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

Explanation:

Collinear Forces

• Collinear forces act along the same line of action.

• Example: Forces on a rope in a tug of war, where all the forces are aligned along the rope.

Coplanar Concurrent Forces 

• These forces lie in the same plane and meet at a common point.

• Example: Forces on a rod resting against a wall, where normal reaction, weight, and friction act in one plane and meet at a point.

Additional InformationNon-Coplanar Concurrent Forces 

• These forces act in different planes but still intersect at a single point.

• Example: A tripod carrying a camera, where the three legs exert forces in different directions but all converge at the camera mount.

Non-Coplanar Parallel Forces → 4

• Forces that are parallel but not in the same plane.

• Example: The weight of benches in a classroom, where each bench’s weight acts vertically but at different locations and planes.

Concept:

The block is acted upon by two horizontal forces:

F₁ = 80 N: F₂ = 70 N

Net Force: F_n = F₁ - F₂ = 80 - 70 = 10 N

The block is on a rough horizontal surface.

Friction force $F_f=\mu N$

where $\mu =0.4$

Force $N=mg=20\times 9.81=196.2N$

So, $F_f=0.4\times 196.2=78.48N$

Calculation:

• Compare F_net​ with friction:

  F_net​=10N
  F_f​=78.48N

• Since F_net < F_f the friction will completely oppose the motion — the block will not move.​

• Net acceleration = 0 m/sec²,  because friction prevents any movement.

Explanation:

Varignon's Theorem:

• It states that the moment of a resultant force about any point is equal to the algebraic sum of the moments of its component forces about the same point.
• This is applicable in both 2D and 3D force systems and is fundamental in statics.
• It is valid for coplanar as well as spatial force systems and is widely used in structural and mechanical analysis.
• It supports the concept of equilibrium by ensuring that replacing a system of forces with their resultant maintains the same turning effect.

 Additional Information

Lami's Theorem:

• Applies only to a particle in equilibrium acted upon by three concurrent, coplanar, and non-parallel forces.
• It relates the magnitudes of the forces to the sine of angles between them.
•  

Cauchy's Theorem:

• Generally used in the context of stress analysis and tensor calculus, not in basic statics or moment calculation.
• This is primarily used in the theory of elasticity and stress tensors.
• It explains how the stress vector on a plane within a body can be expressed using the stress tensor.

Euler's Theorem:

• Refers to various results in mechanics and mathematics. In mechanics, it may relate to Euler's equations of motion or the rotation of rigid bodies, not directly to moments of force systems.
• In mechanics, Euler’s Theorem relates to rotation of rigid bodies and the Euler angles.
• It also refers to Euler's equations of motion in rotational dynamics and stability analysis.

Concept:

To find the resultant of collinear forces, algebraic sum is used. Forces in the same direction are added, and those in the opposite direction are subtracted.

Given:

Three forces: 250 N → right, 150 N → left (opposite direction), 350 N → right

Calculation:

Net Resultant Force = 250 + 350 - 150 = 450 N

Hence, the resultant force is: 450 N

Explanation:

Resultant Force in the Analytical Method:

• When multiple forces act in the same direction, the analytical method simplifies the calculation of the resultant force. In such cases, the forces are additive because they all act along the same line of action, contributing to a cumulative effect in the same direction. The resultant force is determined by summing up all the individual forces acting in that direction.

Correct Option Analysis:

The correct option is:

Option 2: By adding all the forces together.

This option is correct because, when all forces act in the same direction, the resultant force is simply the sum of all the individual forces. This is mathematically expressed as:

R = F₁ + F₂ + F₃ + ... + F_n

Here:

• R = Resultant force
• F₁, F₂, F₃, ... F_n = Individual forces acting in the same direction

The principle of superposition applies, where the net effect of all forces is their summation. This method is straightforward and applicable only when all forces are aligned in the same direction. No vector resolution or trigonometric calculations are necessary in this scenario, making it computationally simple.

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,

$\cos \theta =\frac{3}{5}$

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:

$\Rightarrow R_N=\frac{100}{0.2}=500N$

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 =  $mg\sin {30}^{\circ }=5\times 10\times \frac{1}{2}\Rightarrow 25N$

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

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]

$\Rightarrow \left[a\right]=\frac{\left[ML{T}^{-2}\right]}{\left[T\right]}=\left[ML{T}^{-3}\right]$

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 = $\frac{Distancetravelledbytheeffort}{Distancetravelledbytheload}$

Mechanical Advantage of pulley system:

• Mechanical Advantage = efficiency × Velocity ratio

Calculation:

Given:

Efficiency, η = 60 %

Velocity ratio = $\frac{Distancetravelledbytheeffort}{Distancetravelledbytheload}$ = $\frac{12}{3}$ = 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