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7/29/2019 Equilibruim of forces and how three forces meet at a point
1/32
Facebook page : https://www.facebook.com/MrSherifYehiaAlMaraghy
Mr . Sher if Yehia Al Maraghy https://twitter.com/Mr_Sheri f_yehia
0100998883601009988826 Email : Rooshery@hotmai l.com
Static 3rdsecondary Chapter Two Equil ibri um of forces-2-
SO:To prove that a system of coplanar forces are in equilibrium Stable , you have to
Prove that the algebraic sum in any direction VANISHES
Here Vanishes in any direction means that R = 0 X = 0 and Y = 0
o o o
2 2
X 8 3 Cos30 6 Sin30 14 Sin30 4 Sin30 0
Y 6 Cos30 14Cos30 8 3 Sin30 4Cos30 12 3 0
R X Y 0
The forces are in equilibrium
o30
6
14
12 3
4
o30
o30
8 3
Equil ibrium of Coplanar forces meeting at a point
Introduction: A particle is said to be in equilibriumwhen its acted upon two or more forcesand motion doesnt take place , This means that the resultant of the forces is
zero R = 0 Or stationary as it is not subject to acceleration
---------------------------------------------------------------------------------------------------------------------Example (1)
Five forces 8 3 ,6 ,4 ,12 3 ,14 kg.wt act at a point , the first acts at30north of east , the
Second acts at30 west of north, the third acts at30west of south , the fourth is towardssouth, And the fifth is towards 60north of west . prove that the forces are in Equilibrium.
Answer
---------------------------------------------------------------------------------------------------------------------
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Static 3rdsecondary Chapter Two Equil ibri um of forces--
2 2
ABCD is a rectangle , AD and DC are perpendicular
In HDC : DC 6 cm And HD 2 cm and m( D ) 90
By using Pythagoras : HC 2 6 2 10 cm
6 3Let m( DCH ) Cos
2 10 102 1
Sin2 10 10
In ABC : AB 6 cm BC 8 cm and m( B
2 2
1
1 1 1
2
) 90
By using Pythagoras : AC 8 6 10 cm
Let m( ACB )
8 4 6 3Cos And Sin
10 5 10 5Forces are in Equilibrium R 0 " X 0 and Y 0 "
X 6 10 Sin 5Cos F
1 46 10 5 F 0 2 F 0 F 2 kg.wt
510
Y F 6 10 Cos 5 S
2 2 2
in
3 3
F 6 10 5 0 F 18 3 0 F 15 kg.wt510
A
B C
D2cm
6cm
H
6cm
2F
1F
6 105
Example (2)
ABCD is a rectangle where AB = 6 cm , BC = 8 cm , A point H AD is taken such thatAH = 6 cm , four forces of magnitude
1F , 5 ,6 10 ,
2F kg .wt. act Along (c) in the directions
CB,CA,HC,CD respectively, If the forces are in equilibrium,Then find 1 2F and F
Answer
---------------------------------------------------------------------------------------------------------------------
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Static 3rdsecondary Chapter Two Equil ibri um of forces--
We use triangle of forces if and only if(1) There are three forces act at a point
(2)The sides of the triangles are in the same
cyclic order .(3) If you can find the sidesof the triangle of
forces .
A
B C
D
EF
AB
2DA
AC
AEAF
Let O is the centre of the hexagon.
AB AE AD
AC AF AD
AB AE AC AF 2AD
AB AE AC AF 2DA 2AD 2DA 0
The set of the givenforces are i
n equilibrium
There are another methods to solve Coplanar forces
LAMI 'S rule Tr iangle of forces
We use Lamis rule if three forces act at aPoint . and you can find the anglesbetween
Each two forces .
1
2 1F
2F
W
1
2
3
W
1F
2F
W
1
2
1F
2F
A
B
C1 2
F F W
= =B C A C A B
1 2
1 2 3
F F W
Sin Sin sin
1 2
1 2 3
F F W= =Sin Sin Sin
Example (3)
ABCDEF is a regular hexagon , prove that the forces represented by the lengths AB , AC , 2DA ,
AE , AF acting along the directions AB ,AC ,DA ,AE ,AFrespectively are in equilibrium.
Answer
---------------------------------------------------------------------------------------------------------------------
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Static 3rdsecondary Chapter Two Equil ibri um of forces--
10
8
12
Three forces are in Equilibrium , Then any force of them is the resultant
of the other two forces
HereThe measure of the angle between the first
two forces means that the Resultant here is the third
1 2
2 2 2
1 2 1 2
1 o
force
F 8 N F 10 N R 12N ?
R F F 2F F Cos
144 64 100 2 8 10 Cos
144 164 1 1Cos - Cos - 97 11' 160 8 8
A A
BBC Co45
o45
o30
o60
1
1
122
3
AB : BC : AC 1:1: 2 AB : BC : AC 1: 3 : 2
W W
1F
1F
2F
2F
1 2 1 2W F F W F F
= = = =AB BC AC 1 1 2
1 2 1 2
W F F W F F = = = =
AB BC AC 1 23
Very important remarks
(1) from the above , if three forces acted on a particle can be represented by the sides of a
triangle taken in the same cycli c order , Then they are in equilibrium.
(2) To prove that three forces are in equilibrium , Prove that the resultant of any two forces is
equal to magnitude of the third force and opposite in direction .
Conversely : Ifthreeforces meeting at a poin tare in equilibrium, Then the resultantof
any two forces of them is equal to the thirdforceand opposite in direction .
Example
Three forces of magnitude 8 , 10 and 12 Newton act at a point , if the forces are in equilibrium
, what is the measure of the angle between the fi rst two forces.
Answer
(3) Three forces are in equilibrium if the magnitude of the largest one is less than the sum of
the other two forces
(4) if you have a right angled triangle which make a triangle of forces Then :
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Static 3rdsecondary Chapter Two Equil ibri um of forces--
30 cm
50 cm
40 cmT
F
32W kg.wt
B
Another method
Another method
2 2o
:
In ABC : ( B ) 90 AB 50 30 40 cm
The body is in equilibruim :
3T Cos F 0 F T Cos T (1)
5
4T Sin 32 0 T 32
5
32 5T 40 kg.wt
43
Substitute in (1) F 40 24 kg.wt 5
:
A
C
Examples
Example (1)
A lamp of weight 32 kg.wt. is suspended by a string of length 50 cm , the other end of the stringis fixed at a point on the vertical wall, the body is pulled by a horizontal force (F) until it
becomes 30 cm , apart the wall , find the horizontal force and the tension in the string in state
of equilibrium .
Answer
2 2o
In ABC : ( B ) 90 AB 50 30 40 cm
The body is in equilibruim :
ABC is the triangle of forces :
W T F 32 T F
AB AC BC 40 50 30
50 32T 40kg.wt.
40
30 32F 24kg.wt.
40
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Static 3rdsecondary Chapter Two Equil ibri um of forces--
3 units
4 unitsT
F
B
Another method :
2 2oIn ABC : ( B ) 90 AC 3 4 5 cm
The body is in equilibruim :
3T Cos F 0 F T Cos T (1)
5
4T Sin 60 0 T 60
5
60 5T 75 gm.wt
4
3Substitute in (1) F 75 45 gm.wt 5
5 units
A
C
60W gm.wt
Example (2)
A weight of 60 gm.wt. is suspended by a string , its other end is fixed at a point on a vertical
wall , the body is pulled by a force (F) perpendicular to the wall until the string inclines to the
wall by an angle of measure where Tan = 34
, Find the force and the tension in the string
in state of equilibrium .
Answer
2 2o
In ABC : ( B ) 90 AC 3 4 5 Units
The body is in equilibruim :
ABC is the triangle of forces :
W T F 60 T F
AB AC BC 4 5 3
60 5T 75 kg.wt.
4
60 3F 45 kg.wt.
4
Another method:
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Static 3rdsecondary Chapter Two Equil ibri um of forces--
1
B
2
65 cm
85 25 60 cm 25 cm
A
C
26 gm.wt
1
T2T
1
2
2 2 2 2 2 2
2 2 2
2 1
(AB) (65 ) 4225 And ( BC ) ( AC ) (60 ) ( 25 ) 4225
(AB) ( BC ) ( AC )
ABC is a right angled triangle at C
60 12 25 5
Sin And Sin65 13 65 13
By using the Tiangle of forces method inthe opposite figure:
by lami
1 2
2 1
1 2
2 1
s rule
T T 26
Sin Sin sin90
T T26
12 5
13 135 12
T 26 10 gm .wt & T 26 24 gm .wt 13 13
1T
2T
26
1
2
1
B
2
25 3 gm.wt
A
C
50 gm.wt
2T
o o
2 1
2 1
1 o
2 2
1
By using the Tiangle of forces method inthe opposite figure:
by lamis rule
25 3 25 50
sin90Sin 180 Sin 180
25 3 25 50
Sin Sin sin90
25 3 sin90 3 3Sin Sin 60
50 2 2
25 sin90 1Sin
50 2
1 o
2
1Sin 30
2
25 3
25
50
1
2
25 gm.wt
Example (3)
A weight of 50 gm.wt is suspended by two perpendicular strings , if the tensions in the two
strings are 25 , 25 3 gm.wt. , find the measure of the angle at which each of the two strings
inclines to the vertical .Answer
---------------------------------------------------------------------------------------------------------------------
Example (4)
A light string of length 85 cm , its two ends are fixed at A and B where AB is horizontal lineand AB = 65 cm , A body of weight 26 gm.wt. is suspended from C where C is a point in the
string and AC = 25 cm , find the tension in each part .
Answer
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Static 3rdsecondary Chapter Two Equil ibri um of forces-23-
2 3
2 3 2 3
2 3
2 3 2 3
2 3 2 3
3 3
The body is in equilibruim :
10 T Sin60 T Cos60 0
3 1T Sin60 T cos60 10 T T 10 " 2"
2 2
3T T 20 (1)
T Cos60 T Sin60 0 T Cos60 T Sin60
1 3T T T 3 T ( 2 )
2 2
Substitute (2) in (1) : 3 T T 20 3 3 2
4 T 20 T 5 N and T 5 3 N
o60
2T
1
T 10 N
3T
o60
Example (5)
A point is pulled by three strings , the first is towards east , the second in direction makes an
angle of measure 60 west of north , and the third in direction makes an angle of measure 60
south of west , if the tension in the first string equals 10 Newton , then find the tension in eachof the other two strings in state of equilibrium .
Answer
Another solution :
By using lami`s rule
2 3
2 3
10 T T
sin90 sin120 sin150
10sin120 10sin150T 5 3 Newton & T 5 Newton
sin90 sin90
---------------------------------------------------------------------------------------------------------------------
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Static 3rdsecondary Chapter Two Equil ibri um of forces-2-
B
C
T
F
5 gm.wt
o60
Note
four forces act the particle , Then we can not use tri angle of forces
: Smooth ring Slides on the string means that the tension on both parts of the string is equal
So Forces are in Equilibrium " R 0 ,
o o
o o
X 0 and Y 0"
X T Cos 60 F T Cos 30
1 3 3 1T F T 0 F T T (1)2 2 2 2
Y T Sin 60 T Sin 30 5
1 3T T 5 " 2 " T 3 T 10 T 1 3 10
2 2
10T 5 3 5 gm.wt 5 3 1 gm.wt
1 3
3 1Substitute in (1) : F 5 3 5 5 3 5 10 5 3 gm.wt. 5 2 3 gm.wt 2 2
T
o30
o30o60
B
C
68
1T
A17cm
15cm
2T
8cm
2 2
2 2 2 2
2 2 2
1 2
1 21
The three forces must meet at point C
In ABC: (AB) (17 ) 289
and ( BC ) ( AC ) (15 ) ( 8 ) 289
( AB ) ( BC ) ( AC )
ABCis a right angled at C.
W T Tby lamis Rule:
sin90 sin(90 ) sin(90 )
68 T T 6 T
sin90 cos cos
2
8cos 1568 60 Newton
sin90 1768cos 8
T 68 32 Newtonsin90 17
Example (6)
The two ends of a string are fixed at two points A and B in the same horizontal plane , a smooth
ring of weight 5 gm.wt slides on the string . A horizontal force F acts at the ring is in
Equilibriumwhen the two parts of the string inclines to AB at 30 ,60 ,then find the tension inThe string and magnitude of F.
Answer
---------------------------------------------------------------------------------------------------------------------
Example (7)
A body of weight 68 Newton is suspended by two string with the lengths 8 cm, 15 cm ,the other
two ends of the strings are fixed at the points A and B on a horizontal line where AB=17 cm ,Find in state of equilibrium the magnitude of the tension in each string.
Answer
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Static 3rdsecondary Chapter Two Equil ibri um of forces-2-
o60
2T 2T
1T 3T30
F 20kg.wt.
A B
C D
o60
2T
3T
20kg.wt.
Do120
o150
2 1 1
1
C is in equilibrium , By lami`s rule we find that:
T F T 20 3 F T
Sin150 Sin120 Sin90 Sin150 Sin120 Sin90
20 3 Sin120 20 3 Sin90F 60 kg.wt. & T 40 3 kg.wt.
Sin150 Sin150
2 3
2 3
Dis in equilibrium , By lami`s rule we find that:
20 T T
Sin150 Sin120 Sin90
20Sin120 20Sin90T 20 3 kg.wt. & T 40 kg.wt.Sin150 Sin150
C
1T
2T
F
30
o120
o150
Example (8)
AB is a light string , its two ends are fixed in two points in a horizontal straight line , From C
and D on the string two bodies of weights F and 20 kg.wt. are suspended , the set is in
equilibrium when CD is horizontally and the inclination ofAC andBD with the vertical are30 and 60 , find the tension in each part of the string .
Answer
1 2 3Let the tensions in AC,CD and DB are T ,T and T respectively as in the figure
---------------------------------------------------------------------------------------------------------------------
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Static 3rdsecondary Chapter Two Equil ibri um of forces-22-
1T
30F
20 N
A
B
C
D
30F
20 N
A
B
C
D
2T2T
3T
60
60
1T
20 N
B6030
2T
2 1
2 1
B is in equilibrium , By lami`s rule we find that:
T T 20
Sin150 Sin120 Sin90
20Sin150 20Sin120T 10 N & T 10 3 N
Sin90 Sin90
3
3
C is in equilibrium , By lami`s rule we find that:
T F 10
Sin150 Sin120 Sin90
10Sin150 10Sin120T 5 N & F 5 3 N
Sin90 Sin90
1 2 3Let the tensions in AB,BC and CD are T ,T and T respectively as in the figure
30F
C
3T
2T
Example (9)
In the opposite : ABCD is a light string , A and D
are fixed on the line AD , from B , a body of
weight 20 newton is suspended , from CA force of magnitude F pulls the string until the
partBCbecomes vertically and the inclination of
CD with the horizontal is 30 andAB BC , Then
find F and the tension in each part of the string .
Answer
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Static 3rdsecondary Chapter Two Equil ibri um of forces-2-
W
R
If a body of weight W is placed on a smooth inclined plane which
inclines by with the horizontal , Then the body will be under the
Action of two forces :
1 The weight force W acting vertically dow
nwards .
2 The reaction force R of the inclined plane and it acts
In the direction perpendicular to the plane except
External influences act at the body " hinge , rough ground , ...... "
These two forces can not be in equilibrium because their line of actions are not the same .So , we have to get a third force to act on the body
300
R
o30
F
o
o
o
1Tan m( ) 30 and by lami`s rule
3
F F 300 R
sin150 sin60 sin150sin 180 30
300 Sin30F R 100 3 gm.wt.
Sin60
300
R
o30
F
300 Cos
300 Sin
oF Cos30
oF Sin 30
o o
oo o
o
o o
The forces are in equilibrium R 0
F Cos30 300 Sin30 0
300Sin30F Cos30 300 Sin 30 F 100 3 gm.wt
Cos30
R F Sin30 300Cos30 0
R 50 3 150 3 0 R 100 3 gm.wt
Another method
Another shape of Equilibrium .
Equilibrium of a body on a smooth inclined Plane
---------------------------------------------------------------------------------------------------------------------
Example (1)
A body of weight 300 gm .wt. is placed on a smooth plane inclined to the horizontal at an angle
where1
Tan .3
The body is kept in equilibriumby a force inclines to the line of the
greatest slope of the plane at an angle of measure 30upwards .Find this force and the
Reaction of the plane.Answer
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Static 3rdsecondary Chapter Two Equil ibri um of forces-2-
5 3
R
5 kg.wt
o
5 5 3 R
Sin( 90 ) Sin( 90 2 )Sin 180
5 5 3 R
Sin Cos Cos2Sin 5 1
m( ) 30Cos 5 3 3
5 RR 5 kg.wt
Sin30 Cos60
5 3
R
5
5 3 Cos
5 3 Sin
5 Cos5 Sin
o
o o
The forces are in equilibrium R 0
5Cos 5 3 Sin 0
5 Sin5Cos 5 3 Sin Tan
Cos5 3
1Tan 30
3
R 5Sin 5 3 Cos 0
R 5 Sin 30 5 3 Cos30 0 R 5 gm.wt
Another method
Example (2)
A body of weight5 3 kg .wt. is placed on a smooth plane inclined to the horizontal at an angle
of measure . If the body is kept in equilibriumby a force 5 kg .wt. inclines to the line of
The greatest slope of the plane at an angle of measure also upwards .Find and the reactionOf the plane .
Answer
By lami`s ru le
---------------------------------------------------------------------------------------------------------------------
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Static 3rdsecondary Chapter Two Equil ibri um of forces-2-
W
R
27 N
27 R w
sin( 180 ) sin 90 sin( 90 )
27 R w
sin 1 cos
cos 40w 27 27 cot 27 120 Newton
sin 9
1 41R 27 27 123 Newton.
sin 9
W
R
W Cos
W Sin
27 Cos
The forces are in equilibrium R 0
27Cos W Sin 0
40 9 40W Sin 27 W 27 41 41 41
W 120 N
R 27 Sin W Cos 0
9 40R 27 120 0 R 123 0 R 123 N
41 41
Another method
27 N
27 Sin
9Tan
40
9
40
41
Example (3)A body of weight (W) Newton is placed on a smooth plane inclined to the horizontal at Angle of
measure where9
Tan
40
. The body is kept in equilibriumby a horizontal force of
Magnitude 27 Newton .find (W) and the reaction of the plane.
Answer
By using lami`s rule
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Static 3rdsecondary Chapter Two Equil ibri um of forces-2-
52 kg.wt
R
Let is the angle of inclination of the plane .
5 12Sin , Cos
13 13
Also Sin 0.8 Cos 0.6
By lami s rule
F R 52sin(180 ) sin(90 ) sin(90 )
F R 52(1)
sin cos( ) cos
1cos( ) cos cos sin sin
2 6 5 8 32 16
13 10 13 10 130 65
By substitution in ( 1)
F R 52
5 16 0.6
13 65
5 5 100 16 5 64F 52 kg.wt & R 52 kg .wt
13 3 3 65 3 3
F
5
13
12
8 4
Sin 0.8 10 5
4
3
5
Example (4)
A body of weight 52 kg. wt. is placed on a smooth plane of length 13 decimeter and height 5
decimeter . the body is kept in equilibriumby pulling it by a force inclines to the line of the
Greatest slope of the plane at an acute angle where sin=0.8 .Find F and the reaction of thePlane .
Answer
---------------------------------------------------------------------------------------------------------------------
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Static 3rdsecondary Chapter Two Equil ibri um of forces-2-
o30
o60
Example (5)
A body of weight 8 kg .wt. is placed on a smooth inclined plane whose inclination angle is
30 to the horizontal . The body is in equilibrium under a forceF.FindFand the reaction of
the plane if:(i) Facts in the direction of the plane upwards.
(ii) Fmakes with the horizontal an angle 60 upwards
Answer
---------------------------------------------------------------------------------------------------------------------
( i )By u sing Lami s rule
F N 8
Sin150 Sin120 Sin90
F N 8
1 13
2 2
1 3F 8 4 kg. wt. , N 8 4 3 kg. wt.
2 2
( ii )By u sin g Lami s rule
F N 8
Sin150 Sin150 Sin60
F N 81 1 3
2 2 2
1 2 8 3 8 3F 8 kg. wt. , N kg.
2 3 33
wt.
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Static 3rdsecondary Chapter Two Equil ibri um of forces-2-
RuleIf a bodyis in equilibrium under the action ofthr ee forcessuch that two lines of actions of
any two forces meet at a point , Then you can extend the line of action of the third force tomeet the other two forces at the previous point .
Conclusion: The three forces must intersect at one point
How to sketch
STEP BY STEP...
1. Identify a body with 3 relevant forces.
2. Get all the information you can. Especially ANGLES. Look for gravity (270
o
),special reactions like cables (Angle= cable angle & tensile) no fri ction(90o to Plane)
etc. See table below. (Support Reactions)
3. You must know the direction of 2 forces. Where these intersect is the concurrency point.The third force (unknown angle) must also go through this point.
4. If you drew the relevant details to scale, you now have ALL 3 ANGLES. You have finished
the sketch.
Braced Frame
A weight force of 150kN hangs
from the end of this beam. Thebeam is supported by a strut.
ote:
The strut is a 2 force body The beam is a 3 force body
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Static 3rdsecondary Chapter Two Equil ibri um of forces-3-
j
i
A
B
C
60o
When you wish
upon a star....
h
h/2
h/4
D
E80o
j
i
A
B
C
h
h/2
3h/4 D
E
C
RAy REy
FBD FBD
FBD FBD
R(1/3)Cx
R(1/3)CyR(1/3)
Cy
R(1/3)Cx
(1)
(2)
(3)
W
Mickey Mouse of weight W stands on a step-ladder as shown in the figure. Friction, and theweight of all components in the ladder, may be neglected
So the reactionis perpendicularto the plane floor---------------------------------------------------------------------------------------------------------------------
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Static 3rdsecondary Chapter Two Equil ibri um of forces--
RCos
RSin
T Cos
T Sin
24kg.wt
80cm
30cm 30cm
50cm
50cm
50cm40cm
R
T
o
2 2
The three forces must meet at a point of contact
So Since W and T intersect at M , Therefore
the reaction R passes through M
In ABC , m( A) 90
BC 80 60 100cm
And D is a mid - point of AB and DM / / AC
M
is the mid - point of BC BM MC 50cm
1MD AC 40 cm
2
1Also AM BC " Median from a right angled "
2
MAB is an Isosceles triangle So m( MAB) M( MBA)
Rod is in Equilibrium " R 0 , X 0 and Y 0"
R Cos T Cos 0 R Cos
T Cos
R T (1)
R Sin T Sin 24 0 T Sin T Sin 24 0
24 402T Sin 24 T Sin 12 T 12
2 50
12 50T 15 kg.wt
40
R 15 kg.wt
A B
C
D
M
Example (1)
AB is a uniform rod of length 60 cm . and weight 24 kg .wt. acting at (D) the middle point of
the Rod . Its end (A) is attached to a hingefixed at a vertical wall , the other end (B) is
attached to a string fixed its end at a point (C) on the wall lies vertically above The point (A)such that AC = 80 cm . if the rodequilibriumwhen it is in a horizontal position,Then findthe tension in the string and the reaction of the hinge at A.
Answer
---------------------------------------------------------------------------------------------------------------------
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Static 3rdsecondary Chapter Two Equil ibri um of forces--
24kg.wt
80cm
30cm 30cm
R
T
A B
C
o
2 2
In ABC , m( A) 90
BC 80 60 100cmThe three forces must meet at a point of contact
So Since W and T intersect at M , Therefore the
reaction R passes through M ,So the triangle
AMC is the triangle o
f forces.
M
D
D is a mid - point of AB and DM / / AC
M is the mid - point of BC BM MC 50cm
1MD AC 40 cm2
1Also AM BC 50 " Median from a right angled "
2
24kg.wt
80 cm
40cm40cm
R
T
A
B
C
he didn't say Horizontal in the problem , so draw the rod inclinedNote:
E
D
40 3 cm
The three forces must meet at a point of contact
SoSince T and W intersect at E , Therefore R
also passes through E
DE / / AC and D is the mid point of AB
E is the mid-point of BC , and AC AB
AE BC EC 40 3 cm
22
In AEC : By using pythagorasTheorem :
AE= 80 40 3 40 cm
R T 24
AEC is the triangle of forces 40 8040 3
24 24R 40 12 kg.wt. And T 40 3 12 3 kg.wt.
80 80
40 cm
40 3 cm
T R 24 50 24T 15 kg .wt.
50 50 80 80
Another method :
And R = 15 kg. wt. ,
---------------------------------------------------------------------------------------------------------------------
Example (2)
AB is a uniform rod of length 80 cm . and weight 24 kg .wt. acting at its mid-point (D) .Its
End (A) is attached to a hingefixed at a vertical wall, the other end (B) is attached to a stringof Length 80 3 cm, fixed its end at a point (C) on the wall lies vertically above (A) such that
AC = 80 cm . If the rod is in equilibrium, then find the tension in the string and the reaction of
the Hinge.
Answer
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Static 3rdsecondary Chapter Two Equil ibri um of forces-2-
80 cm
40 cmDA
C
B
M
T R
16
20 cm
o
2 2
In ABC , m( A) 90
BC 80 60 100cm
The three forces must meet at a point of contact
So Since W and T intersect at M , Therefore the
reaction R passes through M ,So the triangle
AMC is the triangle o
f forces.
2
2
2 2
BM BD BM 40DM / / AC ,
BC BA 100 60
40 100 200 100BM cm CM cm
60 3 3
200 160In MDB : MD 40 cm.
3 3160 20 73
In MAD : AM ( 20 ) ( ) cm.3 3
AMC is the triangle of forces
R T 16 4R 73 kg.wt
100 80 320 7333
20 2
and T= 6 kg.wt.3 3
Rod is in Equilibrium " R 0 , X 0 and Y 0"
R Cos T Cos 0 R Cos T Cos
20 3 6 73R T R T (1)
10 520 73
160 3 73 8R Sin T Sin 16 T T 16
5 1020 73
38 4 12 16 5 20 73 20 4 73
T T 16 T 16 T kg.wt R kg.wt 5 5 5 12 3 5 3 3
80 cm
40 cmA
C
B
M
T R
16
20 cm
160
3
D
Example (3)
AB is a rod of length 60 cm , and weight 16 gm .wt . acting at a point (D) on the rod where
AD=20 cm , its end (A) is attached to a hingefixed at a vertical wall and the other end (B) is
Attached to a string fixed its ends at a point (C) on the wall lies vertically above the point (A)Such that AC = 80 cm . if the rod is in equilibrium when it is in a hori zontal positi on, then findthe tension in the string and the reaction of the hinge.
Answer
Another method :
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Static 3rdsecondary Chapter Two Equil ibri um of forces--
2 2o
1 2
he didn't say horizontal .... So the rod is not horizontal
In ABC : m(
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Static 3rdsecondary Chapter Two Equil ibri um of forces--
2 2
The three forces must meet at a point of contact
SoSince W and T intersect at M , Therefore the reaction R
passes through M
MAB is the triangle of forces where MA 30 20 50 cm.
BM 30 cm.
AB ( 50 ) ( 30 ) 40 cm
.
By the rule of the triangle of forces
R T 200 R T 200
MB MA AB 30 50 40
R 150 gm.wt , T 250 gm .wt .
30 cm
20 cm
200 gm.wt
A
MB
30 cm
40 cm
T
R
200 gm.wt
A
MB
T
R
T Cos
o
30 3 40 4In ABM : Sin And Cos50 5 50 5
By lami s rule
T R W
sin 90 sin(90 ) sin 90
T R 200 200 5T 200 250 gm.wt
1 cos cos cos 4
3200
200 cos 5R 150 gm.wt4cos
5
Rod is in Equil
Another method :
Thi rd Solution :
ibrium " R 0 , X 0 and Y 0"
3R T Cos 0 R T Cos R T (1)
5
4T Sin 200 0 T Sin 200 T 200
5
200 5 3T 250 gm.wt R 250 150 gm.wt
4 5
MR
T Cos
T SinT
Example (5)
A homogenous sphere of radius 30 cm and weight 200 gm.wt rests on a smoothvertical wall
And suspended from a point on its surface by a string of length 20 cm, the other end of the string
Is fixed at a point on the wall vertically above the point of contactof the sphere with the wall.Find tension in the string and the reaction of the wall on the sphere .
Answer
Note : Point of contact means point of intersection
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Static 3rdsecondary Chapter Two Equil ibri um of forces--
1 2
1 2
o o
Note: Any reaction is perpendicular to its plane .
The three forces must meet at a point of contact
SoSince R and R intersect at M , Therefore W
also passes through M
By Lami' s Rule :
W R R
Sin150 Sin 90
o
1 2
o o o
o
1 o
o
2 o
Sin120
15 R R
Sin150 Sin 90 Sin120
15Sin90R ( reaction of the inclined plane ) 30 kg.wt
Sin150
15Sin120And R ( reaction of the vertical plane ) 15 3 kg .wt.
Sin150
M
B
1R
2R
15 kg.wt
o30
o60
o30
Example (6)
A metallic sphere of weight 15 kg.wt is placed so that it touches two smoothplanes , one of
Them is vertical and the other inclines to the verticalat angle of measure 30 .Find the reactions
Of the two planes.Answer
o
2 1 2 1
o
1 1
1 2
: The sphere is in equilibrium
3R R Cos 30 0 R R
2
1R Sin 30 15 0 R 15
2
3
R 30 kg.wt R 30 15 3 kg.wt2
Another method
---------------------------------------------------------------------------------------------------------------------
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Static 3rdsecondary Chapter Two Equil ibri um of forces--
1 2
1 2
Note: Any reaction is perpendicular to its plane .
The three forces must meet at a point of contact
SoSince R and R intersect at M , Therefore W
also passes through M
By Lami' s Rule :R R
Sin 90 Sin(180
1 2
1
2
W
) Sin 90
R R W
Sin 90 Sin Cos
W W 5R W kg.wt
Cos 3 5 3
4 WW Sin 45R W kg.wtCos 3 5 3
M
B
1R
2R
4
3
5
W
Example (7)
A metallic sphere of weight (W) kg.wt is placed so that it touches two smoothplanes , one
of them is vertical and the other inclines to the horizontalat an angle of measure where
3Cos5
, if the sphere is stable, find the reaction of the two planes .
Answer
o o
2 1 2 1 1
o
1 1 1
2
: The sphere is in equilibrium
4R R Sin 0 R R Sin R
5
3 5R Cos W 0 R W R W
5 34 5 4
R W W kg.wt5 3 3
Another method
---------------------------------------------------------------------------------------------------------------------
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Static 3rdsecondary Chapter Two Equil ibri um of forces--
45 cm
CA
D
B
M
T
RW
30 cm
50 cm40 cm
15
The rod is in equilibrium by 3 forces : The weight (W)
which acts vertically at M (the mid-point of AB )
The tension (T) in the string CD
and the reaction (R) of the hinge at AAs the two forces (W) and (
2 2
T) meet at N .
( R ) passes through N
the two triangles CAD and CMN are similar .
CA AD CD
CM MN CN
30 40 50MN 20 cm and CN=25 cm.
15 MN CN
DN DC CN 50 25 75 cm.
In AMN : AN ( 20 ) ( 45 ) 2425 5 97 cm.
AND is the
triangle of ofrces .
R T W
AN 75 40
97R W kg.wt
8
15T W kg.wt
8
N
Example (8)
AB is a uniform rod of length 90 cm . and weight (W) kg .wt. acting at the middle of the rod .
Its end (A) is attached to a hingefixed at a vertical wall ,From a point (C) AB where
AC = 30 Cm , the rod is attached to a stringCD where CD=50 cm and D lies on the wallvertically above The point (A) where AD = 40 cm .If the rod is in equilibrium when it is in a
horizontalposition , Find the tension in the string and the reaction of the hinge at (A).
Answer
---------------------------------------------------------------------------------------------------------------------
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Static 3rdsecondary Chapter Two Equil ibri um of forces--
Note: Any reaction is perpendicular to its plane .
The three forces must meet at a point of contactSoSince W and T intersect at O , Therefore R
also passes through M
In AOB : OA AB
OA r r 1Sin 3
OB r r 2r 2
o0
So m( OBA) the measure of the inclination of the plane
to the horizontal ,So it is in the horizontal state .
O
1R
T
o30
312
r r
A
B
o3 Cos 3012o3 Sin3012
T Cos
T Sin
o o
o
o
o
R 12 3 T
Sin 90 Sin 120 Sin150
12 3R 24 kg.wt
Sin 120
12 3 Sin150T 12 kg.wt
Sin 120
Example (9)
A sphere of radius (r) and weight12 3 kg .wt. acting at its centre is placed on an inclined
Smoothplane which inclines to the horizontal at 30, the sphere is prevented from slipping by a
String is fixed at a point on its surface and the end of the string is fixed at a point on the inclinedPlane . if the length of the string equals r and the string lies in the vertical plane which passes
Through the line of the greatest slope , prove the string will be horizontal in state of equilibrium
And find the tension in the stringwhich make the sphere doesnt move downward .and the reaction of the plane on the sphere
Answer
---------------------------------------------------------------------------------------------------------------------
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Static 3rdsecondary Chapter Two Equil ibri um of forces-3-
o
The two lines of action of T and the weight meet at N , So the reaction R passes through N.
C is the mid-point of AB
C is the mid-point of NM
" Side opposite to 30 equals half the hypotunes "
In AMC : Cos
o
2 2
MA30AC
3MA AC Cos30 60 30 3 cm
2
1MN 2MC 2 AC sin30 2 60 60 cm.
2
From ANM : AN ( 30 3 ) (60 ) 6300 30 7 cm
AMN is the triangle of forces
T R 180 T R 180
MA AN NM 6030 3 30 3
T 90 3 gm.wt , R 90 7 gm.wt
R
180
60 cm
o30
A
B
C
N
M60 cm
T
o30
Example (10)
A uniform rod of length 120 cm and weight 180 gm.wt rests at one of two ends on a rough
Horizontal ground, and its other end is pulled by a horizontal string makes 30 with the rod .
Prove that when the rod is in equilibrium the reaction of the ground will be equal to 90 7 gm.wt. and the tension in the string equals 90 3gm .wt.
Answer
---------------------------------------------------------------------------------------------------------------------
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Static 3rdsecondary Chapter Two Equil ibri um of forces--
o
2
As the line of action of the weight and the reaction of the wall intersect at M
The tension in the string passes through M .
MAD is the triangle of forces
1MN AN 20 cm " side opposite to 30 "
2
AM 40 20
2
2 2
20 3 cm
MN // AD CMN CDA
MN CN CM 20 1
DA CA CD 60 3
AD 3MN 60 cm
In AMD : m( A ) 90
MD (60 ) ( 20 3 ) 40 3 cm
60 R T 60 R T
DA AM MD 60 20 3 40 3
R 20 3 Newton And T 40 3 Newton.
o60
o60
There is no fricition
So reaction is to wall
Wrong Graph
Example (11)
AB is a uniform rod of lenght 80 cm and weight 60 Newton.The rod rests at A on a smooth
vertical wall and is tied from C AB ,where BC 20 cm , by a string CD where D lies on the
wall over A , the inclin
ation angle of the rod on the wall is 60 in the position of equilibrium.
Find the tension in the string and the reaction of the wall.
Answer
----------------------------------------------------------------------------------------------------------------------
R
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Static 3rdsecondary Chapter Two Equil ibri um of forces--
ACE is the triangle of forces .
1join CD CD AB AD `Median from right angle `
2
In ABC , ED // AC & D is a mid - point of AB
E is a mid point of BC
1ED AC (1)
2
In BCD , E is a mid point & ED BC
BCD is an iso
2 2
2 2
sceles triangle
CD AD and m( A) 60 ,
m( A) m( ACD ) m( ADC ) 60
AC CD AD ( 2 )
1substitute ( 2 )in (1) : ED AD2
1 3In CDE :CE ( AD ) ( AD ) AD
2 2
3 7In ACE : AE ( AD ) ( AD ) AD
2 2
ACE is th triangle of forces :
T R3 7
AD2
50ADAD
2
350 AD
2T 25 3 kg. wt.AD
750AD
2R 25 7 kg. wt.
AD
o301
AD2AD
AD
A
C B
o60
o60
o60
o30
E
D
R
50
T
Example (12)
AB is a uniform rod of weight 50 Newton. Its end A is fixed at a hingein a vertical wall .The
rod is in equilibrium when a horizontal force acts at its end B If the rod inclines the vertical at
60, Find the magnitude of this force and the reaction of the hinge.Answer
----------------------------------------------------------------------------------------------------------------------
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Static 3rdsecondary Chapter Two Equil ibri um of forces-2-
o30
o60
Example (13)
AB is a uniform rod of length 2L and weight 8 Kg. wt acting at its mid-point. Its end A is hinged
at a point in a vertical wall where as its end B is attached to a light string and the other end of
the string being fixed to a point C on the wall situated vertically above A.If AB = AC = BC , And the rod is equilibrium in a plane perpendicular to the wall, find theintensity of the tension In the string as well as the reaction of the hinge at A.
Answer
The rod is in equilibrium under the effect of three forces which are :
(i) Its weight acting at M.
(ii)The tension in the string (T) acting in BC
and these two forces , the weight and (T) passes
through the point N (middle of BC ).
( iii )The action line of R must passes through (N).
Applying Lamis Rule:
8 T R
sin90 sin150 120
8 T RT 4 kg.wt and R 4 3 kg. wt.
11 3
2 2:
By using ANC
Another solution
as a triangle of forces :
8 T RT 4 kg.wt and R 4 3 kg. wt.
2L L 3L
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8
4R
A
B
2 1
C
D
1 2
m( ACB ) m( DAC ) m( CBD ) 90
ABCD is a rectan gle.
R AC And R BC And the two reactions meet at D.
The line of action of the weight of the rod passes through D.
DM (the line of action of weight) passes through
1 2
1 2
1 2
C
" Diagonals of rectangle "
let the measure of the angle of inclination of the
plane at A to the horizontal is , the plane at B is
by using Lami`s rule:
R 4 8
sin(90 ) sin(90 ) sin90
R 4
cos cos
2 2 1
8
1Cos m( ) 60 And m( ) 30
2
R 8cos30 4 3 Newton
The pressure on the plane at A 4 3 Newton
12
D
Example (14)
A uniform rod AB of weight 8 Newton acting at its midpoint is placed on two smooth
perpendicular planes that are inclined to the horizontal ,such that the vertical plane of the rod
and the two lines of greatest slope of the two inclined planes is perpendicular to theintersection line of two planes . If the magnitude of the pressure on the plane at the end B is 4Newton ,find the magnitude of the pressure on the other plane and measures of two inclination
angles of the planes to the horizontal , in the state of equilibrium .
Answer
He didnt say that the rod is in horizontal position
---------------------------------------------------------------------------------------------------------------------