Static Friction
Static Friction (Statics)
The limiting value of the frictional force is:
\[F_{FR} = \mu_{s}N\]
Where,
\(F_{FR}\) = Frictional Force,
\(\large \mu _{s}\) = Coefficient of static friction
N = Normal reaction between surface and the object

If the applied force is less than this value, there is no motion.

If the applied force is equal to this value, there is impending motion, and the object is about to move.

If the applied force is more than this value, frictional force does not increase, the difference in applied force and limiting value of friction is applied towards acceleration of the body.
Solved Example: 27101
A body of weight W is required to move up on rough inclined plane whose angle of inclination with the horizontal is $\theta$. The effort applied parallel to the plane is given by(where $\mu = \tan \theta$ = Coefficient of friction between the plane and the body.)
A. P = W ($\sin \theta$ + $\mu \cos \theta$)
B. P = W ($\cos \theta$ + $\mu \sin \theta$)
C. P = W ($\sin \theta$  $\mu \cos \theta$)
D. P = W $\tan \theta$
Correct Answer: A
Solved Example: 27102
In the analysis of friction, the angle between the normal force and the resultant force __________ the angle of friction.
A. May be greater than or less than
B. Is greater than
C. Is less than
D. Is equal to
Correct Answer: C
Solved Example: 27103
When a block is place on an inclined plane, its steepest inclination to which the block will be in equilibrium is called:
A. Angle of friction
B. Angle of reaction
C. Angle of normal
D. Angle of repose
Correct Answer: D
Solved Example: 27104
The coefficient of friction depends on: (NPCIL Stipendiary Trainee ME 2018)
A. Area of contact
B. Shape of surfaces
C. Strength of surfaces
D. Nature of surface
Correct Answer: D
Solved Example: 27105
Least force required to draw a body up the inclined plane is W $\sin$(plane inclination + friction angle) applied in the direction:
A. Along the plane
B. Horizontally
C. Vertically
D. At an angle equal to the angle of friction to the inclined plane
Correct Answer: D
Solved Example: 27106
The ratio of limiting friction and normal reaction is known as: (SJVNL JE Mech 2018)
A. Coefficient of friction
B. Angle of friction
C. Angle of repose
D. Sliding friction
Correct Answer: A
Solved Example: 27107
Which one of the following statements is not correct?
A. The tangent of the angle of friction is equal to coefficient of friction
B. The angle of repose is equal to angle of friction
C. The tangent of the angle of repose is equal to coefficient of friction
D. The sine of the angle of repose is equal to coefficient to friction.
Correct Answer: D
Solved Example: 27108
On a ladder resting on smooth ground and leaning against vertical wall, the force of friction will be: (AFCAT EKT Mech Paper 3 Set I/2015)
A. Towards the wall at its upper end
B. Away from the wall at its upper end
C. Upwards at its upper end
D. Downwards at its upper end
Correct Answer: C
Solved Example: 27109
Frictional force encountered after commencement of motion is called:
A. Limiting friction
B. Kinematic friction
C. Frictional resistance
D. Dynamic friction
Correct Answer: D
Solved Example: 27110
Coefficient of friction is the:
A. Angle between normal reaction and the resultant of normal reaction and the limiting friction
B. Ratio of limiting friction and normal reaction
C. The friction force acting when the body is just about to move
D. The friction force acting when the body is in motion
Correct Answer: B
Solved Example: 27111
Pick up wrong statement about friction force for dry surfaces. Friction force is:
A. Proportional to normal load between the surfaces
B. Dependent on the materials of contact surface
C. Proportional to velocity of sliding
D. Independent of the area of contact surfaces
Correct Answer: C
Solved Example: 27112
The angle of inclined plane of a screw jack is also known as:
A. Angle of thread
B. Angle of lead
C. Angle of friction
D. Angle of pitch
Correct Answer: B
Solved Example: 27113
Limiting force of friction is the:
A. Tangent of angle between normalreaction and the resultant of normal reaction and limiting friction
B. Ratio of limiting friction and normal reaction
C. The friction force acting when the body is just about to move
D. The friction force acting when the body is in motion
Correct Answer: C
Solved Example: 27114
Coulomb friction is the friction between: (RRB JE ME CBT Aug 2019)
A. Bodies having relative motion
B. Two dry surfaces
C. Two lubricated surfaces
D. Solids and liquids
Correct Answer: B
Solved Example: 27115
Dynamic friction as compared to static friction is: (SJVNL JE Mech 2018)
A. Same
B. More
C. Less
D. May be less of more depending on nature of surfaces and velocity
Correct Answer: C
Solved Example: 27116
Tangent of angle of friction is equal to:
A. Kinetic friction
B. Limiting friction
C. Angle of repose
D. Coefficient of friction
Correct Answer: D
Solved Example: 27117
A coin with mass 20g is place on the smooth edge of a 25 cmradius phonograph record as the record is brought up to its normal rational speed of 45 rpm. What must be the coefficient of friction between the coin and the record if the coin is not to slip off?
A. 0.45
B. 0.56
C. 0.64
D. 0.78
Correct Answer: B
Solved Example: 27118
A ladder 6 m long has a mass of 18 kg and its center of gravity is 2.4 m from the bottom. The ladder is placed against a vertical wall so that it makes an angle of 60$^\circ$ with the ground. How far up the ladder can a 72kg man climb before the ladder is on the verge of slipping? The angle of friction at all contact surfaces is 15$^\circ$.
A. 3.15 m
B. 3.88 m
C. 4.05 m
D. 5.05 m
\[\mu = \tan 15^\circ\] Amount of friction at contact surfaces, \[f_A = \mu N_A = N_A \tan 15^\circ\] \[f_B = \mu N_B = N_B \tan 15^\circ\] \begin{align*} \Sigma F_V &= 0\\ N_A + f_B &= 18 + 72\\ N_A &= 90  f_B = 90  N_B \tan 15^\circ \end{align*} \begin{align*} \Sigma F_H &= 0\\ f_A &= N_B\\ N_A \tan 15^\circ &= N_B \end{align*} \[(90  N_B \tan 15^\circ) \tan 15^\circ = N_B\] \[90 \tan 15^\circ  N_B \tan^2 15^\circ = N_B\] \[90 \tan 15^\circ = N_B + N_B \tan^2 15^\circ\] \[N_B(1 + \tan^2 15^\circ) = 90 \tan 15^\circ\] \[N_B = \dfrac{90 \tan 15^\circ}{1 + \tan^2 15^\circ}\] \[N_B = 22.5 \, \text{ kg}\] \[f_B = 22.5 \tan 15^\circ = 6.03 \, \text{ kg}\] \[\Sigma M_A = 0\] \[N_B (6 \sin 60^\circ) + f_B (6 \cos 60^\circ)= 18 (2.4 \cos 60^\circ) + 72(x \cos 60^\circ)\] \[N_B (6 \tan 60^\circ) + 6f_B = 18(2.4) + 72x\] \[6(22.5) \tan 60^\circ + 6(6.03) = 43.2 + 72x\] \[72x = 226.81\] \[x = 3.15 m\]
Correct Answer: A
Solved Example: 27119
A block R of mass 100 kg is placed on a block S of mass 150 kg as shown in the figure. Block R is tied to the wall by a massless and inextensible string PQ. If the coefficient of static friction for all surfaces is 0.4, the minimum force F (in kN) needed to move the block S is: (GATE ME 2014 Shift I)
A. 0.69
B. 0.88
C. 0.98
D. 1.37
$F = 0.4 \times 100 \times 9.81 + 0.4 \times 250 \times 9.81 = 1.37 kN$
Correct Answer: D
Solved Example: 27120
A block of negligible mass rests on a surface that is inclined at 30$^\circ$ to the horizontal plane as shown in the figure. When a vertical force of 900 N and a horizontal force of 750 N are applied, the block is just about to slide. The coefficient of static friction between the block and the surface is: (GATE Mechanical 2021)
A. 0.17
B. 0.23
C. 0.24
D. 1.27
\begin{align*} N &= 900 \cos \theta + 750 \sin \theta\\ &= 900 \cos 30^\circ + 750 \sin 30^\circ\\ &= 1154.42\ N \end{align*} \[F_{\mathrm{max}} + 900 \sin 30^\circ = 750 \cos 30^\circ\] \[\mu N = 199.52\] \[\mu = \dfrac{199.52}{1154.42} = 0.17\]
Correct Answer: A