Chapter 1 Friction

(1)

Facebook page : https://www.facebook.com/MrSherifYehiaAlMaraghy

Mr . Sherif Yehia Al Maraghy https://twitter.com/Mr_Sherif_yehia

01009988836 – 01009988826 Email : [email protected]

Static – 3rd secondary Chapter One – Friction-1 -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 , ... "

The line of the greatest slope 2 2 2 1 2 1 2 2 2 1 2 1 2 2 1 2 R F F 2F F cos R F F 2F F Cos

The direction of the resultant : F Sin

Tan = , Where is the angle between R and the first force

F F Cos             

Where : R : The resultant force : The angle between the two forces

Where 0 180      2 F 1 F R  F F Cos F Sin 1  ₂ 2 F W 1  _₂

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

1 2

1 2 3

F

W

=

Sinθ

3



1

F

Introduction

Resultant force between Two forces

Rule:

---Resolution of a force into two

perpendicular Components

---Equilibrium of a body on an inclined Plane

(2)

Movement

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

F_F

---The opposite figure represents a body

rests on a rough horizontal plane, this body is acted by a force .

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

---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.

Friction F

F

(3)

Static – 3rd secondary Chapter One – Friction-3

-Properties of friction

(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.

---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 F_F 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 F_F , so usually F_F R

And from that, we can say that  1 for most applications and problems we will discuss

---Study

(4)

01009988836 – 01009988826 Email : [email protected] λ F F R' R 1 F F 

 

 F 1

i F R where Coefficient of friction and R the normal reaction

Note use this rule when the body is , this is called a limiting friction ii

 

Equilibrium rule : About to move rule :

about to move Not about





 

  



  F

F R friction is not limiting

2

you can use instead of and to use lami's rule

3 Study

It is the angle between th

 2 2 F F to move rule : Resultant reaction R' = F + R Imp. R' F R Angle of friction λ





 

F e resultant reaction R' the resultant between F and R

and the normal to the plane on which body is about to move

Then:

don't use the resultant reaction R'

F F Tan λ = = μ R Imp.

 

except when is given 4

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 

λ Special case on an inclined rough plane :





 

F only Then Tan Tan

If the direction of movement is not mentioned:

least force to move the body means the movement is down F and F are together largest force to mo    Imp





 

F 1 2

ve the body means the movement is up F and F are opposite 5

If two horizontal forces F and F are acted on the body rests on a rough horizontal plane,

Special case on a horizontal rough plane :

 

  



2 ₂ ₂

F 1 2 1 2

F

and the angle between them is , then:

F F F 2F F Cos Where F R    

If the body is not about to move

(5)

- 

 

F F T 1.5 kg.wt F T 1.5 kg.wt 1 1 and R 6 kg.wt So R 6 2 2 3

Then from 1 and 2 : F R

the body is in Equilibruim

Then the friction is not limiting and not about to move

                  

 

                    o o F F o F F 2Sin 30 2.5 F 2.5 2Sin 30 1.5 kg.wt 1 and R 2Cos 30 3 kg.wt R 0.9 3 1.559 2

Then from 1 and 2 : F R

the body is in Equilibruim Then the



friction is not limiting and not about to move

Movement

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. If the coefficient of friction between the block and the table is 1

3 , state whether or not the body is about to move.

Answer

---Example (2)

A wooden block of weight 2 kg.wt rests in equilibrium on a plane, inclined at 30o 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

---6

(6)

01009988836 – 01009988826 Email : [email protected] 16 R F F F Cos

 

F F F F

The body is about to move Then the friction is limiting F R 1 F R 1 4 3 a F F Cos F F 5 4 and R 16 F Sin R 16 F 5 3 1 4 3 1 from 1 : F 16 F F 4 F 5 4 5 5 5                           _  _     

 

2

 

2 3 1 F F 4 F 5 kg.wt 5 5 4 b R' R 1 where R 16 5 12 kg.wt 5 1 1

R' 12 1 3 17 kg.wt and also Tan

4 4                   _{ }      F  F Sin  4 3 5  R' Example (3)

A body of weight 16 kg.wt is placed on a horizontal rough plane. and the friction coefficient between them is 1

4 . Find:

 

a The least force acting on the body inclined to the horizontal plane at an angle whose 3

Cos is so that the motion is about to start. 5

b The magnitude and the direction of the resultant reaction.

Answer

(7)

-26 o 60 R F F 7 8

 

   

  

1 2 2 ₂ ₂ F 1 2 1 2 2 2 _o F

If two horizontal forces F and F are acted on the body rests on a rough horizontal plane, and the angle between them is , then:

F F F 2F F Cos F 7 8 2 7 8 Cos60 13 gm.wt          





 

F And R 26 gm.wt

And the body is about to move friction is limting 13 1

F R 13 26 Tan Angle of friction

26 2               



 







 

F F F

First case: The body is about to slide Then the friction is limiting F R F R So, F 30 Cos and R 60 30 Sin

30 Cos 60 30 Sin divide by 30 Cos 2 Sin 1 Second case:                            



 







 

  





 



 

F o

F 60 Cos and R 60 60 Sin

60 Cos 60 60 Sin divide by 60 Cos 1 Sin 2

From 1 and 2 : 2 Sin 1 Sin divide by

2 Sin 1 Sin 2 Sin 1

1 Sin 30 2 2 Sin 2 From 1 : Cos                                                   Sin 30_o o 3 Cos 30 3  _ 60 R F F 30 Cos 30  60 R F F 60 Cos 60  30 Sin 60 Sin 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

---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 , find the angle of friction between the body and the plane.o

(8)

01009988836 – 01009988826 Email : [email protected] 6 o 120 R F F 2 4

 

   

  

        1 2 2 ₂ ₂ F 1 2 1 2 2 2 _o F

If two horizontal forces F and F are acted on the body rests on a rough horizontal plane, and the angle between them is , then:

F F F 2F F Cos F 4 2 2 4 2 Cos120 2 3 Newto  





 

                st F 1 o nd n And R 6 Newton

1 case : If the body is in equilibrium friction is not limting Angle of friction Tan

And the body is in equilibrium F R 2 3 6 Tan

2 3 2 3 Tan Tan 30 6 6 2 case : If the b        





 

     

  

                      o F F 2 ₂ ₂ F 1 2 1 2 2 2 2 _o 2 2

ody is about to move friction is limting

Angle of friction Tan Tan 45 1

And the body is about to move F R where R 6 Newton F 1 6 6 Newton

And F F F 2F F Cos 6 4 F 2 4 F Cos120 36 16     









                _ _ 2 2 2 2 2 2 2 o 1 o 2 1 F 4F 4 96

F 4F 20 0 then by using formula: F

2 F 2 1 6 Newton

F Sin 4Sin120 2

To get the direction of this force: Tan

F F Cos _{2 1} ₆ _4Cos120 2       1  o 1 2  Tan 35 16' with F 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 being120 . If the body is in equilibrium, o prove that the angle of friction  is not less than 30 . If o  45oand 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

.

(9)

Static – 3rd secondary Chapter One – Friction-9 -R o 30 o 30 112 o 150 Sin30 F F 150 o 150 Cos 30





o F o F

first case: F 112 150 Sin 30 38 gm.wt downward

3

And R 150 Cos 30 75 3 gm.wt and

5 3

R 75 3 45 g.wt F R

5

The body is not about to move Then the friction is not limiting Second case:

When the body

                 o F F o is about to move Then the friction is limiting

F R 45 gm.wt where F F 150 Sin 30

F 45 150 Sin 30 120

The force must increase to 120 gm.wt in order to make the body about to move

           Movement R o 30 o 30 F o 150 Sin30 F F 150 o 150 Cos 30 Movement Example (7) o

A body of 150 gm.wt is placed on a rough plane inclines at angle 30 to the horizontal and the 3

friction coefficient between them is . 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 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.

Answer

(10)

01009988836 – 01009988826 Email : [email protected] R o 30 o 30 o 30 o T Sin30 T o T Cos 30 o 40 Sin30 F F 40 o 40 Cos 30

 

F F o o F F

The body is about to move Then the friction is limiting F R

1

F R 1 4

The least force means that the body is about to move down

So, F T Cos 30 40 Sin 30

F              

 

o o 3 20 T 2

and R 40 Cos 30 T Sin 30

1 R 20 3 T 2 3 1 1 from 1 : 20 T 20 3 T 2 4 2 3 1 3 1 20 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            _  _             _  _  _         _ Movement Example (8)

A body of weight 40 Newtons rests on a rough plane which is inclined to the horizontal at an angle 30o . The body is pulled up by a string which makes an angle 30owith the plane. The string lies in the vertical plane which contains the body and the line of greatest slope. If the coefficient of friction is 1

4. Find the least force in the string which prevents the load from moving down .

Answer

(11)

Static – 3rd secondary Chapter One – Friction-11 -R   F W Sin F F W 130 Cos Movement  12 13 5 F F st F 5 R 130 Cos 130 50 Newton 13

And the body is about to move the friction is limiting 2

F R F 50 20 Newton

5

1 case: If movement is downwards 12 F 130 Sin F 130 20 100 Newton 13 "This is the                       nd F

least force needed" 2 case: If movement is upwards

12

F F 130 Sin 20 130 140 Newton

13 "This is the greatest force needed"

       R   F W Sin F F W 130 Cos Movement Example (9)

A 130 Newton weight is placed on a rough plane which is inclined to the horizontal by an angle whose cosine is 5

13. A force is applied to the weight parallel to the line of greatest slope upwards. If the coefficient of friction between the weight and the plane is 2

5, then find the limits 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.

(12)

01009988836 – 01009988826 Email : [email protected] R   20 Sin F F 20 Movement 1 F R F F 20 Movement 20 Cos 2 F 20 Sin 20 Cos                         F 1 F F 1 F 1

the body is about to move the friction is limiting F R 3 R 20 Cos 20 12 Newton 5 4 F 20 Sin F 20 F 5 F 16 F 16 R F 16 12     st

1 case : If movement is downwards

 









                      _  _        _   _         2 2 2 2 F 2 F 2 2 1 2 1 3 4 4 R 20 Cos F Sin 20 F 12 F 5 5 5 F Cos 20 Sin F 3 3 4 F 16 F 16 R F 16 12 F 5 5 5 3 4 When F F : 16 12 16 12 16 12 5 5 48 36 64 48 16 12 5 5 5 5              nd

2 case : If movement is downwards

 











                           _{ }   2 2 2 2 1 2 48 88 32 0 5 48 88 32 0 8 5 5 5 6 11 32 0 3 4 2 1 0 4 1 1 refused Or F F 16 12 10 Newton 3 2 2            2 F Cos 2 F Sin  5 3 4 Example (10)

A body whose weight is 20 Newton is placed on a rough plane inclined to the horizontal by an angle whose tangent is 4

3. If F1 is the least force when applied along the line of greatest slope 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

(13)

-38 F F Cos  F Sin R F F  4 3 5 st nd

1 case: Inclined plane The body is moving on an inclined plane 1

by angle whose tangent is 5 1 1 Tan Tan 5 5 2 case: horiz            

under its own weight only

F

ontal plane

the body is by a force F inclined by angle

F R

3 4

So, F F Cos F and R 38 F Sin 38 F

5 5 3 1 4 So, F R F 38 F Multiply 5 5 5                    _  _   about to move





 

F

 

by 5 4 4 19 3F 38 F 3F F 38 F 38 F 10 Newton 5 5 5 3

Then from 1 : F 10 6 Newton and from 2 : R 30 Newton

5

           

   

Example (11)

A body of weight 38 Newton is about to move under its own weight when placed on a rough plane inclined to the horizontal at an angle whose tangent is 1

5. If this body is placed on a horizontal plane which is as rough as the inclined plane, and is acted on by a force inclined to the horizontal at an angle whose sine is 4

5 so that the body is about to move. Find the force and the normal reaction.

Answer

(14)

01009988836 – 01009988826 Email : [email protected] R o 30 o 30 F o 2 Sin30 F F 2 o 2 Cos 30 Movement st o o nd

1 case: Inclined plane The body is moving on an inclined plane by angle 30

3 Tan 30

3

2 case: When a force acted on the p



 



    

o F o o F o o F lane

the body is by a force F

inclined by angle 60 to the horizontal F R

3

where F F Cos 30 2 Sin 30 F 1

2 1

and R 2Cos 30 F Sin 30 3 F

2 3 3 1 So, F R F 1 3 F 2 3 2                   _  about to move

 

F

 

3 3 3 3 2 3 F 1 1 F F F 2 F 2 F 3 kg.wt 2 6 2 6 3 3 1 Then from 1 : F 3 1 kg.wt 2 2                  o 30 o 30 o F Sin30 o F Cos 30 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 to the horizontal. If the weight is o then attached to a string which is pulled in a direction inclined at 60 to the horizontal so that o 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.

(15)

Static – 3rd secondary Chapter One – Friction-15 -R o 60 o 60 o 3 Sin60 F F 3 o 3Cos60 Movement st o o nd

1 case: Inclined plane The body is sliding on an inclined plane by angle 30

3 Tan 30

3

2 case: when the angle of the plane increas



 



    

under its weight





 

o o F F o F ed to be 60 3 3

Before the force acts: F 3 Sin60 kg.wt

2 After the force acted:

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

F R 1 3 where F 3 Sin60 F           

 

o rd F o o F o o 3 F 2 3 and R 3Cos60 kg.wt 2 3 3 3 3 from 1 F F 3 kg.wt 2 3 2

3 case: when the force is horizontal

F R 1

3 3 1

where F 3 Sin60 F Cos 60 F

2 2

3 3

and R F Sin60 Cos60 F

2 2 3 3 1 from 1 2 2                          F 3 F 3 3 3 3 3 F 1F 3 2 2 2 3 2 F 3 kg.wt         o F Sin60 R o 60 o 60 o 3Sin60 F F 3 o 3Cos60 o 60 _F o F Cos60 R o 60 o 60 o 3 Sin60 F F 3 o 3Cos60 Movement F 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 , then it is about to slide down. If the inclination of the plane to the horizontal o increased to become 60 , find the force of friction. Then find the magnitude of the least force o 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.

(16)

01009988836 – 01009988826 Email : [email protected] R   F W Sin F F W W Cos Movement F

The body is moving on an inclined plane by angle

Tan Tan Tan

To find the least force to make the body about to move up: Then the friction is limiting F



    



     

  

 





F

R 1

where F F W Sin and R W Cos

Then from 1 : F W Sin Tan W Cos

Sin F W Sin Cos                     



W Cos



2 2 2 W Sin F 2W Sin

To get the resultant in case the body is limiting:

R' R 1 R' W Cos 1 Tan W Cos Sec W Cos Sec

                    1 W Cos W Cos      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 to2W Sin . Prove also that the

resultant reaction is equal to W.

Answer

(17)

Static – 3rd secondary Chapter One – Friction-17 -R   3 Sin F F 3 3Cos Movement





Any body is about to move by its weight when 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 b          _      1 elow the body of the greater friction

2   _       4Cos 4 4 Sin T T

When the two bodies are about to move





 

st F 1 body 1 3 F 3Sin T R 3Cos 1 3Sin T 3Cos 3 T 3Sin Cos 1                    





 

nd F 2 body 1 2 F 4Sin T R 4 Cos 1 4Sin T 4 Cos 2 T 2Cos 4Sin 2                     

 





Then from 1 and 2 : 3Sin Cos 2Cos 4Sin

3

7Sin 3Cos Divide by Cos 7Tan 3 Tan

7                      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 coefficie



1 1

nts 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

(18)

01009988836 – 01009988826 Email : [email protected] R   F Cos W Sin F F W Cos Movement W F F We can solve this problem by two methods

The least tension required to move the rock upward

Tan and R W Cos F Sin

F F Cos W Sin

The body is about to move F R

F Cos W Sin W Cos

                    First method :

















F Sin Tan Sin

F Cos W Sin W Cos F Sin

Cos After multiplying both side by Cos

F Cos Cos W Sin Cos W Cos Sin F Sin Sin

F Cos Cos F Sin Sin W Cos Sin W Sin Cos

F Cos Cos Sin Sin W Cos Sin Sin Cos

                                                































1 o F Cos W Sin W Sin

F Then the minimum tension occurs when Cos is maximum

Cos

The maximum value of Cos 1 Cos 1 0

1000 Sin35 When 35 and W 1000 F 574 kg.wt 1                                               Another solutio

 





































F o o o o

when appears, you may use R' instead of F and R Then we can use Lami's rule between the three force R' , F' ,W

F W R'

Sin 180 Sin 90 Sin 90

F W Sin Sin 90 W Sin F W F Sin Cos                                        n





Cos Then continue...   F  F Sin R   F Cos W Sin F F W Cos Movement R' _ F  F Sin 90 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.

(19)

Static – 3rd secondary Chapter One – Friction-19 -R   F W Sin F F W Cos Movement W F F

R W Cos Tan F W Sin F

The body is about to move

F R W Sin F W Cos Tan

Sin W Sin F W Cos Cos Sin F W Sin W Cos Cos Multiply both sides by Cos :

F Cos W Sin Cos W Cos Sin F Co

                                          









_

_

 

F o s W Sin W Sin F F W Sin Sec Cos

when appears on an inclined plane, you may use R' instead of F and R Then we can use Lami's rule under the three force R' , F' ,W

F Sin 180      _{ } _             Another solution

































_

_

o o o W R' Sin 90 Sin 90 F W F W

Cos Sin Cos

Sin 180 W Sin F F W Sin Sec Cos                                    R   F W Sin F F W Cos Movement W R'   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

(20)

01009988836 – 01009988826 Email : [email protected] R   W 500 Sin 500 500 Cos  4 5 3 W 17525 W F F Movement st F F

1 case: When the least weight attached is T W 175 25 200 gm.wt The least weight means that the movement is down

The body is in equilibrium, F R

3 Where R 500 Cos 500 300 gm.wt 5 And F 500 Sin W 5                  F nd F 4 00 200 200 gm.wt 5 F 200 2 R 300 3

2 case: To get the maximum weight

The maximum weight means that the movement is up

2

The body is in equilibrium, F R where

3 3 And R 500 Cos 500 300 g 5                   

















F F m.wt And F T 25 500 Sin 4 T 25 500 F T 25 400 5 2 T 25 400 300 T 25 600 3 T 575 gm.wt                         R   W 500 Sin 500 500 Cos





W  T 25 W F F Movement Example (18)

A body of 500 gm.wt is placed on a rough plane which is inclined to the horizontal by an angle of measure , such that Tan 4

3

  , the body is then attached to a string passing over a smooth 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.