Friction MCQ Quiz - Objective Question with Answer for Friction - Download Free PDF

Explanation:

Friction is independent of the surface area in contact.

• Friction depends primarily on the normal force (the weight or force pressing the two surfaces together) and the nature of the materials in contact, but not on the surface area in contact.
• Increasing the surface area does not directly affect the frictional force.

 Additional Information

• Static friction is usually greater than kinetic friction.
• Static friction resists the initial motion of an object, while kinetic friction acts once the object is in motion, and is generally lower than static friction. 
• Friction depends on the nature of the surfaces in contact (roughness, material type, etc.). The coefficient of friction varies with the materials involved.
• Friction is directly proportional to the normal reaction force (the force pressing the surfaces together). The greater the normal force, the greater the friction.

Explanation:

Tribology:

• Tribology is a multidisciplinary field that studies the principles of friction, lubrication, and wear of interacting surfaces in relative motion. It is an essential aspect of engineering that impacts the efficiency, reliability, and longevity of machinery and mechanical systems.
• Tribology encompasses three main phenomena: friction, lubrication, and wear. Each of these has significant implications for the performance and maintenance of mechanical systems.

Friction:

• Friction is the resistance to motion that occurs when two surfaces slide against each other. It is a critical factor in mechanical design because it affects the energy efficiency of machines. While friction can be beneficial in certain contexts (e.g., providing traction for vehicles), it often leads to energy losses and wear in mechanical systems.

There are two primary types of friction:

• Static Friction: The resistance to the initiation of motion between two stationary surfaces.
• Kinetic Friction: The resistance encountered when two surfaces are already in relative motion.

Reducing friction through proper material selection and surface treatments can significantly enhance the performance and durability of mechanical components.

Lubrication:

Lubrication is the application of a substance (lubricant) between two surfaces to reduce friction and wear. Lubricants can be liquids, solids, or gases, and they create a thin film that separates the interacting surfaces, minimizing direct contact and, consequently, friction and wear.

There are different types of lubrication:

• Hydrodynamic Lubrication: A thick film of lubricant completely separates the surfaces, preventing direct contact. This type is common in fluid film bearings.
• Boundary Lubrication: A thin film of lubricant exists, and surface asperities may come into contact. This type is typical in conditions where full fluid film lubrication cannot be maintained.
• Mixed Lubrication: A combination of hydrodynamic and boundary lubrication.
• Solid Lubrication: Solid materials (e.g., graphite, molybdenum disulfide) are used as lubricants, especially in high-temperature or vacuum environments.

Effective lubrication is crucial for reducing wear, preventing overheating, and extending the lifespan of mechanical components.

Wear:

Wear is the progressive loss of material from the surface of a solid body due to mechanical action. It is a significant concern in tribology because it affects the reliability and longevity of mechanical systems. Different types of wear mechanisms include:

• Abrasive Wear: Material removal due to hard particles or asperities sliding over a surface.
• Adhesive Wear: Material transfer between surfaces due to strong adhesion forces at the contact points.
• Fatigue Wear: Material removal due to cyclic loading and unloading, leading to crack initiation and propagation.
• Corrosive Wear: Material removal due to chemical or electrochemical reactions.

Understanding wear mechanisms and implementing appropriate countermeasures (e.g., surface treatments, material selection) are essential for enhancing the durability and performance of mechanical systems.

Importance of Tribology:

Tribology plays a vital role in various industries, including automotive, aerospace, manufacturing, and energy. Its importance can be summarized as follows:

• Energy Efficiency: Reducing friction losses improves the energy efficiency of machines and systems, leading to cost savings and reduced environmental impact.
• Reliability and Performance: Proper lubrication and wear management enhance the reliability and performance of mechanical components, reducing downtime and maintenance costs.
• Longevity: Minimizing wear extends the lifespan of machinery, leading to longer service intervals and lower replacement costs.
• Safety: Effective tribological practices ensure the safe operation of critical mechanical systems, preventing accidents and failures.

Concept:

A block subjected to a time-varying force experiences acceleration when the applied force exceeds friction. The final velocity is calculated using Newton’s second law and kinematic equations.

Given:

Weight of block, W = 2 kN = 2000 N

Coefficient of friction, μ = 0.2

g = 10 m/s², Time = 3 s

Force applied, P = 600 N (from graph)

Calculation:

Normal reaction, N = W = 2000 N

Frictional force, \( f = \mu N = 0.2 \times 2000 = 400~N \)

Net force, \( F_{\text{net}} = 600 - 400 = 200~N \)

Mass of the block, \( m = \frac{W}{g} = \frac{2000}{10} = 200~kg \)

Acceleration, \( a = \frac{F_{\text{net}}}{m} = \frac{200}{200} = 1~m/s^2 \)

Using the kinematic equation, \( v = u + at = 0 + 1 \times 3 = 3~m/s \)

Explanation:

Static Friction

Static friction is the force that resists the initiation of sliding motion between two surfaces that are in contact and at rest. It adjusts to be exactly equal and opposite to the applied force until the limit of static friction is reached. The following statements describe various aspects of static friction:

• When the applied force is less than the limiting friction, the body remains at rest. This statement is correct because static friction will adjust to balance the applied force, preventing motion.

• The force of friction depends upon the area of contact between the two surfaces. This statement is incorrect. Static friction depends on the nature of the surfaces and the normal force, but not on the contact area.

• The force of static friction is a self-adjusting force; its magnitude is exactly equal and opposite to the applied force, which tends to produce motion. This statement is correct because static friction adjusts to match the applied force up to its maximum limit.

• If the surfaces are dry (not lubricated), then friction between surfaces is known as solid friction. This statement is correct because the term "solid friction" refers to friction between dry surfaces.

Explanation:

To determine the net work done on the object, we need to consider the work done by the applied force and the work done against friction.

Step 1: Calculate the force of friction

The force of friction (f_k) is given by the formula:

f_k = μ_k × N

where μ_k is the coefficient of kinetic friction and N is the normal force. The normal force for a horizontal surface is equal to the weight of the object (N = mg).

Given:

• Mass (m) = 5 kg
• Acceleration due to gravity (g) = 10 m/s²
• Coefficient of kinetic friction (μ_k) = 0.2

Therefore,

N = m × g = 5 kg × 10 m/s² = 50 N

So,

f_k = μ_k × N = 0.2 × 50 N = 10 N

Step 2: Calculate the work done by the applied force

The work done (W) by the applied force (F) is given by:

W = F × d

where F is the applied force and d is the distance over which the force is applied.

Given:

• Applied force (F) = 20 N
• Distance (d) = 10 m

Therefore,

W_applied = F × d = 20 N × 10 m = 200 J

Step 3: Calculate the work done against friction

The work done against friction (W_friction) is given by:

W_friction = f_k × d

Given:

• Force of friction (f_k) = 10 N
• Distance (d) = 10 m

Therefore,

W_friction = f_k × d = 10 N × 10 m = 100 J

Step 4: Calculate the net work done on the object

The net work done (W_net) on the object is the difference between the work done by the applied force and the work done against friction:

W_net = W_applied - W_friction

Therefore,

W_net = 200 J - 100 J = 100 J

The net work done on the object is 100 J.

Concept:

The friction force is given by:

f = μN

where μ is the coefficient of friction between the surfaces in contact, N is the normal force perpendicular to friction force.

Calculation:

Given:

μ = 0.1, m = 1 kg, F = 0.8 N

Now, we know that

From the FBD as shown below

Normal reaction, N = mg = 1 × 9.81 = 9.81 N

Limiting friction force between the block and the surface, f = μN =  0.1 × 9.81 = 0.98 N

But the applied force is 0.8 N which is less than the limiting friction force.

∴ The friction force for the given case is 0.8 N.

Concept:

The resting on between any frictional floor and a vertical wall will always satisfy all the static equilibrium condition i.e.

∑ F_x = ∑ F_y = ∑ M_{at any point} = 0

Calculation:

Given:

Length of ladder (AB) = 5 m, OB = 3 m

Let W will be the weight of the ladder, N_B and N_A will be support reaction, θ is the angle between ladder and floor and μ is the friction coefficient between ladder and floor.

Free body diagram of the ladder;

OA² = AB² - OB² , OA2 = 52 - 32 

OA2 = 16, OA = 4 m

From Δ OAB,

\(\cos θ = \frac{3}{5}\)

Now apply ∑ Fy = 0

N_B = W 
Now take moment about point A, which should be equal to zero

∑ MA = 0

\(\;\left( {\mu {N_B} \times 4} \right) + \left( {W \times \frac{5}{2} \times \cos \theta } \right) = {N_B} \times 3\)

\(\;\left( {\mu {N_B} \times 4} \right) + \left( {{N_B} \times \frac{5}{2} \times \frac{3}{5}} \right) = {N_B} \times 3\)

\(\left( {\mu \times 4} \right) + \left( {\frac{3}{2}} \right) =~3\)

\(\mu = \frac{3}{8}\)

Hence the value of the coefficient of friction between ladder and floor will be 3/8

Concept:

If the cylinder is kept symmetrically between the V block we will get equal normal reactions at both the surfaces.

Calculation:

Given:

The angle between V block inclined surfaces = 2α

Considering the V block of symmetrical shape and the system is under equilibrium we have,

2 × N × cos (90 - α) = W

2 × N × sin α = W

N = \(\frac{W}{{2\sin \alpha }}\)

Hence, the normal reaction at point A is \(\frac{W}{{2\sin \alpha }}\).

Concept

The equation of motions are 

v = u + at 

v² = u² + 2as                

\(s = ut + \frac{1}{2}a{t^2}\)

Calculation:

Given:

Mass of the block, m = 5 kg, Inclination angle of the plane, θ = 30°, Initial velocity of the block, u = 0 m/s, Distance travelled by the block, s = 3.6 m

Force acting on the block along with the inclination of plane =  \(mg\sin 30^\circ = 5 \times 10 \times \frac{1}{2}\Rightarrow 25\;N\)

Acceleration of the block along the inclined plane, a = \(g\sin 30^\circ = 5~m/{s^2}\) 

Applying equation of motion along the inclined plane.

v² = u² + 2as

v² = 0 + 2 × 5 × 3.6

∴ v = 6 m/s.

Concept:

As the force is applied to the body the friction force acts on the body which will be opposite to the applied force.

As the value of P is going on increasing, at some stage the solid body will be on the verge of motion. 

The friction force corresponding to this stage is called the limiting force of friction.

The frictional force is given by fL = μN; where N is normal force.

If P > fL, then the frictional force acting will be fL;

If P < fL, then the frictional force acting will be P;

Calculation:

Given:

m = 2 kg, μ = 0.1; P = 1 N;

N = mg ⇒ 2 × 9.81 = 19.62 N

Now the limiting frictional force will be  

fL = μN = 0.1 × 19.62 = 1.962 N;

Here P < fL, 

Therefore, frictional force acting will be P = 1 N. 

Concept:

To determine the magnitude of the friction force (\( F_s \)) when an external force of 1 N is applied to a 1 kg block, we need to use the formula for frictional force:

Given:

• \( \mu \) is the coefficient of friction
• \( N \) is the normal force

Values:

• \( \mu = 0.3 \)
• Mass of the block \( m = 1 \, \text{kg} \)
• Gravitational acceleration \( g = 10 \, \text{m/s}^2 \)
• Applied force \( F = 1 \, \text{N} \)

​

First, we calculate the normal force \( N \):

\( N = m \cdot g = 1 \, \text{kg} \times 10 \, \text{m/s}^2 = 10 \, \text{N} \)

Next, we calculate the frictional force \( F_s \):

\( F_s = \mu \cdot N = 0.3 \times 10 \, \text{N} = 3 \, \text{N} \)

Since the frictional force is 3 N, and the applied force is 1 N, the block does not move because the applied force is less than the maximum static friction force.

Hence, the actual friction force will be equal to the applied force (since the block is not moving):

\( F_s = 1 \, \text{N} \)

Therefore, the magnitude of the friction force is:

4) 1 N

Explanation:

Coulomb Law of Friction

• Coulomb’s law of sliding friction can represent the most fundamental and simplest model of friction between dry contacting surfaces.

• When sliding takes place, the coulomb law states that the tangential friction force is proportional to the magnitude of the normal contact force.
• It is independent of relative tangential velocity.

• When contacting bodies slide or tend to slide relative to each other tangential forces are generated.
• These forces are usually referred to as friction forces.
• Three basic principles have been established namely:
  • The friction force acts in a direction opposite to that of the relative motion between the two contacting bodies.
  • The friction force is proportional to the normal load on the contact.
  • The friction force is independent of a normal area of contact.

Explanation:

• When a ladder rests on a smooth horizontal surface and leans against a rough vertical wall.
• As seen from the diagram, the lower end of the ladder will start moving horizontally due to the gravitational pull of earth towards the ladder.
• As the surface is smooth, there would be no frictional force on the smooth horizontal surface.
• But as the ladder moves horizontally, the upper end of the ladder starts moving down and this motion of ladder down will be opposed by a frictional force in the upward direction as the wall is rough.

Concept:

Friction: Friction is an opposing force that comes into play when the body is actually moving (slide or roll) or even tries to move over the surface of another body.

• Static friction(Fs): The opposing force that is active when the body tends to move over the surface, but the actual motion is yet not started.
• Kinetic friction(Fk): Kinetic or dynamic force is an opposing force that is active when the body is actually moving on another surface.
  • Fk < Fs​

μs (coefficient of Static Friction) is effective when the body is at rest and motion of the body is completely opposed by the Frictional Force.

μk (coefficient of kinetic friction) is effective when the body is in motion and its motion is opposed to some small fraction by Friction Force.

The coefficient of static friction is greater than the coefficient of kinetic friction but not always.

Because the static friction is greater than kinetic friction only at its maximum value.

1. The ratio of static friction at its maximum value to dynamic friction is always greater than one.
2. The ratio of static friction to dynamic friction is greater than one, equal to one also.

Explanation:

• The coefficient of friction depends on the material and roughness of the two surfaces in contact.
• In general μs (coefficient of static friction) is slightly greater than μk (coefficient of kinetic friction).
• As long as the normal force is constant, the maximum possible friction does not depend on the area of the surfaces in contact.
• Law of limiting friction for dry surfaces:
  • When two bodies are in contact, the direction of the force of friction on one of them is opposite to the direction in which this body has the tendency to move.
  • The frictional force is independent of the area of contact of the surfaces.
  • When one body is just on the point of sliding over the other body, the maximum force of friction is being exerted; This is called the limiting friction.
  • The force of friction is dependent upon the types of materials of the two bodies in contact. The coefficient of friction depends on the material and roughness of the two surfaces in contact.
  • The limiting frictional force is proportional to the normal force that acts between the bodies in contact.
  • The ratio of frictional force to the normal reaction is constant and is known as the coefficient of friction i.e. μ = F/N.
  • Limiting static friction is somewhat greater than the kinetic friction.
  • The kinetic friction is independent of the relative velocity of the bodies in contact