Physics

Session Opener

A recent space shuttle accident occurred because of failure of heat protecting devices.

How was this heat generated ?

Session Objectives

1. Friction & frictional force

2. Bodies connected in frictional surface

3. Kinetic friction

4. Static friction

5. Angle of friction

6. Coefficient of friction

7. Laws of friction

Session Objective

Friction and Frictional Force

Whenever an object moves or tends to move, contact surfaces of object resist the force tending to generate motion.

This property of the surface is friction

The resistive force is the friction force

An equal and opposite force f acts

Consider the experiment :

1. F applied.

‘A’ does not move.

2. ‘A’ still at rest on increased F.

f adjusts to match F

MB

F MA

A

Bodies Connected in Frictional Surface

f

3. increased further F

A has acceleration. f has a maximum limiting value

f during the motion is lower than at rest.

F

4. is reducedA moves uniformly

Bodies Connected in Frictional Surface

MB

F fF fF fF fF fF fF fF fF fF f

Friction retards motion

f adjusts to stay equal and opposite to F

f has a limiting value

Bodies Connected in Frictional Surface

moving friction force is less than friction force at rest.

Friction force (f) is zero if no external force ( ) existsF

Friction force is a contact force, independent of area of contact.

Bodies Connected in Frictional Surface

Friction force is a non-conservative force as mechanical energy changes to heat during friction.

That is why the space shuttle overheated and exploded

Ff

Kinetic Friction

appears when two objects in contact have relative motion.

Directed opposite the direction of motion

fk is a constant.

fk on B may start the motion of B.

F fk

A

B

F fk

fk

Static Friction

F has component along contact surface, but there is no motion.

f = Fcos till limiting friction.

At limiting friction f(=fs) is constant

fs > fk

f acts along the surface and opposite the force component

F N

mg

fsFcos

Class Exercise

Class Exercise - 1

Blocks A and B are pressed against a smooth wall and are in equilibrium by a horizontal force F as shown. Then friction force due to A on B

(a) upward

(d) system cannot be in equilibrium

(c) dependent on relative mass of A and B

(b) downward

Solution

A B

Wall

F

As wall is smooth, A and B will keep slipping down with acceleration g andcannot be in equilibrium.

Hence answer is (d).

Coefficients of Friction

Limiting static friction

ss s

f: cons tant f N

N

Kinetic friction

kk k

f: cons tant f N

N

are coefficient of friction

s k

0

f

F

Static region

Ffs

Graphically

Kinetic region

Nf kk

Nf smax,s Limiting friction

Coefficients of Friction

Laws of Friction

1. Bodies slip over each other

k kf N

2. Direction of kinetic friction is opposite to the velocity

3. Bodies do not slip over each other

s sf NN is called the limiting friction. s

do not depend upon the area of contact.

k sf or f4.

Class Exercise

Class Exercise - 4

m

Wall

F

An object of mass m is pressed against a wall with a force F and is in equilibrium. The coefficient of friction between the wall and the object is . Then F must be equal to

(c) mg (d) cannot be found

mg(a) mg

(b)

Solution

As equilibrium exists, wall presses against the object by a force of F (equal and opposite to force exerted by object against the wall)

FF

f

mg

f F

mg f F

mgF

At equilibrium

Class Exercise - 5

Fm a

mb m c

Smooth surface

Three blocks of masses ma,mb and mc are arranged on a smooth horizontal surface as shown. Surface between ma and mb is smooth and the coefficient of friction between mb and mc is .Then the minimum force F required to keep mb from sliding down is:

a b

c

m m gm

a a b c b

c

m m m m g

m

b

a b cm m m g

c

a b c b

c

m m m m g

m

d

Solution

F m a

m b m c

Smooth surface

a

Let the force F give an acceleration a to the system

To give an acceleration a to mc, mb presses against mc with force mca, so mc gives a normal reaction mc a (to left) on mb.

a b cF m m m a

b cm g m a

b

c

m ga

m

a b c b

c

m m m m gF

m

for mb not slipping down

Class Exercise - 6

A block of mass M is kept in a lift. Coefficient of friction between the block and the lift is . The force required to initiate horizontal motion of the block is maximum when

(a) Lift is moving up with constant acceleration

(b) Lift is moving down with constant acceleration

(c) Lift is stationary

(d) Lift is in free fall

Solution

F is largest when N is maximum

N is maximum when lift moves up with constant acceleration a [N = m(g + a)].

So F is maximum at that condition.

Condition of motion of M: F N

f

N

W

F

Class Exercise - 8

A 70 Kg box is pulled along a horizontal surface by a 400 N force at an angle of 30° above horizontal. If the coefficient of friction is 0.50, what is the acceleration of the box? (g = 10 m/s2)

Solution

f

N

mg

F cos 30°

F sin F

Vertical: N = mg – F sin30

Horizontal: F cos 30 – f = ma

F cos30 N ma

F cos30 sin30 mg ma

3 0.5400 0.5 70 10 70 a

2 2

Solving for a: a = 1.37 m/s2

Angle of Friction

R

max,sf

f along surface (surface property)

N

N normal to surface

Angle of friction

ftan

N

Static > kinetic

Class Exercise

Class Exercise - 2

The angle between the resultant cont-act force and the normal reaction force exerted by a body on the other when they are one on top of the other is . Then, what is the coefficient of friction,

(a) = tan (b) > tan

(c) < tan (d) and are not related

Solution :

Hence answer is (a)

ftan

N

Class Exercise - 3

F

w

A pushing force F is applied to a body of weight W, placed on a horizontal table, at an angle . The angle of friction is . The magnitude of F to initiate motion in the body is

cos

sinWa.

cos

cosWb.

sin

sinWd.

cos

tanWc.

Solution

F cos

f

F sin N

W

Vertical: sinFWNfcosF Horizontal:

(a = 0 at initiation of motion)

f N tan W F sin tan

F cos W F sin tan

sinF cos sin . W tan

cos

cos cos sin sin WsinF

cos cos

F cos W sin Wsin

Fcos

Angle of Repose

Object at rest

sf

tanN

Object slips

increased .

is angle of repose

Nfs

mg

Angle of Repose

reduced

At , object moves uniformly.k

: Angle of repose (kinetic friction)k

kk

ftan

N

k s

Nfk

mg

Class Exercise

Class Exercise - 9

A block ‘A’ slides from rest down an incline with a 30° inclination with horizontal. It covers 3 m in 5 seconds. What is the value of coefficient of kinetic friction? (g = 10 m/s2)

A

30°

Solution

mg sin

mg cos

x

mg

fN y

Along y: N = mg cos

Along x : mg sin – f = ma

g sin cos a

2 21 1x ut at at

2 2

22x 1

gsin .cos .gt

10 2 3 2.

2 25 3 10

0.53

Class Exercise - 10

M

30°

A block of mass 2kg lies on a rough inclined plane of inclination 30° to the horizontal. Coefficient of friction between the block and the plane is 0.75. What minimum force will make the block move up the incline? (g = 10 m/s2)

Solution

mg cos

mg

mg sin

y N xF

Along y: N = mg cos

F is up the incline (along x)

F – (mg sin + f) = ma

To start the motion a is put equal to zero

mg sin mg cos

mg sin cos

1 3 32 10 . 22.98 N

2 4 2

minF mg sin f

Thank you