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

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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.

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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.

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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(θ2)" id="MathJax-Element-59-Frame" role="presentation" style="position: relative;" tabindex="0">R=2Fcos(θ2)R=2Fcos(θ2) $R=2Fcos(\frac{\theta}{2})$

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 F1x=F" id="MathJax-Element-60-Frame" role="presentation" style="position: relative;" tabindex="0">F1x=FF1x=F $F_{1x} = F$ .
• The y-component of F1 is F1y=0" id="MathJax-Element-61-Frame" role="presentation" style="position: relative;" tabindex="0">F1y=0F1y=0 $F_{1y} = 0$ .
• The x-component of F2 is F2x=Fcos⁡(θ)" id="MathJax-Element-62-Frame" role="presentation" style="position: relative;" tabindex="0">F2x=Fcos(θ)F2x=Fcos⁡(θ) $F_{2x} = F \cos(θ)$ .
• The y-component of F2 is F2y=Fsin⁡(θ)" id="MathJax-Element-63-Frame" role="presentation" style="position: relative;" tabindex="0">F2y=Fsin(θ)F2y=Fsin⁡(θ) $F_{2y} = F \sin(θ)$ .

To find the resultant force, we sum the components:

Resultant x-component Rx" id="MathJax-Element-64-Frame" role="presentation" style="position: relative;" tabindex="0">RxRx $R_x$ :

 

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

Resultant y-component Ry" id="MathJax-Element-66-Frame" role="presentation" style="position: relative;" tabindex="0">RyRy $R_y$ :

 

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

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 Rx" id="MathJax-Element-69-Frame" role="presentation" style="position: relative;" tabindex="0">RxRx $R_x$ and Ry" id="MathJax-Element-70-Frame" role="presentation" style="position: relative;" tabindex="0">RyRy $R_y$ :

 

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

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 cos2⁡(θ)+sin2⁡(θ)=1" id="MathJax-Element-73-Frame" role="presentation" style="position: relative;" tabindex="0">cos2(θ)+sin2(θ)=1cos2⁡(θ)+sin2⁡(θ)=1 $\cos^2(θ) + \sin^2(θ) = 1$ :

 

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

Combining the like terms:

 

$$R = \sqrt{2F^2 + 2F^2 \cos(θ)}$$

Factoring out 2F2" id="MathJax-Element-76-Frame" role="presentation" style="position: relative;" tabindex="0">2F22F2 $2F^2$ :

 

$$R = \sqrt{2F^2(1 + \cos(θ))}$$

We know that cos⁡(θ)=2cos2⁡(θ2)−1" id="MathJax-Element-78-Frame" role="presentation" style="position: relative;" tabindex="0">cos(θ)=2cos2(θ2)−1cos⁡(θ)=2cos2⁡(θ2)−1 $\cos(θ) = 2 \cos^2(\frac{θ}{2}) - 1$ , 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=2Fsin⁡(θ2)" id="MathJax-Element-84-Frame" role="presentation" style="position: relative;" tabindex="0">R=2Fsin(θ2)R=2Fsin⁡(θ2) $R=2F\sin(\frac{\theta}{2})$

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=F1+F2" id="MathJax-Element-85-Frame" role="presentation" style="position: relative;" tabindex="0">R=F1+F2R=F1+F2 $R=F_1+F_2$

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=F1−F2" id="MathJax-Element-86-Frame" role="presentation" style="position: relative;" tabindex="0">R=F1−F2R=F1−F2 $R=F_1-F_2$

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=2Fcos⁡(θ2)" id="MathJax-Element-87-Frame" role="presentation" style="position: relative;" tabindex="0">R=2Fcos(θ2)R=2Fcos⁡(θ2) $R = 2F \cos(\frac{θ}{2})$ , 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.

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Concept:

When two forces act at a right angle (90°), their resultant is found using vector addition based on the Pythagorean theorem.

If:

• Force 1 = F1
• Force 2 = F2
• Angle between them = 90°

Resultant force formula:

This formula gives the magnitude of the resultant of two perpendicular vectors.

Explanation:

Resolution of a Force

Definition: The resolution of a force is a process in which a single force is split into two or more component forces which, when combined, have the same effect as the original force. This is done without changing the overall effect of the original force. These components are usually perpendicular to each other, commonly referred to as the horizontal and vertical components.

Concept and Importance: The resolution of a force is a fundamental concept in physics and engineering, especially in the field of mechanics. It allows us to analyze the effects of a force in different directions, which is crucial for understanding and solving problems involving forces acting at angles. By resolving a force into its components, we can simplify the analysis of structures and mechanisms, making it easier to understand how different parts of a system interact and respond to various forces.

Working Principle: To resolve a force, we typically use trigonometric functions based on the angles involved. For example, if a force F is acting at an angle θ to the horizontal, we can resolve it into two components: F_x (horizontal component) and F_y (vertical component). These components can be calculated using the following equations:

• Horizontal Component (F_x): F_x = F × cos(θ)
• Vertical Component (F_y): F_y = F × sin(θ)

Here, cos(θ) and sin(θ) are the trigonometric functions cosine and sine, respectively, which are based on the angle θ.

Example: Consider a force of 100 N acting at an angle of 30° to the horizontal. To find the components of this force, we use the following calculations:

• Horizontal Component (F_x): F_x = 100 N × cos(30°) ≈ 100 N × 0.866 = 86.6 N
• Vertical Component (F_y): F_y = 100 N × sin(30°) ≈ 100 N × 0.5 = 50 N

Thus, the force of 100 N acting at 30° to the horizontal can be resolved into two components: 86.6 N horizontally and 50 N vertically. These components help in analyzing the effects of the force in different directions.

Applications: The resolution of a force is widely used in various fields, including:

• Structural Engineering: To determine the stresses and strains in different parts of a structure due to applied loads.
• Mechanical Engineering: To analyze the forces acting on machine components and predict their behavior under different loading conditions.
• Physics: To solve problems related to motion, equilibrium, and dynamics by breaking down complex force systems into simpler components.

Analysis of Other Options:

Option 1: Changing the direction of a force without altering magnitude

This option describes the process of changing the direction of a force while keeping its magnitude constant, which is not the same as resolving a force. Resolution of a force specifically involves breaking down a single force into its components, which can then be analyzed separately. Changing the direction of a force might be relevant in vector addition or manipulation but does not align with the concept of force resolution.

Option 2: Combining multiple forces to form a single resultant

This option describes the process of vector addition, where multiple forces acting on a body are combined to find a single resultant force that has the same effect. While this is an important concept in mechanics, it is the opposite of force resolution. Force resolution involves breaking down a single force into components, whereas vector addition combines multiple forces into one.

Option 4: Reducing the magnitude of a force in a given direction

This option describes the process of diminishing the magnitude of a force in a particular direction. However, it does not pertain to the resolution of a force. The resolution of a force involves splitting it into components that can be analyzed separately, without changing the overall effect of the force. Reducing the magnitude of a force could be related to damping or attenuation but is not related to force resolution.

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:

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 = 

Mechanical Advantage of pulley system:

• Mechanical Advantage = efficiency × Velocity ratio

Calculation:

Given:

Efficiency, η = 60 %

Velocity ratio =  =  = 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:

And

                            

                            = × moment

∴ Moment = Twice the area of a triangle

Concept:

Conditions for the system to be in equlibrium

ΣF_x = 0, ΣF_y = 0, ΣM = 0

Calculation:

Given:

m = 2 kg, Assume g = 10 m / s²

Lager will be the moment, smaller will be the force required to lift the rod. Hence, Applying Moment about 0 cm point we get.

w × 50 = F × 100

m × g × 50 = F × 100

2 × 10 × 50 = F × 100

F = 10 N

Explanation:

Type of Connection	Reaction	Number of Unknowns
Weight legs link		One - The reaction is a force that acts in the direction of the link
Rollers		One - The reaction is a force that act perpendicular to the surface at point of contact.
Pin or Hinge		Two - The reaction are two force components
Guided rollar/ Fixed connected collar		Two - The reactions are a force and a moment
Fixed support		Three - The reactions are two forces and a moment
Pin connected collar		One - The reaction is a force that acts perpendicular to the surface at the point of contact

Concept:

Lami's Theorem: It is an equation that relates the magnitude of the three co-planner, concurrent and non-collinear forces that keeps a body in equilibrium. It states that each force is proportional to the sine of the angle between the other two forces.

Calculation:

T₁ = 500 × sin 120° and  T₂ = 500 sin 150°

T₁ = 433 N and T₂ = 250 N

Explanation:

To determine the couple force exerted by the blade edges on the screw slot, we need to calculate the force applied by the screwdriver blade and then multiply it by the lever arm.

The formula for torque is given by:

Torque = Force x Lever Arm

In this case, the torque is 4 Nm and the width of the screwdriver blade is 5 mm. However, we need to convert the width of the blade to meters before proceeding with the calculation. 1 mm is equal to 0.001 meters.

Width of the screwdriver blade = 5 mm = 5 x 0.001 m = 0.005 m

Now we can rearrange the formula for torque to solve for force:

Force = Torque / Lever Arm

Force = 4 Nm / 0.005 m = 800 N

Therefore, the couple force exerted by the blade edges on the screw slot is 800 N.

So, the correct answer is option 2.

Explanation:

Given:

∠BAC = 60° 

∠BCA = 30° 

Let F_AB = compressive 

F_BC = compressive

F_AB = ?

Considering joint B,

∑F_H = 0 

F_AB cos 60° = F_BC cos 30° 

∑F_V = 0 

F_AB sin 60° + F_BC sin 30° = 10

Multiplying both sides by 

Compressive (As direction assumed comp)

Concept:

Lami's theorem:

Lami's theorem is an equation relating the magnitudes of three coplanar, concurrent and non-collinear forces, which keeps an object in static equilibrium, with the angles directly opposite to the corresponding forces. According to the theorem:

Calculation:

Given:

From the given figure we have 

By solving the above equation we have,

 and 

CONCEPT:

Law of Parallelogram of forces: This law is used to determine the resultant of two coplanar forces acting at a point.

• It states that “If two forces acting at a point are represented in magnitude and direction by two adjacent sides of a parallelogram, then their resultant is represented in magnitude and direction by the diagonal of the parallelogram which passes through that common point.”

Let two forces F1 and F2, acting at the point O be represented, in magnitude and direction, by the directed line OA and OB inclined at an angle θ with each other.

Then if the parallelogram OACB be completed, the resultant force R will be represented by the diagonal OC.

CALCULATION:

Given F1 = P, F2 = √2P, θ = 135^∘ 

Then the resultant force is given by