Chapter 1 Friction

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Introduction

Resultant force between Two forces Rule:

F1

R 2  F12  F2 2  2F1 F2 cos 

Where :

R  F12  F2 2  2F1 F2Cos 

 : The angle between the two forces Where 0    180

R : The resultant force

R

 F2

The direction of the resultant :

F2 Sin  , Where  is the angle between R and the first force F1  F2 Cos  ---------------------------------------------------------------------------------------------------------------------F Sin F Tan  =

Resolution of a force into two perpendicular Components



F Cos  ----------------------------------------------------------------------------------------------------------------------

Equilibrium of a body on an inclined Plane If a body of weight W is placed on a smooth inclined plane which

R

inclines by  with the horizontal , Then the body will be under the Action of two forces :

The line of the

1 The weight force W  acting vertically downwards .



 2  The reaction force  R  of the inclined plane and it acts

greatest slope



In the direction perpendicular to the plane except External influences act at the body " hinge , rough ground , ...... " W ----------------------------------------------------------------------------------------------------------------------

Lami’s rule We use Lami’s rule if three forces act at a point . and you can find the angles between each two forces

F1 F2 W = = Sinθ1 Sinθ2 Sinθ3

F2

3 1 2

F1

1  2

W rd

Static – 3 secondary

-1-

Chapter One – Friction

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Friction FF

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Friction is "the resistance an object encounters in moving over another" . It is easier to drag an object over glass than sandpaper. The reason for this is that the sandpaper exerts more frictional resistance. In many problems, it is assumed that a surface is "smooth", which means that it does not exert any frictional force. In real life, however, this wouldn't be the case. A "rough" surface is one which will offer some frictional resistance The friction force is denoted by FF

---------------------------------------------------------------------------------------------------------------------The opposite figure represents a body rests on a rough horizontal plane, this body is acted by a force .

Movement

Notes: as weight of the body increases then the friction increases and the normal reaction increases. ----------------------------------------------------------------------------------------------------------------------

Limiting friction Imagine that you are trying to push a book along a table with your finger. If you apply a very small force, the book will not move. This must mean that the frictional force is equal to the force with which you are pushing the book. If the frictional force were less that the force produced by your finger, the book would slide forward. If you push the book a bit harder, it would still remain stationary. The frictional force must therefore have increased, or the book would have moved. If you continue to push harder, eventually a point is reached when the frictional force increases no more. When the frictional force is at its maximum possible value, here, friction is said to be limiting. If friction is limiting, yet the book is still stationary, it is said to be in limiting equilibrium. If you push ever so slightly harder, the book will start to move. If a body is moving, friction will be taking its limiting value. Static – 3rd secondary

-2-

Chapter One – Friction

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Properties of friction

Study

(1) The force of friction always acts in the direction opposite to the direction the body is tending to move. (2) The force of friction increases with the tangential force that tends to move the body so that the two force are equal provided that the body is in a state of equilibrium. (3) The magnitude of the force of friction increases up to a certain limit which it does not exceed. At this value, the motion is just about to begin and the friction is then called the limiting friction. When motion takes place the magnitude of the force of friction is nearly equal to its maximum value in other words during motion the friction is a limiting friction. (4) The magnitude of the limiting friction bears a constant ratio to the normal reaction. This ratio depends on the nature of the two surfaces in contact and is independent of their shapes and masses. ----------------------------------------------------------------------------------------------------------------------

Coefficient of friction



The coefficient of friction is a number which represents the friction between two surfaces, it is the ratio between the magnitude of the limiting friction FF and the magnitude of the normal reaction R and it is denoted by  

FF R

Note: for most applications,   1 “except for rare materials” So let’s understand what does this mean : A very simple example: which of the following is easier ? To lift a disk or to push it I think pushing it is much easier than lifting it So lifting here tends to R and pushing it tends to FF , so usually FF  R And from that, we can say that   1 for most applications and problems we will discuss ----------------------------------------------------------------------------------------------------------------------

Static – 3rd secondary

-3-

Chapter One – Friction

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1 Equilibrium rule :  i  About to move rule :

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Rules of friction

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FF   R where  Coefficient of friction and R the normal reaction

Note use this rule when the body is about to move , this is called a limiting friction

 ii 

Not about to move rule : FF   R

 2  Resultant reaction

  friction is not limiting 

R' = FF 2 + R 2

Imp. you can use R' instead of FF and R to use lami's rule

 3

Angle of friction  λ 

 Study 

It is the angle between the resultant reaction R'

the resultant between

FF and R



and the normal to the plane on which

Then: Tan λ =

λ

FF

body is about to move FF =μ R

R

R'

Imp. don't use the resultant reaction  R'  except when  λ  is given

4

Special case on an inclined rough plane : If a body is placed on a rough inclined plane with the horizontal by angle  and the body is about to move by its own weight only Then Tan  Tan    Imp If the direction of movement is not mentioned: least force to move the body means the movement is down largest force to move the body means the movement is up

5

F F

F

F

 are opposite 

and F are together and F

Special case on a horizontal rough plane : If two horizontal forces F1 and F2 are acted on the body rests on a rough horizontal plane, and the angle between them is  , then:

 FF 

2

F1

 F12  F2 2  2F1 F2 Cos 

Where FF   R If the body is not about to move Or

FF   R If the body is about to move

Static – 3rd secondary

-4-

 F2 Chapter One – Friction

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Example (1) A wooden block of weight 6 kg.wt. is placed on a horizontal table, and is connected by a string passing over a smooth pulley at the edge, to a weight of magnitude 1.5 kg.wt. which is hanging freely. Given that the block is in equilibrium, find the force of friction and the normal reaction. 1 If the coefficient of friction between the block and the table is , state whether or not the body 3 is about to move. Answer T  1.5 kg.wt   FF  T  1.5 kg.wt    1 and

1 R  6 kg.wt So  R   6  2     2  3

Then from  1 and  2  :

FF   R

 the body is in Equilibruim Then the friction is not limiting and not about to move

6

---------------------------------------------------------------------------------------------------------------------Example (2) A wooden block of weight 2 kg.wt rests in equilibrium on a plane, inclined at 30 o to the horizontal, under the action of a force whose magnitude is 2.5 kg.wt and whose direction is that of the line of greatest slope upwards. If the coefficient of friction is 0.9, find the force of friction. State whether or not the motion is about to begin. Answer o FF  2Sin30  2.5   FF  2.5  2Sin30 o  1.5 kg.wt    1 and

R  2Cos 30 o  3 kg.wt

Movement

  R  0.9 3  1.559     2  Then from  1 and  2  :

FF   R

 the body is in Equilibruim Then the friction is not limiting and not about to move ----------------------------------------------------------------------------------------------------------------------

Static – 3rd secondary

-5-

Chapter One – Friction

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Example (3) A body of weight 16 kg.wt is placed on a horizontal rough plane. and the friction coefficient 1 between them is . Find: 4  a  The least force acting on the body inclined to the horizontal plane at an angle whose 3 so that the motion is about to start. 5  b  The magnitude and the direction of the resultant reaction. Answer The body is about to move  Then the friction is limiting   FF   R R 1 R'  FF  R    1 4 Cos is

F Sin

a

FF  F Cos  and

3 F 5



FF

F Cos 



4 R  16  F Sin    R  16  F 5

 from  1 :

16

3 1 4  3 1 F  16  F    F  4  F 5 4 5  5 5

3 1  F  F 4 5 5

b

  FF 

F

  F  5 kg.wt

R'  R 1   2 where R  16 

5 4

4  5   12 kg.wt 5

 3

2

1 1  R'  12 1     3 17 kg.wt and also Tan     4 4 ----------------------------------------------------------------------------------------------------------------------

Static – 3rd secondary

-6-

Chapter One – Friction

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Example (4) A body of mass 60 kg is placed on a horizontal rough plane, a force of 30 kg.wt acted on it in a direction inclined to the horizontal at an angle   so that it is about to slide, and then a force of a 60 kg.wt acted in the opposite direction of the first force, so that it is about to slide also, find the coefficient of friction and find the angle   . Answer First case: The body is about to slide  Then the friction is limiting   FF   R

 FF   R So,

FF  30 Cos  and R  60  30 Sin 

 30 Cos   60  30 Sin    

 divide by 30  Cos    2  Sin       1

R

Second case: FF  60 Cos  and R  60  60 Sin 

30 Sin

 60 Cos   60  60 Sin   

From



FF

 divide by 60   Cos   1  Sin        2  1 and  2  :  2  Sin     1  Sin     divide by  

30 30 Cos 

60 R

 2  Sin   1  Sin    2 Sin   1 60 Cos  FF 1 o  Sin       30 2  o 2  Sin  2  Sin30 3 60 Sin From  1 :     o 60 Cos  Cos 30 3 60 ---------------------------------------------------------------------------------------------------------------------Example (5) A body of mass 26 gm is placed on a horizontal rough plane, the body is about to move under the action of the of two forces of magnitudes 7 and 8 gm.wt acting horizontally on the body and the angle between them is 60 o , find the angle of friction between the body and the plane. Answer If two horizontal forces F1 and F2 are acted on the body rests on a rough horizontal plane, and the angle between them is  , then: R 7

 FF   FF  And

7 

2

2

 F12  F2 2  2F1F2 Cos 

  8   2 7  8  Cos60  13 gm.wt 2

o

And R  26 gm.wt the body is about to move  friction is limting 

FF

60 o

8

26 13 1  FF   R  13  26        Tan   Angle of friction    26 2 -7rd Static – 3 secondary Chapter One – Friction

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Example (6) A body of weight 6 Newton rests on a rough horizontal plane. Two forces 2 and 4 Newton act horizontally on the body, the angle between them being 120o . If the body is in equilibrium, prove that the angle of friction  is not less than 30 o . If   45o and the two forces act in their previous direction and the 4 Newton force remains constant, find the least value for the other force to move the body and determine the direction of motion. Answer If two horizontal forces F1 and F2 are acted on the body rests on a rough horizontal plane, and the angle between them is  , then: R 2

 FF   FF 

4

2

2

 F12  F2 2  2F1 F2 Cos 

  2   2  4  2  Cos120  2 3 Newton 2

o

FF

120 o

And R  6 Newton

4

1 case : If the body is in equilibrium  friction is not limting  st

6

Angle of friction     Tan    And

the body is in equilibrium   FF   R   2 3  6 Tan 

 Tan  

2 3 2 3    Tan 1    30 o 6 6

2 nd case : If the body is about to move  friction is limting  Angle of friction       Tan   Tan45 o  1 And

the body is about to move   FF   R

where R  6 Newton  FF  1  6  6 Newton

 FF   F12  F2 2  2F1F2 Cos  2 2 2 6    4    F2   2  4  F2  Cos120 o

And 

2

  36  16  F2 2  4F2

 F2 2  4F  20  0  then by using formula: F2 





4  96 2

 F2  2 1  6 Newton To get the direction of this force:    Tan 1 Static – 3rd secondary

F1 Sin 4Sin120 o 2 Tan     F2  F1 Cos  2 1  6  4Cos120 o 2





2  35 o16' with F1 2 -8-

Chapter One – Friction

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Example (7) A body of 150 gm.wt is placed on a rough plane inclines at angle 30 o to the horizontal and the 3 . A force of 112 gm.wt acts on the body parallel to the 5 line of the greatest slope upwards and the body is still in equilibrium, find the magnitude and friction coefficient between them is

the direction of the force of friction in this case, and determine is it the limiting magnitude or not. Also mention the change that should happen to the magnitude of the force so that the body is about to move. first case: And

Answer FF  112  150 Sin30  38 gm.wt  downward  o

R  150 Cos 30 o  75 3 gm.wt

and



112

R

3 5 Movement

FF

3  R  75 3  45 g.wt   FF   R 5  The body is not about to move

150 Sin30o

30 o 30 o

Then the friction is not limiting

150 Cos 30o

Second case:

150

When the body is about to move  FF   R  45 gm.wt

F

R

Then the friction is limiting where FF  F  150 Sin30 o

 F  45  150 Sin30 o  120

Movement

FF

 The force must increase to 120 gm.wt

150 Sin30o

30 o

in order to make the body about to move 30 o

150 Cos 30o

150 ----------------------------------------------------------------------------------------------------------------------

Static – 3rd secondary

-9-

Chapter One – Friction

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Example (8) A body of weight 40 Newtons rests on a rough plane which is inclined to the horizontal at an angle 30 o . The body is pulled up by a string which makes an angle 30 o with the plane. The string lies in the vertical plane which contains the body and the line of greatest slope. If the 1 coefficient of friction is . Find the least force in the string which prevents the load from 4 moving down . Answer The body is about to move  Then the friction is limiting   FF   R 1 R     1 4 The least force means that the body is about to move down  FF 

So,

FF  T Cos 30 o  40 Sin30 o

40 Sin30 o

R  40 Cos 30 o  T Sin30 o

T Cos 30 o

FF 30 o

Movement 30 o

30 o

1  R  20 3  T 2  from  1 : 20 

R T Sin30 o

3  FF  20  T 2 and

T

40

40 Cos 30o

3 1 1  T   20 3  T  2 4 2 

3 1 3 1 T 5 3 T  T  T  20  5 3 2 8 2 8  3 1 20  5 3  T    20  5 3  T    T  15.3 Newton 2 8 3 1    2 8 --------------------------------------------------------------------------------------------------------------------- 20 

Static – 3rd secondary

- 11 -

Chapter One – Friction

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Example (9) A 130 Newton weight is placed on a rough plane which is inclined to the horizontal by an angle 5 whose cosine is . A force is applied to the weight parallel to the line of greatest slope 13 2 upwards. If the coefficient of friction between the weight and the plane is , then find the limits 5 between which the applied force lies, so as to make the weight about to move. Answer limits which the force apply means to get the least and the largest force which will make the body about to move up and down. R  130 Cos   130  And

5  50 Newton 13

F

R FF

the body is about to move   the friction is limiting

2  FF   R   FF   50  20 Newton 5

Movement 

W Sin 

1st case: If movement is downwards 12  20  100 Newton 13 "This is the least force needed"

F  130 Sin  FF  130 

130 Cos

13

W

12



2 nd case: If movement is upwards

5

F

R

12 F  FF  130 Sin  20  130   140 Newton 13 "This is the greatest force needed"

FF

Movement 

W Sin 

130 Cos

W

Static – 3rd secondary

- 11 -

Chapter One – Friction

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Example (10) A body whose weight is 20 Newton is placed on a rough plane inclined to the horizontal by an 4 angle whose tangent is . If F1 is the least force when applied along the line of greatest slope 3 upwards, the body is about to move downwards. F2 is the least force if applied horizontally, the body is about to move downwards. If F1  F2 then find the coefficient of friction of the rough plane and the magnitude of any of the two forces. Answer the body is about to move   the friction is limiting  FF   R 1 case : If movement is downwards R  20 Cos   20 

3  12 Newton 5

F1  20 Sin  FF  20 

Movement

5

st

4

R

F1

FF

 3

4  FF 5

20 Sin

 F1  16  FF  16   R   F1  16  12      1





2 nd case : If movement is downwards

20 Cos

20

3 4 4 R  20 Cos   F2 Sin   20   F2  12  F2 5 5 5 F2 Cos   20 Sin  FF 

Movement

3 3 4   F2  16  FF  16   R   F2  16   12  F2   5 5 5  

3 4 16  12    16   12  16  12     5 5   20 Sin 48 36 64 48 2     16  12      5 5 5 5  48 2 88 32 2      0  5    48   88   32  0   8  5 5 5  6  2  11  32  0   3   4  2   1  0

F2 Cos

R FF



When F1  F2 :



4  refused  3

Or

Static – 3rd secondary





F2

F2 Sin 20 Cos

20

1 1  F1  F2  16  12    10 Newton 2 2

- 12 -

Chapter One – Friction

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Example (11) A body of weight 38 Newton is about to move under its own weight when placed on a rough 1 plane inclined to the horizontal at an angle whose tangent is . If this body is placed on a 5 horizontal plane which is as rough as the inclined plane, and is acted on by a force inclined to 4 the horizontal at an angle whose sine is so that the body is about to move. Find the force and 5 the normal reaction. Answer st 1 case: Inclined plane The body is moving on an inclined plane under its own weight only by angle  whose tangent is  Tan 

1 5

1 1  Tan        5 5

R FF

F

F Sin 

F Cos 



2 nd case: horizontal plane the body is about to move by a force F inclined by angle 

38

 FF   R 3 So, FF  F Cos   F and 5

4 R  38  F Sin   38  F 5

5 4

 3 1 4  FF   R   F  38  F   Multiply by 5  3 5 5 5  4 4 19  3F  38  F   3F  F  38  F  38   F  10 Newton 5 5 5 3 Then from  1 : FF   10  6 Newton and from  2  : R  30 Newton 5 ---------------------------------------------------------------------------------------------------------------------So,

Static – 3rd secondary

- 13 -

Chapter One – Friction

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Example (12) A 2 kg.wt is placed on a horizontal rough plane. The plane is tilted gradually, so the weight is about to slide down when the inclination of the plane is 30 o to the horizontal. If the weight is then attached to a string which is pulled in a direction inclined at 60 o to the horizontal so that the body is about to move upwards. Given that the string is in the vertical plane through the line of greatest slope, calculate the tension in the string and the friction force. Answer 1st case: Inclined plane The body is moving on an inclined plane under its own weight only by angle   30 o Movement

 Tan30 o      

3 3

F

2 nd case: When a force acted on the plane

F Sin30 o

the body is about to move by a force F

FF

inclined by angle 60 o to the horizontal

30 o

30 o

where FF  F Cos 30 o  2 Sin30 o 

So,

30 o

30 o

2 Sin30 o

 FF   R

and

F Cos 30 o

R

3 F 1 2

2

2 Cos 30o

1 R  2Cos 30 o  F Sin30 o  3  F 2 FF   R  

3 3 1  F 1 3  F  2 3  2 

3 3 3 F 11 F  F 2 6 2 3 Then from  1 : FF  3 1 2



 

Static – 3rd secondary

3 2 3 F 2  F  2   F  3 kg.wt 6 3 1 kg.wt 2

- 14 -

Chapter One – Friction

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Example (13) A body of 3 kg.wt is placed on a rough plane. when the plane is is inclined to the horizontal at an angle 30 o , then it is about to slide down. If the inclination of the plane to the horizontal increased to become 60 o , find the force of friction. Then find the magnitude of the least force acting in the direction of the line of the greatest slope that prevents the body to move downwards. And if this force is replaced by another horizontal force, prove that its magnitude is equal to the first force. Answer st 1 case: Inclined plane The body is sliding on an inclined plane under its weight by angle   30 o  Tan30 o      

FF

R

3 3

Movement

2 nd case: when the angle of the plane increased to be 60 o Before the force acts: FF  3 Sin60 o 

3 3 kg.wt 2

3 Sin60 o 60 o 60 o

After the force acted: 3Cos60 o

To find the least force which prevents the body to move down  movement is down 

3 R

 FF   R    1 where FF  3 Sin60 o  F  and

3 3 F 2

Movement 3 Sin60 o

3 R  3Cos60 o  kg.wt 2

60 o

60 o

3 3 3 3 F     F  3 kg.wt 2 3 2 3rd case: when the force is horizontal

from  1 

3Cos60 o

3

FF   R     1 o

3Sin60

3 3 R  F Sin60  Cos60  F 2 2 o

from  1 

FF 60 o

3 3 1 where FF  3 Sin60  F Cos 60   F 2 2 and

F Cos60 o

R

o

F

FF

F

o

60 o

o

F Sin60 o

60 o

3 3 1 3 3 3 3 3 3 1  F F     F F 2 2 3 2 2 2 3 2

3Cos60 o

3

 F  3 kg.wt rd

Static – 3 secondary

- 15 -

Chapter One – Friction

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Example (14) When a weight W is placed on a rough plane inclined at an angle  to the horizontal, it is found that the weight is about to slide down. Prove that the least force along the line of greatest slope which makes the weight about to move upwards is equal to 2W Sin . Prove also that the resultant reaction is equal to W. Answer The body is moving on an inclined plane under its own weight only by angle   Tan  Tan        Tan To find the least force to make the body about to move up: Then the friction is limiting

 FF   R     1



F

R

where FF  F  W Sin and R  W Cos  Then from 1 :   F  W Sin   Tan  W Cos  Sin  F  W Sin  W Cos  Cos 



  W Sin



FF

Movement 

W Sin 

 F  2W Sin 

W Cos

To get the resultant in case the body is limiting:

W

R'  R 1   2   R'  W Cos  1  Tan 2  W Cos  Sec 2  W Cos  Sec  1 W Cos  --------------------------------------------------------------------------------------------------------------------- W Cos  

Static – 3rd secondary

- 16 -

Chapter One – Friction

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Example (15) Two bodies of weights 3 and 4 kg.wt are placed on a plane inclined to the horizontal at angle  the two bodies are connected with a light string coincide to the line of the greatest slope, the 1 1 coefficients of friction between the bodies and the plane respectively are and . 3 2 If the inclination of the plane increased gradually. Determine which body must placed below the other so that the two bodies start to move together, give reason. 3 Then prove that Tan  when the two bodies are about to slide together. 7 Answer FF Any body is about to move by its weight when    R 4 Sin  the angle of inclination  the angle of friction  In this problem, the two bodies will move when    , 1  then we must put the body of smaller friction     3  1  below the body of the greater friction     2 

T





3 Sin

4Cos

 3Cos

3

4

When the two bodies are about to move 1st body



2nd body

1 3

1 2 FF  4Sin  T



FF  3Sin  T

R  4Cos 

R  3Cos   3Sin  T 

Movement

T

1  3Cos  3



 4Sin  T 

1  4Cos  2



 T  2Cos   4Sin      2 

 T  3Sin  Cos      1

Then from  1 and  2  :  3Sin  Cos   2Cos   4Sin 3 7 --------------------------------------------------------------------------------------------------------------------- 7Sin  3Cos   Divide by Cos     7Tan  3   Tan 

Static – 3rd secondary

- 17 -

Chapter One – Friction

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Example (16) (*Excellent*) A rock of weight W is placed on a rough road inclined to the horizontal at angle  , if a horse pulled the rock upward by a string which makes an angle  with the road, so that the rock is about to move, given that the angle of friction between the rock and the road is  . Then prove that the least tension of the string to make the rock about to move upward occurs when    , then find the magnitude of this force when     35o , and the mass of the rock is 1000 kg. Answer We can solve this problem by two methods First method : The least tension required to move the rock upward

  Tan  and

R  W Cos   F Sin 

R

F F Cos 

F Sin

FF  F Cos   W Sin 



The body is about to move   FF   R FF Movement W Sin   F Cos   W Sin   W Cos   F Sin   Tan   Sin   F Cos   W Sin   W Cos   F Sin     Cos   After multiplying both side by  Cos   W Cos  F Cos  Cos   W Sin Cos   W Cos  Sin   F Sin  Sin  W F Cos  Cos   F Sin  Sin   W Cos  Sin   W Sin Cos  F  Cos  Cos   Sin  Sin    W  Cos  Sin   Sin  Cos     F Cos      W Sin      F

W Sin     Then the minimum tension occurs when Cos     is maximum Cos    

The maximum value of Cos      1       Cos 11  0      When     35o and W  1000  F 

1000 Sin35  574 kg.wt 1

Another solution when    appears, you may use R' instead of FF and R Then we can use Lami's rule between the three force R' , F' ,W F W R' R    o o o F Sin Sin  180        Sin  90      Sin  90      R' 90     F W   Sin     Sin  90 o       FF 

F

Sin    



W

Cos    

  F

W Sin     Cos    

W Sin



F F Cos 

Movement



Then continue................

W Cos

W rd

Static – 3 secondary

- 18 -

Chapter One – Friction

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Example (17) A body whose weight is W is placed on a rough plane inclined at angle  to the horizontal and the angle of friction is  . A force F acts on the body parallel to the plane to prevent the body from slipping. Prove that F  W Sin     Sec  Answer F R R  W Cos    Tan   FF  W Sin   F FF The body is about to move  FF   R   W Sin   F  W Cos  Tan  Sin  W Sin   F  W Cos   Cos 

Movement 

W Sin 

Sin   F  W Sin   W Cos   Cos 

W Cos

W

Multiply both sides by Cos  :

F Cos   W Sin  Cos   W Cos  Sin    F Cos   W Sin      F

W Sin      F  W Sin     Sec  Cos 

Another solution when    appears on an inclined plane, you may use R' instead of FF and R Then we can use Lami's rule under the three force R' , F' ,W 

F W R'   Sin  180 o        Sin  90 o    Sin  90 o   



F W F W     Sin     Cos  Sin  180 o       Cos  F 

W Sin      F  W Sin     Sec  Cos 

R' R



FF

F



Movement

W Sin





W Cos W ----------------------------------------------------------------------------------------------------------------------

Static – 3rd secondary

- 19 -

Chapter One – Friction

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Example (18) A body of 500 gm.wt is placed on a rough plane which is inclined to the horizontal by an angle 4 of measure  , such that Tan  , the body is then attached to a string passing over a smooth 3 pulley at the top of the plane and a scale pan of 25 gm mass is attached to the other end of the string, if the least weight to be added to the plane to keep the body in equilibrium is 175 gm.wt. Find the coefficient of friction, then prove that the maximum weight that can be added to the pan without disturbing equilibrium is 575 gm.wt. Answer 1st case: When the least weight attached is T  W  175  25  200 gm.wt Movement

The least weight means that the movement is down

R

The body is in equilibrium,   FF   R Where R  500 Cos   500 

FF

3  300 gm.wt 5

500 Sin

4 And FF  500 Sin   W  500   200  200 gm.wt 5  2

nd

W

W



W  175  25



500 Cos

FF 200 2   R 300 3

500

case: To get the maximum weight

5 4



The maximum weight means that the movement is up The body is in equilibrium,   FF   R where  

3

2 3

R

W

3 And R  500 Cos   500   300 gm.wt 5

W FF

And FF  T  25   500 Sin 

500 Sin

4  T  25   500    FF  T  25   400 5 2  T  25   400   300   T  25  600 3  T  575 gm.wt

Static – 3rd secondary

Movement

- 21 -



W  T  25 



500 Cos

500

Chapter One – Friction