First : staticsFirst : statics

Two forces of magnitude F , 2FTwo forces of magnitude F , 2FNewton act at particle ,Newton act at particle ,

magnitude of their resultant is R .magnitude of their resultant is R .If the mag. of the 1 - isIf the mag. of the 1 - is

increased by 4 N increased by 4 N & the mag. of the 2 – is doubled,& the mag. of the 2 – is doubled,

stst

ndnd

then the measure ofthen the measure ofthe angle betweenthe angle between

the resultant the resultant & the 1 – force does& the 1 – force does

not change ,not change , Find the value of F, 2FFind the value of F, 2F

stst

If R = 4If R = 47 N , Find the measure of7 N , Find the measure ofthe angle between the two forcesthe angle between the two forces

let let be the angle between be the angle betweenthe 2 force ,the 2 force ,

If R = 4If R = 47 N , Find the measure of7 N , Find the measure ofthe angle between the two forcesthe angle between the two forces

let let be the angle between be the angle betweenthe 2 force ,the 2 force ,

SolutionSolution

The angle between The angle between the the

Resultant & the 1 - Resultant & the 1 - force force

does not changedoes not change

stst

Tan Tan for 1 - = Tan for 1 - = Tan for 2 - for 2 - stst ndnd

2F sin 2F sin 4 F sin 4 F sin F + 2F cos F + 2F cos F + 4 + 4F F + 4 + 4F

cos cos 2F + 4F cos 2F + 4F cos = F + 4 + 4Fcos = F + 4 + 4Fcos

2F = F + 42F = F + 4

F = 4F = 4

=

64 Cos 64 Cos = 32 = 32

= 60= 60

R = F + 4 F + 2 F (2F) Cos R = F + 4 F + 2 F (2F) Cos 22 22 22

(4(47) = 16 + 64 + 64 Cos 7) = 16 + 64 + 64 Cos 22

Cos Cos = =2211

A B C D H O is a regularA B C D H O is a regularHexagon of side length 10 cm.Hexagon of side length 10 cm.forces of mag. 3 , 2forces of mag. 3 , 23 , 4 , 33 , 4 , 33 ,3 ,

2 N act at the vertex D2 N act at the vertex Din directions DC , DB , DA , DO,DHin directions DC , DB , DA , DO,DH

respectively respectively

A B C D H O is a regularA B C D H O is a regularHexagon of side length 10 cm.Hexagon of side length 10 cm.forces of mag. 3 , 2forces of mag. 3 , 23 , 4 , 33 , 4 , 33 ,3 ,

2 N act at the vertex D2 N act at the vertex Din directions DC , DB , DA , DO,DHin directions DC , DB , DA , DO,DH

respectively respectively

Find the magnitude & dir. ofFind the magnitude & dir. ofthe resultant & the distancethe resultant & the distance

from B to the linefrom B to the lineof action of the resultant .of action of the resultant .

Find the magnitude & dir. ofFind the magnitude & dir. ofthe resultant & the distancethe resultant & the distance

from B to the linefrom B to the lineof action of the resultant .of action of the resultant .

SolutionSolution

OO AA

BB

XXCCDD

HH

yy

22

33

44

2233

3333

3030ºº3030ºº3030ºº

3030ºº

F = ( 3 , 0 º ) F = ( 3 , 0 º ) F = (2F = (23 ,30)3 ,30)F = ( 4 , 60 º)F = ( 4 , 60 º)

F = ( 3F = ( 33 , 90 º )3 , 90 º ) F = ( 2 , 120 º ) F = ( 2 , 120 º )

F = ( 3 , 0 º ) F = ( 3 , 0 º ) F = (2F = (23 ,30)3 ,30)F = ( 4 , 60 º)F = ( 4 , 60 º)

F = ( 3F = ( 33 , 90 º )3 , 90 º ) F = ( 2 , 120 º ) F = ( 2 , 120 º )

221 1

3344

55

OO AA

BB

XXCCDD

HH

yy

22

33

44

2233

3333

3030ºº3030ºº3030ºº

3030ºº

X = 3 + 2X = 3 + 23 Cos 303 Cos 30ºº + 4 Cos 60 + 4 Cos 60º º + 3+ 33 Cos 90º + 2 Cos 120º = 73 Cos 90º + 2 Cos 120º = 7

X = 3 + 2X = 3 + 23 Cos 303 Cos 30ºº + 4 Cos 60 + 4 Cos 60º º + 3+ 33 Cos 90º + 2 Cos 120º = 73 Cos 90º + 2 Cos 120º = 7

OO AA

BB

XXCCDD

HH

yy

22

33

44

2233

3333

3030ºº3030ºº3030ºº

3030ºº

Y = 3 Sin 0Y = 3 Sin 0º + 2º + 23 sin 3 sin 30º30º

+ 4 sin 60º + 3+ 4 sin 60º + 33 sin3 sin 9090ºº

+ 2 sin 120º = 7+ 2 sin 120º = 733

Y = 3 Sin 0Y = 3 Sin 0º + 2º + 23 sin 3 sin 30º30º

+ 4 sin 60º + 3+ 4 sin 60º + 33 sin3 sin 9090ºº

+ 2 sin 120º = 7+ 2 sin 120º = 733

OO AA

BB

XXCCDD

HH

yy

22

33

44

2233

3333

3030ºº3030ºº3030ºº

3030ºº

Notice that both X , Y areNotice that both X , Y arePositive then Positive then lies in the 1- quant lies in the 1- quant

Notice that both X , Y areNotice that both X , Y arePositive then Positive then lies in the 1- quant lies in the 1- quantstst

OO AA

BB

XXCCDD

HH

yy

22

33

44

2233

3333

3030ºº3030ºº3030ºº

3030ºº

R = X + y = 49 + 147 = 196R = X + y = 49 + 147 = 196 R = 14 N ,R = 14 N ,

Tan Tan = = = = 33 = 60º= 60º

R = X + y = 49 + 147 = 196R = X + y = 49 + 147 = 196 R = 14 N ,R = 14 N ,

Tan Tan = = = = 33 = 60º= 60º

22 22 22

XXYY

OO AA

BB

XXCCDD

HH

yy

22

33

44

2233

3333

3030ºº3030ºº3030ºº

3030ºº

R acts along DA ,R acts along DA ,Rup = 14 sin 60º = 7Rup = 14 sin 60º = 73 cm.3 cm.

R acts along DA ,R acts along DA ,Rup = 14 sin 60º = 7Rup = 14 sin 60º = 73 cm.3 cm.

A particle moves from pointA particle moves from pointA ( -3 ,1) to point B ( 3 , 9 )A ( -3 ,1) to point B ( 3 , 9 )

under the action of the forceunder the action of the forceF = 3 i + 4 j .F = 3 i + 4 j .

Find the alg. Component &Find the alg. Component &the vector component of the vector component of F F

in dir . Of in dir . Of ABAB

A particle moves from pointA particle moves from pointA ( -3 ,1) to point B ( 3 , 9 )A ( -3 ,1) to point B ( 3 , 9 )

under the action of the forceunder the action of the forceF = 3 i + 4 j .F = 3 i + 4 j .

Find the alg. Component &Find the alg. Component &the vector component of the vector component of F F

in dir . Of in dir . Of ABAB

^̂^̂

SolutionSolution

AB = B – A = ( 6 , 8 )AB = B – A = ( 6 , 8 )alg . comp. =alg . comp. = = =

= = = = = = 5 units 5 units

AB = B – A = ( 6 , 8 )AB = B – A = ( 6 , 8 )alg . comp. =alg . comp. = = =

= = = = = = 5 units 5 units

F ABF ABABAB

(3,4) (6,8)(3,4) (6,8) 36 + 6436 + 64

....

18 + 3218 + 32

1010 5050 1010

Vector comp. =Vector comp. =

= (6 i + 8 j )= (6 i + 8 j )

= 3 i + 4 j = 3 i + 4 j

Vector comp. =Vector comp. =

= (6 i + 8 j )= (6 i + 8 j )

= 3 i + 4 j = 3 i + 4 j

ABAB F ABF AB

ABAB

..22

5050

100100^̂ ^̂

^̂ ^̂

FF , F , F , F , F are three coplanar are three coplanarforces where F = 3 i – 5 j ,forces where F = 3 i – 5 j ,

F = - 7i + 2 j , F = 4 i + 3 j act resp.F = - 7i + 2 j , F = 4 i + 3 j act resp.in A ( 2, 41 ) , B ( 5 , -1 ) , C ( -2 , 1 ) in A ( 2, 41 ) , B ( 5 , -1 ) , C ( -2 , 1 )

FF , F , F , F , F are three coplanar are three coplanarforces where F = 3 i – 5 j ,forces where F = 3 i – 5 j ,

F = - 7i + 2 j , F = 4 i + 3 j act resp.F = - 7i + 2 j , F = 4 i + 3 j act resp.in A ( 2, 41 ) , B ( 5 , -1 ) , C ( -2 , 1 ) in A ( 2, 41 ) , B ( 5 , -1 ) , C ( -2 , 1 )

1 1 2 2 3 3 1 1

2 2 3 3

^̂ ^̂^̂ ^̂ ^̂ ^̂

Prove that the system is Prove that the system is equivalent to a couple & findequivalent to a couple & find

the norm of its moment .the norm of its moment .

Prove that the system is Prove that the system is equivalent to a couple & findequivalent to a couple & find

the norm of its moment .the norm of its moment .

SolutionSolution

Firstly Find the resultant ofFirstly Find the resultant ofF , F for exampleF , F for example

Firstly Find the resultant ofFirstly Find the resultant ofF , F for exampleF , F for example2 2 3 3

R R = F = F + F + F= (-7 i + 2 j)+ (4i + 3j)= (-7 i + 2 j)+ (4i + 3j)

R R = - 3 i + 5j = - 3 i + 5j R R = - F = - F The system is equivalentThe system is equivalent

to a couple .to a couple .

R R = F = F + F + F= (-7 i + 2 j)+ (4i + 3j)= (-7 i + 2 j)+ (4i + 3j)

R R = - 3 i + 5j = - 3 i + 5j R R = - F = - F The system is equivalentThe system is equivalent

to a couple .to a couple .

2 2 3 3

1 1 ^̂

^̂ ^̂ ^̂ ^̂^̂

M = MA = AB M = MA = AB F + F + AC AC F F

= ( 3i – 5j) = ( 3i – 5j) (-7i + 2j) (-7i + 2j)+ (- 4i – 3j) + (- 4i – 3j) ( 4i + 3j) ( 4i + 3j)= - 29 k + O = - 29 K= - 29 k + O = - 29 K

M M = 29 units = 29 units

M = MA = AB M = MA = AB F + F + AC AC F F

= ( 3i – 5j) = ( 3i – 5j) (-7i + 2j) (-7i + 2j)+ (- 4i – 3j) + (- 4i – 3j) ( 4i + 3j) ( 4i + 3j)= - 29 k + O = - 29 K= - 29 k + O = - 29 K

M M = 29 units = 29 units

^̂

^̂ ^̂ ^̂ ^̂^̂^̂^̂^̂^̂

AB is a uniform rod of lengthAB is a uniform rod of length60 cm , its weight is 400 gm. wt,60 cm , its weight is 400 gm. wt,rests on a horizontal position onrests on a horizontal position on

a support 20 cm from A &a support 20 cm from A &is kept in equilibrium by means is kept in equilibrium by means

of a vertical light stringof a vertical light stringat its end B .at its end B .

AB is a uniform rod of lengthAB is a uniform rod of length60 cm , its weight is 400 gm. wt,60 cm , its weight is 400 gm. wt,rests on a horizontal position onrests on a horizontal position on

a support 20 cm from A &a support 20 cm from A &is kept in equilibrium by means is kept in equilibrium by means

of a vertical light stringof a vertical light stringat its end B .at its end B .

Find :Find :

The tension in the string &The tension in the string &The reaction of the support .The reaction of the support .

Find :Find :

The tension in the string &The tension in the string &The reaction of the support .The reaction of the support .

Find :Find :

The mag. of the weight that The mag. of the weight that should be suspended at A should be suspended at A

so that the tension inso that the tension inthe string is about to bethe string is about to be

vanished .vanished .

Find :Find :

The mag. of the weight that The mag. of the weight that should be suspended at A should be suspended at A

so that the tension inso that the tension inthe string is about to bethe string is about to be

vanished .vanished .

SolutionSolution

TT

400400

RR

AACC

10cm10cm30cm30cmBB

WW20cm20cm

T + R = 400T + R = 400, M , M BB = 0 = 0 40 R – 400 40 R – 400 30 = 0 30 = 0R = 300 gm. wt R = 300 gm. wt T = 100 gm . wtT = 100 gm . wt

T + R = 400T + R = 400, M , M BB = 0 = 0 40 R – 400 40 R – 400 30 = 0 30 = 0R = 300 gm. wt R = 300 gm. wt T = 100 gm . wtT = 100 gm . wt

TT

400400

RR

AACC

10cm10cm30cm30cmBB

WW20cm20cm

When T = 0 When T = 0 MC = 0 MC = 0 400 400 10 – 20 w = 0 10 – 20 w = 0 W = 200 gm . Wt.W = 200 gm . Wt.

When T = 0 When T = 0 MC = 0 MC = 0 400 400 10 – 20 w = 0 10 – 20 w = 0 W = 200 gm . Wt.W = 200 gm . Wt.

A B C D is a rectangle inA B C D is a rectangle inWhich AB = 6 cm , BC = 8 cm.Which AB = 6 cm , BC = 8 cm.

Forces of mag.15 , 20 , 3 , 4 , FForces of mag.15 , 20 , 3 , 4 , FNewton act along BA , CB , DCNewton act along BA , CB , DC

AD , AC resp. If the system isAD , AC resp. If the system isEquivalent to a couple ,Equivalent to a couple , find F & the momentfind F & the momentnorm of the couplenorm of the couple

A B C D is a rectangle inA B C D is a rectangle inWhich AB = 6 cm , BC = 8 cm.Which AB = 6 cm , BC = 8 cm.

Forces of mag.15 , 20 , 3 , 4 , FForces of mag.15 , 20 , 3 , 4 , FNewton act along BA , CB , DCNewton act along BA , CB , DC

AD , AC resp. If the system isAD , AC resp. If the system isEquivalent to a couple ,Equivalent to a couple , find F & the momentfind F & the momentnorm of the couplenorm of the couple

SolutionSolution