Laws of Friction

Laws of friction : Force of friction between two surfaces obey the following laws :
1. Friction is an opposing force and comes into play, only, if the body is in motion or tending to move over the surface of another body.
2. Force of friction  acts along the tangent to the two surfaces at the point of contact.
3. The force of friction acts in a direction opposite to the direction of motion of the body.
4. The force of friction is independent of the areas of the surfaces in contact.
5. The force of friction is directly proportional to the normal reaction between the two surfaces in contact i.e., f ∝ R

Co-efficient of friction : The ratio of the limiting (maximum) force of friction to the  normal reaction is called co-efficient of friction.

The force of limiting friction which comes into play when a body is just to move  over the surface of another body is directly proportional to the normal reaction i.e.,
                           f ∝ R
                          f = µR
where µ is constant of proportionality and is called the co-efficient of friction.

                         µ =f/R   = Force of limiting friction/ Normal Reaction

It is a unites quantity and has different values for different pair  of surfaces but obeys the inequality i.e., 1 >µ > 0 for all practical purposes.

Friction

Friction : The opposing force which comes into play in between the surfaces of two bodies, when one is moved over the surface of another is called force of friction or friction.

Whenever there is a relative motion between the two surfaces in contact, an opposing force comes into play to prevent the relative motion. This opposing force is called force of friction.

When a body is at rest on a surface and no external force is applied, no friction exists. In this case the normal reaction R balances the weight Mg i.e. for vertical equilibrium R = Mg and frictional force =0

            

                  No friction

Causes of Friction : The surfaces of solid bodies are never perfectly smooth. Friction is produced by the interlocking of irregularities of one surface with that of the other surface in contact.

Static Friction : The friction offered by the surfaces in contact subjected to external forces until there is no motion between them is called static friction.

When an external force (F) is applied to move the body and the body does not move, then the frictional force acts opposite to applied force F and is equal to the applied force i.e., F – f =0 frictional force ,f = applied force F. When the body remains at rest, the frictional force is called the static friction. Static friction is self adjusting force.
                           

Limiting Friction: The maximum value of friction which comes into play when a body just starts moving over the surface of another body is called limiting  friction.
The force of friction goes on increasing as the pull on the body is increased till it just starts moving over the surface of another body.

                       
Consider a block of wood placed over the surface of a table. When no force is applied on the block, there is no force of friction between the surfaces in contact and the block is under the action of the following two forces:
(i) Weight of the block W = mg acting vertically downwards.
(ii) Normal Reaction are acting vertically upward.

When some force is applied to pull the body, friction comes into play and the body does not move. If the pulling force is gradually increased, the force of friction also increases. When the body just starts moving or is at the verge of motion the opposing force attains a maximum value and is called the force of limiting friction. The various forces acting on the block when it just starts moving are:
(i) weight of the block W = mg acting vertically downwards
(ii) Normal reaction R acting vertically upward
(iii) pulling force F
(iv) Force of friction f acting opposite to F

Since the body is in equilibrium under these forces , we have 
W = R, and F = f

i.e.,                   Weight of the body = Normal reaction
and,                       pulling force (F) = Force of friction (f)

When the pulling force slightly increases the force of limiting friction, the body just start moving.

courtesy: sciencehq.com

Types of Work

Positive Work : From the mathematical expression of work

                            W = F.S Cos θ

Where θ is the smallest angle between F and the displacement S.

When θ is acute angle i.e., less than 90°, then Cos θ is +ve 
i.e., if θ < 90, Cos θ = +ve and work done will also be positive

i.e.,                      W = F.S Cos θ  =  +ve

Work done is said to be +ve if the angle θ between applied force F and the displacement S is a acute angle i.e., θ < 90°.

Examples of positive work done : 
(i) When a body falls freely under gravitational pull, the work done by gravity is positive.
(ii) When a horse pulls a cart on a level road, the work done is positive.
(iii) When a string is stretched, the stretching force acts in the direction of displacement, here work done by stretching force is positive.

In all the above examples F and S are in the same direction hence work done is +ve.

Negative Work : From the mathematical expression of work

                            W = F.S Cos θ
When θ > 90°, Cos θ is –ve, hence work is also negative. Thus work done is said to be negative if angle(θ) between applied force (F) and the displacement (S) is an obtuse angle i.e., θ > 90°.

Examples of negative work done : 
(i) When a body is made to slide over a surface, the work done by the frictional forces is negative.
(ii) When two similar poles of a magnet are brought close to each other, the work done by the magnetic force of repulsion is negative.
(iii) When two similar charges either both positive or both negative are brought close to each other, they repel each other hence the work done by electrostatic force of repulsion acting between them will be negative.

In all the above examples F and S are in the opposite directions, hence work done is –ve.

Zero Work : From the mathematical expression of work

                            W = F.S Cos θ
Work done by a force will be zero when either θ = 90° or  S = 0 i.e., if there is no displacement. 
Hence, work done will be zero if either angle (θ) between applied force (F) and the displacement (S) is 90° or the body does not move under the application of force.

Examples of zero work done : 
(i) When a person travels on a platform with a load on his head, work done by him is zero.
(ii) when a body moves along a circular path with the help of a string, the work done by tension in the string is zero.
(iii) When a man tending to push a wall, he is doing no work. though the force is being applied, because the point to application of force i.e., wall does not move.

In the above example, displacement is zero hence work done is also zero, while in the example (i) and (ii), angle between F and S. i.e.,  θ = 90°. Hence work is zero.

Practice Problems:
1. What is the work done by a man in carrying a suitcase weighing 30 kg over his head when he travels a distance  of 10 m in :
(i) the direction direction;
(ii) vertically upward direction.
Sol. (i)  Work done = F Cos θ × S
Here,  F = force exerted by the man = mg = 30 × 9.8 N
θ  = 90°, Since the body moves in horizontal direction and force acts vertically upwards.
∴     Cos θ = Cos 90° = 0;
             S = 10 m
∴        W.D = 30 × 9.8 × 0 × 10
                =  0
( No work is done when the suitcase of 30 kg is moved by a man on his head in the horizontal direction)
(ii) Work done in moving in vertical upwards direction = mgh
Here,    m = 30 Kg;
             g = 9.8 m/s²
∴        W.D = 30 × 9.8 × 10 J
                = 2940 J (Ans.)
2940 J work is done by a man in carrying a suitcase of 30 Kg on his head in the vertical upward direction.

2. A man weighing 50 Kg caries a load of 10 kg on his head. Find the W.D. when he goes (i) 15 m  vertically up
(iii) 15 m on a leveled path on the ground.
Sol. Here, Mass of the man, m1 = 50 Kg
   Mass carried by the man , m2 = 10 kg
       Total mass, M = m1 + m2
                            = 50 + 10 
                            = 60 kg.
(i) When the man goes vertically up,
Height through which rises, h = 15 m.
∴     W.D. = Mgh = 60 × 9.8 × 15
                        = 8820 J (Ans.)
(ii) When the man goes on a leveled path on the ground.
       W.D. =  F Cos θ × S
Here,             θ  = 90°
∴     Cos θ = Cos 90° = 0;
             S = 15 m. ∴     W.D, = F × 0 × 15 = 0
(No work done in moving on a leveled path i.e., horizontally)
3. A body moves through a horizontal distance of 20 m under the action of a force of 1.5 N. Calculate the work done if the force makes an angle of
(i)  30° with the horizontal
(ii) 60° with the horizontal.
Sol. (i)  Here,
                     Force, F = 15 N and distance, S = 20 m
Angle with horizontal,  θ  = 30°
∴    work done is given by;
                W = F Cos θ × S

                    = 15 N × cos 30° × 20 m

                    = 300 × √3/2 Nm

                    = 259.8 joules (Ans.)

(ii) Here,    F = 15 N, S = 20 m, θ = 60°
Therefore, work done is given by,
                W = F Cos θ × S
                    = 15 N ×  cos 60° ×  20 m
                    = 15 N ×  1/2 ×  20 m 
                    = 150 J (Ans.)

4. Calculate the amount of work done when 1 kg water falls through the height 10 m.
Sol. Work done = mgh = 1×  9.8×  10
                                  =  98 J  (Ans.)

5. A body of mass 20 kg initially at rest is moved by a horizontal force of 5N on a smooth frictionless table. What will be work done by the force in 8 sec. ?
Sol.       Given,  u = 0 , t = 8
                        a = F/m  = 5/20

We know that,   S= ut + (1/2) at² 
                          = 0 × 8 + (1/2) ×  (5/20) × (8)² 
                       S = 1/2 × 5/20 × 64 
                          = 8 m
Work done = FS = 5 × 8  = 40 J. (Ans.)

Check your understanding with Multiple choice questions:

1. Work is done when force
(a) is applied
(b) produces motion
(c) is normal to displacement
(d) is strong

2. When the angle between force and displacement is obtuse, the work done is
(a) 1                         (b)  Zero 
(c) +ve                     (d)  –ve

3. When we lower a bucket into a well to fetch the water. The work done is
(a) 1                         (b) 0
(c) +ve                     (d)–ve

4. Work is measured in 
(a) Newton                (b) Joule
(c) Watt                    (d) Newton-cm

5. A stone of mass 0.1 Kg is whirled in a horizontal circle of radius 21cm with a force of 100 dyne work done to complete one cycle is
(a) 0                         (b) 100 J
(c) 100 dyne-cm        (d) 220 dyne-cm

6. The sun's gravitational force is keeping the earth in its orbit. The work done by this force is
(a) 0                         (b) +ve
(c) –ve                      (d) Infinite

7. Which of the following is not related with work ?
(a) F.S                       (b)Nm
(c) Joule                    (d)Hz.

8. Which of the following is not a unit of work?
(a) Joule                    (b) Kg m
(c) g cm                    (d) All are units of work.

9. Choose the incorrect statement .........
(a) Work done by centripetal force is Zero
(b) When a body slides on a horizontal surface, no work is done either by weight (mg) or normal reaction (R)
(c) Work done by frictional force is always negative
(d) A person walking on a horizontal road with a load on his head is doing work.

10. Work is
(a) Scalar quantity
(b) Vector quantity
(c) Neutral quantity
(d) F × S

11. A man pushes a wall but fails to displace it; he does
(a) negative work
(b) positive work 
(c) maximum work 
(d) No work at all

12. The work performed on an object does not depend on-
(a) applied Force
(b) displacement 
(c) angle at which force is inclined to displacement 
(d) initial velocity of the object.

13. A body carrying a box on his head is walking on level road from one place to another on a straight road is doing no work. This statement is -
(a) correct
(b) incorrect
(c) partly correct
(d) insufficient data

Work, Power and Energy

Work 
Work is said to be done by a force if the point of application of force moves in the direction of force.

                                                OR

Work is said to be done by a force when it actually moves the body through  a certain distances.


                                      
 Mathematically,          

Work done = Force (in the direction of motion)  × Distance through which Force acts.
                                                                                           
Let a force F   displaces a body placed on a horizontal surface through displacement  R,  in the direction of force  (F), then work done is given by,

 Work done = Force (in the direction of motion)  × Horizontal displacement
i.e.,        W = F.R

But when force (F) acts along a direction making an angle θ  with the direction of displacement (R), then work done is found by resolving the force.  

Since force is a vector quantity and can be resolved into two mutually perpendicular components-

(i)                        Fx = F Cos θ  
(ii)                       Fy = F Sin θ

Since  Fy component is perpendicular to the direction of motion of the body, hence it has no effect on the motion of the body i.e., No work is done by Fy component, it just balances the weight of the body. Only Fx component does work and moves the body (horizontally),

hence,       W =   Fx .R

              =   F Cos θ.R

OR            W =    F.R

Hence work may also be defined as the dot product of applied force (F) and the displacement (R) produced by force in the body.

Work is a scalar quantity and its dimensional formula is [M¹L²T^-2].

S.I. Unit of Work : The unit of work depend upon the units of force and the units of distance.
In S.I. system the unit of force is newton and unit of distance is metre.

∴         Unit of work  = newton × meter
                              = Nm
                              = Joule
The S.I. unit of work is newton metre or joule.

One Joule (or 1 Nm) is the work done by a constant force of 1 N, if it moves the body on which it acts through a distance of 1m in the direction of force.

C.G.S Unit of Work : In C.G.S system, unit of work is erg.

                                  1 erg = 1 dyne × 1 cm

One erg work is said to be done by a constant force of 1 dyne, if it moves the body on which it acts through a distance of 1 cm in the direction of force.

 S.I. units  and C.G.S units of work are related to each other by the relation.

                                  1 Joule = 10⁷ ergs.

courtesy:engineeringarchives.com
               

Electric Flux

The total number of electric lines of force that cut through a given surface area held perpendicular to the direction of electric lines of force is called electric flux.

It is represented by the Greek letter Ø . The surface through which the electric lines of force cross may be closed or open.

For closed surfaces electric flux is positive if the electric lines of force flow outwards and is negative if the electric lines of force flow inwards. Electric flux is positive for surface  around charge +q and negative for surface around charge q2.

For an open surface electric lines of force of a uniform electric field (E) are closing a surface of area A perpendicular to the electric field. Therefore, Electric flux  Ø, through the surface is given as
                               
                                           Ø = EA                   ......(1)

If the surface makes an angle θ with  the direction of electric lines of force, then the electric flux is given as
                                               Ø = EA cos θ

                                     
Units of Electric flux:

From the units of E and A the SI units of Electric flux are 

  N. m² C⁻¹  or  JmC^−1.   

The number of lines of force coming out of a positively charged body determines the charge on the body in coulomb.
Therefore,units of electric flux are also expressed as coulomb. Thus if a positively charged body has a charge of q coulomb then, Electric flux due to charged body,  Ø  =  q coulomb.

Electric flux through an area :
Electric flux through an area is defined as the number of electric lines of force passing through that area normally. It is a scalar quantity and is denoted by  Ø.

 Let we have an area  A in an electric field E, and let dA be small area vector element, then small flux (dØ) passing through small area (dA) is given by;

                        dØ = E.dA                         ......(2)

                         dØ = E dA cos θ                  .......(3)

where θ is the angle between E and dA.

If        θ = 0°,      i.e., E || dA, then flux is maximum and 

when  θ = 90°,     i.e., E ⊥ dA, then flux is zero.

To find flux through the whole closed surface A, we integrate eqn. (2) & (3) i.e.,

              

                          Ø = ∮A  E.dA

                              = ∮A  E  dA cos θ 

Electric flux is a scalar quantity as it is dot product of two vector quantities E and dA.

courtesy:vias.org  
hyperphysics.phy -astr.gsu.edu                              

Electric field Distribution due to Different Charges

Some typical electric field due to different charges are given below:

(i) Isolated Charged Sphere : In an isolated positively charged sphere, the electric lines of force leaving the surface are radially outward i.e., they appear to proceed from the centre of the sphere.
           

(ii) A positive and a negative charge : The electric field distribution due to two equally and oppositely charged spheres is as shown in fig.. 
                              
The electric lines of force start from the positive charge and end on the negative charge, behaving as if they were bonds pulling the charge towards the other.

(iii) Two positive charges : Consider two positively charged spheres placed at some distance apart as shown in fig.
                     

The electric field distribution will be as shown. The electric lines of force start from both the charges and behave as if they were bonds pulling apart one charge from the other.

There will be some point on the line joining the two spheres where electric field intensity is zero i.e. if a unit positive charge is placed at the centre of line joining two spheres, it will experience no force. Such a point is called neutral point.

(iv) Charged parallel plates :Consider two metal plates equally and oppositely charged and held at some distance apart as shown in fig. Such an arrangement forms parallel plate capacitor.

           
The electric field distribution is uniform between the plates i.e.,electric lines of force are straight and parallel in this region. However, near the edges of the plates, the field distribution is non-uniform as the lines of force are not equally spaced.

Courtesy:chem.ox.ac.uk
itacanet.org

Electric Lines Of Force

An electric line of force is defined as the path traced by a test charge when placed in the electric field.

According to coulomb's Laws, a positive charge repels a positive charge and attracts a negative charge, therefore, an electric line of force always starts from the positive charge and ends on the negative charge.
                   

Properties of electric lines of force :Following are the important properties of electric lines of force :

1. An electric line of force starts from the positive charge and ends on the negative charge.

2. An electric line of force does not form a closed loop like a magnetic line of force. There is no electric field inside a charged body and, therefore, the electric line of force ends on the negative charge.
                                 
3. Two electric lines of force never cross each other. The crossing of two lines of force means that there are two directions of the force on a test charge at a point which is impossible.It is well known that a body can move in one and only one direction under the action of force. Hence two electric lines of force can never cross each other.

4. The electric lines of force in the same direction repel each other and those in the opposite direction attract each other.
                                 
5. The tendency of electric lines of force is to taken easy electric path.