Tag: laws of motion

Questions Related to laws of motion

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

friction laws of motion physics

  1. $\mu^2h$

  2. $\mu/h$

  3. $\mu h$

  4. $\mu^2h^2$

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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

friction laws of motion physics

  1. $9.8\ N$

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

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

  4. $0.7\times 9.8\ N$

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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:

friction laws of motion physics

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

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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:

friction laws of motion physics

  1. 20 N

  2. 20 $\sqrt{3}$ N

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

  4. zero

Show answer
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, $mg=20N, \mu=\cfrac{1}{\sqrt 3}$
$\Rightarrow F _{min}=\cfrac{20}{\sqrt 3}N$