Equilibruim of forces and how three forces meet at a point

7/29/2019 Equilibruim of forces and how three forces meet at a point

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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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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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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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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


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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3 units

4 unitsT

F

B

Another method :

2 2oIn ABC : ( B ) 90 AC 3 4 5 cm


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


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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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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2 3

2 3 2 3

2 3

2 3 2 3

2 3 2 3

3 3


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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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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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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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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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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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


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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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


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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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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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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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


18/32





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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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

---------------------------------------------------------------------------------------------------------------------


20/32





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


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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80 cm

40 cmDA

C

B

M

T R

16

20 cm

o

2 2

In ABC , m( A) 90

BC 80 60 100cm


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 :


22/32





2 2o

1 2

he didn't say horizontal .... So the rod is not horizontal

In ABC : m(


23/32





2 2


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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1 2

1 2

o o

Note: Any reaction is perpendicular to its plane .


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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1 2

1 2



SoSince R and R intersect at M , Therefore W


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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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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The three forces must meet at a point of contactSoSince W and T intersect at O , Therefore R


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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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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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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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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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

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Equilibruim of forces and how three forces meet at a point

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