Equivalent Force System and Free Body Diagram MCQ Quiz - Objective Question with Answer for Equivalent Force System and Free Body Diagram - Download Free PDF

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.

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:

\( R = \sqrt{F_1^2 + F_2^2} \)

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

Explanation:

Rotational Equilibrium

• Rotational equilibrium refers to the state of a rigid body where the sum of all torques (moments) acting about any axis is zero.
• In other words, for a body to be in rotational equilibrium, there must be no net torque causing it to start rotating or change its rotational motion.

Condition for Rotational Equilibrium:

• For a lamina or any rigid body to be in rotational equilibrium, the vector sum of all the torques acting on it must be zero. Mathematically, this is expressed as:

Στ = 0

where Στ represents the sum of all torques acting on the body.

Concept:

Lami's Theorem

If a body is in equilibrium under the action of three forces, then each force is proportional to the sine of the angle between the other two forces.

\({F_1\over \sin(α) }={F_2\over \sin(β)}={F_3\over \sin(γ )}\)

where  F1, F2, and F3 are three forces on equilibrium body, α = angle between force F2 and F3, β = angle between force F1 and F3, γ = angle between force F1 and F2.

Calculation:

Given:

in equilibrium under the action of three forces

W = Weight of the roller, F = Pull force, and R = the reaction force

From the geometry of the figure α = 120°, β = 150°, and γ = 90°.

Applying Lami's Theorem 

\({W\over \sin(α) }={F\over \sin(β)}={R\over \sin(γ )}\)

\({W\over \sin(120) }={F\over \sin(150)}={R\over \sin(90)}\)

\({F}={W\;\times\; sin (150)\over \sin(120)}=\frac{W}{\sqrt{3}}\)

\(F= \frac{W}{\sqrt{3}}=\frac{\sqrt{3}W}{3}\)

Explanation:

Principle of Force Transmissibility:

• The principle of transmissibility of force states that the condition of motion of a rigid body remains unchanged if a force F of a given magnitude, direction, and sense acts anywhere along the same line of action on the rigid body.
• For example, a force F acting at a₂ along the line of action a₁a₂ is equivalent to a force F acting at a₁ along the line of action a₁a₂.

​

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:

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:

\(\frac{A}{{\sin \alpha }} = \frac{B}{{\sin\beta }} = \frac{C}{{\sin\gamma }}\)

Calculation:

Given:

From the given figure we have 

\(\frac{T_1}{sin~(120)}~=~\frac{T_2}{sin~(150)}~=~\frac{T_3}{sin~(90)}\)

By solving the above equation we have,

\(\frac{T_1}{T_2}~=~\sqrt3\) and \(\frac{T_1}{T_3}~={\frac{\sqrt{3}}{2}}\)

Explanation:

When the roller is just about to climb the step then, 

• Static friction at point A is zero because when the cylinder is about to make out of the curb, it will lose its contact at A, the only contact will be at point B.
• At point B, the roller will be in a state of pure rolling so even the surfaces are rough there will be no friction at point B.
• Force F, contact force at point B, and weight of the roller W will be concurrent at point C.

Therefore the free body diagram will be as follows

Explanation:

Lami's theorem: 

If three coplanar and concurrent forces acting on a particle, keep it in equilibrium, then each force is proportional to the sine of the angle between the other two and the constant of proportionality is the same.

Consider three forces F1, F2, F3 acting on a particle or rigid body making angles α, β, and γ with each other.

Mathematically, 

\(\frac{{{F_1}}}{{\sin \alpha }} = \frac{{{F_2}}}{{\sin \beta }} = \frac{{{F_3}}}{{\sin \gamma }}\)

Concept:

Lami's theorem:

It states that if three forces acting at a point are in equilibrium, each force is proportional to the sine of the angle between the other two forces. Consider three forces FA, FB, FC acting on a particle or rigid body making angles α, β and γ with each other.

Therefore, \(\frac{{{F_A}}}{{sin\ \alpha }} = \frac{{{F_B}}}{{sin\ \beta }} = \frac{{{F_C}}}{{sin\ \gamma }}\)
Calculation:

Given:

We have angle BAC = 180° - (30° + 60°) = 90°

angle BAC = 90°

By Lami’s Theorem,

\(\frac{{{800}}}{{sin~90°}} = \frac{{{T_{AC}}}}{{sin~150° }}\)

TAC = 400 N

Concept:

Making free body diagram of each block:

Balancing forces in positive x direction

∴ F = μR_N1 + μR_N2 = μ(R_N1 + R_N2)

where, R_N1 = Normal reaction force on block R and  RN₂ = Normal reaction force on block S

Calculation:

Given:

μ = 0.4

R_N1 = 100g = 100 × 9.81 = 981 N

R_N2 = (100 + 150)g = 250 × 9.81 = 2452.5 N

Minimum force (F) is

F = μ(RN1 + RN2)

F = 0.4 × (981 + 2452.5)  = 1373.4 N = 1.37 kN

Concept:

Parallelogram law of forces:

This law states that if two forces are acting simultaneously on a body at a point are represented in magnitude and direction by the two adjacent sides of a parallelogram, their resultant is represented in magnitude and direction by the diagonal of the parallelogram which passes through the point of intersection of the two sides representing the forces.

Net resultant force, \({{R}} = \sqrt {{{{a}}^2} + {{{b}}^2}} \) \(% MathType!MTEF!2!1!+- % feaagKart1ev2aqatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn % hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr % 4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq-Jc9 % vqaqpepm0xbba9pwe9Q8fs0-yqaqpepae9pg0FirpepeKkFr0xfr-x % fr-xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaaeaaaaaaaaa8 % qacaqGsbGaeyypa0ZaaOaaa8aabaWdbiaabggapaWaaWbaaSqabeaa % peGaaGOmaaaakiabgUcaRiaabkgapaWaaWbaaSqabeaapeGaaGOmaa % aaaeqaaaaa!3CE6! {\rm{R}} = \sqrt {{{\rm{a}}^2} + {{\rm{b}}^2}} \)

Calculation:

Given:

\(\rm Let\;{\rm{\vec a}} = 800{\rm{\;N}},{\rm{\;\vec b}} = 200{\rm{\;N}},{\rm{\;\vec c}} = 400{\rm{\;N}},{\rm{\;\vec d}} = 600{\rm{\;N}}\)\(% MathType!MTEF!2!1!+- % feaagKart1ev2aqatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn % hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr % 4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq-Jc9 % vqaqpepm0xbba9pwe9Q8fs0-yqaqpepae9pg0FirpepeKkFr0xfr-x % fr-xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaaeaaaaaaaaa8 % qaceqGHbWdayaalaWdbiabg2da9iaaiIdacaaIWaGaaGimaiaabcka % caqGobGaaiilaiaabckaceqGIbWdayaalaWdbiabg2da9iaaikdaca % aIWaGaaGimaiaabckacaqGobGaaiilaiaabckaceqGJbWdayaalaWd % biabg2da9iaaisdacaaIWaGaaGimaiaabckacaqGobGaaiilaiaabc % kaceqGKbWdayaalaWdbiabg2da9iaaiAdacaaIWaGaaGimaiaabcka % aaa!53CD! {\rm{\vec a}} = 800{\rm{\;N}},{\rm{\;\vec b}} = 200{\rm{\;N}},{\rm{\;\vec c}} = 400{\rm{\;N}},{\rm{\;\vec d}} = 600{\rm{\;}}\)

ΣX = 200 – 600 = -400 N

ΣY = 400 – 800 = -400 N

\({\rm{\theta }} = {\tan ^{ - 1}}\frac{{Σ {\rm{X}}}}{{Σ {\rm{Y}}}} = {\tan ^{ - 1}}\frac{{400}}{{400}} = 45^\circ \), as shown in the figure.

\({\rm{R}} = \sqrt {{{\left( {Σ {\rm{X}}} \right)}^2} + {{\left( {Σ {\rm{Y}}} \right)}^2}} = \sqrt {{{400}^2} + {{400}^2}} = 400\sqrt 2 {\rm{\;N}}\)

ΣM_A = 400 × 1 + 600 × 1 = 1000 Nm.

Let x be the distance from A along the x-axis, where the resultant cuts AB. Then,

\({\rm{x}} = \frac{{\Sigma {{\rm{M}}_{\rm{A}}}}}{{\Sigma {\rm{Y}}}} = \frac{{1000}}{{400}} = 2.5{\rm{\;m}}\)

Explanation:

Principles of equilibrium

1. Two force principle: If only two forces act on a body that is in equilibrium, then they must be equal in magnitude, co-linear and opposite in sense.

2. Three force principle: If a body in equilibrium is acted upon by three forces, then the resultant of any two forces must be equal, opposite and collinear with the third force. If a three-force member is in equilibrium and the forces are not parallel, they must be concurrent. Therefore, the lines of action of all three forces acting on such a member must intersect at a common point; any single force is, therefore, the equilibrant of the other two forces.

If it does not pass through a common point, it will produce a couple.

A solid body applied to three forces whose lines of action are not parallel, is in equilibrium if the three following conditions satisfies:

1. The lines of action are coplanar (in the same plane).
2. The lines of action are meeting at a point.
3. The vector sum of these forces is equal to the zero vector.

Concept:

From the free body diagram:

F – N = Ma     … (i)

And N = ma    … (ii)

Therefore, \(a = \frac{{F}}{{\left( {m + M} \right)}}\)

From equation (ii), we get

\(N = \frac{{mF}}{{\left( {m + M} \right)}}\)

Explanation:

Physical independence of forces:

• The Physical independence principal states that the effect of a force on a body is not affected by the presence of the other forces
• So even if there are a number of forces acting on the same body, each force has it’s own influence as other forces were absent.
• In other words, when number of forces act simultaneously on a particle, then the resultant of these forces will have the same effect as produced by the all the forces

Principle of transmissibility:

• Principle of transmissibility of forces states that the conditions of equilibrium or conditions of motion of a rigid body will remain unchanged if a force acting at a give point of the rigid body is replaced by a force of the same magnitude and same direction, but acting at a different point, provided that the two forces have the same line of action.