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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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Newton’s Laws of Motion 3genius PHYSICS by Pradeep Kshetrapal
(&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
=⇒∝
ordt
pd F = ( 0 in C)8)S) and S)I) units!
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4 Newton’s Laws of Motiongenius PHYSICS by Pradeep Kshetrapal
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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(*! :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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(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
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?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
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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
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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$$ording to Newton’s third law of %otion BA AB F F −=
(9! F@a%ple 4 (i! " booB lying on a table e@erts a for$e on the table whi$h is e-ual to theweight of the booB) /his is the for$e of a$tion)
/he table supports the booB' by e@erting an e-ual for$e on the booB) /his is the for$e of rea$tion)
"s the syste% is at rest' net for$e on it is ero) /herefore for$e of a$tion and rea$tion %ust be e-ual and opposite)
(ii! Swi%%ing is possible due to third law of %otion)
(iii! hen a gun is Ared' the bullet %o&es forward (a$tion!) /he gun re$oils ba$Bward(rea$tion!
R
mg
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(i&! ebounding of rubber ball taBes pla$e due to third law of %otion)
(&! hile walBing a person presses the ground in the ba$Bward dire$tion(a$tion! by his feet) /he ground pushes the person in forward dire$tion withan e-ual for$e (rea$tion!) /he $o%ponent of rea$tion in horiontal dire$tion
%aBes the person %o&e forward)(&i! It is diT$ult to walB on sand or i$e)
(&ii! :ri&ing a nail into a wooden blo$B without holding the blo$B isdiT$ult)
Sample problem based on Newton’s third law
P roblem 22. You are on a fri$tionless horiontal plane) How $an you get o< if no horiontal for$e ise@erted by pushing against the surfa$e
(a! Ey #u%ping (b! Ey splitting or sneeing
($! Ey rolling your body on the surfa$e (d! Ey running on the plane
Solution 4 (b! Ey doing so we $an get push in ba$Bward dire$tion in a$$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$$eleration' plane whi$h is taBing o
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4.1# Im9u/se.
(! hen a large for$e worBs on a body for &ery s%all ti%e inter&al' it is $alled i%pulsi&efor$e)
"n i%pulsi&e for$e does not re%ain $onstant' but $hanges Arst fro% ero to %a@i%u% and
then fro% %a@i%u% to ero) In su$h $ase we %easure the total e o r $ e
/i%e
t
F
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(i&! "n athlete is ad&ised to $o%e to stop slowly after Anishing a fast ra$e) So that ti%e of stop in$reases and hen$e for$e e@perien$ed by hi% de$reases)
(&! China wares are wrapped in straw or paper before pa$Bing)
Sample problem based on Impulse
P roblem 23. " ball of %ass 2Gg %o&ing with an a$$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)
(! "$$ording to this law for a syste% of parti$lesdt
pd F =
In the absen$e of e@ternal for$e G=F then = p $onstant
i.e.' =+++= )))),* p p p p $onstant)
or =+++ →→→
)))),,** v mv mv m $onstant
/his e-uation shows that in absen$e of e@ternal for$e for a $losed syste% the linear%o%entu% of indi&idual parti$les %ay $hange but their su% re%ains un$hanged with ti%e)
(*! Law of $onser&ation of linear %o%entu% is independent of fra%e of referen$e thoughlinear %o%entu% depends on fra%e of referen$e)
(,! Conser&ation of linear %o%entu% is e-ui&alent to Newton’s third law of %otion)
>or a syste% of two parti$les in absen$e of e@ternal for$e by law of $onser&ation of linear%o%entu%)
=+ * p p $onstant) =+ ** v mv m $onstant)
* . 9 =
O *
*
> o r $ e
( ! !
/i%e (s!
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:i
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(a! /hrust on the ro$Bet 4 mgdt
dmuF −−=
Here negati&e sign indi$ates that dire$tion of thrust is opposite to the dire$tion of es$apinggases)
dt
dmuF −= (if e
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Solution 4 (d! Mo%entu% of body for gi&en options are 4
(a! * seH*G*)GGG kgmmv =×== (b! * se$H.)GGGG. , kgmmv =××== −
($! * se$HG,)9G*)G** .9 kgmmE −− ×=××==
(d! * se$H=*)*GG*G*G* ,, kgmg/m =××××== −
So for option (d! %o%entu% is %a@i%u%)
P roblem 2. " ro$Bet with a lift5o< %ass .G2), × kg is blasted upwards with an initial a$$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$h o
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Condition Fiure ?e/ocit< Acce/eration
+eaction Conc/usion
Lift is at rest v 0 G a 0 GR mg 0 G
∴ R 0 mg"pparent weight0 "$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$$eleratingdownward atthe rate ofa(g!
v 0 &ariable a g mg R 0 ma R 0 mg ma R 0 ve
"pparent weightnegati&e %eansthe body will risefro% the Door ofthe lift and sti$B
SpringEalan$e
R
mg
LIF"
Spring
Ealan$e
R
mg
LIF"
a
SpringEalan$e
R
mg
LIF"
g
SpringEalan$e
R
mg
LIF"
a
SpringEalan$e
R
mg
LIF"
g
SpringEalan$e
R
mg
LIF"
a g
SpringEalan$e
R
mg
LIF"
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to the $eiling ofthe lift)
Sample problems based on lit
P roblem 33. " %an weighs )=Gkg He stands on a weighing s$ale in a lift whi$h is %o&ing upwards with a
unifor% a$$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($! g,
*(d! g
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Solution 4 (b!*
,
!(lift%o&ingdownwardin%anaofweight
liftstationaryin%anaofweight=
−=
agm
mg
∴*
,=
− ag
g ⇒ agg ,,* −= or
,
ga =
P roblem 4#. " 9G kg %an stands on a spring s$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
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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$$elerating fra%e' an inertial for$e ( m2! a$ts on it in the opposite
dire$tion)Nor%al rea$tion R 0 mg $osθ O m2 sinθ
and ma 0 mg sin θ 9 m2 $os θ
∴ a 0 g sinθ 2 $osθ
Note 4 /he $ondition for the body to be at rest relati&e to the in$lined plane 4 a 0 g sinθ 2 $osθ 0 G
∴ 2 0 g tanθ
4.1, Motion of 5/oc;s in Contact.
Condition Free 0od< diaram &uation Force and acce/eration
mg $osθ Om2 sinθ
R
θ mg
a
θ
θ
2
mg
R
m F
$osθ
F F
sinθ θ
F
sinθ
R
m F
$osθ
aF
F mg θ
mg $osθ
R
θ mg
a F
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am& F =−* mm
F a
+=
am& *=*
*
mm
F m&
+=
am& =* mm
F a
+=
am& F *=−*
mm
F m&
+=
am& F =−,* mmm
F a
++=
am& & ** =−,*
,*
!(
mmm
F mm&
++
+
=
am& ,* = ,*,
* mmm
F m
& ++=
am& =,* mmm
F a
++=
am& & ** =−,*
mmm
F m&
++=
am& F ,* =−,*
**
!(
mmm
F mm&
++
+
=
Sample problems based on motion o blocks in contact
P roblem 43. /wo blo$Bs of %ass . kg and 9 kg are pla$ed in $onta$t with ea$h other on a fri$tionlesshoriontal surfa$e) If we apply a push of 2 ! on the hea&ier %ass' the for$e on the lighter
%ass will be
(a! 2 !
m
F &
ma
m*
&
m*a
m
&
ma
m*
&
m*a
F
m
&
ma
F
m*a
& m
*
& *
m
&
ma
m*a
& m
*
& *
m,
& *
m,a
F
m
m,F
B
m*
1
mm
*F
A
B
m
m*
F A
B
m
m,
F
A B
m*
1
.kg
9kg
2!
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24 Newton’s Laws of Motiongenius PHYSICS by Pradeep Kshetrapal
(b! . !
($! * !
(d! None of the abo&e
Solution 4 ($! Let kgmkgm .'9 * == and !F 2= (gi&en!
>or$e on the lighter %ass 0*
*
mm
F m
+×
.9
2.
+×
= !*=
P roblem 44. In the abo&e proble%' if a push of 2 ! is applied on the lighter %ass' the for$e e@erted bythe lighter %ass on the hea&ier %ass will be
(a! 2 ! (b! . ! ($! * ! (d! None of the abo&e
Solution 4 (d! >or$e on the hea&ier %ass*
mm
F m
+=
.9
29
+×
= !,=
P roblem 4). In the abo&e proble%' the a$$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"$$elerati = *H2)G
G
2sm==
P roblem 4,. /wo blo$Bs are in $onta$t on a fri$tionless table one has a %ass m and the other * m asshown in Agure) >or$e F is applied on %ass *m then syste% %o&es towards right) Now the
sa%e for$e F is applied on m) /he ratio of for$e of $onta$t between the two blo$Bs will be in
the two $ases respe$ti&ely)
(a! 4 (b! 4 * ($! 4 , (d! 4 .
Solution 4 (b! hen the for$e is applied on %ass *m $onta$t for$e,*
gg
mm
m& =
+=
hen the for$e is applied on %ass m $onta$t for$e ggmm
m&
,
*
*
** =+=
atio of $onta$t for$es *
* =& &
F m*m
F
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4.1- Motion of 5/oc;s Connected 0< Mass Less *trin.
Condition Free 0od< diaram &uation "ension and acce/eration
amT =* mmF a+
=
amT F *=−*
mm
F mT
+=
amT F =−* mm
F a
+=
amT *=*
*
mm
F mT
+=
amT =,* mmm
F a
++=
amT T ** =−,*
mmmF mT ++
=
amT F ,* =−,*
**
!(
mmm
F mmT
++
+=
amT F =−,* mmm
F a
++=
amT T ** =−,*
,*
!(
mmm
F mmT
++
+
=
amT ,* =,*
,*
mmm
F mT
++
=
m
T
ma
m*
F
m*a
T
m
T
ma
F
m
T
ma
m,
F
m,a
T *
T *
m*a
T m
*
m
T
ma
F
T *
m*a
T m
*
m,
m,a
T *
m*
m*a
T
m
m*
F T
B
A
m
m*
F T
B
A
m
m,
F A B
m*
1
T
T *
m,
F A B
m*
1
T
T *
m
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Sample problems based on motion o blocks connected by mass less string
P roblem 4-. " %onBey of %ass *G kg is holding a &erti$al rope) /he rope will not breaB when a %assof *2 kg is suspended fro% it but will breaB if the %ass e@$eeds *2 kg) hat is the %a@i%u%
a$$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$$eleration of *H* sm in
the upward dire$tion in both the blo$Bs) If T and T ′ be the tensions in the two parts of thestring' then [AME '&n.( 2###%
(a! !T =);G= and !T *).;=′
(b! !T =)2== and !T *).;=′
($! !T =);G= and !T =)2==′
(d! !T =);G= and G=′T Solution 4 (a! >ro% >)E):) of %ass . kg gT a .W. −= )R)(i!
>ro% >)E):) of %ass * kg gT T a *W* −−= )R)(ii!
>or total syste% upward for$e
( ) ( )agT F ++== .* ( ) !*=9 += 0 ;G)= N
by substituting the &alue of / in e-uation (i! and (ii!
and sol&ing we get !T *).;W=
P roblem )#. /hree %asses of 2 kg) G kg and 2 kg are suspended &erti$ally as shown in the Ag) If thestring atta$hed to the support breaBs and the syste% falls freely' what will be the tension in
*kg2kg,kg
G! T
T *
T F
*kg
.kg
T ′
.kg
T ′
.g
.a *kg
T ′
*g
*a
T ′
2k g
Gk g
2kg
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the string between G kg and 2 kg %asses) /aBe *G −= msg ) It is assu%ed that the string
re%ains tight during the %otion
(a! ,GG ! (b! *2G ! ($! 2G ! (d! Uero
Solution 4 (d! In the $ondition of free fall' tension be$o%es ero)
P roblem )1. " sphere is a$$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
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, *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
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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*
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3# Newton’s Laws of Motiongenius PHYSICS by Pradeep Kshetrapal
"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
($! *G −ms
(d! *,G −ms
Solution 4 (b! gmm
ma
+
=*
*
*H,G,;
,sm=
+=
m
T
mg
ma
m*
T *
m*g
m*a
MT *
T
Ma
M
m
A
T
m*
B
T
aa
T *
T *
1
B
A
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Newton’s Laws of Motion 31genius PHYSICS by Pradeep Kshetrapal
P roblem )4. /wo %asses m and *m are atta$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$$eleration
m
m*
G kg
9 kg
M
M*
θ
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32 Newton’s Laws of Motiongenius PHYSICS by Pradeep Kshetrapal
gmm
mma
*
* sinW+
−=
θ G)
*
,Gsin*
=+
−= g
mm
mm6"s m* 0 *m7
i.e. the *m will not %o&e)
P roblem ). In the abo&e proble%' gi&en that M* 0 *M and the tension in the string is T ) If thepositions of the %asses are re&ersed' the tension in the string will be
(a! . T (b! T ($! T (d! T *
Solution 4 ($! /ension in the string( )
gmm
mmT
*
* sin
++
= θ
If the position of the %asses are re&ersed then there will be no e
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Newton’s Laws of Motion 33genius PHYSICS by Pradeep Kshetrapal
>ro% (i! and (ii! ,H,H
*
==g
g
a
a
P roblem ,4. In the ad#oining Agure m 0 .m*) /he pulleys are s%ooth and light) "t ti%e t 0 G' the
syste% is at rest) If the syste% is released and if the a$$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 $o%ing to rest positionwill be
m
m*
*Gcm
m*
T
m*g
m*(*a
!
m
*T
mg
ma
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34 Newton’s Laws of Motiongenius PHYSICS by Pradeep Kshetrapal
(a! *G cm (b! .G cm ($! 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!
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Newton’s Laws of Motion 3)genius PHYSICS by Pradeep Kshetrapal
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
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4.2# *9rin 5a/ance and P8
and the needle %ust in %iddle of the bea% i.e. a 0 2)E
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Newton’s Laws of Motion 3-genius PHYSICS by Pradeep Kshetrapal
>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
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3 Newton’s Laws of Motiongenius PHYSICS by Pradeep Kshetrapal
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
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Newton’s Laws of Motion 3genius PHYSICS by Pradeep Kshetrapal
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 "
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(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
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).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$$elerating
&ehi$le until !)mgma s µ < If there had been no fri$tion between body and &ehi$le the body will not%o&e along with the &ehi$le)
hilepedalling
Pedalling isstoped
ma µ smg
a
"$tion
>ri$tion
θ
A
1B
F l
F k
> o r $ e
o f f r i $ t i o n
"pplied for$e4
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42 Newton’s Laws of Motiongenius PHYSICS by Pradeep Kshetrapal
>ro% these e@a%ples it is $lear that without fri$tion %otion $annot be started' stopped ortransferred fro% one body to the other)
Sample problems based on undamentals o riction
P roblem 1. If a ladder weighing *2G! is pla$ed against a s%ooth &erti$al wall ha&ing $oeT$ient of fri$tion between it and Door is G),' then what is the %a@i%u% for$e of fri$tion a&ailable at
the point of $onta$t between the ladder and the Door[AIIM* 2##2%
(a! ;2 ! (b! 2G ! ($! ,2 ! (d! *2 !
Solution 4 (a! Ma@i%u% for$e of fri$tion !RF sl ;2*2G,)G =×== µ
P roblem 2. +n the horiontal surfa$e of a tru$B ( µ 0 G)9!' a blo$B of %ass kg is pla$ed) If the tru$B isa$$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
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Newton’s Laws of Motion 43genius PHYSICS by Pradeep Kshetrapal
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
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44 Newton’s Laws of Motiongenius PHYSICS by Pradeep Kshetrapal
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 $os α α
α
F
mg sin α
mg
S%ooth
ough
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⇒ !$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!
>or the $ondition of e-uilibriu%
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
!($os
!(sin
θ α
λ θ
−+
= .
*
(.! Minimum force on 0od< in downward direction a/on t8e surface of inc/ined 9/ane to
start its motion
Ey esol&ing * in the dire$tion of the plane and perpendi$ular to the plane (as shown in the
Agure!
>or the $ondition of e-uilibriu%
*
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
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λ α $ossin . *R =+
α λ sin$os *. R −=
and λ α sin$os . *F +=
Ey substituting these &alues in RF µ = and sol&ing we get
!($os
!sin(
θ α
λ θ
−−
= .
*
(2! Minimum force to aoid s/idin a 0od< down 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!
>or the $ondition of e-uilibriu%
λ α $ossin . *R =+
∴ α λ sin$os *. R −=
and λ α sin$os . F * =+
∴ α λ $ossin *. F −=
Ey substituting these &alues in RF µ = and sol&ing we get
+−
=!($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
* $osα F
mg
*
α
λ
α *
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Newton’s Laws of Motion 4-genius PHYSICS by Pradeep Kshetrapal
*
*
*
µ
µ
µ
µ
++
+
≥ mg
* * µ
µ
+≥
mg
∴ *%in µ µ +
= mg*
Sample problems based on orce against riction
P roblem . hat is the %a@i%u% &alue of the for$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
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∴m
F *a k
−=