Learning Objectives:

• Understand the difference between frictional value in statics and frictional value in dynamics.
• Understand the concept of limiting friction.
• Calculate free body diagrams for objects on inclined plane.

The limiting value of the frictional force is:

$F_{FR} = \mu_{s}N$       Page 97

Where,
\(F_{FR}\) = Frictional Force,
\(\large \mu _{s}\) = Coefficient of static friction
N = Normal reaction between surface and the object

1. If the applied force is less than this value, there is no motion.
2. If the applied force is equal to this value, there is impending motion, and the object is about to move.
3. 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.

Hanjin Deviasse, CC BY-SA 3.0, via Wikimedia Commons

Solved Example: 27-1-01

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 $\tan \theta$

Normal Reaction: \[N = W \cos \theta\] Frictional Force: \[F_{FR} = \mu (W \cos \theta)\] Effort required: \begin{align*} P &= W \sin \theta + F_{FR}\\ &= W \sin \theta + \mu W \cos \theta\\ &= W (\sin \theta + \mu \cos \theta)\\ \end{align*}

Correct Answer: A

Solved Example: 27-1-02

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 or equal to

D. Is equal to

When applied force increases, force of friction increases upto limiting friction.

Friction is a self-adjusting force.

Let's consider two cases:

Case I: At the point of impending motion
Here, the angle between the normal force and the resultant force is equal to the angle of friction.

Case II: Before the point of impending motion
Here, the angle between the normal force and the resultant force is less than the angle of friction.

Correct Answer: C

Solved Example: 27-1-03

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: 27-1-04

The coefficient of friction depends on:

A. Area of contact

B. Shape of surfaces

C. Strength of surfaces

D. Nature of surface

Correct Answer: D

Solved Example: 27-1-05

The minimum force required to move a body up an inclined plane is three times the minimum force required to prevent it from sliding down the plane. If the coefficient of friction between the body and the inclined plane is $\dfrac{1}{2\sqrt{3}}$, the angle of the inclined plane is:

A. 60$^\circ$

B. 45$^\circ$

C. 30$^\circ$

D. 15$^\circ$

Efforts required while raising the load \begin{align*} P_1 &= W \sin \theta + F_{FR}\\ &= W \sin \theta + \mu W \cos \theta \end{align*} Efforts required while preventing the load from lowering \begin{align*} P_2 &= W \sin \theta - F_{FR}\\ &= W \sin \theta - \mu W \cos \theta \end{align*} Given: \[\mu = \dfrac{1}{2\sqrt{3}} = 0.29\] Also, \[P_1 = 3P_2\] Solving, \begin{align*} W \sin \theta + \mu W \cos \theta &= 3(W \sin \theta - \mu W \cos \theta)\\ \sin \theta + 0.29 \cos \theta &= 3( \sin \theta - 0.29 \cos \theta)\\ \sin \theta + 0.29 \cos \theta &= 3 \sin \theta - 0.87 \cos \theta)\\ 1.16 \cos \theta &= 2 \sin \theta \\ \tan \theta &= 0.58\\ \theta &= 30^\circ \end{align*}

Correct Answer: C

Solved Example: 27-1-06

The ratio of limiting friction and normal reaction is known as:

A. Coefficient of friction

B. Angle of friction

C. Angle of repose

D. Sliding friction

Correct Answer: A

Solved Example: 27-1-07

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: 27-1-08

On a ladder resting on smooth ground and leaning against vertical wall, the force of friction will be:

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

Since it is given that the ground is smooth, there will not be any friction on the ground end of the ladder.

The natural tendancy of the ladder will be to slide down due to the gravity. The wall end of the ladder will try to slide down, hence the frictional force at the wall end will be opposite to motion. Hence, the frictional force will be along the wall, but upwards.

Correct Answer: C

Solved Example: 27-1-09

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: 27-1-10

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: 27-1-11

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: 27-1-13

Limiting force of friction is the:

A. Tangent of angle between normal-reaction 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: 27-1-14

Coulomb friction is the friction between:

A. Bodies having relative motion

B. Two dry surfaces

C. Two lubricated surfaces

D. Solids and liquids

Correct Answer: B

Solved Example: 27-1-15

Dynamic friction as compared to static friction is:

A. Same

B. More

C. Less

D. May be less of more depending on nature of surfaces and velocity

Correct Answer: C

Solved Example: 27-1-16

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: 27-1-17

A coin with mass 20g is place on the smooth edge of a 25 cm-radius 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: 27-1-18

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 72-kg 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: 27-1-19

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:

A. 0.69

B. 0.88

C. 0.98

D. 1.37

Frictional force from upper surface, \[F = 0.4 \times 100 \times 9.81\ \mathrm{N}\] Frictional force from lower surface, \[F = 0.4 \times 250 \times 9.81\ \mathrm{N}\] Total frictional force, \[F = 0.4 \times 100 \times 9.81 + 0.4 \times 250 \times 9.81 = 1.37\ \mathrm{kN}\]

Correct Answer: D

Solved Example: 27-1-20

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:

A. 0.17

B. 0.23

C. 0.24

D. 1.27

Normal to surface: \begin{align*} N &= 900 \cos \theta + 750 \sin \theta\\ &= 900 \cos 30^\circ + 750 \sin 30^\circ\\ &= 1154.42\ N \end{align*} Parallel to surface: \[F_{\mathrm{FR}} + 900 \sin 30^\circ = 750 \cos 30^\circ\] \[\mu N = 199.52\] \[\mu = \dfrac{199.52}{1154.42} = 0.17\]

Correct Answer: A

Solved Example: 27-1-21

Select the FALSE statement for static friction.

A. The force of friction is dependent on the area of contact between the two surfaces

B. The force of friction depends upon the roughness of friction

C. The magnitude of force of friction is exactly equal to the force, which tends to move the body.

D. The force of friction always acts in a direction, opposite to that in which body tends to move, if the force of friction would have been absent

Friction force depends upon the frictional coefficient between two surfaces and the normal reaction offered by the surface. It does not depend on the nature of the surface in contact.

Correct Answer: A

Learning Objectives:

• Calculate force required to raise or lower a particular load given the parameters of screw jack.
• Understand importance of self-locking and its effect on efficiency.

$M = Pr \tan (\alpha \pm \phi)$       Page 97

where,
P = Load on Screw Jack in N.
r = Mean thread radius
$\alpha$ = Pitch angle of the thread
$\phi$ = $\tan^{-1}$ $\mu$
$\mu$ = Frictional Coefficient

Here plus sign is to be used when the load is raised (screw is tightened) and minus sign when the load is lowered (screw is loosened), The efforts in the second case will be lower as the gravity is assisting the efforts.

Interestingly, if $\phi$ is more than $\alpha$, then the effort (Moment) will be negative, means we do not have to put any effort and the load is lowered automatically. Such cases are rare, as we want to hold the load at certain position and NOT return back automatically.

So, in practice, $\alpha$ is greater than $\phi$ and the screw is called Self-Locking.

Pearson Scott Foresman, Public domain, via Wikimedia Commons

Solved Example: 27-1-12

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. Pitch angle of the thread

Correct Answer: D

Solved Example: 27-2-01

Screw jacks generally have:

A. American threads

B. British threads

C. Acme threads

D. Square threads

Correct Answer: D

Solved Example: 27-2-02

In a screw jack, the effort required to lower the load is __________ the effort required to raise the same load.

A. Less than

B. Equal to

C. More than

D. None of these

For a screw-jack, square thread, \[M = Pr \tan (\alpha \pm \phi)\]

Here, plus sign is taken while calculating force(and Moment) to raise the load, whereas negative sign is taken while lowering the load. Hence lowering force requirement is lesser than raising force requirement.

Correct Answer: A

Learning Objectives:

• To be able to calculate efforts required to raise a load supported by pulley or prevent the load from being lowered.

Solved Example: 27-3-02

If T$_1$, T$_2$ are the respective tensions on the tight and slack side of the open belt drive in Newtons and v is the velocity in the belt in m/s, then the power transmitted by the belt drive will be given by:

A. $P = \dfrac{(T_1 - T_2)}{v}$

B. $P = \dfrac{(T_1 + T_2)}{v}$

C. $P = (T_1 - T_2) \cdot v$

D. $P = (T_1 + T_2) \cdot v$

Correct Answer: C

Solved Example: 27-3-03

Which of the following relation is valid for belt drive when the belt is on the point of slipping on the pulley?

T₁= Tension on the tight side,

T₂= Tension on slack side

θ = angle of lap or contact of the belt over pulley

μ = coefficient of friction between belt and pulley

A. $T_1 \times T_2= e^{\mu \theta}$

B. $\dfrac{T_2}{T_1} = e^{\mu \theta}$

C. $\dfrac{T_1}{T_2} = e^{\mu \theta}$

D. $\dfrac{T_2}{T_1} = e^{\dfrac{\mu}{\theta}}$

Correct Answer: C

Solved Example: 27-3-04

Which of the following statements is not correct for the ratio of friction tension in flat-belt \[\left( {\frac{{{T_1}}}{{{T_2}}} = {e^{\mu \theta }}} \right)\]

A. T$_1$= tension on tight side

B. T$_2$= tension on tight side

C. $\theta$ = angle of lap over the pulley

D. $\mu$ = coefficient of friction between belt and pulley

Correct Answer: B