Informative line

Static Friction

Learn static friction definition physics and types of friction with examples, Practice to calculate angle of repose and friction problems.

Visual Idea of Friction

  • When we look at the surface of rigid body at microscopic level, we find that surface of an evenly smooth polished body is not as smooth as it  appears.
  • There are some irregularities on the surface.
  • There are several projections and several depressions making the surface irregular.
  • Whenever two surfaces are in contact with each other, then due to these irregularities of surface, at microscopic level, they get interlocked so that they can't slip over each other. 
  • This phenomenon is known as friction and the force that resists the relative motion of two surfaces is called friction force.

Illustration Questions

Consider two blocks  A and B, which are placed as shown in figure. The block A is moving with velocity \(v_1= \) 5 m/sec and the block B is moving with velocity \(v_2= \) 10 m/sec. What will be the direction of friction force on block A?

A Right 

B Left 

C Upward 

D Downward

×

Relative velocity of block A with respect to B (vAB),

vAB = vA–vB

vAB = 5–10

vAB = –5 m/s

image

Friction always acts opposite to relative motion.

image

Hence, option (A) is correct.

Consider two blocks  A and B, which are placed as shown in figure. The block A is moving with velocity \(v_1= \) 5 m/sec and the block B is moving with velocity \(v_2= \) 10 m/sec. What will be the direction of friction force on block A?

image
A

Right 

.

B

Left 

C

Upward 

D

Downward

Option A is Correct

Types of Friction

  • There are three types of friction:
  1. Static friction 
  2. Kinetic friction 
  3. Rolling friction

1. Static Friction

  • Static friction comes into act when there is no slipping even after applying the force.
  • The maximum value of static friction is  \(f_s = \mu_s N\)

where,  \(\mu_s\) is coefficient of static friction 

N is normal force

  • Consider a block placed on a rough surface and pulled towards right with force F.
  • The block does not move even the applied force is increasing as a counteracting force of friction is also increasing till it reaches its maximum value and balances the applied force.

  • The block does not move upto the maximum limit of fs i.e., \(f_s = \mu_ s\,N\).

2 . Kinetic Friction

  • Kinetic friction comes into play, when a body slips over other surface.
  • When external applied force is sufficient to move a body along a surface then the force which opposes this motion is called as kinetic frictional force.
  • The magnitude of kinetic frictional force, \(f_k = \mu_ k\,N\) (constant).  
  • \(\mu_k\) is the coefficient of kinetic frictional force.
  • N is the net normal force on the  body. 

3. Rolling Friction

  • The  rolling friction comes into act for bodies in rolling motion because the portion of the body which is in contact with the surface is stationary with respect to surface, i.e., the body is not sliding.
  • For Example : 

           Wheel of moving car

  • Molecular bonds are quickly established where the wheel presses against the surface. These bonds have to be broken as the wheel rolls forward and the effort needed to break them causes rolling friction.    

Illustration Questions

After applying a force on a heavy body, body is not moving. Which kind of friction is certainly acting?

A Static Friction 

B Kinetic Friction 

C Rolling Friction 

D None of these

×

Since there is no relative motion so, static friction is acting.

Hence, option (A) is correct.

After applying a force on a heavy body, body is not moving. Which kind of friction is certainly acting?

A

Static Friction 

.

B

Kinetic Friction 

C

Rolling Friction 

D

None of these

Option A is Correct

Calculation of Friction Force

  • Consider a block of mass m which is at rest on a horizontal surface.
  • The coefficient of static friction between blocks and surface is \(\mu_s\).

Case 1: When there is no external force applied on the block

  • If there is no external force applied on the block, then friction force will be zero.

\(f_s = 0\)

 

Case 2: When external force is applied horizontally and the block is still at rest

  • As external force is applied, frictional force comes into the act.

Maximum friction force  = \(\mu _s N\)

  • If body is in equilibrium, it means that friction force is adjusting its value = Fext.

Illustration Questions

A block weighing 20 N rests on a horizontal surface. The coefficient of static friction \(\mu_s = 0.40\). What will be the maximum value of friction force, if horizontal force Fext = 5 N is exerted on the block?

A 10 N 

B 8 N

C 5 N

D 3 N

×

Maximum value of static friction 

fmax =  \(\mu_s N\)N = 20 N 

fmax =  0.40 × 20

fmax = 8 N

image

Since, Fext <  fmax

Block is in static equilibrium, it means body is not slipping.

image

Therefore, applied force of friction = Fext

Frictional force = 5 N

image image

A block weighing 20 N rests on a horizontal surface. The coefficient of static friction \(\mu_s = 0.40\). What will be the maximum value of friction force, if horizontal force Fext = 5 N is exerted on the block?

image
A

10 N 

.

B

8 N

C

5 N

D

3 N

Option C is Correct

Calculation of Force of Friction when Fext  is Applied on Body with some Angle and Body is at Rest

  • Consider a block of mass m which is kept on a rough horizontal surface.
  • External force F is applied on block with some angle \(\theta\) with the horizontal, as shown in figure.

  • Even when the force is applied, the body is still at rest. 
  • To find the force of friction, first draw the FBD of the system.
  • Net force in horizontal direction,

             \(f_s - F \,cos\theta = 0 \)   (As body is at rest)

  • Net force in vertical direction,

           \(N+F\,sin\theta-mg=0\)                 

         Since,  \(\Sigma F_y = 0\)       (As, there is no acceleration along y-axis)

            \(N=mg-F\,sin\theta\)  

            \( F_{max} = \mu_s N \)

             \(f_{s_{max}} = \mu_s (mg - F sin \theta)\)

  • Only static friction will be applied because there is no relative motion between the surface or body is at rest.
  • Thus, \(f_s = F cos \theta\)

Illustration Questions

A block of mass m = 20 kg, kept on rough horizontal surface, is pulled by a force F = 80 N at an angle \(\theta\) = 37º with the horizontal. If coefficient of static friction \(\mu = 0.5\), then calculate the force of friction.   ( Given : sin 37º = 3/5, cos 37 º = 4/5 )

A 85 N 

B 80 N 

C 64 N 

D 76 N 

×

Net force in y - direction : 

 \(N+F\,sin\theta-mg=0\)

 \(N=mg-F\,sin\theta\)

     =  20 × 10 – 80 × sin 37º 

     =  \(200-80\times\dfrac{3}{5}\)

      = 152 N

image

Maximum friction force :

\( f_s = \mu_s N\)

= 0.5 × 152 

 \(f_{max} = 76N\)  

image

Horizontal component of force :

\(F_x = F cos \theta\)

    = 80 × cos 37 º 

\(= 80 × \dfrac{4}{5}\)

= 64 N 

image

As \(F_x < f_{max}\)

Friction force applied is 

\(f = F_x\)

\(f = 64 \,N\)      (\(\because\) Body is at rest )

image

A block of mass m = 20 kg, kept on rough horizontal surface, is pulled by a force F = 80 N at an angle \(\theta\) = 37º with the horizontal. If coefficient of static friction \(\mu = 0.5\), then calculate the force of friction.   ( Given : sin 37º = 3/5, cos 37 º = 4/5 )

image
A

85 N 

.

B

80 N 

C

64 N 

D

76 N 

Option C is Correct

Angle of Repose

  • Consider a block of mass m which is kept on an inclined plane, as shown in figure. The angle of inclination can be varied. 
  • At certain value of \(\theta\), sliding of block just starts.
  • This state is known as impending state of motion.

At this value of \(\theta\)\(f_s= f_{max} = \mu_s \,N\) 

Net force in \(x- \text{direction} \,\,{(\sum f_x)}\)

\(= -mg\,sin\, \theta + \mu_s \,N =0 \) ..........(i)

Net force in \(y- \text{direction} \,\,{(\sum f_y)}\),

\(= -mg \,cos \,\theta +N =0 \) ...........(ii)

Putting the value of N from equitation (ii) in (i)

\(tan\,\theta = \mu_s\)

\(\theta = \theta _R = tan^{-1} \, \mu_s\)

This angle \(\theta _R\) is known as angle of Repose.

Illustration Questions

A block of mass m = 50 kg, is placed on an inclined plane whose inclination angle can be varied. At which angle the block will begin to slide if coefficient of static friction is \(\mu_s = \) 0.5?

A tan -1 (4)

B tan -1 (2)

C tan -1 (0.5)

D tan -1 (0.4)

×

Let \(\theta\) be the angle of repose.

image

Net force in x-direction 

\(\Sigma F_x = f_s - mg \,sin\theta\)

\(0 = \mu_s N - mg \,sin\theta\)     ..............(i)

image

Net force in y-direction 

\(\Sigma F_y = - mg \,cos\theta + N\)

\(0= - mg \,cos\theta +N\)

\(N = mg \,cos\theta\)    ..............(ii)

image

Eliminating N from equation (i) and (ii),

\(tan\theta = \mu_s\)

\(\theta = tan^{-1} \mu_s\)

\(\theta = tan^{-1} (0.5)\)

image

A block of mass m = 50 kg, is placed on an inclined plane whose inclination angle can be varied. At which angle the block will begin to slide if coefficient of static friction is \(\mu_s = \) 0.5?

image
A

tan -1 (4)

.

B

tan -1 (2)

C

tan -1 (0.5)

D

tan -1 (0.4)

Option C is Correct

Calculation of Maximum and Minimum Value of M2 for which M1 Remain at Rest

  • Consider a block of mass m1, which is kept on a fixed inclined plane and attached to a block of mass m2 by a rope of negligible mass, as shown in figure.
  • The inclined plane makes angle \(\theta\) with horizontal.
  • The coefficient of friction is \(\mu\).

Case 1:

Calculation of minimum value of m2 for which m1 remains at rest

  • If m2 is smaller than m1, then m1 will tends to move downward on the inclined plane for the smallest possible value of m2.

  • To remain at equilibrium, m1 will experience maximum friction in upward direction.

Free body diagram of both the blocks 

  • For the block of mass m2,

 T = mg  ..............(i)

For equilibrium of m1,

  • Net force along the inclined plane,

\(T + \mu_s N - m_1 g \,sin\theta =0\)   ............(ii)

  • Net force perpendicular to the inclined plane,

\(m_1 \,g \,cos \theta - N =0\)

\( N = m_1 g \,cos\theta \)  ..............(iii)

Putting the value of N and T in equation (ii) from (iii) and (i)

\(m_2 g + \mu_s (m_1 g \,cos\theta) - m_1 g \, sin \theta = 0 \)

\(m_2 = m_1 (sin \theta - \mu_s cos\theta)\)

  • This is the minimum value of m2 for which mass m1 remains at rest.

\(\dfrac{m_2}{m_1} = sin \theta - \mu_s cos\theta\)

 

Case 2:

Calculation of maximum value of m2 for which mass mremains at rest

  • Now if m2 is increased, then m1 will tends to move in upward direction on the inclined plane. The force of friction will be directed in downward direction on the inclined plane.

Free body diagrams for this case 

  • For equilibrium of m2

T = m2 g    ...........(i)

For equilibrium of m1,

  • Net force along the inclined plane,

\(T - \mu_s N - m_1 g \,sin\theta = 0 \)     ...............(ii)

  • Net force perpendicular to the inclined plane,

\(m_1\, g \,cos \theta - N = 0\)

\(N = m_1 \,g \, cos\theta\)    ...............(iii)

Putting the value of N and T in equation (ii) from (iii) and (i)

\(m_2 = m_1 (sin\theta + \mu_s cos\theta)\)

\(\dfrac{m_2}{m_1} = sin\theta+ \mu_s cos\theta\)

  • Two different values of m2 are obtained for two different impending states of motion of m1.

Illustration Questions

Consider a block of mass m1 = 5 kg, which is kept on a fixed inclined plane and attached to a block of mass m2 by a rope, as shown in figure. The inclined plane makes an angle \(\theta = 37^\circ\) with the horizontal and coefficient of friction \(\mu= 0.5\). Find the minimum value of m2 for which m1 remains at rest. Given : [cos 37º = 4/5, sin 37º = 3/5]

A 3 kg

B 2 kg

C 1 kg

D 4 kg

×

Calculation of minimum value of m2 for which m1 remains at rest

If m2 is smaller than m1, then m1 will tends to move downward on the inclined plane for the smallest possible value of m2.

To remain at equilibrium, m1 will experience maximum friction in upward direction.

Free body diagram of both the blocks

For the block of mass m2,

 T = mg  ..............(i)

image

For equilibrium of m1,

Net force along the inclined plane, 

\(T + \mu_s N - m_1 g \,sin\theta = 0 \)     ...............(ii)

Net force perpendicular to the inclined plane,

\(m_1 \,g \,cos \theta - N =0\)

\(\Rightarrow N = m_1 g \,cos\theta \)  ..............(iii)

Putting the value of N and T in equation (ii) from (iii) and (i),

\(m_2 g + \mu_s (m_1 g \,cos\theta) - m_1 g \, sin \theta = 0 \)

    \(m_2 = m_1 (sin \theta - \mu_s cos\theta)\)

This is the minimum value of m2 for which mass m1 remains at rest.

\(\dfrac{m_2}{m_1} = sin \theta - \mu_s cos\theta\)

 

image

\(m_2 = 5 (sin37º - 0.5× cos37º )\)

\(m_2 = 5 \left(\dfrac{3}{5} - 0.5 × \dfrac{4}{5} \right)\)

m2 = 1 kg 

Consider a block of mass m1 = 5 kg, which is kept on a fixed inclined plane and attached to a block of mass m2 by a rope, as shown in figure. The inclined plane makes an angle \(\theta = 37^\circ\) with the horizontal and coefficient of friction \(\mu= 0.5\). Find the minimum value of m2 for which m1 remains at rest. Given : [cos 37º = 4/5, sin 37º = 3/5]

image
A

3 kg

.

B

2 kg

C

1 kg

D

4 kg

Option C is Correct

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