Tag: friction

Questions Related to friction

A light ladder is supported on a rough floor and lens against a smooth wall, touching the wall at height 'h' above the floor. A man climbs up the ladder until the base of the ladder is on the verge of slipping. The coefficient of statice friction between the foot of the ladder and the floort is $\mu$. The horizontal distance moved by the man is

  1. $\mu^2h$

  2. $\mu/h$

  3. $\mu h$

  4. $\mu^2h^2$


Correct Option: C

A block of mass $2\ kg$ rests on a rough inclined plane making an angle of ${30}^{o}$ with the horizontal. The coefficient of static friction between the block and the plane is $0.7$. The frictional force on the block is

  1. $9.8\ N$

  2. $0.7\times 9.8\times \sqrt { 3 } N$

  3. $9.8\times \sqrt { 3 } N$

  4. $0.7\times 9.8\ N$


Correct Option: A
Explanation:

The force applied on the body that is on the inclined plane is given as,

$F = mg\sin \theta $

$F = 2 \times 9.8 \times \sin 30^\circ $

$ = 9.8\;{\rm{N}}$

The limiting friction force between the block and the inclined plane is given as,

$f = \mu mg\cos \theta $

$f = 0.7 \times 2 \times 9.8\cos 30^\circ $

$ = 11.88\;{\rm{N}}$

Since the limiting friction force is greater than the force that tends to slide the body.

Thus, the body will be at rest and the force of friction on the block is $9.8\;{\rm{N}}$.

An object is placed on the surface of a smooth inclined plane of inclination $\theta$. It takes time $t$ to reach the bottom. If the same object is allowed to slide down a rough inclined plane of same inclination $\theta $, it takes times nth to reach the bottom where $n$ number greater than $1$. The coefficient of friction $\mu$ is given by:

  1. $\mu =\tan { \theta \left( 1-1/{ n }^{ 2 } \right) } $

  2. $\mu =\cot { \theta \left( 1-1/{ n }^{ 2 } \right) } $

  3. ${ \mu =\tan { \theta \left( 1-1/{ n }^{ 2 } \right) } }^{ 1/2 }$

  4. ${ \mu =\cot { \theta \left( 1-1/{ n }^{ 2 } \right) } }^{ 1/2 }$


Correct Option: A
Explanation:

$\dfrac { { V } _{ a } }{ n } =\sqrt { 2Lg\left( \sin\theta -\mu \cos\theta  \right)  } \quad \longrightarrow \left( 1 \right) $

${ V } _{ a }=\sqrt { 2Lg\sin\theta  } \quad \longrightarrow \left( 2 \right) $
By putting equ(2) in eq(1) we get
$\dfrac { \sqrt { 2Lg\sin\theta  }  }{ n } =\sqrt { 2Lg\left( \sin\theta -\mu \cos\theta  \right)  } $
$\dfrac { 2Lg\sin\theta  }{ { n }^{ 2 } } =2Lg\left( \sin\theta -\mu \cos\theta  \right) $
$\dfrac { \sin\theta  }{ { n }^{ 2 } } =\sin\theta -\mu \cos\theta $
$\mu \cos\theta =\sin\theta -\dfrac { \sin\theta  }{ { n }^{ 2 } } $
$\mu \cos\theta =\sin\theta \left( 1-\dfrac { 1 }{ { n }^{ 2 } }  \right) $
$\mu =\tan\theta \left( 1-\dfrac { 1 }{ { n }^{ 2 } }  \right) $

A body of  weight 20 N is on a horizontal surface, minimum force applied to pull it when applied force makes an angle $60^0$ with horizontal (angle of friction a = $30^0$) is:

  1. 20 N

  2. 20 $\sqrt{3}$ N

  3. $\dfrac{20}{\sqrt{3}}$ N

  4. zero


Correct Option: C
Explanation:
For minimum force, we will use limiting friction  i.e.
$\Rightarrow \tan\phi=\mu _{s}$
and ATQ, $\phi=30$
$\Rightarrow \tan 30=\mu _{s}$
$\Rightarrow \mu _{s}=\cfrac{1}{\sqrt 3}$
Minimum force can be calculated as $=\mu _{s}mg$
ATQ, $mg=20N, \mu=\cfrac{1}{\sqrt 3}$
$\Rightarrow F _{min}=\cfrac{20}{\sqrt 3}N$

The surface of the head of a match stick and sides of a match box are deliberately made rough to:

  1. increase friction

  2. decrease friction

  3. increase amount of heat

  4. decrease amount of heat


Correct Option: A
Explanation:

The rough surface will increase the friction and so the rough surface of match box and match stick will produce more friction which helps in firing the stick easily.  

Correct air pressure in our vehicle tyres help to reduce:

  1. static friction

  2. sliding friction

  3. rolling friction

  4. all of the above


Correct Option: C
Explanation:

Correct air pressure helps in reducing deformation of the tyre and also provide an air cushion to reduce rolling friction. 


Option C is correct.

Wheels reduce friction due to:

  1. more area of contact

  2. less area of contact

  3. size

  4. all of the above


Correct Option: B
Explanation:

The friction will be more for more area of contact. Since the wheels have less area of contact so the friction will be less for them. 

With increase of temperature, the frictional force acting between two surfaces increases.

  1. True

  2. False

  3. Ambiguous

  4. Data Insufficient


Correct Option: B
Explanation:

Friction decreases with increase in temperature.

Friction is minimised by using

  1. Wheels and balI bearings

  2. More number of tyres in a car

  3. Rough surfaces

  4. None of these


Correct Option: A
Explanation:

Answer is A.

Bearings reduce friction by providing smooth metal balls or rollers, and a smooth inner and outer metal surface for the balls to roll against. These balls or rollers 'bear' the load, allowing the device to spin smoothly.
The concept behind a bearing is very simple. Things roll better than they slide. The wheels on our car are like big bearings. If we had something like skis instead of wheels, our car would be a lot more difficult to push down the road.
That is because when things slide, the friction between them causes a force that tends to slow them down. But if the two surfaces can roll over each other, the friction is greatly reduced.

powder sprinkled on a carrom board

  1. magnifies friction.

  2. reduces friction.

  3. removes friction.

  4. does not affect friction.


Correct Option: B
Explanation:

The carom board may sometimes be very rough. Sprinkling powder on a microscopic level, the powder particles offer rolling friction to the striker because of which the striker can be used for smooth and fast play.