motion 1)genius PHYSICS by Pradeep Kshetrapal

(i&! "n athlete is ad&ised to $o%e to stop slowly after Anishing a fast ra$e) So that ti%e of stop in$reases and hen$e for$e e@perien$ed by hi% de$reases)

(&! China wares are wrapped in straw or paper before pa$Bing)

Sample problem based on Impulse

P roblem 23. " ball of %ass 2Gg %o&ing with an a$$eleration *H*G sm is hit by a for$e' whi$h a$ts on itfor G) sec) /he i%pulsi&e for$e is

[AFMC 1%

(a! G)2 !-s (b! G) !-s ($! G), !-s (d! )* !-s

Solution 4 ($! I%pulsi&e for$e ti%for$e×= t am ×= )G*G2)G ××= 0 G), !-s

P roblem 24. " for$e of 2G d%nes is a$ted on a body of %ass 2 g whi$h is at rest for an inter&al of ,seconds' then i%pulse is

(a! s! 5G2)G ,−× (b! s! 5G?=)G ,−× ($! s! 5G2) ,−× (d! s! 5G2)* ,−×Solution 4 ($! ti%for$epulseI% ×= ,G2G 2 ××= − s5G2) ,!−×=

P roblem 2). /he for$e5ti%e (F t ! $ur&e of a parti$le e@e$uting linear %otion is as shown in the Agure) /he %o%entu% a$-uired by the parti$le in ti%e inter&al fro% ero to = second will be

[CPM" 1%

(a! * !-s

(b! O . !-s

($! 9 !-s

(d! Uero

Solution 4 (d! Mo%entu% a$-uired by the parti$le is nu%eri$ally e-ual to the area en$losed between theF 5t $ur&e and ti%e "@is) >or the gi&en diagra% area in a upper half is positi&e and in lowerhalf is negati&e (and e-ual to the upper half!) So net area is ero) Hen$e the %o%entu%a$-uired by the parti$le will be ero)

4.11 Law of Conseration of Linear Momentum.

If no e@ternal for$e a$ts on a syste% ($alled isolated! of $onstant %ass' the total %o%entu%of the syste% re%ains $onstant with ti%e)

(! "$$ording to this law for a syste% of parti$lesdt

pd F =

In the absen$e of e@ternal for$e G=F then = p $onstant

i.e.' =+++= )))),* p p p p $onstant)

or =+++ →→→

)))),,** v mv mv m $onstant

/his e-uation shows that in absen$e of e@ternal for$e for a $losed syste% the linear%o%entu% of indi&idual parti$les %ay $hange but their su% re%ains un$hanged with ti%e)

(*! Law of $onser&ation of linear %o%entu% is independent of fra%e of referen$e thoughlinear %o%entu% depends on fra%e of referen$e)

(,! Conser&ation of linear %o%entu% is e-ui&alent to Newton’s third law of %otion)

>or a syste% of two parti$les in absen$e of e@ternal for$e by law of $onser&ation of linear%o%entu%)

=+ * p p $onstant) =+ ** v mv m $onstant)

* . 9 =

O *

*

> o r $ e

( ! !

/i%e (s!


16/65

1, Newton’s Laws of Motiongenius PHYSICS by Pradeep Kshetrapal

:i


17/65

Newton’s Laws of Motion 1-genius PHYSICS by Pradeep Kshetrapal

(a! /hrust on the ro$Bet 4 mgdt

dmuF −−=

Here negati&e sign indi$ates that dire$tion of thrust is opposite to the dire$tion of es$apinggases)

dt

dmuF −= (if e


18/65


Solution 4 (d! Mo%entu% of body for gi&en options are 4

(a! * seH*G*)GGG kgmmv =×== (b! * se$H.)GGGG. , kgmmv =××== −

($! * se$HG,)9G*)G** .9 kgmmE −− ×=××==

(d! * se$H=*)*GG*G*G* ,, kgmg/m =××××== −

So for option (d! %o%entu% is %a@i%u%)

P roblem 2. " ro$Bet with a lift5o< %ass .G2), × kg is blasted upwards with an initial a$$eleration of )HG *sm /hen the initial thrust of the blast is

[AI&&& 2##3%

(a! !2G;2) × (b! !2G2), × ($! !2GG); × (d! !2GG). ×

Solution 4 ($! Initial thrust on the ro$Bet !( agmF += !GG(G2), . +×= !2GG); ×=

P roblem 3#. In a ro$Bet of %ass GGG kg fuel is $onsu%ed at a rate of .G kgs) /he &elo$ity of the gasese#e$ted fro% the ro$Bet is smHG2 .× ) /he thrust on the ro$Bet is

[MP PM" 14%

(a! !,G*× (b! !.G2× ($! !9G*× (d! !?G*×

Solution 4 ($! /hrust on the ro$Betdt

udmF = !.G(G2 .×= !9G*×=

P roblem 31. If the for$e on a ro$Bet %o&ing with a &elo$ity of ,GG ms is *G !' then the rate of $o%bustion of the fuel is

(a!G); kgs (b! ). kgs ($! G)G; kgs (d! G); kgs

Solution 4 (a! >or$e on the ro$Betdt

udm= ∴ ate of $o%bustion of fuel )H;)G,GG

*Gskg

u

F

dt

dm===

P roblem 32. " ro$Bet has a %ass of GG kg) ?GV of this is fuel) It e#e$ts fuel &apours at the rate of

kgsec with a &elo$ity of 2GG msec relati&e to the ro$Bet) It is supposed that the ro$Bet isoutside the gra&itational Aeld) /he initial upthrust on the ro$Bet when it #ust starts %o&ingupwards is [NC&+" 1-%

(a! Uero (b! 2GG ! ($! GGG ! (d! *GGG !

Solution 4 (b! 3p thrust for$e

=dt

dmuF !2GG2GG =×=

4.12 Free 5od< =iaram.

In this diagra% the ob#e$t of interest is isolated fro% its surroundings and the intera$tionsbetween the ob#e$t and the surroundings are represented in ter%s of for$es)

Example 4

4.13 A99arent >ei8t of a 5od< in a Lift.

hen a body of %ass m is pla$ed on a weighing %a$hine whi$h ispla$ed in a lift' then a$tual weight of the body is mg) /his a$ts on a

weighing %a$hine whi$h o


19/65


Condition Fiure ?e/ocit< Acce/eration

+eaction Conc/usion

Lift is at rest v 0 G a 0 GR mg 0 G

∴ R 0 mg"pparent weight0 "$tual weight

Lift %o&ingupward ordownwardwith $onstant&elo$ity

v 0 $onstanta 0 G

R mg 0 G

∴ R 0 mg"pparent weight0 "$tual weight

Lifta$$eleratingupward at therate of Wa’

v 0 &ariablea J g

R mg 0 ma

∴R 0 m(g Oa!

"pparent weight "$tual weight

Lifta$$eleratingupward at therate of g’

v 0 &ariablea 0 g

R mg 0 mgR 0 *mg

"pparent weight0 * "$tualweight

Lifta$$elerating

downward atthe rate of a’

v 0 &ariablea J g

mg R 0 ma

∴ R 0 m(g a!

"pparent weight

J "$tual weight

Lifta$$eleratingdownward atthe rate of g’

v 0 &ariablea 0 g

mg R 0 mg R 0 G

"pparent weight0 Uero(weightlessness!

Lifta$$eleratingdownward atthe rate ofa(g!

v 0 &ariable a g mg R 0 ma R 0 mg ma R 0 ve

"pparent weightnegati&e %eansthe body will risefro% the Door ofthe lift and sti$B

SpringEalan$e

R

mg

LIF"

Spring

Ealan$e

R

mg

LIF"

a

SpringEalan$e

R

mg

LIF"

g

SpringEalan$e

R

mg

LIF"

a

SpringEalan$e

R

mg

LIF"

g

SpringEalan$e

R

mg

LIF"

a g

SpringEalan$e

R

mg

LIF"


20/65


to the $eiling ofthe lift)

Sample problems based on lit

P roblem 33. " %an weighs )=Gkg He stands on a weighing s$ale in a lift whi$h is %o&ing upwards with a

unifor% a$$eleration of )H2 *sm hat would be the reading on the s$ale) !HG( *smg = [C5*& 2##3%

(a! .GG ! (b! =GG ! ($! *GG ! (d! Uero

Solution 4 ($! eading of weighing s$ale !( agm += !2G(=G += !*GG=

P roblem 34. " body of %ass * kg is hung on a spring balan$e %ounted &erti$ally in a lift) If the liftdes$ends with an a$$eleration e-ual to the a$$eleration due to gra&ity g’' the reading on

the spring balan$e will be

(a! * kg (b! (. × g! kg ($! (* × g! kg (d! Uero

Solution 4 (d! !( agmR −= G!( =−= gg 6be$ause the lift is %o&ing downward with a 0 g7

P roblem 3). In the abo&e proble%' if the lift %o&es up with a $onstant &elo$ity of * msec' the reading onthe balan$e will

Ee (a! * kg (b! . kg ($! Uero (d! kg

Solution 4 (a! mgR = !e#tog*= or kg* 6be$ause the lift is %o&ing with the ero a$$eleration7

P roblem 3,. If the lift in proble%' %o&es up with an a$$eleration e-ual to the a$$eleration due togra&ity' the reading on the spring balan$e will be

(a! * kg (b! (* × g! kg ($! (. × g! kg (d! . kg

Solution 4 (d! !( agmR += !( ggm += 6be$ause the lift is %o&ing upward with a 0g7

mg*= !gR **×= !g.= or kg.

P roblem 3-. " %an is standing on a weighing %a$hine pla$ed in a lift' when stationary' his weight is

re$orded as .G kg) If the lift is a$$elerated upwards with an a$$eleration of 2smH* ' then the

weight re$orded in the %a$hine will be !HG( *smg = [MP PM" 14%

(a! ,* kg (b! .G kg ($! .* kg (d! .= kg

Solution 4 (d! !( agmR += !*G(.G += !.=G= or kg.=

P roblem 3. "n ele&ator weighing 9GGG kg is pulled upward by a $able with an a$$eleration of *2 −ms )

/aBing g to be *G −ms ' then the tension in the $able is[Mani9a/ M&& 1)%

(a! 9GGG ! (b! ?GGG ! ($! 9GGGG ! (d! ?GGGG !

Solution 4 (d! !( agmT += !2G(9GGG += !T GGG'?G=

P roblem 3. /he ratio of the weight of a %an in a stationary lift and when it is %o&ing downward withunifor% a$$eleration a’ is , 4 *) /he &alue of a’ is (g5 "$$eleration due to gra&ity on the

earth! [MP P&" 1-%

(a! g*

,(b! ,

g($! g,

*(d! g


21/65


Solution 4 (b!*

,

!(lift%o&ingdownwardin%anaofweight

liftstationaryin%anaofweight=

−=

agm

mg

∴*

,=

− ag

g ⇒ agg ,,* −= or

,

ga =

P roblem 4#. " 9G kg %an stands on a spring s$ale in the lift) "t so%e instant he Ands' s$ale reading has$hanged fro% 9G kg to 2G kg for a while and then $o%es ba$B to the original %arB) hatshould we $on$lude

(a! /he lift was in $onstant %otion upwards

(b! /he lift was in $onstant %otion downwards

($! /he lift while in $onstant %otion upwards' is stopped suddenly

(d! /he lift while in $onstant %otion downwards' is suddenly stopped

Solution 4 ($! >or retarding %otion of a lift !( agmR += for downward %otion

!( agmR −= for upward %otion

Sin$e the weight of the body de$rease for a while and then $o%es ba$B to original &alue it%eans the lift was %o&ing upward and stops suddenly)

Note 4 8enerally we use !( agmR += for upward %otion!( agmR −= for downward %otion

here a0 a$$eleration' but for the gi&en proble% a0 retardation

P roblem 41. " bird is sitting in a large $losed $age whi$h is pla$ed on a spring balan$e) It re$ords aweight pla$ed on a spring balan$e) It re$ords a weight of *2 !) /he bird (%ass 0 G)2kg! Dies

upward in the $age with an a$$eleration of *H* sm ) /he spring balan$e will now re$ord a

weight of [MP PM" 1%

(a! *. ! (b! *2 ! ($! *9 ! (d! *; ! Solution 4 (b! Sin$e the $age is $losed and we $an treat bird $age and air as a $losed (Isolated! syste%) In

this $ondition the for$e applied by the bird on the $age is an internal for$e due to thisreading of spring balan$e will not $hange)

P roblem 42. " bird is sitting in a wire $age hanging fro% the spring balan$e) Let the reading of the springbalan$e be . ) If the bird Dies about inside the $age' the reading of the spring balan$e is

*. ) hi$h of the following is true

(a! * . . = (b! * . . >

($! * . . < (d! Nothing deAnite $an be predi$ted

Solution 4 (b! In this proble% the $age is wire5$age the %o%entu% of the syste% will not be $onser&edand due to this the weight of the syste% will be lesser when the bird is Dying as $o%pared to

the weight of the sa%e syste% when bird is resting is * . . < )

4.14 Acce/eration of 5/oc; on @orionta/ *moot8 *urface.

(! >8en a 9u// is 8orionta/

R 0 mg

and F 0 ma

∴ a 0 F m

(*! >8en a 9u// is actin at an an/e ' ( to t8e 8orionta/ 'u9ward(

mg

R

m F a


22/65


R O F sin θ 0 mg

⇒ R 0 mg F sinθ

and F $osθ 8 ma

∴ m

F a θ $os=

(,! >8en a 9us8 is actin at an an/e ' ( to t8e 8orionta/ 'downward(

R 0 mg O F sinθ

and F $osθ 8 ma

m

F a

θ $os=

4.1) Acce/eration of 5/oc; on *moot8 Inc/ined P/ane.

(! >8en inc/ined 9/ane is at rest

Nor%al rea$tion R 0 mg $osθ

>or$e along a in$lined plane

F B mg sinθ

ma 8 mg sinθ

∴ a 8 g sinθ

(*! >8en a inc/ined 9/ane ien a 8orionta/ acce/eration b’

Sin$e the body lies in an a$$elerating fra%e' an inertial for$e ( m2! a$ts on it in the opposite

dire$tion)Nor%al rea$tion R 0 mg $osθ O m2 sinθ

and ma 0 mg sin θ 9 m2 $os θ

∴ a 0 g sinθ 2 $osθ

Note 4 /he $ondition for the body to be at rest relati&e to the in$lined plane 4 a 0 g sinθ 2 $osθ 0 G

∴ 2 0 g tanθ

4.1, Motion of 5/oc;s in Contact.

Condition Free 0od< diaram &uation Force and acce/eration

mg $osθ Om2 sinθ

R

θ mg

a

θ

θ

2

mg

R

m F

$osθ

F F

sinθ θ

F

sinθ

R

m F

$osθ

aF

F mg θ

mg $osθ

R

θ mg

a F


23/65


am& F =−* mm

F a

+=

am& *=*

*

mm

F m&

+=

am& =* mm

F a

+=

am& F *=−*

mm

F m&

+=

am& F =−,* mmm

F a

++=

am& & ** =−,*

,*

!(

mmm

F mm&

++

+

=

am& ,* = ,*,

* mmm

F m

& ++=

am& =,* mmm

F a

++=

am& & ** =−,*

mmm

F m&

++=

am& F ,* =−,*

**

!(

mmm

F mm&

++

+

=

Sample problems based on motion o blocks in contact

P roblem 43. /wo blo$Bs of %ass . kg and 9 kg are pla$ed in $onta$t with ea$h other on a fri$tionlesshoriontal surfa$e) If we apply a push of 2 ! on the hea&ier %ass' the for$e on the lighter

%ass will be

(a! 2 !

m

F &

ma

m*

&

m*a

m

&

ma

m*

&

m*a

F

m

&

ma

F

m*a

& m

*

& *

m

&

ma

m*a

& m

*

& *

m,

& *

m,a

F

m

m,F

B

m*

1

mm

*F

A

B

m

m*

F A

B

m

m,

F

A B

m*

1

.kg

9kg

2!


24/65


(b! . !

($! * !

(d! None of the abo&e

Solution 4 ($! Let kgmkgm .'9 * == and !F 2= (gi&en!

>or$e on the lighter %ass 0*

*

mm

F m

+×

.9

2.

+×

= !*=

P roblem 44. In the abo&e proble%' if a push of 2 ! is applied on the lighter %ass' the for$e e@erted bythe lighter %ass on the hea&ier %ass will be

(a! 2 ! (b! . ! ($! * ! (d! None of the abo&e

Solution 4 (d! >or$e on the hea&ier %ass*

mm

F m

+=

.9

29

+×

= !,=

P roblem 4). In the abo&e proble%' the a$$eleration of the lighter %ass will be

(a! *2)G −ms (b!*

.

2 −ms ($! *9

2 −ms (d! None of the abo&e

Solution 4 (a!systeof the%ass /otal

systetheonfor$eNeton"$$elerati = *H2)G

G

2sm==

P roblem 4,. /wo blo$Bs are in $onta$t on a fri$tionless table one has a %ass m and the other * m asshown in Agure) >or$e F is applied on %ass *m then syste% %o&es towards right) Now the

sa%e for$e F is applied on m) /he ratio of for$e of $onta$t between the two blo$Bs will be in

the two $ases respe$ti&ely)

(a! 4 (b! 4 * ($! 4 , (d! 4 .

Solution 4 (b! hen the for$e is applied on %ass *m $onta$t for$e,*

gg

mm

m& =

+=

hen the for$e is applied on %ass m $onta$t for$e ggmm

m&

,

*

*

** =+=

atio of $onta$t for$es *

* =& &

F m*m

F


25/65


4.1- Motion of 5/oc;s Connected 0< Mass Less *trin.

Condition Free 0od< diaram &uation "ension and acce/eration

amT =* mmF a+

=

amT F *=−*

mm

F mT

+=

amT F =−* mm

F a

+=

amT *=*

*

mm

F mT

+=

amT =,* mmm

F a

++=

amT T ** =−,*

mmmF mT ++

=

amT F ,* =−,*

**

!(

mmm

F mmT

++

+=

amT F =−,* mmm

F a

++=

amT T ** =−,*

,*

!(

mmm

F mmT

++

+

=

amT ,* =,*

,*

mmm

F mT

++

=

m

T

ma

m*

F

m*a

T

m

T

ma

F

m

T

ma

m,

F

m,a

T *

T *

m*a

T m

*

m

T

ma

F

T *

m*a

T m

*

m,

m,a

T *

m*

m*a

T

m

m*

F T

B

A

m

m*

F T

B

A

m

m,

F A B

m*

1

T

T *

m,

F A B

m*

1

T

T *

m


26/65


Sample problems based on motion o blocks connected by mass less string

P roblem 4-. " %onBey of %ass *G kg is holding a &erti$al rope) /he rope will not breaB when a %assof *2 kg is suspended fro% it but will breaB if the %ass e@$eeds *2 kg) hat is the %a@i%u%

a$$eleration with whi$h the %onBey $an $li%b up along the rope !HG( *smg =

[C5*& 2##3%(a! *HG sm (b! *H*2 sm ($! *H2)* sm (d! *H2 sm

Solution 4 ($! Ma@i%u% tension that string $an bear (T max ! 0 *2 × g ! 0 *2G !

/ension in rope when the %onBey $li%b up ( )agmT +=

>or li%iting $ondition %a@T T = ⇒ *2G!( =+ agm ⇒ ( ) *2GG*G =+ a ∴ *H2)* sma=

P roblem 4. /hree blo$Bs of %asses * kg' , kg and 2 kg are $onne$ted to ea$h other with light string andare then pla$ed on a fri$tionless surfa$e as shown in the Agure) /he syste% is pulled by a

for$e 'G!F = then tension =T [Drissa 7&& 2##2%

(a! ! (b! 2 ! ($! = ! (d! G !

Solution 4 ($! Ey $o%paring the abo&e proble% with general e@pression)( )

,*

,*

mmm

F mmT

+++

= ( )2,*

G2,

+++

=

!e#to==

P roblem 4. /wo blo$Bs are $onne$ted by a string as shown in the diagra%) /he upper blo$B is hung by

another string) " for$e F applied on the upper string produ$es an a$$eleration of *H* sm in

the upward dire$tion in both the blo$Bs) If T and T ′ be the tensions in the two parts of thestring' then [AME '&n.( 2###%

(a! !T =);G= and !T *).;=′

(b! !T =)2== and !T *).;=′

($! !T =);G= and !T =)2==′

(d! !T =);G= and G=′T Solution 4 (a! >ro% >)E):) of %ass . kg gT a .W. −= )R)(i!

>ro% >)E):) of %ass * kg gT T a *W* −−= )R)(ii!

>or total syste% upward for$e

( ) ( )agT F ++== .* ( ) !*=9 += 0 ;G)= N

by substituting the &alue of / in e-uation (i! and (ii!

and sol&ing we get !T *).;W=

P roblem )#. /hree %asses of 2 kg) G kg and 2 kg are suspended &erti$ally as shown in the Ag) If thestring atta$hed to the support breaBs and the syste% falls freely' what will be the tension in

*kg2kg,kg

G! T

T *

T F

*kg

.kg

T ′

.kg

T ′

.g

.a *kg

T ′

*g

*a

T ′

2k g

Gk g

2kg


27/65


the string between G kg and 2 kg %asses) /aBe *G −= msg ) It is assu%ed that the string

re%ains tight during the %otion

(a! ,GG ! (b! *2G ! ($! 2G ! (d! Uero

Solution 4 (d! In the $ondition of free fall' tension be$o%es ero)

P roblem )1. " sphere is a$$elerated upwards with the help of a $ord whose breaBing strength is A&eti%es its weight) /he %a@i%u% a$$eleration with whi$h the sphere $an %o&e up without

$ord breaBing is

(a! .g (b! ,g ($! *g (d! g

Solution 4 (a! /ension in the $ord 0 m(g O a! and breaBing strength 0 2 mg

>or $riti$al $ondition ( ) mgagm 2=+ ⇒ ga .= /his is the %a@i%u% a$$eleration with whi$h the sphere $an %o&e up with $ord breaBing)

4.1 Motion of Connected 5/oc; Der a Pu//e


28/65


, *T T =,*

,* 7!6(

mmm

gmmma

++−+=

Condition Free 0od< diaram &uation "ension andacce/eration

hen pulley ha&e aAnite %ass M andradius R then tension intwo seg%ents of stringare di


29/65


T gmam −= ** gmm

mmT

*

* !sin(

++

= θ

amgmT sin =− α g

mm

mma

*

* !sinsin(

+−

= α β

T gmam −= β sin** gmm

mmT

*

* !sin(sin

++

= β α

Condition Free 0od< diaram &uation "ension andacce/eration

amT gm sin =−θ *

sin

mm

gma

+=

θ

amT *= gmm

mmT

*

*

.

*

+=

amT =*

*.

*

mm

gmaa +==

*

**

. mm

gma

+=

m*aT

m*g

sinβ

m*

β

m

A

*

m*

aT

θ B

m

ma

T

θ

mg sinθ

m*

*T

m*g

m*

(a

*!

m

T

ma

m*

T

m*a

m

ma

T

α

mg

sinα a

m

m

*

T T a

α β A

B

*

B

T T

a*

a

m

A

m*


30/65


"s**

* !(

dt

x d

*

* !(

*

dt

x d=

*

*a

a =∴

=a a$$eleration of blo$B A

=*a a$$eleration of blo$B B

*

*

.*mmgmmT

+=T gmam ** ** −=

T gmam −= gMmm

mma

76

!(

*

*

++−

=

gmT am *** −= gMmm

MmmT

76

!*(

*

* ++

+=

MaT T =− * gMmm

MmmT

76

!*(

*

*** ++

+=

Sample problems based on motion o blocks over pulley

P roblem )2. " light string passing o&er a s%ooth light pulley $onne$ts two blo$Bs of %asses m and *m(&erti$ally!) If the a$$eleration of the syste% is g/ = then the ratio of the %asses is

[AI&&& 2##2%

(a! = 4 (b! ? 4 ; ($! . 4 , (d! 2 4 ,

Solution 4 (b! gmm

mma

+−

=*

* 0

=

gQ by sol&ing ;H?

* =m

m

P roblem )3. " blo$B A of %ass ; kg is pla$ed on a fri$tionless table) " thread tied to it passes o&er a

fri$tionless pulley and $arries a body B of %ass , kg at the other end) /he a$$eleration of the syste% is (gi&en g 0 G !*−ms

[era/a '&n.( 2###%

(a! *GG −ms

(b! *, −ms

($! *G −ms

(d! *,G −ms

Solution 4 (b! gmm

ma

+

=*

*

*H,G,;

,sm=

+=

m

T

mg

ma

m*

T *

m*g

m*a

MT *

T

Ma

M

m

A

T

m*

B

T

aa

T *

T *

1

B

A


31/65


P roblem )4. /wo %asses m and *m are atta$hed to a string whi$h passes o&er a fri$tionless s%ooth

pulley) hen 'G kgm = '9* kgm = the a$$eleration of %asses is[Drissa 7&& 2##2%

(a! *G *H sm

(b! *H2 sm

($! *)2 *H sm

(d! *HG sm

Solution 4 ($! gmm

mma

*

*

+−

= G9G

9G

+−

= *H2)* sm=

P roblem )). /wo weights . and *. are suspended fro% the ends of a light string passing o&er as%ooth A@ed pulley) If the pulley is pulled up with an a$$eleration g' the tension in the stringwill be

(a!*

*.. . . .

+(b!

*

**. . . . +

($!*

*

. . . . +

(d!!(* *

*

. . . . +

Solution 4 (a! hen the syste% is at rest tension in string( ) gmm

mmT

*

**

+=

If the syste% %o&es upward with a$$eleration g then ( )ggmm

mmT +

+=

*

** gmm

mm

*

*.

+= or

*

*.

##

##T

+=

P roblem ),. /wo %asses M and M* are atta$hed to the ends of a string whi$h passes o&er a pulley

atta$hed to the top of an in$lined plane) /he angle of in$lination of the plane in θ ) /aBe g 0G ms*)

If M 0 G kg' M* 0 2 kg' θ 0 ,Go' what is the a$$eleration of %ass M*

(a! *G −ms (b! *2 −ms

($!*

,

* −ms (d! Uero

Solution 4 (d! "$$eleration gmm

mm

*

* sin

+−

= θ

gG2

,Gsin)G2

+−

= gG2

22

+−= 0 G

P roblem )-. In the abo&e proble%' what is the tension in the string

(a! GG ! (b! 2G ! ($! *2 ! (d! Uero

Solution 4 (b!( )

gmm

mmT

*

* sin

++= θ ( )

2G

G),Gsin2G

++×= !2G=

P roblem ). In the abo&e proble%' gi&en that M* 0 *M and M* %o&es &erti$ally downwards witha$$eleration a) If the position of the %asses are re&ersed the a$$eleration of M* down thein$lined plane will be

(a! * a (b! a ($! a* (d! None of the abo&e

Solution 4 (d! If * *mm = ' then *m %o&es &erti$ally downward with a$$eleration

gmm

mm

a *

* sin

+

−

=

θ

gmm

mm

*

,Gsin*

+

−

= 0 g/ *

If the position of %asses are re&ersed then *m %o&es downward with a$$eleration

m

m*

G kg

9 kg

M

M*

θ


32/65


gmm

mma

*

* sinW+

−=

θ G)

*

,Gsin*

=+

−= g

mm

mm6"s m* 0 *m7

i.e. the *m will not %o&e)

P roblem ). In the abo&e proble%' gi&en that M* 0 *M and the tension in the string is T ) If thepositions of the %asses are re&ersed' the tension in the string will be

(a! . T (b! T ($! T (d! T *

Solution 4 ($! /ension in the string( )

gmm

mmT

*

* sin

++

= θ

If the position of the %asses are re&ersed then there will be no e


33/65


>ro% (i! and (ii! ,H,H

*

==g

g

a

a

P roblem ,4. In the ad#oining Agure m 0 .m*) /he pulleys are s%ooth and light) "t ti%e t 0 G' the

syste% is at rest) If the syste% is released and if the a$$eleration of %ass m is a' then thea$$eleration of m* will be

(a! g

(b! a

($!*

a

(d! *a

Solution 4 (d! Sin$e the %ass *m tra&els double distan$e in $o%parison to %ass m therefore

its a$$eleration will be double i.e) *a

P roblem ,). In the abo&e proble% (9.!' the &alue of a will be

(a! g (b!*

g($!

.

g(d!

=

g

Solution 4 ($! Ey drawing the >E: of m and *m

T gmam * −= R))(i!

( ) gmT am ** * −= R))(ii!

by sol&ing these e-uation .Hga =

P roblem ,,. In the abo&e proble%' the tension T in the string will be

(a! m*g (b!

*

*gm ($! gm*,

*(d! gm*

*

,

Solution 4 (d! >ro% the solution (92! by sol&ing e-uation

gmT **

,=

P roblem ,-. In the abo&e proble%' the ti%e taBen by m in $o%ing to rest position will be

(a! G)* s (b! G). s ($! G)9 s (d! G)= s

Solution 4 (b! /i%e taBen by %ass *m to $o&er the distan$e *G cm

2)*

*)G*

.H

*)G** ×=×==ga

/t se$.)G=

P roblem ,. In the abo&e proble%' the distan$e $o&ered by m* in G). s will be(a! .G cm (b! *G cm ($! G cm (d! =G cm

Solution 4 (a! Sin$e the *m %ass $o&er double distan$e therefore S 0 * × *G 0 .G cm

P roblem ,. In the abo&e proble%' the &elo$ity a$-uired by m* in G). se$ond will be

(a! GG cms (b! *GG cms ($! ,GG cms (d! .GG cms

Solution 4 (b! elo$ity a$-uired by %ass *m in G). se$

>ro% at uv += 6"s *H2*

G*H smga === 7

.)G2G ×+=v )H*GGH* seccmsm ==

P roblem -#. In the abo&e proble%' the additional distan$e tra&ersed by m* in $o%ing to rest positionwill be

m

m*

*Gcm

m*

T

m*g

m*(*a

!

m

*T

mg

ma


34/65


(a! *G cm (b! .G cm ($! 9G cm (d! =G cm

Solution 4 (a! hen *m %ass a$-uired &elo$ity *GG $%se$ it will %o&e upward till its &elo$ity be$o%es

ero)

( )cm

g

u; *G

GG*

*GG

*

**

=

×

==

P roblem -1. /he a$$eleration of blo$B B in the Agure will be

(a!!.( *

*

mm

gm

+

(b!!.(

*

*

*

mm

gm

+

($!!.(

*

*

mm

gm

+

(d!!(

**

mmgm

+

Solution 4 (a! hen the blo$B *m %o&es downward with a$$eleration a' the a$$eleration of %ass m will

be a* be$ause it $o&ers double distan$e in the sa%e ti%e in $o%parison to *m )

Let T is the tension in the string)

Ey drawing the free body diagra% of A and B

amT *= RR))(i!

amT gm ** * =− RR))(ii!

by sol&ing (i! and (ii!

( )*

*

. mm

gma

+

=

m

m

*

A

B

*T

m

*

m*g

m*am

T

m(*a!


35/65


4.1 Motion of Massie *trin.

Condition Free 0od< diaram &uation "ension and

acce/eration

=T for$e applied bythe string on the blo$B

amMF !( +=

MaT =mM

F a

+=

!(

mM

F MT

+=

F mM

mMT

!(*

!*(* +

+=

=*T /ension at %idpoint of the rope

am

MT

+=

*

*

m 0 Mass of string

T 0 /ension in string ata distan$e x fro% theend where the for$e isapplied

maF =

mF a H=

F L

x LT

−=

aL

x LmT

−=

M 0 Mass of unifor%rod

L 0 Length of rod

L

MxaT F =−

M

F F a *

−=

MaF F =− *

+

−=

L

x F

L

x F T *

Mass of seg%ent B1

x LM =

F L

x LT

−= F

L

x LT

−=

T

M

a

F m

M

T *

m*M

F m

aF

LT

x

m 6(L x !L7

aT

F

F *

L

x B A

LT

T

x

1

B

F

A

T

B

A

F

T

(M/L! x A B

a

F *

F

a

M


36/65


4.2# *9rin 5a/ance and P8

and the needle %ust in %iddle of the bea% i.e. a 0 2)E


37/65


>or rotational e-uilibriu% about point 4’

2. > a. = !(!( * +=+ RR(iii!

>ro% (i!' (ii! and (iii!

/rue weight *. . . =

(b! If the bea% of physi$al balan$e is not horiontal (when the pans are e%pty! and the

ar%s are e-ual

i.e. > = > and 2a =

In this physi$al balan$e if a body of weight is pla$ed in = Pan then to balan$e it)

e ha&e to put a weight in > Pan

>or e-uilibriu% . > . = +=+ R))(i!

Now if pans are $hanged then to balan$e the bodywe ha&e to put a weight *. in X Pan)

>or e-uilibriu% . > . = +=+ * R))(ii!

>ro% (i! and (ii!

/rue weight*

* . . . +

=

Sample problems 'iscellaneous(

P roblem -2. " body weighs = gm' when pla$ed in one pan and =gm' when pla$ed in the other pan of a false balan$e) If the bea% is horiontal (when both the pans are e%pty!' the true weight of the body is

(a! , gm (b! * gm ($! 2)2 gm (d! 2 gm

Solution 4 (b! >or gi&en $ondition true weight 0 *. . ==×= 0* gm)

P roblem -3. " plu%b line is suspended fro% a $eiling of a $ar %o&ing with horiontal a$$eleration of a) hat will be the angle of in$lination with &erti$al

[Drissa 7&& 2##3%

(a! !H(tan ga− (b! !H(tan ag− ($! !H($os ga− (d! !H($os ag−

Solution 4 (a! >ro% the Agure

g

a=θ tan

( )ga Htan −=θ

P roblem -4. " blo$B of %ass kg2 is %o&ing horiontally at a speed of )2 ms) " perpendi$ularfor$e of 2 ! a$ts on it for . sec) hat will be the distan$e of the blo$B fro% the point wherethe for$e started a$ting [P0 PM" 2##2%

(a! G m (b! = m ($! 9 m (d! * m

=

>

AB

4

a 2

a

gθ

a

θ


38/65


Solution 4 (a! In the gi&en proble% for$e is worBing in a dire$tion perpendi$ular to initial &elo$ity) So thebody will %o&e under the eor $riti$al $ondition m(g a! 0 *, mg ⇒ mgmamg

,

*=− ∴

,

ga =

So' this is the %ini%u% a$$eleration by whi$h a Are%an $an slides down on a rope)

T T

θ θ

*T sin θ


39/65


P roblem -. " $ar %o&ing at a speed of ,G kilo mete"s per hour’s is brought to a halt in = met"es byapplying braBes) If the sa%e $ar is %o&ing at 9G km) per hour' it $an be brought to a haltwith sa%e braBing power in

(a! = met"es (b! 9 met"es ($! *. met"es (d! ,* met"es

Solution 4 (d! >ro% asuv *** −= asu *G * −=

a

us

*

*

= ⇒ *us∝ (if a 0 $onstant!

.,G

9G**

*

* =

=

=

u

u

s

s ⇒ =.. * ×== ss ,*= met"es)

).1 Introduction.

If we slide or try to slide a body o&er a surfa$e the %otion is resisted by a bonding between

the body and the surfa$e) /his resistan$e is represented by a single for$e and is $alled fri$tion)

/he for$e of fri$tion is parallel to the surfa$e and opposite to the dire$tion of intended

%otion)

).2 "


40/65


(ii! :ire$tion of the for$e of li%iting fri$tion is always opposite to the dire$tion in whi$h one

body is at the &erge of %o&ing o&er the other

(iii! CoeT$ient of stati$ fri$tion 4 (a! s µ is $alled $oeT$ient of stati$ fri$tion and deAned as

the ratio of for$e of li%iting fri$tion and nor%al rea$tion R

F s = µ

(b! :i%ension 4 76 GGG T LM

($! 3nit 4 It has no unit)

(d! alue of s µ lies in between G and

(e! alue of µ depends on %aterial and nature of surfa$es in $onta$t that %eans whether

dry or wet Q rough or s%ooth polished or non5polished)

(f! alue of µ does not depend upon apparent area of $onta$t)

(,! inetic or d


41/65


).3 ra98 5etween A99/ied Force and Force of Friction.

(! Part 4A of the $ur&e represents stati$ fri$tion !( sF ) Its &alue in$reases linearly with the

applied for$e

(*! "t point A the stati$ fri$tion is %a@i%u%) /his

represent li%iting fri$tion !( lF )

(,! Eeyond A' the for$e of fri$tion is seen to de$rease

slightly) /he portion B1 of the $ur&e therefore represents

the Bineti$ fri$tion !( k F )

(.! "s the portion B1 of the $ur&e is parallel to x 5a@is

therefore Bineti$ fri$tion does not $hange with the applied

for$e' it re%ains $onstant' whate&er be the applied for$e)

).4 Friction is a Cause of Motion.

It is a general %is$on$eption that fri$tion always opposes the %otion) No doubt fri$tion

opposes the %otion of a %o&ing body but in %any $ases it is also the $ause of %otion) >or

e@a%ple 4

(! In %o&ing' a person or &ehi$le pushes the ground

ba$Bwards (a$tion! and the rough surfa$e of ground rea$ts and

e@erts a forward for$e due to fri$tion whi$h $auses the %otion)

If there had been no fri$tion there will be slipping and no %otion)

(*! In $y$ling' the rear wheel %o&es by the for$e $o%%uni$ated to it by pedalling while front

wheel %o&es by itself) So' when pedalling a bi$y$le' the for$e e@erted by rear wheel on ground

%aBes for$e of fri$tion a$t on it in the forward dire$tion (liBe walBing!) >ront wheel %o&ing by

itself e@perien$e for$e of fri$tion in ba$Bward dire$tion (liBe rolling of a ball!) 6Howe&er' if

pedalling is stopped both wheels %o&e by the%sel&es and so e@perien$e for$e of fri$tion in

ba$Bward dire$tion)7

(,! If a body is pla$ed in a &ehi$le whi$h is a$$elerating' the for$e of fri$tion is the $ause of

%otion of the body along with the &ehi$le ( i.e.' the body will re%ain at rest in the a$$elerating

&ehi$le until !)mgma s µ < If there had been no fri$tion between body and &ehi$le the body will not%o&e along with the &ehi$le)

hilepedalling

Pedalling isstoped

ma µ smg

a

"$tion

>ri$tion

θ

A

1B

F l

F k

> o r $ e

o f f r i $ t i o n

"pplied for$e4


42/65


>ro% these e@a%ples it is $lear that without fri$tion %otion $annot be started' stopped ortransferred fro% one body to the other)

Sample problems based on undamentals o riction

P roblem 1. If a ladder weighing *2G! is pla$ed against a s%ooth &erti$al wall ha&ing $oeT$ient of fri$tion between it and Door is G),' then what is the %a@i%u% for$e of fri$tion a&ailable at

the point of $onta$t between the ladder and the Door[AIIM* 2##2%

(a! ;2 ! (b! 2G ! ($! ,2 ! (d! *2 !

Solution 4 (a! Ma@i%u% for$e of fri$tion !RF sl ;2*2G,)G =×== µ

P roblem 2. +n the horiontal surfa$e of a tru$B ( µ 0 G)9!' a blo$B of %ass kg is pla$ed) If the tru$B isa$$elerating at the rate of 2m/sec* then fri$tional for$e on the blo$B will be

[C5*& PM" 2##1%

(a! 2 ! (b! 9 ! ($! 2)== ! (d! = !

Solution 4 (a! Li%iting fri$tion !mgR ss ==)2=)?9)G =××=== µ µ

hen tru$B a$$elerates in forward dire$tion at the rate of *H2 sm a pseudo for$e !(ma of

2! worBs on blo$B in ba$B ward dire$tion) Here the %agnitude of pseudo for$e is less than

li%iting fri$tion So' stati$ fri$tion worBs in between the blo$B and the surfa$e of the tru$B

and as we Bnow' stati$ fri$tion 0 "pplied for$e 0 2!)

P roblem 3. " blo$B of %ass * kg is Bept on the Door) /he $oeT$ient of stati$ fri$tion is G).) If a for$e F of

*)2 ! is applied on the blo$B as shown in the Agure' the fri$tional for$e between the blo$Band the Door will be [MP P&" 2###%

(a! *)2 !

(b! 2 !

($! ;)=. !

(d! G !

Solution 4 (a! "pplied for$e 0 *)2 ! and li%iting fri$tion 0 µ mg 0 G). 1 * 1 ?)= 0 ;)=. !

"s applied for$e is less than li%iting fri$tion) So' for the gi&en $ondition stati$ fri$tion will

worB)

Stati$ fri$tion on a body 0 "pplied for$e 0 *)2 !)

P roblem 4. " blo$B A with %ass GG kg is resting on another blo$B B of %ass *GG kg) "s shown in Agurea horiontal rope tied to a wall holds it) /he $oeT$ient of fri$tion between A and B is G)*

while $oeT$ient of fri$tion between B and the ground is G),) /he %ini%u% re-uired for$e F

to start %o&ing B will be [+P&" 1%

(a! ?GG !

(b! GG !

($! GG !

(d! *GG !

Solution 4 ($! /wo fri$tional for$e will worB on blo$B B)

B' AB & & F += gmmgm B AB'a AB !( ++= µ µ 0 G)* 1 GG 1 G O G), (,GG! 1 G

0 *GG O ?GG 0 GG!) (/his is the re-uired %ini%u% for$e!

P roblem ). " *G kg blo$B is initially at rest on a rough horiontal surfa$e) " horiontal for$e of ;2 ! isre-uired to set the blo$B in %otion) "fter it is in %otion' a horiontal for$e of 9G ! is re-uired

F

A

B F

& AB

A

B F

& B' 8roun

d


43/65


to Beep the blo$B %o&ing with $onstant speed) /he $oeT$ient of stati$ fri$tion is [AME 1%

(a! G),= (b! G).. ($! G)2* (d! G)9G

Solution 4 (a! CoeT$ient of stati$ fri$tion ,=)G=)?*G

;2=

×==R

F l

S

µ )

P roblem ,. " blo$B of %ass M is pla$ed on a rough Door of a lift) /he $oeT$ient of fri$tion between the

blo$B and the Door is µ ) hen the lift falls freely' the blo$B is pulled horiontally on the Door)hat will be the for$e of fri$tion

(a! µ Mg (b! µ Mg* ($! * µ Mg (d! None of theseSolution 4 (d! hen the lift %o&es down ward with a$$eleration WaW then e


44/65


R

F =θ tan

∴ tan θ 0 µ 6"s we Bnow µ =R

F 7

or !(tan

µ θ

−

=Hen$e $oeT$ient of li%iting fri$tion is e-ual to tangent of the angle of fri$tion)

). +esu/tant Force &Gerted 0< *urface on 5/oc; .

In the abo&e Agure resultant for$e ** RF S +=** !(!( mgmgS += µ

* += µ mgS

when there is no fri$tion !G( = µ S will be %ini%u% i.e)' S 0 mg

Hen$e the range of S $an be gi&en by' * +≤≤ µ mgSmg

). An/e of +e9ose.

"ngle of repose is deAned as the angle of the in$lined plane with horiontal su$h that a body

pla$ed on it is #ust begins to slide)

Ey deAnition α is $alled the angle of repose)

In li%iting $ondition α sinmgF =

and α $osmgR =

So α tan=R

F

∴ α θ µ tantan ===RF

6"s we Bnow θ µ tan==RF

7

/hus the $oeT$ient of li%iting fri$tion is e-ual to the tangent of angle of repose)

"s well as θ α = i.e. angle of repose 0 angle of fri$tion)

Sample problems based on angle o riction and angle o repose

P roblem -. " body of 2 kg weight Bept on a rough in$lined plane of angle ,Go starts sliding with a $onstant&elo$ity) /hen the $oeT$ient of fri$tion is (assu%e g 0 G m/s*!

[7IPM&+ 2##2%

(a! ,H (b! ,H* ($! , (d! ,*Solution 4 (a! Here the gi&en angle is $alled the angle of repose

So',

,Gtan == o µ

P roblem . /he upper half of an in$lined plane of in$lination θ is perfe$tly s%ooth while the lower half isrough) " body starting fro% the rest at top $o%es ba$B to rest at the botto% if the

$oeT$ient of fri$tion for the lower half is gi&en[P0 PM" 2###%

(a! µ 0 sin θ (b! µ 0 $ot θ ($! µ 0 * $os θ (d! µ 0 * tan θ

Solution 4 (d! >or upper half by the e-uation of %otion asuv *** +=*H!sin(*G** lgv θ += θ singl= 6"s 7sin'*H'G θ galsu ===

>or lower half

!$os(sin*G * θ µ θ −+= gu l * 6"s !$os(sin'*H'G θ µ θ −=== galsv 7

R

mg $os α α

α

F

mg sin α

mg

S%ooth

ough


45/65


⇒ !$os(sinsinG θ µ θ θ −+= glgl 6"s Anal &elo$ity of upper half will be e-ual to the initial&elo$ity of lower half7

⇒ θ µ θ $ossin* = ⇒ θ µ tan*=

).1# Ca/cu/ation of Necessar< Force in =iHerent Conditions.

If 0 weight of the body' θ 0 angle of fri$tion' == θ µ tan $oeT$ient of fri$tionthen we $an $al$ulate ne$essary for$e for dior the $ondition of e-uilibriu%

α $os*F = and α sin*. R −= Ey substituting these &alue in RF µ =

!sin($os α µ α *. * −=

⇒ !sin($ossin$os α

θ θ α *. * −= 6"s θ µ tan= 7

⇒ !($os

sin

θ α

θ

−= .

*

(*! Minimum 9us8in force P at an an/e from t8e 8orionta/

Ey esol&ing * in horiontal and &erti$al dire$tion (as shown in the Agure!

>or the $ondition of e-uilibriu%

α $os*F = and α sin*. R += Ey substituting these &alue in RF µ =

⇒ !sin($os α µ α *. * +=

⇒ !sin($ossin$os α

θ

θ α *. * += 6"s θ µ tan= 7

⇒ !($os

sin

θ α

θ

+= .

*

(,! Minimum 9u//in force P to moe t8e 0od< u9 an inc/ined 9/ane

Ey esol&ing * in the dire$tion of the plane and perpendi$ular to the plane (as shown in the

Agure!


λ α $ossin . *R =+∴ α λ sin$os *. R −=

and α λ $ossin *. F =+∴ λ α sin$os . *F −=

Ey substituting these &alues in RF µ = and sol&ing we get

!($os

!(sin

θ α

λ θ

−+

= .

*

(.! Minimum force on 0od< in downward direction a/on t8e surface of inc/ined 9/ane to

start its motion


Agure!


*

α

R

* sinα

* $osα F

*

α

R

* sinα

* $osα F

λ

*

α

λ

R O * sinα

$osλ λ

F O sinλ

* $osα

λ

*

α

λ

F R O * sinα

* $osα O

sinλ

$osλ

λ


46/65


λ α $ossin . *R =+

α λ sin$os *. R −=

and λ α sin$os . *F +=


!($os

!sin(

θ α

λ θ

−−

= .

*

(2! Minimum force to aoid s/idin a 0od< down an inc/ined 9/ane


Agure!


λ α $ossin . *R =+

∴ α λ sin$os *. R −=

and λ α sin$os . F * =+

∴ α λ $ossin *. F −=


+−

=!($os

!(sin

α θ

θ λ . *

(9! Minimum force for motion and its direction

Let the for$e * be applied at an angleα with the horiontal)Ey resol&ing * in horiontal and &erti$al dire$tion (as shown in Agure!

>or &erti$al e-uilibriu%

mg*R =+ α sin

∴ α sin*mgR −= R)(i!

and for horiontal %otion

F * ≥α $osi.e. R* µ α ≥$os R)(ii!

Substituting &alue of R fro% (i! in (ii!

!sin($os α µ α *mg* −≥

α µ α

µ

sin$os +≥

mg* R)(iii!

>or the for$e * to be %ini%u% !sin($os α µ α + %ust be %a@i%u% i.e.

G7sin6$os =+ α µ α α d

d ⇒ G$ossin =+− α µ α

∴ µ α =tan

or fri$tionofangle!(tan == − µ α

i.e. Fo" %ini%u% &alue of * its angle fro% the horiontal should be e-ual to angle of fri$tion

"s µ α =tan so fro% the Agure *sin

µ

µ α

+= and *

$os

µ α

+=

Ey substituting these &alue in e-uation (iii!

λ

F O *

$osα

R O * sinα

sinλ λ $osλ

R O * sinα

* $osα F

mg

*

α

λ

α *


47/65


*

*

*

µ

µ

µ

µ

++

+

≥ mg

* * µ

µ

+≥

mg

∴ *%in µ µ +

= mg*

Sample problems based on orce against riction

P roblem . hat is the %a@i%u% &alue of the for$e F su$h that the blo$B shown in the arrange%ent'

does not %o&e ( ,*H= µ !

[II"67&& '*creenin( 2##3%

(a! *G !

(b! G !

($! * !

(d! 2 !

Solution 4 (a! >ri$tional for$e R& µ =

⇒ !9Gsin(9G$os F . F += µ

⇒ ( )9Gsin,,*

9G$os F gF +=

⇒ !F *G= )

P roblem 1#. " blo$B of %ass m rests on a rough horiontal surfa$e as shown in the Agure) CoeT$ient of

fri$tion between the blo$B and the surfa$e is µ ) " for$e F 0 %g a$ting at angle θ with the&erti$al side of the blo$B pulls it) In whi$h of the following $ases the blo$B $an be pulled

along the surfa$e

(a! µ θ ≥tan

(b! µ θ ≥$ot

($! µ θ ≥*Htan

(d! µ θ ≥*H$ot

Solution 4 (d! >or pulling of blo$B & * ≥⇒ Rmg µ θ ≥sin ⇒ !$os(sin θ µ θ mgmgmg −≥

⇒ !$os(sin θ µ θ −≥

⇒

≥

*sin*

*$os

*sin* *

θ µ

θ θ ⇒ µ

θ ≥

*

$ot

).11 Acce/eration of a 5/oc; Aainst Friction.

(! Acce/eration of a 0/oc; on 8orionta/ surface

hen body is %o&ing under appli$ation of for$e *' then Bineti$ fri$tion opposes its %otion)

Let a is the net a$$eleration of the body

>ro% the Agure

k F *ma −=

α

µ

* µ +

R

*F k

mg

ma

9Go

F m 8

√ ,kg

θ m

mg 0 F

+

F $os 9G&

F sin 9G

+m

cos mg sin θ

0 p

&

mg


48/65


∴m

F *a k

−=

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