stat,dinamic,2,2009
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First : staticsFirst : statics
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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
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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
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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 ,
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SolutionSolution
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The angle between The angle between the the
Resultant & the 1 - Resultant & the 1 - force force
does not changedoes not change
stst
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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
=
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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
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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
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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 .
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SolutionSolution
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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
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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
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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
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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
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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
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OO AA
BB
XXCCDD
HH
yy
22
33
44
2233
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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.
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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
^̂^̂
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SolutionSolution
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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
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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^̂ ^̂
^̂ ^̂
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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
^̂ ^̂^̂ ^̂ ^̂ ^̂
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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 .
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SolutionSolution
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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
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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 ^̂
^̂ ^̂ ^̂ ^̂^̂
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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
^̂
^̂ ^̂ ^̂ ^̂^̂^̂^̂^̂^̂
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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 .
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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 .
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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 .
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SolutionSolution
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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
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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.
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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
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SolutionSolution