Chapter 4 General Equilibrium

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General General equilibrium equilibrium

Introduction

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In previous year, we have taken the equilibrium of a body under the action of three force. In this chapter, we will deal with a body in equilibrium under the action of more than three forces placed on smooth or rough planes. ---------------------------------------------------------------------------------------------------------------------Definition : A set of coplanar forces is said to be in " equilibrium " if both the vector sum of the forces and the vector sum of the moments of the forces about any point vanish. Also when a body is acted on by such a set of coplanar forces, the body is said to be in equilibrium.

Conditionsofofequilibrium equilibrium Conditions ( 1) ∑ M = 0

( 2) ∑ X = 0

( 3 ) ∑Y = 0

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

Some notes which will make your work easier

( 1) Your drawing must be in a large scale ( 2 ) If you didn't find the word "horizontal position", then your body is inclined to the horizontal ( 3 ) The reaction of any body is always perpendicular to the plane R ( 4 ) We are dealing with two planes : Smooth planes R

( 5)

Rough planes FF

R

Equilibrium rule : About to move rule : FF = µ R where µ Coefficient of friction and R the normal reaction This is called a limiting friction

( 6 ) Resultant reaction

Static – 3rd secondary

R' = FF 2 + R 2

1

Chapter four – General equilibrium

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(7)

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Use the following methods to solve your problem quickly A

A

60 o 2x C

45o 2x

x

30 o

C

B

3x

A

45o

45o x B

x

C

A x C

α xCos α

10

Example →

45o ?

α Example →

x Sin α B

C

10

B ? = 10

2

A 10 Cos α

10 Sin α

αα

?

B

A If DE // BC AD AE DE = = AB AC BC

αα

D

E

θθ

C B C B ---------------------------------------------------------------------------------------------------------------------

Static – 3rd secondary

2

Chapter four – General equilibrium

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Problemsof ofequilibrium equilibriumon onsmooth smoothplanes planes Problems Example (1) A uniform rod AB of 80 cm long of weight 3 kg.wt is hinged at B to a vertical wall, A lamp of 12 kg.wt is hung from a point on the rod 10 cm from the end A, this end is tied with a light rope with its end C fixed on the wall at point vertically above B and far from it 60 cm so that the rod is kept horizontal, find the tension of the string and the reaction of the hinge and its direction. Answer Notes ( 1) The problem mentioned that your rod is horizontal C

( 2 ) No "smooth" or "rough" mentioned in the problem ( 3 ) When there is a hinge in the problem, then its reaction has no fixed direction, so we must make

( )

a resolution in the x - axis ( Rx ) and the y - axis R y Q AC =

T

60 cm

( 60 ) 2 + ( 80 ) 2 = 100 cm

T Sinθ

R

Ry

Rx

α

θ T Cos θ

B

∑ M B = 0 : " try to vanish the most unknown forces"

10 cm

30 cm

40 cm 3

A

12

∴ -3 × 40 − 12 × 70 + T Sinθ × 80 = 0

60 = 12 ⇒ ∴ T = 20 kg.wt 100 80 ∑ X = 0: ( → ) : Rx − T Cos θ = 0 ⇒ ∴ Rx = T Cos θ ⇒ ∴ Rx = 20 × 100 = 16 kg.wt ∴ 80T Sinθ = 960 ⇒ T Sinθ = 12 ⇒ ∴ T ×

∑Y = 0: ( ↑ )

: RY + T Sinθ = 15 ⇒ RY = 15 − 12 = 3 kg.wt

Then the reaction of the hinge is: R = Rx 2 + R y 2 = Ry

( 16 ) 2 + ( 3 ) 2 =

265 kg.wt

3 Rx 16 ---------------------------------------------------------------------------------------------------------------------And its direction is Tan α =

Static – 3rd secondary

=

3

Chapter four – General equilibrium

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Example (2) AB is a rod of 6 m long of weight 5 kg.wt and its center of gravity is at 4.5 m from A and hinged at A to a fixed horizontal hinge, the end B is tied to a light string passing over a smooth pulley lies vertically above A and at 4.5 m away from it, the string carries at its other end a body of weight W. If a body of weight 3 kg.wt is hung from B so that the rod became horizontal, then find the magnitude of W and find the pressure of the hinge. Answer Notes ( 1) The problem mentioned that your rod is horizontal

( 2 ) No "smooth" or "rough" mentioned in the problem ( 3 ) When there is a hinge in the problem, then its reaction has no fixed direction, so we must make

C

( )

a resolution in the x - axis ( Rx ) and the y - axis R y Q BC =

( 4.5 )

2

W Sinθ

W =T 4.5 m W

Ry

α

+ ( 6 ) = 7.5 m 2

T Sinθ

R

Rx

A

First : we have to know that W = T

θ T Cos θ

4.5 m

1.5 m 5

∑ M A = 0 : " try to vanish the most unknown forces"

B 3

∴ -5 × 4.5 − 3 × 6 + T Sinθ × 6 = 0

4.5 = 6.75 ⇒ ∴ T = W = 11.25 kg.wt 7.5 6 ∑ X = 0: ( → ) : Rx − T Cosθ = 0 ⇒ ∴ Rx = T Cos θ ⇒ ∴ Rx = 11.25 × 7.5 = 9 kg.wt ∴ 6 T Sinθ = 40.5 ⇒ T Sinθ = 6.75 ⇒ ∴ T ×

∑Y = 0: ( ↑ )

: " Don' t take ( W ) here as we know that W = T"

Q RY + T Sinθ = 8 ⇒ RY = 8 − 6.75 = 1.25 kg.wt 1321 kg.wt 4 ---------------------------------------------------------------------------------------------------------------------Then the reaction of the hinge is: R = Rx 2 + R y 2 =

Static – 3rd secondary

4

( 9 ) 2 + ( 1.25 ) 2 =

Chapter four – General equilibrium

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Example (3) A uniform rod AB of 160 cm long and weight 300 gm.wt is hung by a fixed nail C by the mean of two strings attached to its end A and B, a weight of 600 gm.wt is hung to a point N on the rod. If the rod is in equilibrium in a horizontal position and the two strings AC and BC are inclined to the rod at angles 60 o ,30 o respectively, find the length of AN and the magnitude of the tension in the two strings Answer Notes ( 1) The problem mentioned that your rod is horizontal C ( 2 ) No "smooth" or "rough" mentioned in the problem T1 T1 Sin60

A

T2

o

60 o

T1 Cos60

N

o

T2 Sin30 o 30 o

T2 Cos 30 80 cm

∑ M B = 0 : " try to vanish the most unknown forces"

x 300

600

o

B

80 − x

∴ 600 ( 80 − x ) + 300 × 80 − T1 Sin60 o × 160 = 0

∴ 48000 − 600x + 24000 − 80 3 T1 = 0 ⇒ ∴ 80 3 T1 + 600x = 7200 ( ÷ 40 ) ∴ 2 3 T1 + 15x = 1800 − − − ( 1) 1 3 : T1 Cos60 o − T2 Cos 30 o = 0 ⇒ ∴ T1 − T2 = 0 ⇒ ∴ T1 = 3 T2 − − − ( 2 ) 2 2 3 1 ∑Y = 0: ( ↑ ) : T1 Sin 60o − 300 − 600 − T2 Sin30 o = 0 ⇒ ∴ 2 T1 + 2 T2 = 900

∑ X = 0: ( → )

∴

3 T1 + T2 = 1800 − − − ( 3 )

Then from ( 2 ) and ( 3 ) : 3T2 + T2 = 1800 ⇒ ∴ T2 = 400 gm.wt Then from ( 2 ) : T1 = 400 3 gm.wt Then from ( 1) : 2 3 × 400 3 + 15x = 1800 ⇒ 15x = -900 ⇒ x = -60 cm ⇒ ∴ AN = 20 cm ----------------------------------------------------------------------------------------------------------------------

Static – 3rd secondary

5

Chapter four – General equilibrium

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Example (4) A uniform rod AB of 160 cm long and weight 30 kg.wt , its lower end A rests on a vertical smooth wall AD, and is tied from point H by one end of a string whose other end is attached to point D so that the rod reached its equilibrium state in a position inclined to the vertical at an angle 60 o . If the length AH = 60 cm, then find the angle ADH and also find the magnitudes of the tension of the string and the reaction of the wall. Answer Note when a body is placed on a smooth plane, then the D θ T Cos θ B reaction is perpendicular to the plane

∑ M H = 0 : " try to vanish the most unknown forces" ∴ -3 × 10 3 + R × 30 = 0 ⇒ 30R = 30 3 ⇒ ∴ R = 3 kg.wt

T Sinθ

∑ X = 0 : R − T Sin θ = 0 ⇒ ∴ T Sin θ = R = ∑ Y = 0: T Cosθ = 3 − − − ( 2 )

60 cm

Then divide ( 1) by ( 2 ) : ∴ Tan θ =

o

3 − − − ( 1) A

3 ⇒ ∴ θ = 30 o 2

θ 20

80 cm

C

30 o

10

30 H 10 3 15 cm

60 o o 30

R

30 3 cm

Smooth

3

Then from ( 1) : T Sin 30 o = 3 ⇒ ∴ T = 2 3 kg.wt ---------------------------------------------------------------------------------------------------------------------Example (5) A uniform ladder AB of weight 40 kg.wt and 12 m length rests with its end A on a smooth horizontal plane and its end B on a smooth vertical wall, the ladder is kept in equilibrium by a string tied to one of its ends A, and its other end is attached to a point on the horizontal plane vertically below B, if the ladder is inclined at 45o to the horizontal, given that the string can not bear a tension of more than 50 kg.wt, prove that a man of weight = the weight of the ladder can not ascend more than 9m without cutting the string Answer Smooth R Note ( 1) When a body is placed on a smooth plane, then the B o 3 x reaction is perpendicular to the plane 45

( 2 ) We have to prove that if a man of weight 40 kg.wt ascends a ddistance 9 m from A, then the maximum tension of the string will be 50 kg.wt 6 9 12 ∑ M A = 0 : ∴ 40 × 2 + 40 × 2 − Rx × 2 = 0 2

( )

∴ 240 + 360 − 12Rx ⇒ ∴ Rx = 50 kg.wt X = –0:3 ( secondary → ) : Rx = T ∑Static rd

⇒ ∴ T = 50 kg.wt

12 2

3 2

3 Ry

6 2

6 45 o

C

T

40

6

A

40

Smooth

Chapter four – General equilibrium

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Example (6) A uniform rod AB of lenght 8 m and weight 20 kg.wt, its end A is attached to a hinge fixed to a vertical wall, and its end B being joined by a light string 6 m length whose other end is fixed to the wall at C vertically above A and 10 m away from it, if a weight of 15 kg.wt is hung from the end B. Find the tension in the string, and the magnitude and the direction of the reaction of the hinge Answer Note ( 1) When a body is placed on a smooth plane, then the C reaction is perpendicular to the plane

6 cm

( 2 ) When there is a hinge in the problem, then its reaction has no fixed direction, so we must make

( )

a resolution in the x - axis ( Rx ) and the y - axis R y

And Q

( AB )

2

α α

Ry

+ ( BC ) = ( 6 ) + ( 8 ) = 100 2

2

2

2

2

⇒ ∴ m ( < B ) = 90

Not Smooth

o

∑ M B = 0 : ∴ 20 ( 4 Sinα ) − Ry ( 8 Sinα ) + Rx ( 8Cos α ) = 0

B

4

2

∴ ( AC ) = ( AB ) + ( BC ) 2

Smooth

T Cos α

10 cm

In ∆ ABC: Q ( AC ) = 10 = 100 2

T Sin α

T

4 Sinα

α α

4 cm

20

Rx

A

15 8 Cos α

8 Sinα

6 6 8 − 8 × R y + 8 × Rx = 0 ⇒ ∴ 48 − 4.8 R y + 6.4 Rx = 0 − − − ( 1) 10 10 10 8 4 ∑ X = 0 : Rx = T Cos α ⇒ Rx = 10 T ⇒ ∴ Rx = 5 T − − − ( 2 )

∴ 80 ×

∑ Y = 0:

3 R y + T Sin α = 20 + 15 = 35 ⇒ ∴ R y = 35 − T − − − ( 3 ) 5

3   4  Substitute ( 2 ) and ( 3 ) in ( 1) : ∴ 48 − 4.8  35 − T ÷+ 6.4  T ÷ = 0 5   5  72 128 ∴ 48 − 168 + T + T = 0 ⇒ ∴ 8T = 120 ⇒ ∴ T = 15 kg.wt 25 25 4 3 Then Rx = × 15 = 12 kg.wt and R y = 35 − × 15 = 26 kg.wt 5 5 Q R = Rx 2 + R y 2 =

( 12 ) 2 + ( 26 ) 2 Ry

= 2 205 kg.wt

26 13 = Rx 12 6 ---------------------------------------------------------------------------------------------------------------------Then Tanθ =

Static – 3rd secondary

=

7

Chapter four – General equilibrium

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Example (7) A uniform rod AB of weight 150 kg.wt and 54 cm rests with its lower end A on a vertical smooth wall, and it is fixed at point H on a horizontal smooth support parallel to the wall, if the length AH equals 12 cm and the rod is in equilibrium when it is perpendicular to the wall, prove that 2 the sin angle of the inclination between the rod and the wall is , then calculate the magnitude 3 of the reaction of the support and that of the wall. Answer Note when a body is placed on a smooth plane, then the reaction is perpendicular to the plane

(

∑MH = 0 :

)  − 150 ( 15 Sinα ) = 0

R2

∴ R1 12 Sin 90 o − α  ∴ 12 R1 Cos α − 250 Sin α = 0

R2 Sin α

( ÷ 6 Cos α ) ∴ 2 R1 − 375Tan α = 0 − − − ( 1)

Smooth 15

R2 Cos α Smooth

α

A

R Sin α 150 Divide ( 3 ) and by ( 2 ) : 2 = R2 Cos α R1

α

(

12 Sin 90 o − α

12 90 − α

R1

(

27 Cos 90 o − α

)

)

150

 150  150 − − − ( 4 ) , then substitute in ( 1) : 2 R1 − 375  ÷ = 0 ( × R1 ) R1  R1 

∴ 2R12 − 56250 = 0 ⇒ R12 = 28125 ⇒ ∴ R1 = 75 5 gm.wt From ( 4 ) : Tan α = ∴ Sin α =

B

H α 15 Sin α

∑ X = 0 : R1 = R2 Cos α − − − ( 2 ) ∑ Y = 0: R2 Sinα = 150 − − − ( 3 )

∴ Tan α =

27

2 and 3

Cos α =

150 2 =  → 75 5 5

3 2

α 5

5 2 ⇒ then from ( 3 ) : R2 × = 150 ⇒ ∴ R2 = 225 gm.wt 3 3

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

Static – 3rd secondary

8

Chapter four – General equilibrium

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Example (8) A uniform ladder of length 5 m and weight 20 kg.wt rests with one end on a smooth vertical wall and the other end on a smooth horizontal floor, if the lower end of the ladder is 3m away from the wall, the ladder is kept from slipping by a string attached to a point on it and the other end of the string is fixed to a point on the line of intersection of the floor and the wall, knowing that the direction of the string is perpendicular to that of the ladder, find the tension of the string and the reaction of each of the wall and the floor. Answer Smooth Note ( 1) When a body is placed on a smooth plane, then the B

reaction is perpendicular to the plane BE =

16 Cos α 5 F

( 5) − ( 3) = 4 m 2

2

Rx

α

2.5 1.5

BC =

16 m 5

D

1 4m Ry 2.5 Q DF = AE = 1.5 m and Q BE ⊥ AE T Cos α C 2 O α 16 Sinα And EC ⊥ AB , then by using Euclid's theorem 5 T T Sin α A 16 2 E Q ( EB ) = BC × BA ⇒ ∴ 16 = 5BC ⇒∴ BC = m 20 Smooth 5 3m 16 16 ∑ M B = 0 : ∴ - 20 × 1.5 − T Cos α × 5 Cos α − T Sin α × 5 Sin α + R y × 3 = 0 4 16 4 3 16 3 16 ∴ -30 − T × × × − T × × × + 3R y = 0 ⇒ ∴ 3R y − T = 30 − − − ( 1) 5 5 5 5 5 5 5 4 ∑ X = 0: ( → ) : Rx = T Cos α = 5 T − − − ( 2 ) 3 ∑ Y = 0: ( ↑ ) : Ry = 20 + T Sin α = 20 + 5 T − − − ( 3 ) 3  16 9 16 150  Substitute ( 3 ) in ( 1) : 3  20 + T ÷− T = 30 ⇒ 60 + T − T = 30 ⇒ ∴ T = kg.wt 5  5 5 5 7  4 150 120 From ( 2 ) : Rx = × = kg.wt 5 7 7 3 150 230 From ( 3 ) : R y = 20 + × = kg.wt 5 7 7

Static – 3rd secondary

9

Chapter four – General equilibrium

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Example (9) A uniform ladder rests with its end A on a smooth vertical wall and the its other end B on a smooth horizontal plane, the ladder is kept from slipping by a string whose one of its ends fixed to a point at the bottom of the wall vertically below A and the other end is fixed to one stair at a 1 distance from B equals of the ladder length, if the pressure of the ladder on the wall is ( P ) 4 and on the floor is ( Q ) . If A,B are at distance 60 cm, 80 cm respectively from the bottom of the P 4 wall, then prove that: = Q 5 Answer Note ( 1) When a body is placed on a smooth plane, then the P A reaction is perpendicular to the plane BE =

( 5)

2

α

− ( 3) = 4 m 2

30

1 Q DF = BE = 40 cm ⇒ ∴ AF = 30 cm 2 In ∆ ∆ AFD, AHC : Q FD // HC AF AD FD 30 50 40 ∴ = = ⇒ = = AH AC HC AH 75 HC ∴ AH = 45 cm ⇒ FH = 15 cm ⇒ HC = 60 cm In ∆ CEH: Q EC =

( 15 ) 2 + ( 60 ) 2

We don't need ( W ) , so we will take

F 15 H

50 40 cm

60 cm T Cos α 15 17

15 E

D

25

α

T

C

Q 25

T Sin α

20

40

20

B

W

= 15 17 cm

∑MD = 0 :

∴ - P ( 30 ) − T Sin α ( 20 ) − T Cos α ( 15 ) + Q ( 40 ) = 0 ∴ -30P − T ×

15 60 20 60 × 20 − T × × 15 + 40Q = 0 ⇒ ∴ 40Q − 30P − T− T =0 15 17 15 17 17 17 ∴ 40Q − 30P −

∑ X = 0: ( → )

: P = T Cos α =

80 T = 0 − − − ( 1) 17

60 17 T= T − − − ( 2) 17 15 17

80 17 × P =0 4 17 P 4 ∴ 40Q − 30P − 20P = 0 ⇒ ∴ 40Q = 50P ⇒ ∴ = rd Static – 3 secondary 10 Q 5 Substitute ( 2 ) in ( 1) : 40Q − 30P −

Chapter four – General equilibrium

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Example (10) AB is a uniform rod of mass 20 kg rests with its upper end B on a smooth inclined plane that's to the horizontal at an angle of measure 60 o , and it is resting with its lower end A on a smooth horizontal plane, it is kept from slipping by a string AC which is attached with one end to A and with the other end to C, so that the plane ABC is perpendicular to the line of intersection of the two planes. Find the magnitude of the reaction of each of the plane on the rod, and find the tension on the string when the angle of inclination of the rod to the horizontal is 30 o Answer Note when a body is placed on a smooth plane, then the reaction is perpendicular to the plane

∑MB = 0 :

RB Sin30 o RB

1  ∴ -RA  x 3  + 20  3 x ÷+ T x = 0 2 

RB Cos 30 o

Divide by ( x ) ⇒ ∴ - 3RA + 10 3 + T = 0 − − − ( 1)

∑ X = 0 : T − RB Cos 30 o = 0 ⇒ T = ∴ RB =

∑ Y = 0:

3 RB 2

x

2 T − − − ( 2) 3

x

30 o

RA

1 3x 2

x

30 o

A

R A + RB Sin30 = 20 − − − ( 3 ) o

T

120 60 C

30 o B 60 o 30 o30 o 1 x 2

1 x 2

o

20 x 3

1 2 3 Substitute ( 2 ) in ( 3 ) : ∴ R A + × T = 20 ⇒ R A = 20 − T − − − ( 4) 2 3 3  3  Substitute ( 4 ) in ( 1) : ∴ - 3 20 − T  + 10 3 + T = 0 ⇒ - 20 3 + T + 10 3 + T = 0 3   ∴ 2T = 10 3 ⇒ ∴ T = 5 3 kg.wt , then from ( 2 ) : ∴ RB = then from ( 4 ) : R A = 20 −

2 × 5 3 = 10 kg.wt 3

3 × 5 3 = 15 kg.wt 3

---------------------------------------------------------------------------------------------------------------------Static – 3rd secondary

11

Chapter four – General equilibrium

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Example (11) A uniform rod LM of length 150 cm and weight 5 kg.wt tied by its two ends L,M by means of two strings LN and MH whose lenghts are 150, 300 cm respectively, and the two free ends of the two strings are fixed at points N, H on a horizontal straight line, NH = 300 cm, then a horizontal force ( F ) acted on the rod and its end ( L ) so the rod became in equilibrium position as the string LN is vertical and the whole system of forces lie in the same vertical plane, determine

( F)

and find the magnitude of the tension in each string. Answer

In ∆ HLN : HL =

( 300 ) 2 + ( 150 ) 2 = 150

5 cm N

300 cm

x H α 150 1 α 2 And Q Sin = = α 2 2 150 5 5 150 5 300 150 T1 α 300 2 And Cos = = T2 F α 2 2 150 5 5 O L α α 1 2 4 T2 Sin α F 90 cm ∴ Sin α = 2Cos Sin = 2 × × = 75 α 2 2 5 5 5 75 M T Cos α 2 3 5 4 ∴ Cos α = 5 5 α 3 In ∆ MHX MX 4 4 Q Sin α = = ⇒ ∴ MX = 300 × = 240 cm ⇒ ∴ MO = 240 − 150 = 90 cm 300 5 5 HX 3 3 Also Q Cos α = = ⇒ ∴ HX = 300 × = 180 cm ⇒ ∴ NX = OL = 300 − 180 = 120 cm 300 5 5 ∴ LF = 60 cm

∑ML = 0 :

∴ -5 ( 60 ) + T2 Sin α ( 120 ) + T2 Cos α ( 90 ) = 0

4 3 ∴ -300 + T2 × × 120 + T2 × × 90 = 0 ⇒ ∴ 150T2 = 300 ⇒ ∴ T2 = 2 kg.wt 5 5 3 6 ∑ X = 0: ( → ) : T2 Cos α = F ⇒ ∴ F = 2 × 5 = 5 kg.wt 4 17 ∑ Y = 0: ( ↑ ) : T1 = 5 − T2 Sin α = 5 − 2 × 5 = 5 kg.wt Static – 3rd secondary

12

Chapter four – General equilibrium

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Problemsof ofequilibrium equilibriumon onrough roughplanes planes Problems Example (12) A uniform ladder of weight 30 kg.wt rests with one of its ends on a smooth vertical wall and with its other end on a rough horizontal floor, so that it is inclined at angle 45o to the horizontal, if a man whose weight is 60 kg.wt is ascending the ladder slowly, and the ladder was about to slip 3 when the man ascended of the length of the ladder, find the coefficient of friction between 4 the the floor and the ladder, and if the man wanted to continue to ascend the top of the ladder, then find the least horizontal force which acts at the lower end at this case. Note ( 1) when a body is placed on a smooth plane, then the Answer Smooth R reaction is perpendicular to the plane 1 A x ( 2 ) when a body is placed on a rough plane, then the 45o reaction is perpendicular to the plane and FF appears let the length of the ladder be 4x 3 1st case when the man ascended the length of the ladder: 4  2  4x   3x   2x  M = 0 : ∴ -R + 60 + 30 = 0  ÷ ∑ B 1 ÷  ÷  ÷  2  2  2  x 

x 2

4x 2

x R2

2x 2

2x 45o

C

FF 60

2x 2

30

∴ 4R1 = 240 ⇒ ∴ R1 = 60 kg.wt

∑ X = 0: ( → ) : R1 = FF = 60 kg.wt ∑Y = 0: ( ↑ ) : R2 = 60 + 30 = 90 kg.wt And Q FF = µ R ⇒ ∴ µ =

FF 60 2 = = R 90 3

4x 2

nd

2 case when the man reaches the top of the ladder:

∑MB = 0 :

 4x   4x   ∴ -R1  ÷+ 60  ÷+ 30   2  2 

 2C  2x  = 0  ÷ ÷ 2  2x 

∴ 2R1 = 150 ⇒ ∴ R1 = 75 kg.wt X = –0:3 ( secondary → ) : F + R1 = FF ∑Static rd

45o

Rough

3x 2

R1

A 60

B

2x

R2

2x 2

2x 45o FF 30

B

F

2x 2

⇒ F = FF − R1 =1360 − 75 = -15Chapter kg.wt four – General equilibrium

∴ F = 15 kg.wt in the ( ¬

)

direction

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Example (13) A uniform ladder of length 4 m rests on a horizontal rough floor and a smooth vertical wall. If the ladder is about to slide under the action of a weight ( W ) when it is inclined to the horizontal 3 at an angle of tangent , find the coefficient of friction, and if the ladder was inclined to the Note ( 1) when a body4is placed on a smooth plane, then the horizontal at an angle 45 o , find the farthest distance on the ladder a man whose weight is 4 times reaction is perpendicular to the plane the weight ascend ladder about to slide. when a body is can placed on abefore roughthe plane, thenis the ( 2 ) ladder's reaction is perpendicular to the plane andAnswer FF appears Smooth R 1 A AC 3 AC 12 Q Sinθ = ⇒∴ = ⇒ ∴ AC = m 2 4 5 4 5 12 5

BD 4 BD 8 And Q Cos θ = ⇒∴ = ⇒ ∴ BD = m 2 5 2 5 1st case If the ladder is about to slide under weight ( W ) :  12  8  5 ∑ M B = 0 : ∴ -R1  5 ÷ + W  5 ÷ = 0  × 4 ÷ 2 ∴ 3R1 = 2W ⇒ ∴ R1 = W − − − ( 1) 3 2 ∑ X = 0: ( → ) : R1 = FF = 3 W ∑ Y = 0: ( ↑ ) : R2 = W

R2 2

C 5

θ A 60

FF 8 5

W

3

B

Rough

4 R1

45

o 4−x 2

4 2

2 2 And Q FF = µ R2 ⇒ ∴ W = µ W ⇒ ∴ µ = 3 3

θ

D

4−x E x 2 2

R2

2 o

45 2 nd case when θ = 45o : C FF B F  4   x   2  2 4W W ∑ M B = 0 : ∴ -R1  2 ÷ + 4W  2 ÷ + W  2 ÷ = 0 × 2 2 1 ∴ 4R1 = 4W x + 2W ( ÷ 4 ) ⇒ ∴ R1 = W x + W − − − ( 1) x 2 2 1 ∑ X = 0: ( → ) : R1 = FF = W x + 2 W And ∑Y = 0: ( ↑ ) : R2 = 5W Static – 3rd secondary 14 1 2 1 10Chapter four –17General equilibrium and Q FF = µ R ⇒ W x + W = × 5W ( ÷W ) ⇒ x + = ⇒ ∴ x= m 2 3 2 3 6

(

)

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Example (14) A uniform ladder AB of length 5 m and weight 20 kg.wt rests with its end A on a smooth vertical wall and with its other end on a rough horizontal floor, the coefficient of friction between the 1 ground and the ladder is . If the point B is 3 m away from the wall, prove that the ladder can 4 not be in equilibrium in this case, and then find the magnitude of the least weight could be placed 1 at B to prevent sliding, given that the coeficient of friction between this weight and the ground is . 5 Note ( 1) when a body is placed on a smoothAnswer plane, then the Smooth R 2 A reaction is perpendicular to the plane

( 2 ) when a body is placed on a rough plane, then the

2.5

reaction is perpendicular to the plane and FF appears let the length of the ladder be 4x 1st case

∑MB = 0 :

4

2.5

∴ 20 ( 1.5 ) − R2 ( 4 ) = 0

B

C

∴ 4R2 = 30 ⇒ ∴ R2 = 7.5 kg.wt

1 × 20 = 5 kg.wt 4 FF1 > µ R1 ⇒ Then the ladder is not in equilibrium

So Q FF1 = 7.5 and µ R1 =

2 nd case to get the least weight to prevent the ladder from sliding ⇒ ∴ FF1 must be equal to µ R1 ∴ FF1 = µ R1 = 5 kg.wt

A

R3

2.5

∑ M B = 0 : ∴ -R3 ( 4 ) + 20 ( 1.5 ) = 0 ⇒ ∴ 4R3 = 30 ∑ X = 0: ( → ) : R3 − FF − FF = 0 ⇒ 7.5 − 5 − FF

2

R2

4

2.5

C

And Q the weight has another friction FF2

2

Rough

3m

1

1

FF1

20

∑ X = 0: ( → ) : R2 = FF = 7.5 kg.wt ∑Y = 0: ( ↑ ) : R1 = 20 kg.wt Then

R1

⇒ ∴ R3 = 7.5 kg.wt

FF2 20 3m

R1 FF1

B W

⇒ ∴ FF2 = 2.5 kg.wt

1 R = 2.5 ⇒ ∴ R2 = 12.5 kg.wt Static – 3 secondary 5 2 15 Chapter four – General equilibrium ∑Y = 0: ( ↑ ) : R1 + R2 = 20 + W ⇒ 20 + 12.5 = 20 + W ⇒ ∴ W = 12.5 kg.wt ∴ µ2 Rrd2 = 2.5 ⇒

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Example (15) A uniform inclined ladder AB rests with the end A on a rough horizontal floor, and the other end B 1 on a rough vertical wall, if the coefficient of the friction between the ladder and the floor is , and 2 1 that between the ladder and the wall is , and the ladder was about to slide, find the tangent of the 4 angle of the inclination of the ladder to the horizontal. Answer Note when a body is placed on a rough plane, then the reaction is perpendicular to the plane and FF appears 1st case

FF2

B

µ=

1 4

rough

∑M A = 0 :

∴ -FF2 ( 2xCos θ ) − R2 ( 2x Sin θ ) + W xCos θ = 0

( ÷ x)

R2 x

2x Sinθ

∴ -2FF2 Cos θ − 2R2 Sinθ + W Cos θ = 0

∴ -2 ( µ 2 R2 ) Cos θ − 2R2 Sinθ + W Cos θ = 0 1  ∴ -2  R2 ÷Cos θ − 2R2 Sin θ + W Cos θ = 0 4 

FF1

C W

( ÷ Cos θ )

2x Cos θ

1 R2 + 2R2 Tanθ = W − − − ( 1) 2 1 ∑ X = 0: ( → ) : R2 = FF1 = µ1 R1 = 2 R1 − − − ( 2 ) : FF2 + R1 = W ⇒ µ2 R2 + R1 = W ⇒

θ

A

x Cos θ

∴

∑ Y = 0: ( ↑ )

R1

x

Rough

µ=

1 2

1 R2 + R1 = W − − − ( 3 ) 4

From ( 2 ) : R1 = 2R2 − − − ( 4 ) Then substitute ( 4 ) in ( 3 ) : ∴

1 9 4 R2 + 2R2 = W ⇒∴ R2 = W ⇒ ∴ R2 = W − − − ( 5 ) 4 4 9

Then substitute ( 5 ) in ( 1) : ∴

1 4   4   W ÷+ 2  W ÷Tan θ = W 29   9 

8 2 7 7 ∴ Tanθ rd= 1 − = ⇒ ∴ Tanθ = Static 9 – 3 secondary 9 9 8

16

( ÷W )

⇒

2 8 + Tan θ = 1 9 9

Chapter four – General equilibrium

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Example (16) A uniform rod AB of length 130 cm, and weight 60 kg.wt rests with its ends A on a rough vertical wall and its other end B on a horizontal floor, if the coefficient of the friction between the wall and 2 1 the rod is , and that between the rod and the floor is , find the magnitude of the least force which 5 3 acts at B and sufficient to move this end towards the wall when B is 50 cm away from the wall Answer 2 Note when a body is placed on a rough plane, then the µ= A 5 R1 reaction is perpendicular to the plane rough

( 130 ) 2 − ( 50 ) 2 = 120 cm ∑ M B = 0 : ∴ -R1 ( 120 ) + 60 ( 25 ) + FF1 ( 50 ) = 0 ( ÷ 10 ) ∴ -12 R1 + 150 + 5FF1 = 0 ⇒ ∴ 12 R1 = 150 + 5 ( µ1 R1 ) Q AC =

2  ∴ 12 R1 = 150 + 5  R1 ÷ ⇒ 10R1 = 150 5 

FF1

120

R2

65

B

F

C

FF2

25 cm 60

∴ R1 = 15 kg.wt

∑ X = 0: ( → )

65

Rough µ =

1 3

50 cm

: R1 + FF2 = F ⇒ R1 + µ2 R2 = F

1 ∴ F = 15 + R2 − − − ( 2 ) 3

∑ Y = 0: ( ↑ )

: R2 = FF1 + 60 ⇒ R2 = µ1 R1 + 60 ⇒ ∴ R2 =

2 ( 15 ) + 60 = 66 kg.wt 5

1 ( 66 ) = 37 kg.wt 3 ---------------------------------------------------------------------------------------------------------------------From ( 2 ) : F = 15 +

Static – 3rd secondary

17

Chapter four – General equilibrium

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Example (17) A uniform rod AB of length 300 cm, and weight 5 kg.wt rests with its ends A on a rough horizontal plane, and on a horizontal support at a point C on the rod, the support is at 125 cm above the horizontal plane, if the rod was in a vertical plane that's perpendicular to the support, and the rod was about to slide when its angle of inclination to the horizontal is 30 o , find the reaction of the support and the coefficient of the friction between the rod and the plane. Answer ∑ M B = 0 : ∴ 5 75 3 − R2 Cos60o ( 125 ) − R2 Sin60 o 125 3 = 0

(

∴ 375 3 −

)

(

125 375 R2 − R2 Sin60 o = 0 2 2

∴ 250 R2 = 375 3 ⇒ ∴ R2 =

∑ X = 0: ( → ) ∴ FF =

)

R2

R2 Sin60 o

B 50

3 3 kg.wt 2

C

60 o 30 o

60 o

o

: R2 Cos60 = FF

125

3 3 1 3 3 × = kg.wt 2 2 4

R2 Cos60 o 100 60 o

30 o

∑ Y = 0: ( ↑ ) : R2 Sin60 + R1 = 5 o

∴ R1 = 5 −

R1

150

75 3

FF

A

5

3 3 3 11 × = 2 2 4

125 3

3 3 11 3 3 =µ× ⇒ ∴µ= 4 4 11 ---------------------------------------------------------------------------------------------------------------------But FF = µ R1 ⇒

Static – 3rd secondary

18

Chapter four – General equilibrium

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Example (18) A uniform rod of length 100 cm, rests with one end on a rough horizontal floor, and rests by one of its points that is 80 cm away from its other end on the surface of a smooth hemi - sphere whose radius is 60 cm that rests with its plane surface on the horizontal floor, if the rod is about to slip, find the coefficient of friction between the rod and the floor. Answer R2

R2 Cos α α

20

B

R2 Sin α

R1

30 50

80 Sin α

60

50Cos α

W

α

FF

A

100 cm 80 Cos α

∑MA = 0 :

∴ W ( 50 Cos α ) − R2 Sin α ( 80 Sin α ) − R2 Cos α ( 80Cos α ) = 0 2

2

80  60   80  ∴ 50 × W − R2  − 80R ÷ 2 ÷ =0 100  100   100  144 256 1 R2 − R2 = 0 ⇒ ∴ 80R2 = 40W ⇒ ∴ R2 = W kg.wt 5 5 2 1 60 3 ∑ X = 0: ( → ) : R2 Sinα = FF = 2 W × 100 = 10 W 1 80 2 ∑ Y = 0: ( ↑ ) : R2 Cos α + R1 = W ⇒ ∴ 2 W × 100 + R1 = W ⇒ ∴ 5 W + R1 = W 2 3 ∴ R1 = W − W ⇒ ∴ R1 = W kg.wt 5 5 ∴ 40W −

But FF = µ R1 ⇒

3 3 1 W =µ× W ⇒ ∴ µ = 10 5 2

Static – 3rd secondary

19

Chapter four – General equilibrium

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

Specialcases cases Special Example (19) A uniform ladder rests on a horizontal floor and on a vertical wall, if the ladder is about to slide, prove that the two resultant reactions of the wall and the floor are perpendicular when the angle of friction between the ladder and the wall equals the angle of friction between the ladder and the floor, and if the measure of the angle of friction is 30 o , prove that the angle of inclination of the ladder to the vertical equals 60 o . Answer Note ( 1) Don' t use the resultant reaction ( R' ) in any

R2 '

FF1

problem except when the angle of friction is mentioned.

( 2)

B

In last year, we have taken that if three forces act

θ

R1 '

C

34

λ=2

R1

D

R2

on a body, then they must meet at a point C Q The angle of friction ( λ1 ) between the ladder and the wall

λ 1

M

= The angle of friction ( λ2 ) between the ladder and the floor

W

A

FF2

∴ m ( < 1) = m ( < 2 ) and Q m ( < 1) = m ( < 4 ) " alt. angle" ⇒ ∴ m ( < 2 ) = m ( < 4 )

In ∆ BCD, Q m ( < 2 ) + m ( < 3 ) = 90 o ⇒ ∴ m ( < 4 ) + m ( < 3 ) = 90 o ⇒ ∴ R1 ' ⊥ R2 ' When λ1 = λ2 = 30 o , and Q CM is a median from a right angle ⇒ ∴ CM = MB ∴ m ( < 3 ) = m ( < CBM ) = 60 o but m ( < 2 ) = 30 o ⇒ ∴ m ( < MBD ) = 30 o ⇒ θ = 90 o − 30 o = 60 o. Note ( 1) Don' t use the resultant reaction ( R' ) in any ---------------------------------------------------------------------------------------------------------------------problem except when the angle of frictionExample is mentioned. (20) a uniform ladder weight W rests onact a smooth vertical wall, and the other end on 2 ) Inend lastofyear, we have takenofthat if three forces (One a floor to the θ ,D if the ladder is about to slide, prove that it is on ainclined body, then theyhorizontal must meetatatangle a point inclined the vertical at and an angle tangent 2Tan ( α − θ ) where α is the angle of friction. Q M is ato midpoint of AB ME //ofBC Answer ∴ E is a midpoint of AC, so let ( < ABC ) = β AC 2 AE ⇒ ∴ Tan β = − − − ( 1) BC DE B DE AE 1 β In ∆ AED, Q Tan 90 o − α + θ = ⇒∴ = AE DE Tan 90 o − α + θ

R2 '

In ∆ ABC, Q Tan β =

(

Static – 3rd secondary

∴ Tan β =

)

(

2

Tan 90 o − α + θ

)

(

(

)

)

D M

( 90

o

−α

)

R2 20 o equilibrium = 2Cot 90 − ( α − θ ) Chapter = 2Tanfour ( α −– θGeneral )

A

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θ

E C

Static – 3rd secondary

21

α

W

Chapter four – General equilibrium