IEMDaily - Video Lecture Notes

Example

Example of maximum static friction in non-inertial frame

Example:
A block of metal is lying on the floor of a bus. Find the maximum acceleration which can be given to the bus so that the block may remain at rest.
Solution:Let

a

be the acceleration of the bus. The contact between the metal block and the bus floor is due to the frictional force. The forces acting on the block are the inertia force and the frictional force. For equilibrium we have

m a = μ m g \Rightarrow a = μ g

where

μ

is the friction coefficient in between block and the floor of bus.

Example

Retardation due to friction

A car of mass

1000

k g

moving with a velocity of

10 m s^{- 1}

is acted upon by a forward force of

1000

N

due to engine and retarding force of

500

N

due to friction. It's velocity after

10

s

is:

For a body in equilibrium,
Equating forces along Y-axis,
Normal reaction

N = m g

Equating forces along X-axis,

m a + μ N = F

m a + μ m g = F

m a = F - μ m g

a = \frac{F - μ m g}{m}

a = \frac{1000 - 500}{1000}

∴ a = 0.5 m / s^{2}

Now,

v = u + a t

∴ v = u + a \times t

∴ v = 10 + 0.5 \times 10

∴ v = 15 m / s

Example

Friction acting at one fixed horizontal surface

The maximum value of the force

F

such that the block shown in the arrangement, does not move is given by:
Block does not move till the horizontal force on it becomes more than the maximum static frictional force.

F_{m a x} \cos \frac{π}{3} = μ m g N μ (m g + F_{m a x} \sin \frac{π}{3}) = \frac{1 \times (\sqrt{3} \times 10 + \frac{{\sqrt{3} F}_{m a x}}{2})}{2 \sqrt{3}} \Rightarrow F_{m a x} = 20 N

Definition

Friction at fixed inclined surface

A plane surface is inclined at an angle of

60^{0}

. A body of mass

10

kg is placed on it. If the value of coefficient of friction

μ_{K}

, between the body and the inclined surface is

0.2

, calculate the downward acceleration of the body, along the inclined plane surface. (Take

g = 10 m s^{- 2}

) At angles greater than the the critical angle of inclination, the block slides down the incline with uniform acceleration

a

.The frictional force is

μ_{K} N

. Here

N

is the normal reaction.
The net force acting on the body in a direction along the plane is:

F_{x} = m g s i n θ - μ_{K} m g c o s θ = m a

Hence, the acceleration

a

of the body is related to

θ

μ_{K}

by the equation:

a = g (s i n θ - μ_{k} c o s θ)

On substituting the respective values:

a = 10 (\frac{\sqrt{3}}{2} - 0.2 \times \frac{1}{2})

a = 7.66 m s^{- 2}

Example

Friction

1. A horizontal force of F N is necessary to just hold a block stationary against a wall. The coefficient of friction between the block and the wall is . The weight of the block is
Solution: Let, W - weight of body, R = Reaction force(F), Weight downward=W
For no movement
Frictional force = Weight downward

∴ W = μ F

2. A block of Mass M is moving with a velocity v on straight surface. What is the shortest distance and shortest time in which the block can be stopped if is coefficent of frictionSolution:
Force of friction=

μ N = μ m g

Therefore retardation =

μ m g / m = μ g

From

v^{2} = u^{2} + 2 a s

S = \frac{v^{2}}{2 g}

from v=u+at
or

t = \frac{v}{μ g}

Definition

Friction that involve multiple forces

A block of mass

m

is pressed against a vertical wall by applying a force

P

at an angle

θ

to the horizontal as shown in the figure. As a result, if that block is prevented from falling down and

μ

is the coefficient of static friction between the block and the wall, the value of

P

will be given as:
From the FBD of block, we have

P c o s θ = N

where

N

is normal reaction.

μ N - P s i n θ = m g

Or,

μ P c o s θ - P s i n θ = m g

P = \frac{m g}{μ c o s θ - s i n θ}

Example

Friction on Multiple horizontal surfaces

A block A of mass

2

kg rests on another block B of mass

8

kg which rests on a horizontal floor. The coefficient of friction between A and B is

0.2

, while that between B and the floor is

0.5

. When a horizontal force of

25

N is applied on B, the force of friction between A and B is:
Here

F = μ m g = 0.5 \times 10 \times 10

50 N

To move block of

8

kg along

2

kg,

50 N

force is required
But only

25 N

is applied.

∴

Block 'B' will not move.
Hence relative motion between A and B is not there hence friction

= 0

Definition

Kinetic friction in accelerated frame

Block A of mass

35

kg is resting on a frictionless floor. Another block B of mass

7

kg is resting on it as shown in the figure. The coefficient of friction between the blocks is

0.5

while kinetic friction is

0.4

. If a horizontal force of

100

N. is applied to block B, the acceleration of the block A will be

(g = 10 m s^{- 2})

f = 0.4 \times 7 \times 10 = 28

a = \frac{f}{35} = \frac{28}{35} = 0.8 m / s^{2}

Example

Problems on friction where magnitude of force is variable

Example: A variable force F

= 10 t

is applied to block B placed on a smooth surface. The coefficient of friction between A and B is

0.5

. (

t

is time in seconds. Initial velocities are zero). At what instant block A starts sliding on B?

Solution:

f_{m a x} = μ \times 3 g = 0.5 \times 3 \times 10 N = 15 N

where

f_{m a x}

is max force of static friction
Block A starts sliding when friction becomes

f_{m a x}

\Rightarrow 15 = 3 a

\Rightarrow a = 5 m / s^{2}

Both the blocks will move with same acceleration at this instant

F - 15 = 7 a

10 t - 15 = 7 a = 7 \times 5

\Rightarrow 10 t = 50

\Rightarrow t = 5 s

Work done

= \int F . d s = \int 10 t . d s (a = \frac{F}{m} = \frac{10 t}{10} = t)

= \int_{0}^{5} 10 t . V d t (d s = v d t)

= \int_{0}^{5} 10 t . \frac{t^{2}}{2} d t (v = \int a d t = \int t d t = \frac{t^{2}}{2})

= \int_{0}^{5} 5 t^{3} d t = 5 {[\frac{t^{4}}{4}]}_{0}^{5} = \frac{5}{4} [625 - 0] = \frac{625 \times 5}{4} = 781.25 J

.
This is the amount of heat produced due to friction.

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