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This is one of our many quizzes on Friction. Click on this to access more such competitive quizzes on Dynamics.

Friction Quiz 2

physics

What will be the frictional force acting when we paint the walls using a painting roller and rubbing one person’s hands together? 

 Rolling friction is a kind of kinetic friction and sliding friction will be acting where the surfaces are slipping over each other. 

  When a paint roller is rolling over a wall, rolling friction will be acting in between the roller and the wall. And also when a person is rubbing his or her hands together, the heat will be generated between the hands. As we are rubbing our hands, they will be sliding between each other, hence it will be categorised as the sliding friction.

Meenu was applying a force over a load of mass $2\text{ }kg$ which has been moving with an acceleration, $g$. The frictional force acting between the body and the ground is found to be equivalent to the force experienced on a car moving with an energy of $20\text{ }J$ to a distance of $50\text{ m}$. Find out the angle between the normal and the force applied. 

 The coefficient of friction will be the ratio of the frictional force to the force acting over the load. 

  $\text{The equation of frictional force,} \\ f=\mu {{F}_{L}}..........(1) \\ \text{The force over the load can be found by equation,} \\ {{F}_{L}}=mg \\ \text{Substituting the values,} \\ {{F}_{L}}=2\times 10 \\ \Rightarrow {{F}_{L}}=20\text{ }N \\ \text{The frictional force,} \\ f=\dfrac{W}{d} \\ \text{Substituting the values,} \\ \Rightarrow f=\dfrac{20}{50} \\ \therefore f=0.4\text{ }N \\ \text{Rearranging the equation (1),} \\ \mu =\dfrac{f}{{{F}_{L}}} \\ \text{Substituting the values,} \\ \Rightarrow \mu =\dfrac{0.4}{20} \\ \therefore \mu =0.02 \\ \text{The angle between the normal and the force applied,} \\ \mu =\tan \theta  \\ \Rightarrow \theta ={{\tan }^{-1}}\left( \mu  \right) \\ \text{Substituting the values,} \\ \theta \text{=}{{\tan }^{-1}}\left( 0.02 \right) \\ \therefore \theta =1.146{}^\circ$

Appu was riding a bicycle of total mass $120\text{ }kg$ through a rough road with the coefficient of rolling friction mentioned as $0.8$.  What will be the rolling resistance of the bicycle? 

 The rolling frictional force will be the product of the coefficient of rolling friction and the weight of the body. 

 $\text{The equation of rolling frictional force,} \\ {{f}_{r}}={{\mu }_{r}}W \\ \text{Weight of the body,} \\ W=mg \\ \text{Substituting this in the equation,} \\ \Rightarrow {{f}_{r}}={{\mu }_{r}}\times mg \\ \text{Substituting the values,} \\ \Rightarrow {{f}_{r}}=0.8\times 120\times 10 \\ \therefore {{f}_{r}}=960\text{ N}$

Heena was going to a park by riding her car on a circular banked road having a circumference of the road mentioned as $1\text{ }km$. The coefficient of friction has been given as $1.732$. What will be the ratio of the safest speed she should maintain in order to avoid wear and tear and the force of friction if the mass of the car is $100\text{ }kg$? 

 The safest speed will be the square root of the product of the radius, acceleration due to gravity and the tangent of the angle. 

 $\text{The equation of safest speed,} \\ v=\sqrt{Rg\tan \theta } \\ \text{The radius of the park,} \\ \text{R=}\dfrac{\text{Circumference}}{\text{2 }\!\!\pi\!\!\text{ }} \\ \text{Substituting the values in it,} \\ R=\dfrac{1000}{2\pi } \\ \Rightarrow R=159.15\text{ m} \\ \text{The angle,} \\ \text{tan}\theta \text{=}\mu  \\ \text{Substituting and rearranging,} \\ \theta ={{\tan }^{-1}}\left( \mu  \right) \\ \Rightarrow \theta =60{}^\circ  \\ \text{Substituting the values in equation (1),} \\ \Rightarrow v=\sqrt{159.15\times 10\times \tan 60{}^\circ } \\ \Rightarrow v=\sqrt{2756.478} \\ \therefore v=52.50\text{ }m{{s}^{-1}} \\ \text{Force of friction,} \\ f=\mu N \\ \Rightarrow f=1.732\times 100\times 10 \\ \therefore f=1732\text{ }N \\ \text{The ratio,} \\ \dfrac{v}{f}=\dfrac{52.50}{1732} \\ \therefore \dfrac{v}{f}=0.0303\text{ }m{{s}^{-1}}{{N}^{-1}}$

A body of mass $3\text{ }kg$ was placed on an inclined plane with an angle of inclination mentioned as $30{}^\circ $. The coefficient of friction has been given as $\dfrac{1}{\sqrt{5}}$. Calculate the force required to start the body moving upwards and the force required to prevent the body from sliding down the inclined plane. 

 The frictional force acting on a body which is kept on an inclined plane will be the product of coefficient of friction, mass, acceleration due to gravity and the cosine of the angle of inclination. 

 $\text{The minimum force required to make the body move upwards,} \\ {{F}_{1}}=mg\sin \theta +\mu mg\cos \theta  \\ \text{Substituting the values in it,} \\ \Rightarrow {{F}_{1}}=3\times 10\times \sin 30{}^\circ +0.447\times 3\times 10\times \cos 30{}^\circ  \\ \Rightarrow {{F}_{1}}=15+11.613 \\ \therefore {{F}_{1}}=26.6134\text{ }N \\ \text{The minimum force required to prevent the body from sliding down,} \\ {{F}_{2}}=mg\sin \theta -\mu mg\cos \theta  \\ \text{Substituting the values,} \\ \Rightarrow {{F}_{2}}=3\times 10\times \sin 30{}^\circ -0.447\times 3\times 10\times \cos 30{}^\circ  \\ \Rightarrow {{F}_{2}}=15-11.613 \\ \therefore {{F}_{2}}=3.387\text{ }N$