FRICTION NOTES | IIT-JEE | BITSAT | NEET | AIIMS

FRICTION

1.

FRICTION

When two t wo bodies are kept in contact, electromagnetic forces act between the charged particles (molecules) at the surfaces of the bodies. Thus, each body exerts a contact force on the other. The magnitudes of the contact forces acting on the two bodies are equal but their directions are opposite and therefore t he contact forces obey Newton’s Newton’s third law.

The direction of t he contact force acting on a particular body is not necessarily necessarily perpendicular to the contact surface. We can resolv e this contact force into two components, one perpendicular to the contact surface and the other parallel to it (figure. (figur e. The perpendicular component is called the normal contact for ce or normal force ( generally written as N) and the parallel component is called frict ion (generally written as f). Therefore if R is contact force then R= 2.

f 2  N2

R E A S O N S F O R F R IC T I O N

(i) nter-locking of ext ended parts of one object into the extended parts of the other object. (ii) Bonding between the molecules of the two surfaces or objects in contact. 3.

FRICTION FORC ORCE IS OF TWO TYPES.

a. Kinetic (a)

b. Static

K i n et i c Fr Fr i c t i o n Fo Fo r c e

Kinetic friction exists between two contact surfaces only when there is relative relative motio n between the two contact surfaces. It stops acting when relative motion between two surfaces ceases. Direction Direction o f kinetic friction on an object relativ e velocity of the object with respect to the other object i n contact considered. considered. t is opposite to the relative Note that its direction is not op posite to th e force applied applied it i s opposi te to the relative relative motion of th e body con sidered whic h is in contact with t he other surface.

Example 1. Find the direction of kinetic friction f orce

(a) on the block, exerted exert ed by the ground.

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Sol.

(a) (a)

(b) (b) where f 1 and f 2 are the fricti on forces on the block and ground respectiv respectively ely.. Example 2. The correct relation between magnitude of f 1 and f 2 is (A) f 1 > f 2 (B) f 2 > f 1 (C) f 1 = f 2 (D) not possible to decide due to insufficient data. Solution : By Newton‘s Newton‘s third law the above friction forces f orces are action-reaction action-reaction pair and equal but opposite to each each other in direction. Hence (C). Also note that the direction of kinetic friction has nothing to do with with applied force F. F. Example 3. All surfaces as shown shown in the figure are rough. Draw the friction force on A & B A B

10m/s 20m/s

////////////////////////////////////////////////////////

Solution :

Kinetic friction fri ction acts in such a way way so as to reduce relativ e motion. Example 4. Find out the distance trav elled by the blocks shown in the figure before it stops. 10 m/s 10 kg µk=0. 0.5 5

Solution :

N

f k

10g

/////////////////////////////////////////















 S = 10 m

Example 5. Find out the distance travel led by the block on incline before it stops. Initial velocity of the block is 10 m/s and and coefficeint of fr iction between between the block and incline is  = 0.5. Solution : N = mg cos37°  mg sin 37° + µN = ma a = 10 m/s2 down the incline Now v 2 = u2 + 2as 0 = 102 + 2(–10) S  S = 5 m

s s m / 0 1



37° fixed µ

Example 6. Find the time taken in the above case by the block to reach the initial position. Solution :

5 m

a = g sin sin 37° – µg cos 37°  a = 2 m/s2 down the incline  S = ut +  t =

1 2

at2



S=

1 2

× 2 × t2

5 sec.

Example 7. A block is given a velocit y of 10 m/s and a force of 100 N in addition to friction force i s also acting on the block. Find the retardation of the block? 10 m/s 

100 N  µ

Sol.

As there is relative motion  kinetic friction will act to reduce this relativ e motion.















a=

110 10

= 11 m/s2

Example 8. Find out the acceleration accelerati on of the block as shown in the figure.

10 Kg

30 N 

µ=0.5

30 N 10 Kg

Sol.



f = 30 0  f  f max 0  f s  µN  30 – f = 0 Hence a = 0

(b)

STATIC FRICT ION two surfaces when there is tendency of relative m otion but no relative motion m otion t exists between the two along the two contact surface. For example exam ple consider a bed inside a room ; when we gently push the bed with a finger, the bed does not move. mov e. This means that the bed has has a tendency to move in the direction of appli ed force but does not move as there exists static fricti on force acting in the opposite direction of the applied force.

Example 9. What is value of static fri ction force on the block?

Solution :

In horizontal direction as acceleration is zero.















Note : Here Here once again again the static f ricti on is invol ved when there is no relative motion between between two surfaces.

Example 10. In the following foll owing figure an object of m ass M is kept on a rough table as seen seen from abov e. Forces are applied on it as shown. shown. Find the direction of static friction if the object does not move. Solution : In the above problem we fi rst draw the free body diagram of f ind the resultant force.

As the object doe not move this is not a case of limiting friction. The direction of static friction is opposite opposite to the direction of the resultant force FR as shown in figure by f s . Its magnitude is equal to 25 N.

4.

MAGN MAGNIT ITUD UDE E OF OF KIN KINET ETI IC AND AND STA STATIC TIC FRI FRICT CTI ION

Kinetic friction : The magnitude of the kinetic fr iction is proportional to the normal force acting between the two bodies. bodies. We can write f k = k N where N is the normal force. The proportionality constant k is called the coefficient coefficient of ki netic friction and its value depends depends on t he nature of th e two su rfaces in contact. If the surfaces are smooth k will be small, if the surfaces are rough k will be large. It also depends on the materials of the two bodies in contact. Static Static friction : The magnitude of static static friction fricti on is equal and opposite to the external force exerted, till the object at which force is exerted is at rest. This means it is a variable and self adjusting force. However it has a maximum value called limiting friction. f max = sN The actual force of static friction fri ction may be smaller thansN and its value val ue depends on other forces acting on the















Following table gives a rough estimate of the values of coeffi cient of static fri ction between certain pairs pairs of materials. The actual value depends on the degree of smoothness and other environmental factors. For example, wood may be prepared at various degrees of smoothness and the friction coefficient will vary.

Material

s

Material

s

Steel and steel

0.58

Copper and copper

1.60

Steel and brass

0.35

Teflon and teflon

0.04

Glass and glass

1.00

Wood and wood

0.35

Rubber tyre on dry concrete road

1.0

Wood and metal

0.40

Rubber tyre on wet concrete road

0.7

Example 11. Find acceleration of block. Ini tially the block is at rest.

10 Kg

50 N 

µ=0.5 Solution :

zero

Example 12. Find out acceleration of the block. Initially Initiall y the block is at rest.

40 N 37°

10 Kg

















N + 24 – 100 = 0  N = 76 N Now

0 for vertical verti cal direction

0  of µs N 0  f s  76 × 0.5 0  f  38 N 32 < 38 Hence f = 32 acceleration of block is zero.

 

Example 13. Find out acceleration of the bl ock for different ranges of F. m

F

µ Solution : 0  f  µSN 0  f  µSmg a=0 a=

if F  µSmg

F  µMg if F > µMg M

Example 14. Find out acceleration of t he block. Initiall y the block is at rest. 10 kg µS = 0. 0.5 = 0.3 0.3

 51N















f s = 100 . ....... (1) F + 10 g = N  N = 100 + F . .......... (2) Now 0  f S  N 0  f s  µN 100  0.5 N 100  0.5N [100 + F] 200  100 + F F  100 N  Minimum F = 100 N 

Example 16. The angle of inclinat ion is slowly increased. Find out the angle at which the block starts moving.

µ



Solution : f

N s i n g m 



 o s c m g

0  &  µS N mg sin > f smax mg sin  > µN mg sin > µ mg cos 















N

7 5

6 0

8 0

We will put value of f in the last i.e. in the direction opposite to resultant of other forces. f acts down the incline and its val ue is of = 75 – 60 = 1 15 5N Example 18. In the above abov e problem how much force should be added to 75 N f orce so that block starts to mov e up the incline. Sol.  60 + 40 = 75 + f s  f s = 25 N Example 19. In the abvov e problem what is the minimum f orce by which 75 N force should be replaced with so that the block does not move. Solution : In this case the block has a tendency to m ove downwards. Hence friction fricti on acts upwards. upwards.

4 0 6 0  F + 40 = 60  F = 20 N

Example 20. Top view of a bl ock on a table is shown (g = 10 m/s2) .















µ

50

100 kg Sol.

1000

T = 100 g = 1000 N  f = 1000 to keep the block stationary Now f max = 1000

f

µN = 1000 µ=2 Can µ be greater than 1 ? Yes 0 < µ   Example 22. Find out minim um acceleration of block A so that the 10 kg block doesn’t fall. 

A

a

10

µ

Solution : Applying NL in horizontal horizontal direction direction N = 10 a . ........ (1) Applying NL in vertical direction 10 g = µ N . ......(2) 10 g = µ 10 a fro from (1) (1) & (2)  a =

µN N

10

a

10 g

µ

Example 23. Find the tension in the string in situation as shwon in the figure below. Forces 120 N and 100 N start acting















Assuming that the 10 kg block reaches limiti ng friction first then using FBD’s.

Also

120 = T + 90 T = 30 N  T + f = 100  30 + f = 100  f = 70 N which is not possible as the limiting v alue is 60 N for this surface of block.  Our assumption is wrong and now taking the 20 kg surface to be limi ting we have

T + 60 = 100 N  T = 40 N Also f + T = 120 N  f = 80 N This is acceptable as static fricti on at this surface should be less than 90 N. Hence the tension in the string is T = 40 N. Example 24. In the following figur e force F is gradually increased from zero. Draw the graph between applied applied force F and tension T in the string. The coef ficient of static friction fri ction between the block and the ground is s . Solution : As the external external force F is gradually gradually increased increased from zero it is compensated compensated by the friction fricti on and the string bears no no tension. When limiting lim iting friction fricti on is achieved by increasing force F to a value till  mg, the f urther increase in F is transferred to the string. string.

















F cos 37  sN F cos 37º > 0.5 [ 10g – F sin 37º] Hence Fslide >

Fslide >

Flift >

( N = 10 g – F sin sin 37º)

50 cos 37 º 0.5 sin 37 º

500 N 11 500 3

N.

Fslide < Flift Therefore the block will begin to slide before lifti ng.



TWO BLOCK PROBLEMS

Example 26. Find the acceleration of the two blocks. The system is initially at rest and the friction coeff icent are as shown shown in the figure? µ=0.5

10 A

Smooth 10 B

F = 50 N















(ii)

Check friction f or B : f = 10 × 2.5 = 25 25 N is required which is less than availabl e friciton hence they will move together. togeth er. and a A = aB = 2.5 m/s2

Example 27. Find the acceleration of the two blocks. The system is initially initi ally at rest and the friction fricti on coefficent coeffi cent are as shown in the figure? µ=0.5

10 A

101 N

smooth 10 B

//////////////////////////////////////////////////

Solution :

f max = 50 N  f  50 N A

101

f B

f

(i)

If they move together

(ii)

Check friction on B

a=

101 = 5.05 m/s2 20















40 < 50 (therefore (theref ore required < available) together..  will move together Example 29. In above example fi nd maximum F for which two blocks will will mov e together. together. Soution : Observing the critical situation where friction becomes limiting.

fmax = 50

10

fmax = 50

F

20

 F – f max = 10 a . ........ (1)

f max = 20 a . ......... (2)  F = 75 N Example 30. Initially Initiall y the system is at rest. find out minimum value of F for which sliding starts between the two two blocks. µ=0.5

10

Smooth 20

F

////////////////////////////////////

Solution : At just sliding condition limiting friction is acting.















aB =

30 = 1.5 m/s2 ( only friction fric tion force by block A is responsible responsible for producing acceleration in block B) 20

Because 4 > 1.5 m/s2 we can conclude that the blocks do not mov e together. Now drawing drawing the F.B.D. F.B.D. of each bl ock, for finding f inding out indivi dual accelerations.

a A =

aB =

120  30 = 9 m/s2 towards right 10 30 20

= 1.5 m/s2 towards right.

Case (ii) F (ii) F is increased from zero till t he two blocks just just start moving toget her. As the two blocks move togethe togetherr the friction is static in nature nature and its value is is limiting. FBD in this case will be

a A =

120  30 10

= 9 m/s2



aB =

F  30 20

=a A



F  30 20

= 9

 F = 150 N

Hence when 0 < F < 150 N the blocks do not move together and the friction is kinetic. As F increases















PART - I : SUBJECTIVE QUESTIONS

SECTION (A) : KINETIC FRICTION A-1.

Find the direction of f riction forces on each block and the ground (Assume all surfaces are rough and all velocities are with respect to ground).

A-2.

The wheel shown is fixed at ‘O’ and is in contact with a rough surface as shown . The wheel rotates with an angular velocity . W hat is the direction and nature of friction force on the wheel and on the ground. ground.

A-3.

In the following figure, find the direction of f riction on the blocks and ground .

















B-4.

In the figure fi gure shown calculate the angle of friction. The block does not slide. Take Take g = 10 m/s2.

SECTION SECTION (C) : M ISCELLAN ISCELLAN EOUS QUESTIONS QUESTIONS C-1.

C-2.

A block of mass 2.5 kg is kept on a rough horizontal surface . It is found that the block does not slide slide if a horizontal force less than 15 N is applied to it. Also it is found that it takes 5 seconds to slide throughout the first 10 m if a horizontal force of 15 N is applied and the block is gently pushed pushed to start the motion. Taking g= 10 m/s2, calculate the coefficients of static and kinetic friction between the block and the surface. Find the acceleration accelerations s and the friction forces involv ed : µ=0 5kg A

F=15N

µ=0.5 5kg A

30N















B2.

A box 'A' is lying on the horizontal horizo ntal floor floo r of the compartment compa rtment of a train running along horizontal rail s from left to right. At time 't', it decelerates. Then the reaction R by the floor on the box i s given best by :

(A) B3.

(B)

(C)

(D)

A cart of mass M has a blo ck of m ass m attache att ached d to it as shown in t he figure fi gure.. Co-efficient of friction between the block and cart is . What is the minimum acceleration of the cart so that the block m does not fall ? (A)  g

(B) /g

(C) g/ 

(D) none

B4.

A rope so li es on a tabl e that part pa rt of i t lays ov er. The rop e begi ns to sli de when the l ength engt h of ha ngi ng part is 25 % of entire length. The co-effi cient of friction between rope and table is : (A) 0.33 (B) 0.25 (C) 0.5 (D) 0.2

B 5.

A block of mass mass 1 kg kg lies on a horizontal horizontal surface surface in a truck. The coefficient of static static friction between between the block















C 3. 3.

A forc e F = t is appli app lied ed to b lock lo ck A as shown in f ig ure. T he force fo rce i s appl ied at t = 0 secon ds when t he system was was at rest and string is just straight without tension. tension. Whi ch of the foll owing graphs gives gives the friction force between B and horizontal surface as a function of time ‘t’.

(A)

(B)















4.

In the given situation it is known that when released the blocks slide. Find the accelerations of the two blocks. Also find the time when the small block will fall off from the larger block.

5.

A bead of mass ‘m’ is fitted onto a rod with with a length of 2, and can move on i t with friction having the coefficient of f riction . At the initial moment the bead is in the middle of the rod. The rod moves translationally in a horizontal plane with an acceleration ‘a’ in the direction forming an angle  with the rod. Find the time tim e when the bead will leave t he rod : (Neglect the weight weight of the bead).

6.

M A = 3 kg, MB = 4 kg and M C = 8 kg.  between any two two surface is 0.25. Pulley is fricti onless and string is m assless. assless. A is connected to the wall through a massless rigid rod.(g=10m/s2)















3.

4.

A chain of length length L is placed on a horizontal horizontal surface as shown shown in figure. At any instant instant x is the length of chain chain on rough surface and the remaining portion lies on smooth surface. Initial ly x = 0. A horizontal force P is applied to the chain (as shown shown in figure ). In the duration x changes from x = 0 to x = L, for chain to move with constant speed.

(A) (A) the magnitude magnitude of of P should increas increase e with time time (B) the magnitude of P should decrease with time (C) the magnitude of P should increase fi rst and then decrease with time (D) the magnitude of P should decrease first and then increase with time   The upper portion of an inclined plane of inclination  is smooth and the lower portion is rough. A particl e slides down from rest from the top and just comes to rest at the foot. If the rati o of the smooth length to rough length is m : n, the coefficient of friction is : (A) 

m

n

 tan

 m  n  (B)   cot 

 m  n   cot  (C) 

(D)

1

















10.

In the arrangement shown in the fi gure mass of the block B and A are are 2 m, , 8 m respectively. Surface between B and floor is smooth. The block B is connected to block C by means of a pulley. If the whole system is released then the minimum v alue of m ass of the block C so that the block A remains stationary stationary with respect to B is : (Co-effici ent of fr iction between A and B is  and pulley is ideal)















17.

The coefficient of friction between 4kg and 5 kg blocks is 0.2 and between 5 kg block and ground is 0.1 respectiv ely. Choose the correct statements















PART - I : MATCH THE COLUMN















5.

If applied applie d force F = 120 N, then then magnitude of accelerati on of 15 kg block will will be : 2 2 2 (A) 8 m/s (B) 4 m/s (C) 3.2 m/s (D) 4.8 m/s2

6.

Continuing with the situation, if the force F = 80 N is directed vertically as shown, the acceleration of the 10kg

10 kg















PART - IV : TRUE / FALSE 13.

A block B of mass 10 10 kg is placed place d on smoot smoot h horizontal surf ace ove r it another block. A of same mass is placed. A horizontal horizontal force F i s applied on block B. (i) No block will will mov e unless F > 10 N. (ii) Block A will move towards left.













































































FRICTION NOTES | IIT-JEE | BITSAT | NEET | AIIMS

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