Physics

Friction and Inclined Planes

58 Questions

Friction and inclined plane problems focus on calculating forces, coefficients of static and kinetic friction, and motion on rough surfaces. Questions cover blocks on horizontal tables, angled planes, and simple harmonic motion systems. These physics concepts are strictly evaluated in engineering entrance and civil services tests.

Static friction coefficientKinetic friction forceInclined plane blocksDamped vibrations systemHorizontal table motion

Friction and Inclined Planes Questions

Multiple choice

Directions: Carry One Mark Each.

Which of the following statements are TRUE for damped vibrations?

P. For a system having critical damping, the value of damping ratio is unity and the system does not undergo a vibratory motion. Q. Logarithmic decrement method is used to determine the amount of damping in a physical system. R. In case of damping due to dry friction between moving surfaces, resisting force of constant magnitude acts opposite to the relative motion. S. For the case of viscous damping, drag force is directly proportional to the square of relative velocity.

  1. P and Q only

  2. P and S only

  3. P, Q and R only

  4. Q and S only
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

Multiple choice

A wardrobe (mass 100 kg, height 4 m, width 2 m, depth 1 m), symmetric about the Y-axis, stands on a rough level floor as shown in the figure. A force P is applied at mid-height on the wardrobe so as to tip it about point Q without slipping. What are the minimum values of the force (in newton) and the static coefficient of friction $\mu$ between the floor and the wardrobe, respectively?

  1. 490.5 and 0.5

  2. 981 and 0.5

  3. 1000.5 and 0.15

  4. 1000.5 and 0.25
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

Correct option is (1).

Multiple choice friction laws of motion physics

A block released from rest from the top of a smooth inclined plane of angle $\theta _1$ reaches the bottom in time $t _1$. The same block released from rest from the top of another smooth inclined plane of angle $\theta _2$, reaches the bottom in time $t _2$. If the two inclined planes have the same height, the relation between $t _1$ and $t _2$ is 

  1. $\cfrac{t _2}{t _1} = \left(\cfrac{sin \theta _1}{sin \theta _2}\right)^{1/2}$
  2. $\cfrac{t _2}{t _1} = 1$
  3. $\cfrac{t _2}{t _1} = \left(\cfrac{sin \theta _1}{sin \theta _2}\right)$
  4. $\cfrac{t _2}{t _1} = \left(\cfrac{sin^2 \theta _1}{sin^2 \theta _2}\right)$
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation
From the figure
we see that
down the incline acceleration us $a=g\sin \theta$
distance $s=\dfrac {h}{\sin \theta _1}$

using $s=ut+\dfrac {1}{2} at^2 \ ;\ u=0$ (initially rest)

gives $t=\sqrt {\dfrac {2s}{a}}$

For $\theta _1 \quad t _1=\sqrt {\dfrac {2\ h}{\sin \theta _1 \times g\sin \theta _1}}=\dfrac {1}{\sin \theta _1} \sqrt {\dfrac {2\ h}{g}}$

so for $O _2, \ t _2=\dfrac {1}{\sin \theta _2}\sqrt {\dfrac {2\ h}{g}}$

$\Rightarrow \ \dfrac {t _2}{t _1}=\dfrac {\sin \theta _1}{\sin \theta _2}$
Multiple choice friction laws of motion physics

The coefficient of friction between two surfaces is 0.2. The angle of friction is 

  1. sin$^{-1}$(0.2)
  2. cos $^{-1}$(0.2)
  3. tan$^{-1}$(0.1)
  4. cot$^{-1}$(5)
Reveal answer Fill a bubble to check yourself
D Correct answer
Explanation
The correct option is D

We have,

 The coefficient of friction is $0.2$

Since we know that,

The coefficient of friction $= tan \theta$  Where $\theta $  is the angle of friction.

$\dfrac{1}{5}=tan \theta$

$\theta=tan^{-1}\dfrac{1}{5}$

$=cot^{-1}5$
Multiple choice friction laws of motion physics

If angle of repose is ${30}^{o}$, then coefficient of friction will be

  1. $1$
  2. $15$
  3. $\cfrac { 1 }{ \sqrt { 3 } } $
  4. $\cfrac { \sqrt { 3 } }{ 2 } $
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation

$\mu =tan\theta \ =tan30=\frac { 1 }{ \sqrt { 3 }  } \ $


Multiple choice friction laws of motion physics

The coefficient of friction between a chain & a table is m. If a chain is placed on a horizontal table so that a part of it is hanging from one end, the minimum fraction of length of the chain that can be on the table, so that the chain may not slip off is

  1. $\frac{\mu}{\mu + 1}$
  2. $\frac{1}{\mu + 1}$

  3. 1

  4. Zero
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

For the chain not to slip, the friction force on the table portion must equal the weight of the hanging portion. If L is the total length and x is the length on the table, then mu * (x/L) * m * g = ((L-x)/L) * m * g. Solving for the fraction x/L gives mu/(mu+1).