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

Last updated on Jun 7, 2025

Latest Equilibrium and Friction MCQ Objective Questions

Equilibrium and Friction Question 1:

What is a characteristic of a coplanar parallel force system?

  1. Forces act in different planes and are parallel.
  2. Forces act in the same plane but are not parallel.
  3. Forces act in different planes and are not parallel.
  4. Forces act in the same plane and are parallel.

Answer (Detailed Solution Below)

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

Equilibrium and Friction Question 1 Detailed Solution

```html

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.

```

Equilibrium and Friction Question 2:

Which of the following options best describes non-coplanar concurrent forces?

  1. Forces that meet at one point but their lines of action do not lie on the same plane
  2. Forces that do not meet at one point and their lines of action lie on the same plane
  3. Forces that meet at one point and their lines of action lie on the same plane
  4. Forces that do not meet at one point but their lines of action lie on different planes

Answer (Detailed Solution Below)

Option 1 : Forces that meet at one point but their lines of action do not lie on the same plane

Equilibrium and Friction Question 2 Detailed Solution

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.

Equilibrium and Friction Question 3:

Which of the following happens when two equal and opposite forces are applied at a point on a rigid body?

  1. They produce an additional force on the body.
  2. They create rotational motion in the body.
  3. They cancel each other and have no effect.
  4. They change the magnitude of the original force.

Answer (Detailed Solution Below)

Option 3 : They cancel each other and have no effect.

Equilibrium and Friction Question 3 Detailed Solution

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.

Equilibrium and Friction Question 4:

When two equal forces F act at an angle θ, the resultant force is given by which of the following expressions?

  1. \(R=2Fsin(\frac{\theta}{2})\)
  2. \(R=F_1+F_2 \)
  3. \(R=F_1-F_2\)
  4. \(R=2Fcos(\frac{\theta}{2})\)

Answer (Detailed Solution Below)

Option 4 : \(R=2Fcos(\frac{\theta}{2})\)

Equilibrium and Friction Question 4 Detailed Solution

```html

Explanation:

When two equal forces F act at an angle θ, the resultant force R is given by the following expression:

The correct option is:

Option 4: \(R=2Fcos(\frac{\theta}{2})\)

Let's derive and explain why this is the correct option:

Derivation:

When two forces of equal magnitude F act at an angle θ, the resultant force can be calculated using the law of cosines in vector addition. The resultant force R can be determined by breaking down the forces into their components and combining them vectorially.

Consider the two forces F acting at an angle θ:

  • Force F1 acts along the positive x-axis.
  • Force F2 acts at an angle θ relative to F1.

The components of these forces can be resolved as follows:

  • The x-component of F1 is \(F_{1x} = F\).
  • The y-component of F1 is \(F_{1y} = 0\).
  • The x-component of F2 is \(F_{2x} = F \cos(θ)\).
  • The y-component of F2 is \(F_{2y} = F \sin(θ)\).

To find the resultant force, we sum the components:

Resultant x-component \(R_x\):

\[ R_x = F_{1x} + F_{2x} = F + F \cos(θ) \]

Resultant y-component \(R_y\):

\[ R_y = F_{1y} + F_{2y} = 0 + F \sin(θ) \]

Now, the magnitude of the resultant force R is given by the Pythagorean theorem:

\[ R = \sqrt{R_x^2 + R_y^2} \]

Substituting the values of \(R_x\) and \(R_y\):

\[ R = \sqrt{(F + F \cos(θ))^2 + (F \sin(θ))^2} \]

We can simplify this equation:

\[ R = \sqrt{F^2 + 2F^2 \cos(θ) + F^2 \cos^2(θ) + F^2 \sin^2(θ)} \]

Using the Pythagorean identity \(\cos^2(θ) + \sin^2(θ) = 1\):

\[ R = \sqrt{F^2 + 2F^2 \cos(θ) + F^2} \]

Combining the like terms:

\[ R = \sqrt{2F^2 + 2F^2 \cos(θ)} \]

Factoring out \(2F^2\):

\[ R = \sqrt{2F^2(1 + \cos(θ))} \]

We know that \(\cos(θ) = 2 \cos^2(\frac{θ}{2}) - 1\), so:

\[ 1 + \cos(θ) = 1 + 2 \cos^2(\frac{θ}{2}) - 1 = 2 \cos^2(\frac{θ}{2}) \]

Substituting this back into the equation:

\[ R = \sqrt{2F^2 \cdot 2 \cos^2(\frac{θ}{2})} \]

\[ R = \sqrt{4F^2 \cos^2(\frac{θ}{2})} \]

Taking the square root of both terms:

\[ R = 2F \cos(\frac{θ}{2}) \]

Therefore, the resultant force R when two equal forces F act at an angle θ is:

\[ R = 2F \cos(\frac{θ}{2}) \]

This confirms that the correct answer is Option 4.

Additional Information

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

Option 1: \(R=2F\sin(\frac{\theta}{2})\)

This option is incorrect because it describes the resultant force in terms of the sine function. The correct derivation using the law of cosines and vector addition shows that the resultant force involves the cosine function, not the sine function.

Option 2: \(R=F_1+F_2\)

This option is incorrect because it represents the simple scalar addition of forces. When forces act at an angle to each other, their vector nature must be considered. The correct resultant force requires vector addition, not simple scalar addition.

Option 3: \(R=F_1-F_2\)

This option is incorrect because it represents the difference between the magnitudes of the forces. The resultant force of two equal forces acting at an angle requires vector addition, not subtraction.

Conclusion:

Understanding the vector addition of forces is crucial in determining the resultant force when two equal forces act at an angle. The correct expression for the resultant force is \(R = 2F \cos(\frac{θ}{2})\), which takes into account the angle between the forces and their magnitudes. Evaluating the other options helps clarify common misconceptions and emphasizes the importance of proper vector addition in physics.

```

Equilibrium and Friction Question 5:

How is the resultant force calculated if two forces act along the same straight line but in opposite directions?

  1. R=F1+F2
  2. \(R=\sqrt{F_{1}^{2} + F_{2}^{2} } \)
  3. \(R=2F_1F_2cos\theta\)
  4. \(R=F_1 - F_2\)

Answer (Detailed Solution Below)

Option 4 : \(R=F_1 - F_2\)

Equilibrium and Friction Question 5 Detailed Solution

Concept:

When two forces act along the same straight line but in opposite directions, the resultant force is determined by their algebraic sum, considering their directions.

Given:

  • Force 1, \( F_1 \) (acting in the positive direction)
  • Force 2, \( F_2 \) (acting in the negative direction)

Step 1: Assign Directions

Assume \( F_1 \) is positive (e.g., rightward) and \( F_2 \) is negative (e.g., leftward).

Step 2: Calculate Resultant Force

The resultant force \( R \) is the sum of the two forces, accounting for their directions:

\( R = F_1 - F_2 \)

Top Equilibrium and Friction MCQ Objective Questions

A 1kg block is resting on a surface with co effcient of friction, µ = 0.1. A force of 0.8N is applied to the block as shown in figure. The friction force in Newton is 

GATE ME 2011 Images-Q43

  1. 0
  2. 0.98
  3. 1.2
  4. 0.8

Answer (Detailed Solution Below)

Option 4 : 0.8

Equilibrium and Friction Question 6 Detailed Solution

Download Solution PDF

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

F1 Satya Madhu 18.07.20 D3

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.

A 5 m long ladder is resting on a smooth vertical wall with its lower end 3 m from the wall. What should be the coefficient of friction between the ladder and the floor for equilibrium?

  1. \(\frac{1}{2}\)
  2. \(\frac{3}{8}\)
  3. \(\frac{1}{3}\)
  4. \(\frac{3}{5}\)

Answer (Detailed Solution Below)

Option 2 : \(\frac{3}{8}\)

Equilibrium and Friction Question 7 Detailed Solution

Download Solution PDF

Concept:

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

∑ Fx = ∑ Fy = ∑ Mat any point = 0

Calculation:

Given:

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

Let W will be the weight of the ladder, NB and NA 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;

F1 Ashiq 6.11.20 Pallavi D5

OA2 = AB2 - OB, OA2 = 52 - 32 

OA2 = 16, OA = 4 m

From Δ OAB,

\(\cos θ = \frac{3}{5}\)

Now apply ∑ Fy = 0

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

∑ M= 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

A block weighing 981 N is resting on a horizontal surface. The coefficient of friction between the block and the horizontal surface is μ = 0.2. A vertical cable attached to the block provides partial support as shown. A man can pull horizontally with a force of 100 N. What will be the tension, T (in N) in the cable if the man is just able to move the block to the right?

GATE ME 2009 Images-Q10

  1. 176.2
  2. 196.0
  3. 481.0
  4. 981.0

Answer (Detailed Solution Below)

Option 3 : 481.0

Equilibrium and Friction Question 8 Detailed Solution

Download Solution PDF

Concept:

GATE ME 2009 Images-Q10.1

Number of vertical forces:

∑Fy = T + RN - W = 0 

Static friction force is given as,

Fs = μRN 

Calculation:

Given:

W = 981 N, μ = 0.2, Fs = 100 N

Normal reaction:

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

Tension T

⇒ T = 981 - 500 = 481 N

Define free-body diagram.

  1. A figure that represents external forces acting on a body.
  2. A diagram that represents internal forces acting on a body.
  3. A free-hand sketch representing a body.
  4. A diagram that only represents moments acting on a body.

Answer (Detailed Solution Below)

Option 1 : A figure that represents external forces acting on a body.

Equilibrium and Friction Question 9 Detailed Solution

Download Solution PDF

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:

F1 J.S 18.5.20 Pallavi D1 

F1 J.S 18.5.20 Pallavi D2

A short cylinder of circular cross-section and weight N is resting on a V block of angle 2α as shown in the figure. The reaction at point A is

quesOptionImage512

  1. W/2
  2. W/(2 Sin α)
  3. W/(2 cos α)
  4. W sin(α/2) 

Answer (Detailed Solution Below)

Option 2 : W/(2 Sin α)

Equilibrium and Friction Question 10 Detailed Solution

Download Solution PDF

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

F1 Ashik Madhu 31.10.20 D4

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 \(\frac{W}{{2\sin \alpha }}\).

 A block of mass 5 kg slides down from rest along a frictionless inclined plane that makes an angle of 30° with horizontal. What will be the speed of the block after it covers a distance of 3.6 m along the plane? [g = 10 m/s2]

  1. 5 m/s
  2. 6 m/s
  3. 7 m/s
  4. 8 m/s

Answer (Detailed Solution Below)

Option 2 : 6 m/s

Equilibrium and Friction Question 11 Detailed Solution

Download Solution PDF

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

 

F1 Ashik Madhu 14.08.20 D17

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

v= 0 + 2 × 5 × 3.6

∴ v = 6 m/s.

A force F is given by F = at + bt2 where t is time, what are the dimension of a and b.

  1. MLT1, MLT0
  2. MLT3, ML2T4
  3. MLT-4, MLT-4
  4. MLT-3, MLT-4

Answer (Detailed Solution Below)

Option 4 : MLT-3, MLT-4

Equilibrium and Friction Question 12 Detailed Solution

Download Solution PDF

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]

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

Find the mechanical advantage of a pulley system if it has an efficiency of 60%. The load lifts by 3 m when the rope is pulled by 12 m.

  1. 4.8
  2. 3.6
  3. 1.2
  4. 2.4

Answer (Detailed Solution Below)

Option 4 : 2.4

Equilibrium and Friction Question 13 Detailed Solution

Download Solution PDF

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.

A 2 kg block is resting on a rough surface with 0.1 coefficient of friction. A force of 1 N is applied to the block. The friction force is:

  1. 0 N
  2. 1 N
  3. 1.96 N
  4. 1.20 N

Answer (Detailed Solution Below)

Option 2 : 1 N

Equilibrium and Friction Question 14 Detailed Solution

Download Solution PDF

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.

SSC JE Mechanical 13 10Q 25th Jan Morning Part 2 Hindi - Final images Q4

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. 

An external force of 1 N is applied on the block of 1 kg as shown in the figure. The magnitude of the friction force Fs is (where, μ  = 0.3, g = 10 m/s2):

new 16316261022441

  1. 0.3 N
  2. 0.1 N
  3. 3 N
  4. 1 N

Answer (Detailed Solution Below)

Option 4 : 1 N

Equilibrium and Friction Question 15 Detailed Solution

Download Solution PDF

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

F1 Ram Ravi 16.09.21 D1

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

Get Free Access Now
Hot Links: teen patti bliss teen patti master update teen patti yas teen patti joy apk lotus teen patti