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

Concept:

The rod is in equilibrium under the action of three forces: weight W acting at center, normal reaction at A (vertical), and horizontal reaction at B from the smooth wall. Friction at A balances the horizontal reaction.

Equilibrium Equations:

• Vertical: \( R_A = W \)
• Horizontal: \( F_A = R_B \)
• Moment about A: \( R_B \times L \times \sin(30^\circ) = W \times \frac{L}{2} \times \cos(30^\circ) \)

Solving:

\( R_B \times \frac{L}{2} = \frac{W L \cos(30^\circ)}{2} \Rightarrow R_B = W \times \cos(30^\circ) = W \times \frac{\sqrt{3}}{2} \)

Final Answer:

\( \frac{\sqrt{3}}{2} W \)

Concept:

When a body is just about to slip on an inclined plane, the condition of limiting equilibrium applies. The frictional force is at its maximum, and we use the relation:

\( \mu = \tanθ \)

Calculation:

Given:

• Mass m = 1 kg
• Inclination θ = 30^∘

On an inclined plane:

• Normal reaction: \( N = mg \cosθ \)
• Downslope force: \( mg \sinθ \)

At limiting equilibrium, friction balances the downslope force:

\( mg \sinθ = \mu mg \cosθ \)

\( \Rightarrow \mu = \frac{\sinθ}{\cosθ} = \tanθ \)

\( \mu = \tan 30^∘ = \frac{1}{\sqrt{3}} \)

Explanation:

Coplanar Parallel Force System

Definition: A coplanar parallel force system is a system in which all the forces act in the same plane and are parallel to each other. This type of force system is commonly encountered in structural engineering and mechanics, where forces such as loads on beams and columns need to be analyzed.

Characteristics:

• All forces lie in the same plane.
• Forces are parallel to each other, meaning they have the same direction but may have different magnitudes.

Applications: Coplanar parallel force systems are often used in the analysis of structures such as beams, trusses, and frames. They simplify the analysis by reducing the problem to two dimensions and focusing on the effects of parallel forces.

Advantages:

• Simplifies the analysis of structural elements by reducing the problem to two dimensions.
• Allows for straightforward calculations of resultant forces and moments.

Disadvantages:

• Only applicable to systems where forces are truly coplanar and parallel.
• May not fully represent the complexity of real-world force interactions in three-dimensional structures.

Correct Option Analysis:

The correct option is:

Option 4: Forces act in the same plane and are parallel.

This option correctly describes a coplanar parallel force system. All forces are in the same plane and are parallel to each other, which is the defining characteristic of this type of force system.

Additional Information

To further understand the analysis, let’s evaluate the other options:

Option 1: Forces act in different planes and are parallel.

This option is incorrect because it describes forces that are parallel but not coplanar. If forces act in different planes, they cannot be considered part of a coplanar force system.

Option 2: Forces act in the same plane but are not parallel.

This option is incorrect as it describes a coplanar force system, but not a parallel one. The forces are in the same plane but have different directions, which does not fit the definition of a coplanar parallel force system.

Option 3: Forces act in different planes and are not parallel.

This option is incorrect because it describes a situation where forces are neither coplanar nor parallel. This scenario does not fit the definition of a coplanar parallel force system.

Conclusion:

Understanding the characteristics of a coplanar parallel force system is crucial for accurately analyzing structural elements and mechanical systems. A coplanar parallel force system involves forces that lie in the same plane and are parallel, simplifying the analysis and calculation of resultant forces and moments. This fundamental concept is widely used in engineering to ensure the stability and integrity of various structures.

Explanation:

Non-Coplanar Concurrent Forces

• Non-coplanar concurrent forces are forces that meet at a single point, but their lines of action do not lie within the same plane. These forces exist in three-dimensional space and are commonly encountered in engineering problems involving structures, mechanics, or physics.

Key Characteristics:

• Concurrent: All forces meet at one single point.
• Non-Coplanar: The lines of action of the forces do not lie on the same plane, i.e., they are distributed in 3D space.

Importance in Engineering Applications:

Non-coplanar concurrent forces are critical for analyzing structures and systems in three-dimensional space. For instance:

• In truss and frame structures, forces acting at joints can be concurrent but not coplanar.
• In mechanical systems, forces on components such as shafts and gears often act in different planes but converge at specific points.
• In aerospace and automotive engineering, forces acting on vehicles or aircraft may be non-coplanar due to the complex interaction of aerodynamic forces, gravity, and thrust.

Analysis of Forces:

To analyze non-coplanar concurrent forces, vector methods are typically used. These include:

• Vector Addition: All forces are represented as vectors in three-dimensional space, and their resultant can be determined using vector addition.
• Resolution of Forces: Forces can be resolved into components along standard axes (x, y, z) to simplify calculations.
• Equilibrium Analysis: For a system in equilibrium, the sum of all forces and moments (torques) must be zero in all directions.

Explanation:

Understanding the Effect of Equal and Opposite Forces on a Rigid Body

Definition: When two equal and opposite forces are applied at a point on a rigid body, they are known as balanced forces. Balanced forces are forces that are equal in magnitude but opposite in direction. They act along the same line of action and, as a result, they cancel each other out.

Working Principle: In physics, forces are vectors, meaning they have both magnitude and direction. When two forces of equal magnitude but opposite direction are applied at a point on a rigid body, the net force on the body is the vector sum of the two forces. Since the forces are equal and opposite, their vector sum is zero. This means that the forces cancel each other out, resulting in no net force acting on the body.

Analysis of Correct Option (Option 3):

When two equal and opposite forces are applied at a point on a rigid body, they cancel each other and have no effect. This means that the body remains in its state of rest or uniform motion, according to Newton's First Law of Motion, which states that an object will remain at rest or in uniform motion unless acted upon by an external force.

To understand this better, consider the following points:

• Equilibrium: A rigid body is said to be in equilibrium when the net force and net torque acting on it are zero. In this case, since the forces are equal and opposite, the net force is zero, and the body remains in equilibrium.
• Translational Motion: Since the net force is zero, there is no translational motion induced in the body. The body does not accelerate in any direction.
• Rotational Motion: For rotational motion to occur, there must be a net torque acting on the body. In this scenario, the equal and opposite forces do not create a net torque because they act along the same line of action and their moments cancel each other out.

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:

Number of vertical forces:

∑F_y = T + R_N - W = 0 

Static friction force is given as,

F_s = μR_N 

Calculation:

Given:

W = 981 N, μ = 0.2, F_s = 100 N

Normal reaction:

\(\Rightarrow {R_N} = \frac{{100}}{{0.2}} = 500~N\)

Tension T

⇒ T = 981 - 500 = 481 N

Explanation:

Free-Body Diagram: These are the diagrams used to show the relative magnitude and direction of all external forces acting upon an object in a given situation. A free-body diagram is a special example of vector diagram.

Some common rules for making a free-body diagram:

• The size of the arrow in a free-body diagram reflects the magnitude of the force.
• The direction of the arrow shows the direction that the force is acting.
• Each force arrow in the diagram is labeled to indicate the exact type of force.
• It is generally customary in a free-body diagram to represent the object by a box and to draw the force arrow from the center of the box outward in the direction that the force is acting.

Example:

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:

Principle of homogeneity of dimensions:

• According to this principle, a physical equation will be dimensionally correct if the dimensions of all the terms occurring on both sides of the equation are the same.
• This principle is based on the fact that only the physical quantities of the same kind can be added, subtracted, or compared.
• Thus, velocity can be added to velocity but not to force.

EXPLANATION

Given - F = at + bt²

From the principle of dimensional homogeneity, the left-hand side of the equation dimensionally equal to the right-hand side of the equation.

The dimension formula of force (F) = [MLT^-2]

∴ [MLT^-2] = [a] [T]

\(⇒ \left[ a \right] = \frac{{\left[ {ML{T^{ - 2}}} \right]}}{{\left[ T \right]}} = \left[ {ML{T^{ - 3}}} \right]\)

For second term,

⇒ [MLT^-2] = [b] [T²]

\(\Rightarrow \left[ b \right] = \frac{{\left[ {ML{T^{ - 2}}} \right]}}{{\left[ {{T^2}} \right]}} = \left[ {ML{T^{ - 4}}} \right]\)

Concept:

Velocity ratio in a pulley system:

• The ratio of the distance moved by the effort force applied to the object and the distance moved by the object under load is known as the Velocity ratio of the pulley system.

Velocity ratio = \(\frac{Distance~travelled~by~the~effort}{Distance~travelled~by~the~load}\)

Mechanical Advantage of pulley system:

• Mechanical Advantage = efficiency × Velocity ratio

Calculation:

Given:

Efficiency, η = 60 %

Velocity ratio = \(\frac{Distance~travelled~by~the~effort}{Distance~travelled~by~the~load}\) = \(\frac{12}{3}\) = 4

Mechanical Advantage = efficiency × Velocity ratio = 0.6 × 4 = 2.4

Additional InformationEfficiency:

• It is a measure of performance and effectiveness of a system or component.
• The main approach to define efficiency is the ratio of useful output per required input.

Mechanical Advantage:

• Mechanical Advantage is the ratio of load to effort.
• Pulleys and levers alike rely on mechanical advantage.
• The larger the advantage is the easier it will be to lift the weight.
• The mechanical advantage (MA) of a pulley system is equal to the number of ropes supporting the movable load.

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