Hints and Solution LOM

8/18/2019 Hints and Solution LOM

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Laws of Motion

Solutions to Subjective AssignmentsLEVEL – I

1. We find that the each of the blocks A & B experience following forces.

Forces acting on A :(a) Earths gra!itational field force " weight of the block A " #1g

(b) $or%al contact force exerted b the hori'ontal srface " $1.

(c) ontact force exerted b the block B " $

(d) *he external hori'ontal force " F.

Forces acting on B :

(a) Earths gra!itational field force " weight of the block B " #+g

(b) $or%al contact force exerted b the hori'ontal srface " $+.

(c) ontact force exerted b the block A " $

Both the blocks %o!e together. ,ence both will ha!e sa%e acceleration- let s

chose /axis shown. *here is no 0/axial acceleration as the blocks are not co%ing

off the srface. *he F.B.. diagra% of 2A3 as a sste% is shown in figre (i) . *he

F.B.. of 2B3 as a sste% shown in figre (ii).$1

F$

#1g

a

$+

$

#+g

a

Fig. (i) Fig. (ii)

Fro% figre (i)- $1 4 #1g " 5 6(1)

For A- F 4 $ " #1a 6(+)

Fro% figre (ii) $+ 4 #+g " 5 6(7)

For B- $ " #+a 6(8)

*hs sol!ing (1)- (+)- (7) and (8) We obtain

a "+1

+

+1 ##

F#$and

##

F

+=

+

+. Along !ertical direction-

* cos θ " %g 6(1) Along hori'ontal direction

* sinθ " %a 6(+)9sing (1) and (+) tan θ " ag

%g

* sin θ

* cos θ

a

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∴ a " g tan θ " ;.< tan 75o

"+s:%

7

%s+.

7. *he F.B.. of %1 gi!es-

$1 " %1g os θ 6(1)* " %1g ?in θ 6(+)*he F.B.. of %+- gi!es-

$+ " %+g is α 6(7)* " %+g ?in α 6(8)

*

%1 g sin θ %1 g cos θ

$1

$+

%+ g cos α %+ g sin α

*

%1 %+

*aking the ratio of (+) and (8)-

θα

=?in

?in

%

%

+

1" o

o

@5?in

75?in "

7

1

7

+

+

1=× .

∴7

1

%

%

+

1 = .

8. et the block be displaced to the left b x

the forces acting on the block will be kx to the right b both the springs

ths F " %a or a "%

kx+

%

F=

=. $


3/18

" µk%g " 5.1= × 1 × ;.<" 1.8> $.


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@. F " (%1 D %+)a

∴ a "+=

1

%%

F

+1 +=

+ " >1

%s+

if the string breaks- %+ will %o!e with constant !elocit attained b it while %1 will%o!e with a acceleration

a′ "=1 %s+

>. (a) Angle of repose θ " tan−1(µ)∴ θ " tan−1(5.7) " 1@.>°

(b) Frictional force " µs%gcos(θ+)

∴ f fr " 5.7 × cos

+

1@.>%g " 5.18= %g


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15. *aking the differential ele%ent at an anglar

position θ- we obtain*he net tangential force " ∫ (d%)g sin θ

" θ

∫ α

sing)ds.(l

#

5

θ

θ∫ α

sing)Ed.(l

#

5

" ∫ α

θθ5

dsinl

#Eg

" [ ]αθ− 5cosl

#Eg-

where α " l ⇒ . l*R +⇒ *angential acceleration "

)

E

cos1(Eg

− .

θ dθR

(dm)g

dm


RSM79-P1-LOM (S -P!-


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LEVEL – II

1. *he forces acting on the particle are%g (!erticall downward) %a5 (psedoforce !erticall downward) $ nor%al

reaction- nor%al to the inclined plane)esol!ing the forces along theincline- we ha!e

F " % (a5 D g) sin θ⇒ acceleration down the plane

a " (a5 D g) sin θ "

+ g

+

gsin 755 "

8

g7

istance tra!ersed along the incline

sec θ " +at+

1 "

+t<

g7

t " g77

A1@

g7

75secA<

=

+. #otion of the bod p the plane

F " %g sin θ D µ %g cos θ " %g sin θ D µcos θG#otion of the bod down the plane

F" µ%g cos θ / %g sin θ " %g H( µcos θ / sin θ) HIt µ " 5.+- θ " 755- & % " + kg

mg sinθ

µmg cos θ

F'

mg sinθ

µmg cos θ

m

m

F

'ote / Jf F is negati!e alter its ass%ed direction.

7. Acceleration of the %an relati!e to the ele!ator " + %s+ (pwards) Acceleration of the ele!ator relati!e to earth " 7 %s+ (downwards),ence- acceleration of the %ass relati!e to earth " 7 4 + " 1 %s+ (downwards)onsider free bod diagra% of %an :Jf * " tension in the spring K%g 4 * " %a

⇒ =5 × 15 4 * " =5 × 1⇒ * " 8=5 $.

,ence- extension in the spring " +555

8=5

" 5.++= %.

8. %!ds

d! " αcos

7

%gds "

a

dα

∫ !

5

!d! " αα∫ α

dcosa7

g

5


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α= sina7

g

+

!+

! " αsina7

g+.

=. Acceleration of a block sliding down an inclinedplane is gi!en b

a " g ( / µ cos θ)for + kg block µ1 µ+ - hence a1 C a+,ence the blocks %o!e separatel and with neLalaccelerations. Acceleration of + kg block isa " g(sin 75 4 5.+ cos 75)

" 15 (5.= 4 5.1>7+) " 7.+> %s+

Jn case (b)- when µ1 " 5.7 and µ+ " 5.+- the + kg block has less acceleration thanthe 7 kg block.

*hs the blocks %o!e together. When we draw the FB and resol!ed the forces

along the plane- we get

F " (%1 D %+) g sin θ / (%1µ D %+µ+) g cos θ

⇒ a "+1 %%

F

+ " g sin θ /θ

+µ+µ

cosg%%

)%%(

+1

++11

" 15 sin 75 /@

)+.587.5+( ×+× × 15 × cos 75° " +.;> %s+

@. Jnitiall

%g 4 F " %a ↓ . . . (i) After releasing a %ass (∆%)F 4 (%/ ∆%)g " (% / ∆%) a ↑ . . .(ii)Where F " pward thrst of air

B sol!ing (i) and (ii) & ptting %g " w-

obtain the answer.

a

F

mg

a

F

(m-∆m)g

>. et there is no relati!e %otion between

the blocks B and ,ence * " (%B D %)a (1)

And % Ag / * " % Aa (+)

Fro% (1) and (+)- we get

a "+

B A

A s:%7

=

1<

75

%%%

g%==

++⇒ $et force on the block is

F " %a " 15 × (=7) $ " 1@.@ $

mAg

T

F.B.D. of the blocks

NB+mC

f r

f r

NC

mB g

T

NB


RSM79-P1-LOM (S -P!-7


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?ince force acting on the block is onl frictional force.

Jf %axi%% !ale of frictional force acting on is f %ax - then

f (%ax) " µ%Bg " 5.8 × = × 15 " +5 $ F ≤ f %ax ,ence there is no relati!e %otion between the block B and . *herefore

distance %o!ed b is +% onl.


9/18

(c) When F " 185 $

* " >5 $

T

m1

m1g

1

⇒ * / %1g " %1a1 (1)⇒ >5 $ / +5 $ " + × a1 ⇒ a1 " += %s+

yP

y1

m1

y2

m2

F

* / %+g " %+a+ (1)

⇒ >5$ / =5 $ " =a+ ⇒ a+ " 8 %s+ onstraint eLation

I / + D I / 1 " c⇒ +I / 1 / + " c

⇒ 5dt

&d

dt

&d

dt

&d+

+

+

+

+

1

+

I

+

=−−

⇒ aI "+1s:%

+

+;

+

aa+=

+

;. ELation of %otion for %:

ΣFx " %a ⇒ $ " %a 6(1)

ΣF " 5 ?ince % is at rest relation to # at the !erge ofslipping.⇒ f 4 %g " 5

a

$

µ$

%g

ELation of %otion for #:

F 4 µ$′ 4 $ " #a 6(8)

Where $ " µ%g

- a "µ

=µ

ga-

%g and since the block

# does not ha!e !erticall %otion.

⇒ $′ " #g D f Itting !ales of $′- $ and a in (8)- we obtain

F 4 µ (#g D f) 4 $ " #a F / µ (#g D %g) 4 %a " #a⇒ F / µg (# D %) " (# D %)a

$a

F

µ$′

µ$ #g

$′

⇒ a "%#

)%#(gF

++µ−

f ≤ f %ax ⇒ %g ≤ µ$⇒ %g ≤ µ%a




10/18

⇒ g ≤ µ.%#

)%#(gF

++µ−

⇒ (# D %) g ≤ µF / µ+ g (# D %)⇒ µ+g(# D %) / µF D (# D %) g ≤ 5⇒ +µ+ / =µ D + ≤ 5

For critical-+µ+ / =µ D + " 5(µ / +) (+µ / 1) " 5

∴ µ " + or µ "+

1

Iracticall µ does not beco%e +.∴ µ " 5.=.

15. et x1 → distance of the plle (wedge)fro% the wall.

x+ → distance of the bar fro% the plle.*herefore

x1 D x+ " l

+

1+

dt

xd → acceleration of wedge

+

1+

dt

xd "

+

++

dt

xd−

α

ax+

x1ax

&

++

+

dt

xd → acceleration of the bar- w.r.t.

wedge

+

1+

dt

xd " +

++

dt

xd " a (sa).

Acceleration of the bar

a

a cos α

a sin αa

α

$et acceleration of wedge along the

hori'ontal srface " a 6(1)

$et acceleration of bar along the hori'ontal

srface figre (i)

" a 4 co%p. of acceleration of bar w.r.t.

wedge

" a 4 a cos α 6(+)Merticall downward " a sin α 6(7)

$1

*

a

*

$

#g

α

Forces on wedge and bar are shown

separatel in figre (ii) and fig. (iii)

respecti!el (i!) and (!).

Fro% the F.B.. of the wedge

Along the chosen /direction

* D $1 sin α 4 * cos α " #a 6(8)

α

$1

%g

α

a/acos α

a sin α

along 0/direction

For bar


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$1 cos α D * sin α D #g 4 $ " 5 6(=)Fro% the F.B.. of the bar along /direction

cos α 4 $1 sin α " % (a 4 a cos α) 6(@)For Wedge: along 0/direction

%g 4 $1 cos α 4 * sin α " % a sin α 6(>)

$1 cos α * sin α

$1 sin α* cos α

%g

a/a cos α

a sin α

Fro% eLations (8)- (=)- (@) and (>)

eli%inating other Lantities we can find a.*he

proble% can be sol!ed b choosing /0 axis

for the sste% along and nor%al to the

srface of wedge. Nst do b orself.

$

* cos α

#g

$1 cos α

* sin α

$1 sin α*

a




12/18

Solutions to Objective Assignments

LEVEL – I

"+ *+ " %a " %7

F

%7

F=

F 4 *1 " %a

*1 " F / %a

" F /7

F+

7

F=

⇒ .+

1

7:F+

7:F

*

*

1

+ ==

m

a a aZYX

Fm mT1

T1

T2

T2

#+ *he contact force F accelerates the bod

of %ass %1 " + kg with an acceleration

a " s:%1+1

7

%%

F

+1

=+

=+

⇒ F " %1a " 1 × 1 " 1'+

1 kg2 kg

m1F '

m1

m2

F

+ a "%

Fi%pact "%

)dt:d%(! "

+

)1)(=( " "+m*s"+

+ Jn nifor% circlar %otion !F⊥ where F is the centripetal force

⇒ OE " ttancons%!+

1 +

=%+ *he net force F " #1g sinα / #+ g sin β

" (#1 sin α / #+ sin β)gWhen #1 " #+ and α1 " α+ -the net force is 'ero.

⇒ *he acceleration of the sste% " 5

7+ 9sing the expression obtained in Lestion no. 1

We obtain µ " 1/ (1n+)- ptting n " +we obtain µ " &+7

)+ *he acceleration of % along the plane isg sin θ becase the inclined plane is s%ooth⇒ *he co%ponent of the tension * parallelto the inclined plane is 'ero. *he onl

dri!ing force is co%ponent of gra!itational

force %g parallel to the plane- that is eLal

to %g sin θ.*herefore- for the eLilibri% of the bod-

mg cos θ . 0+

θT

m

mg

mg sin θθ mg cosθ


RSM79-P1-LOM (S -P!-1"


13/18

9+ ?ince block has been pshed and is broght to %otion hence kinetic friction will

act

∴ %gsinθ − f r " %a∴ f r " µ+%gcosθ∴ gsinθ − µ+g cosθ " a

⇒ a " g(sinθ − µ+cosθ)

1&+ a "%ass

Fi%pact "[ ]

%

)dt:d%(!

"%

)!(! α

"%

!+α (n%ericall)

11+ Following the pre!ios

procedreF 4 *+ " %7a

⇒ F 4 *+ " %7 7+1 %%%

F

++

⇒ *+ "7+1

+1

%%%

F)%%(

++

+

⇒ *+ " $;7@)+>


14/18

1+ For the eLilibri% of the bod %.

* cosθ " %gFor hori'ontal acceleration of %:

* sin θ " %a

⇒

%g

%a

cos*

sin*=

θ

θ

⇒ θ " t$n 1 ($*g+

T cos θ

T sin θ

T

θθ

mg

a

1%+ For eLilibri% of the bod (block)

f 4 %g " 5

⇒ f " %g " (5.1) (;.


15/18

⇒ µcot θ " 1 /+n

1

⇒ µ " t$nn

11

"

− .

LEVEL – II

1. Acceleration of #-F

aM

= ÷

+1 F .t+ #

=l

+#t

F=

l.

+. #g 4 * " #a 6(i)

* " %a6(ii)?ol!ing (i) and (ii)

Mga

(M m)=

+ N

a

M g

FB of %an#g 4 $ " #a

#%g$(# %)

= +7. #g * #a− = 6(i)

* " #a 6(ii)

#g a(# #)= +# g

a(# #)

=+

ma sin mg cosθ = θa g cot= θ

T

R

M g

M

#ggcot

(# #)θ =

+cot # cot # #θ + θ =




16/18

#cot#

(1 cot )

θ=

− θ* " # a

" #. g cot θ#g

* tan= θ .m a s i n θ

m a

m g s i n θm g c o s θ

θ

+ m a c o s θ

8. Jf the tendenc of relati!e %otion along the co%%on tangent does not exist- thenco%ponent of contact force along co%%on tangent will be 'ero.

=. For the two !ales of F- i.e.- F " 1=5 $ and F " 1+5 $K the tensions in the string are

F*

+= " >= $ and @5 $. Jn the first case- the accelerations of the two %asses are eLal

and opposite- while in the second the accelerations are 'ero.

@. *1 " +µ %g 6(i)F " ; µ%g 6(ii)

∴ (a)- (b).

2 m g

T2 m gµ

F2 m gµ

2 m gµ5 µ m g

>. kx " 1+5 $-

22 12a ! m " s2

−= .

1 2

2 2 k g

;. ?ince there is relati!e acceleration between the fra%es.?1 %a be at rest ?+ is accelerating

?1 %a be acceleration and ?+ is at rest?1 and ?+ both are accelerating with difference acceleration?1 and ?+ both cannot be at rest.

15. 1 1$ % H a H sin %gcos+ θ = θ


RSM79-P1-LOM (S -P!-1%


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m g c o s θθ

N + m # a # s i n1 1 θ

m g s i n + m # a # c o sθ θ1

m # a #1

# a #1

[ ]1 1 N m g cos # a # sin= θ− θ1 1 N sin M # a #θ =

FB of %

m g c o s θ

N + m # a # s i n1 1 θ

m g s i n θ

2 1 N mg N cos= + θ

2 1 N mg m(g cos # a # sin )cos= + θ− θ θ

2 N (M m)g∴ < + .

m g

N s i n1 θ

N2

N c o s1 θ

COMPREHENSION

1.+

?

%!%g

r ≤ µ - since friction is static.

+. $or%al reaction decreases- and so the li%iting friction is lower.7. $et force " f 4 " 5.8. to @.

F.B.. of %an

a

N

T

M g

$ D * 4 #g " #aWith $ " #g

⇒ * " #a 6(i)




18/18

F.B.. of box :

a

N

T

( m + m ) g’

T $ N $ mg $ m′g" (% D %′)a 6(ii)Where # " @5 kg-

% " +5 kg- %′ " 75 kg?ol!ing eLation (i) and (ii)- we get

a " 115 %s 41 * " @@55 $.

MA02! 0!E 3OLLO4I'5

1. When bod is at rest then friction is static and if bod is %o!ing then friction on it iskinetic.

RSM79-P1-LOM (S -P!-1)

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