SOLUTION OF EME-102; CENTROID 2

HOW TO FIND THE CENTROID OF A COMPOSITE AREA

(a composite area consists of several straight or curved lines.)

(i) Draw the figure in a coordinate system. Draw the dimensions too. Every dimensions will be measured with respect to origin of the coordinate system

(ii) Divide the composite area into several parts of basic geometric areas. Lebel them as part-1, part-2, part-3, .......part-n. Let the corresponding areas are A₁, A₂, A₃, .... A_n. Let the centroids are G₁(X₁,Y₁), G₂(X₂,Y₂), G₃(X₃,Y₃), ...... G_n(X_n,Y_n).

(iii) Let the centroid of the composite area be G(Xg,Yg). Hence,
Xg = (A₁X₁ + A₂X₂ +A₃X₃)/(A₁ + A₂ + A₃)

Yg = (A₁Y₁ + A₂Y₂ +A₃Y₃)/(A₁ + A₂ + A₃)

 

(a) Suppose we have certain area of magnitude (A) in a coordinate system. The centroid of the area will be at its mid-point. A centroid is denoted by G.

                       In the figure we have a complex geometrical area composed of three basic geometrical areas. Three rectangles. Let us denote the centroids as G₁, G₂, G₃ for the given areas in the figure.

We shall have to find the Centroid of the entire area composed of  A₁, A₂, A₃.

At first, the composite line is divided into three parts.

Part -1 : The middle rectangle : Let the centroid of the area A₁ be G₁(X₁,Y₁)

Area, A₁ = (100-15-15)x15 mm² = 70x15  mm²  = 1050  mm²                 
          X₁ = 7.5 mm
          Y₁ =0


Part -2 : The lower Rectangle : Let the centroid of the A₂ be G₂(X₂,Y₂)


Area, A₂ = 50 x 15  mm² = 750 mm²                
          X₂ = 25 mm
          Y₂ = - 42.5 mm


Part -3 : The upper Rectangle : Let the centroid of the area Area, A₃ be G₃(X₃,Y₃)


Area, A₃ = 50 x 15  mm² = 750 mm²                
          X₃ = 25 mm
          Y₃ = 42.5 mm






If the centroid of the composite line be G  (X_g,Y_g)

X_g = (∑A_iX_i)/(∑A_i) 

    = (A₁X₁ + A₂X₂ +A₃X₃)/(A₁ + A₂ + A₃)
    = (1050 x 7.5 + 750 x 25 + 750 x 25)/( 1050 + 750 + 750)
    = 17.79 mm
     

Y_g = (∑A_iY_i)/(∑A_i) 

    = (A₁Y₁ + A₂Y₂ +A₃Y₃)/(A₁ + A₂ + A₃)
    = {1050 x 0 + 750 x (-42.5) + 750 x 42.5}/( 1050 + 750 + 750)
    = 0

As the figure is symmetrical about X axis, hence when we take the line of symmetry as our X axis (as we have taken here), we can directly write, Y_g =0

SOLUTION OF EME-102; CENTROID

HOW TO FIND THE CENTROID OF A COMPOSITE AREA

(a composite area consists of several straight or curved lines.)

(i) Draw the figure in a coordinate system. Draw the dimensions too. Every dimensions will be measured with respect to origin of the coordinate system

(ii) Divide the composite area into several parts of basic geometric areas. Lebel them as part-1, part-2, part-3, .......part-n. Let the corresponding areas are A₁, A₂, A₃, .... A_n. Let the centroids are G₁(X₁,Y₁), G₂(X₂,Y₂), G₃(X₃,Y₃), ...... G_n(X_n,Y_n).

(iii) Let the centroid of the composite area be G(Xg,Yg). Hence,
Xg = (A₁X₁ + A₂X₂ +A₃X₃)/(A₁ + A₂ + A₃)

Yg = (A₁Y₁ + A₂Y₂ +A₃Y₃)/(A₁ + A₂ + A₃)

 

(a) Suppose we have certain area of magnitude (A) in a coordinate system. The centroid of the area will be at its mid-point. A centroid is denoted by G.

                       In the figure we have a complex geometrical area composed of three basic geometrical areas. A rectangle, a semi circle and a isosceles triangle. Let us denote the centroids as G₁, G₂, G₃ for the given areas in the figure.

We shall have to find the Centroid of the entire area composed of  A₁, A₂, A₃.

At first, the composite line is divided into three parts.

Part -1 : The semi-circle : Let the centroid of the area A₁ be G₁(X₁,Y₁)

Area, A₁ = (π/2)x(25)² mm² = 981.74 mm²                  
          X₁ = { 25 -  (4x25)/(3xπ)} mm = 14.39 mm
          Y₁ = 25 mm


Part -2 : The Rectangle : Let the centroid of the A₂ be G₂(X₂,Y₂)


Area, A₂ = 100 x 50  mm² = 5000 mm²             
          X₂ = 25 + (100/2) = 75 mm
          Y₂ = 25 mm
Part -3 : The Triangle : Let the centroid of the area Area, A₃ be G₃(X₃,Y₃)


Area, A₃ = (1/2) x 50 x 50 mm² = 1250 mm²             
          X₃ = 25 + 50 + 25 = 100 mm
          Y₃ = 50 + (50/3) =  66.67 mm






If the centroid of the composite line be G  (X_g,Y_g)

X_g = (∑A_iX_i)/(∑A_i) 

    = (A₁X₁ + A₂X₂ +A₃X₃)/(A₁ + A₂ + A₃)
    = (981.74 x 14.39 + 5000 x 75 + 1250 x 100)/( 981.74 + 5000 + 1250)
    = 71.09
     

Y_g = (∑A_iY_i)/(∑A_i) 

    = (A₁Y₁ + A₂Y₂ +A₃Y₃)/(A₁ + A₂ + A₃)
    = (981.74 x 25 + 5000 x 25 + 1250 x 66.67)/ ( 981.74 + 5000 + 1250)
    = 32.20

SOLUTION OF EME-102; EQUILIBRIUM OF FORCES

EQUILIBRIUM OF FORCES IN 2D

A light string ABCDE whose extremity A is fixed, has weights W₁ & W₂ attached to it at B & C. It passes round a small smooth pulley at D carrying a weight of 300 N at the free end E as shown in figure. If in the equilibrium position, BC is horizontal and AB & CD make 150° and 120° with BC, find (i) Tensions in the strings and (ii) magnitudes of W₁ & W₂

Although ABCDE is a single string/rope but still the tensions in the string/rope will be different at different segments like in the segment AB the tensions will be T₁ , but in BC segment it will be different as the weight is attached at a fixed point (point B) on the string, hence it will be T₂ here and in CD it will be T₃  there. As at point D the string is not attached rather passes over a smooth pulley hence the tension in DE and CD will be same ie. T₃  again.

. 

To solve for equilibrium of forces in 2D, follow these steps:

1. Draw a free body diagram: Draw a diagram of the object in question and identify all the forces acting on it. This will help you visualize the problem and identify any unknown forces or angles.

2. Break forces into components: Resolve all forces into their x- and y-components. This is done by using trigonometry to determine the horizontal and vertical components of each force.

3. Apply Newton's second law: For the object to be in equilibrium, the sum of the forces in the x-direction and the sum of the forces in the y-direction must be equal to zero. This gives you two equations to solve simultaneously.

4. Solve the equations: Solve the equations for the unknown forces or angles by using algebraic methods. This will give you the values of the unknown forces or angles required for the object to be in equilibrium.

5. Check for consistency: Check that the forces and angles you have calculated are consistent with the problem and the free body diagram. For example, make sure that the direction of the forces makes sense and that the angles are reasonable.

6. Interpret the results: Interpret the results and explain what they mean in the context of the problem. This might involve calculating the tension in a rope, the force required to lift an object, or the angle required for an object to remain stationary.

Physics of Inertial Forces

Force Effects Caused by Inertia and Accelerating Reference Frames

Inertial forces are not real forces, rather they are often called as "pseudo forces". They produce effects that feel like forces but actually arise from Newton's inertial law in a reference frame that is accelerating.

When a car accelerates forward rapidly, a person inside the car feel pushed back into their seats. When the car turns around a curve, the person feels pulled to the outside of the curve. If the car suddenly comes to a stop, the persons inside the car not wearing seatbelts fly forward, possibly hitting the windshield. The car’s occupants may feel like some force is pushing them around, but in reality there are no forces shoving them in the directions they move inside the car. They feel shoved around the car because the car is accelerating. The occupants, however, follow Newton’s first law, the inertial law, and continue their original motion, as the car accelerates.


Newton’s First Law

Also called the inertial law, Newton’s first law requires that any object with no outside forces acting on it continues to move at a constant velocity. A constant velocity is a constant speed in a straight line because in physics the velocity includes direction. Any change in an object’s velocity (increasing speed, decreasing speed, or changing direction) is called an acceleration and requires an external force to act on the object. This tendency for objects to continue to move at a constant velocity is called inertia.


Inertial Forces

Despite the name, inertial forces are not real forces. Rather they are effects caused by an object’s inertia when the object is in or on something that is accelerating, what physicists call an accelerating reference frame.

For example, consider the occupants of a car rapidly increasing its speed. They feel pushed back into their seats, but not because some force is shoving them in the chest. The only real force acting on them is the back of the car seat accelerating them forward. Because of Newton’s first law, however, these occupants have inertia that tends to keep them at rest. They feel squeezed back into the car seat because the car is accelerating forward while their inertia would tend to keep them at rest.

When the car goes around a curve, it is also accelerating because the direction of the car’s velocity is changing. The occupants’ inertia tends to keep them moving in a straight line. Hence they feel pulled sideways in the car because they experience an inertial force (or effect) caused by the car’s acceleration as it changes direction.

Flying forward into the windshield when a car stops suddenly (Always wear seatbelts!) is a similar inertial effect. The car accelerates to a stop, and the occupants continue moving forward until their seatbelts (or the windshield) exerts a stopping force on the occupants. There is no real force pushing them forward, they just continue their forward motion, as required by Newton’s first law, while the car accelerates (slowing) to a stop.
Circular Motion

To move in a circular path, an object must have a centripetal force acting on it. The centripetal force points inward towards the center of the circle. The outward centrifugal effect is the tendency of the object to continue in a straight line motion. Hence, the centrifugal effect is an example of an inertial force and is not a real force acting on an object moving in a circular path.

Inertial forces feel like forces, but they are not real forces. They are effects caused by an object’s inertia when it is in or on something that is accelerating.


 D'Alembert's principle of inertial forces

D'Alembert showed that one can transform an accelerating rigid body into an equivalent static system by adding the so-called "inertial force" and "inertial torque" or moment. The inertial force must act through the center of mass and the inertial torque can act anywhere. The system can then be analyzed exactly as a static system subjected to this "inertial force and moment" and the external forces. The advantage is that, in the equivalent static system' one can take moments about any point (not just the center of mass). This often leads to simpler calculations because any force (in turn) can be eliminated from the moment equations by choosing the appropriate point about which to apply the moment equation (sum of moments = zero). Even in the course of Fundamentals of Dynamics and Kinematics of machines, this principle helps in analyzing the forces that act on a link of a mechanism when it is in motion. In textbooks of engineering dynamics this is sometimes referred to as d'Alembert's principle.

d’Alembert’s principle,  alternative form of Newton’s second law of motion, stated by the 18th-century French polymath Jean le Rond d’Alembert. In effect, the principle reduces a problem in dynamics to a problem in statics. The second law states that the force F acting on a body is equal to the product of the mass m and acceleration a of the body, or F = ma; in d’Alembert’s form, the force F plus the negative of the mass m times acceleration a of the body is equal to zero: F - ma = 0. In other words, the body is in equilibrium under the action of the real force F and the fictitious force -ma. The fictitious force is also called an inertial force and a reversed effective force.

QUESTION BANKS: Analyse the following Trusses:

 Analyse the following Trusses:

(1)      A cantilever truss has been as shown in the figure. Find the value of W which will produces a force of magnitudes 15 kN  in the member AB.

(2)        A cantilever truss is loaded as shown in the figure. Find the nature and magnitudes of the forces in each link.

(3)      A cantilever truss has been as shown in the figure. Find the value of W which will produces a force of magnitudes 15 kN  in the member AB.

(4)      A truss has been loaded as shown in the figure. Find the nature and the magnitudes of the forces in the links BC, CH and GH by the methods of sections.

(5) A truss has been loaded as shown in the figure. Find the internal forces in each of the beam.

(6)      A truss has been loaded as shown in the figure. Find the forces in each member and tabulate them by any methods.

compiled by Subhankar Karmakar

more content: click the following links for more questions.

THEORETICAL QUESTIONS ON SIMPLE TRUSSES part-3

QUESTION BANK : ENGINEERING MECHANICS PART-2

QUESTION BANK : ENGINEERING MECHANICS

 

THEORETICAL QUESTIONS ON SIMPLE TRUSSES

SHORT QUESTIONS: TOPIC - TRUSS ANALYSIS

1)      What is a truss? Classify them with proper diagrams.

2)      State the differences between a perfect truss and an imperfect truss.

3)      Distinguish between a deficient truss and a redundant truss.

4)      Write the Maxwell’s Truss Equation.

5)      What are the assumptions made, while finding out the forces in the various members of a truss?

6)      What are the differences between a simply supported truss and a cantilever truss? Discuss the method of finding out reactions in both the cases.

Analyse the following Trusses:

 

(1)      Analyse the Truss by the method of Joints.

(2) Find the internal forces on the links 1, 2 and 3 by the method of Sections.

(3) Determine the magnitude and the nature of the forces in the members BC, GC and GF of the given truss.

(4)   A truss of span 10 m is loaded as shown in the figure. Find the forces in all the links by any method.






compiled by Subhankar Karmakar 

PARALLEL AXIS THEOREM AND IT'S USES IN MOI

Moment Of Inertia of an Area.

download pdf file of solved numerical

MOI or MOMENTS OF INERTIA is a physical quantity which represents the inertia or resistances shown by the body against the tendency to rotate under the action external forces on the body. It is a rotational axis dependent function as its magnitude depends upon our selection of rotational axis. Although for any axis, we can derive the expression for MOI with the help of calculus, but still it is a cumbersome process.


Now suppose we take a different issue. We know MOI of an area about its centroidal axis is easily be obtained by integral calculus, but can we find a general formula by which we can calculate MOI of an area about any axis if we know its CENTROIDAL MOI.

We shall here find that we can indeed derive an expression by which MOI of any area (A) can be calculated about any Axis, if we know its centroidal MOI and the distance of the axis from it's Centroid G.


If I_GX be the centroidal moment of inertia of an area (A) about X axis, then we can calculate MOI of the Area about a parallel axis (here X axis passing through the point P) at a distance Ŷ-Y'=Y from the centroid if we know the value of I_GX and Y, then I_PX will be
I_PX = I_GX + A.Y² where Y=Ŷ-Y'


I_XX = I_OX = I_GX + A.Ŷ²

<sup>Where I_XX is the moment of inertia of the area about the co-ordinate axis parallel to X axis and passing through origin O, hence we can say,

I_XX = I_OX

 IMPORTANT: The notation of Moment of Inertia

MOI of an area about an axis passing through a point B will be written as I_BX</sup>

Q: Find the Centroidal Moment of Inertia of the figure given above. Each small division represents 50 mm.

To find out Centroidal MOI

TEST PAPER: EME-102: ENGINEERING MECHANICS

TEST PAPER: EME- EME-303: THERMODYNAMICS

Section A:

      (1)           Attempt All The Questions:                                             5x2 = 10

a)     Define system & surroundings.

b)    What is heat pump & refrigerator?

c)     What is availability?

d)    What is Entropy?

e)  What is triple point of water?

Section B:

    (2)          Attempt any three questions                                                          3x5 = 15

    

(a)    Distinguish between microscopic & macroscopic approaches of thermodynamics.

(b)   What are the limitations of First law of thermodynamics? Explain the statements of Second law of thermodynamics.

(c)    2 kg of a gas at 10 bar expands adiabatically and reversibly till the pressure drops to 5 bar. During the process 120 kJ of non-flow work is done by the system, and the temperature falls from 377° centigrade to 257°C. Calculate the value of the index of expansion and the characteristics gas constants.

Let the equation of expansion be P1-γ.Tγ = constant Hence, P1(1-γ)T1γ = P2(1-γ)T2γ

(d)   Derive the Tds equations.

(e)    Steam at a pressure of 4 bar absolute and having dryness fraction of 0.75 is heated at constant volume to a pressure of 5 bar absolute. Find the final condition of the steam and the heat absorbed by 1 kg of steam.

Section C:

Attempt part (a) or part (b) of the following questions                           5x5=25

(3) (a) Explain thermodynamic equilibrium and quasi-static process.

     (b) A steam turbine developing 110 kW is supplied steam at 17.5 bar with an internal energy of 2600 kJ/min and specific volume of 0.155 m³/kg and velocity of 100 m/s. exhaust from turbine is at 0.1 bar with internal energy of 2093 kJ/min and sp. Volume = 15.5 m³/kg and velocity of 275 m/s. heat loss from the steam turbine 37.6 kJ/kg neglecting potential energy changes, determine steam flow rate in kg/hr.

(4)(a) Prove the equivalence of Kelvin-Planck statement & Clausius statement.

     (b) A reversible engine takes 2400 kJ/min from a reservoir at 750 K develops 400 kJ/min of work during the cycle. The engine rejects heat to two reservoirs at 650 K & 550 K. Find the heat rejected to each sink.

(5)(a) Explain the principle of entropy increase.

    (b) Explain the Gibbs Function & Gibbs free energy

           

(6)(a) Distinguish between Universal gas constant and characteristics gas constant with proper example.

    (b) Explain the causes of internal and external irreversibility.

(7)(a) A gas having a moleculer mass of 28 occupies 0.13 m³ at a pressure of 1.5 bar and a temperature 21°C. Find the mass of gas and the volume as well as the density at 0°C and 1 bar pressure.

(b)    One kg of an ideal gas is heated from 18.3°C to 93.4°C. Assuming R=287 J/kg-K and

 γ = 1.18 for the gas. Find out (i) specific heats, (ii) change in internal energy and

(iii) change in enthalpy

T H E R M O D Y N A M I C S

 T H E R M O D Y N A M I C S


(1) Explain briefly what you understand about Microscopic and Macroscopic approaches to study Thermodynamics. 

or

Differentiate between them. Also state which approaches is considered in studying Engineering thermodynamics.


ANSWER: There are two approaches to study thermodynamic problem. They are known as

• (i) Microscopic approach and
• (ii) Macroscopic approach.

(i) If we try to analyse a system by considering it as comprising of discrete particles which are its atoms and molecules, we say that the approach is microscopic here. Here mass is regarded as a macro object which are composed of billions of billions microscopic particles known as atoms and molecules.


Where as when we analyse a system by its gross or time averaged effects of molecules we say the approach is macroscopic. Here matter is assumed to be continuous not discrete. It is regarded as a continuum, just like a physical field is taken as continuum.


(ii) As no. of molecules are very large hence its not possible to study individual molecules, hence the analysis in microscopic approach is done by statistical methods with the help of the theory of probability and the concept is known as Statistical Mechanics.


Where as in macroscopic approach the analysis is done on the basis classical or Newtonian Mechanics.


(iii) In Microscopic approach Statistical Mechanics and different probability distribution theories like Maxwell's velocity distribution theory have been employed.


Where as in Macroscopic approach is based on classical mechanics and calculus.




(iv) In Microscopic approach the value of the system parameters are indirectly calculated as most of them cant be directly measured.


Where as in macroscopic approach most of properties are not only measurable but also sensible too.






(2) What do you understand by the term Temperature? What is Thermal Equilibrium?


Here I want to write my view points about the term "equilibrium" with it's precise definition as an physical real events.


We all know that although we comprehend matter as a continuous distribution of masses that means that we can take infinitesimally and arbitrarily small volume of mass, but in reality, matter is composed of tiny particles called molecules. These molecules in gases are almost free of intermolecular forces, and always move randomly which is named as Brownian Motion. Gas molecules posses kinetic energy, so whenever they collide with the wall of the container within which the gas has been kept. Every collision is responsible for the momentum transfer to the wall which the wall resisted due to it's elastic properties. The change in momentum produces a thrust to the wall and we call it the pressure of the gas which means total force per unit surface area of the wall of the container. So, gross kinetic energy of the molecules due to their random Brownian motion has two effects on the wall of the container, one is due to the thrust on the wall named as Pressure of the enclosed gas. And the average Kinetic Energy of a molecules is the basis of stored energy of the gas molecules and we perceive it as temperature.


     When a body at certain temperature, T₁ is kept in contact with another body having a different temperature T₂, it means that there exists a difference in average kinetic energy of the molecules between the bodies. 




(3) Explain the term Thermodynamic Equilibrium. Explain the conditions of Thermodynamic Equilibrium. Also explain the conditions of Thermodynamic Equilibrium.                            (5) 

Ans: Equilibrium is a state or condition of a system, when there is no change in the value of properties with respect to time. In equilibrium condition, there exists no driving force inside the system and absence of driving force ensures that there is no change in the properties of the system. Basically, changes occur due to the existence of either (a) a temperature gradient, (b) a pressure gradient or (c) chemical potential in the system or between system and surroundings.  

Based on these equilibrium conditions are of three types. 

1. Thermal Equilibrium
2. Mechanical Equilibrium
3. Chemical Equilibrium 

(i) Thermal Equilibrium : If there is not any temperature difference between a system and its surroundings, there will not be any kind of heat exchange between the system and the surroundings. This state or condition of a system is known as Thermal Equilibrium of the system.

(ii) Mechanical Equilibrium: If there doesn't exist any pressure difference between a system and its surroundings, then the system is in Mechanical Equilibrium and  it tells that there will not be any work interactions between a system and its surroundings.

(iii) Chemical Equilibrium: If in a system that contains multi-components working fluid/substance and if there is not any chemical potential between them, then there will not be any chemical reaction inside a system and this condition is called as chemical equilibrium.

When a system is in thermal, mechanical and chemical equilibrium, then the system is called is in Thermodynamic equilibrium.






(4) Explain the statements of Second law of thermodynamics.

Answer: Kelvin Planck and Clausius statements of the second law of Thermodynamics.

The second law of thermodynamics can also be stated using Clausius, Kelvin and Planck statements also. Each statement is based on an irreversible process. The Clausius and the Kelvin and Planck statements of the second law of Thermodynamics are given below:


Clausius statement:
 
Clausius statement states “it is impossible for a self acting machine working in a cyclic process without any external force, to transfer heat from a body at a lower temperature to a body at a higher temperature. It considers transformation of heat between two heat reservoirs.


Kelvin – Planck statement:
 
Kelvin – Planck statement states “it is impossible to construct an engine, which is operating in a cycle produces no other effect except to external heat from a single reservoir and do equivalent amount of work.
It considers the transformation of heat into work.


Equivalence of Clausius statement to the Kelvin – Planck statement
 
Consider a reservoir having temperature T₁ and another reservoir at temperature T₂. The temperature T₁ is higher than the temperature T₂. Consider a heat pump which requires no work and transfers an amount of Q₂ from low temperature to a higher temperature reservoir, which is violating the Clausius statement. Consider an amount of heat Q₁ (greater than Q₂) be transferred from higher temperature reservoir to a heat engine which develops a net work, W = Q₁ – Q₂ and rejects Q₂ to the low temperature reservoir.
Since there is no heat interaction with the low temperature, it can be eliminated. The combined system of the heat engine and heat pump acts then like a heat engine exchanging heat with a single reservoir, which violates the Kelvin – Planck statement.

 

 

(5) Explain the concepts of continuum with the help of density.explain whether we can term density as an intensive properties of the system,justify your answer.


Ans: "Continuum" is a concept.

(6)What is a thermodynamic process? Explain the differences between reversible and irreversible process? What are the common causes of irreversibility.


(7)"....heat and work done are actually different forms of energy...and we termed them as energy in transition..." justify the statement.

or

Compare Heat Transfer with Work transfer...also explain why they are termed as path function? Differentiate between path and point function of a thermodynamic system.



(8) Discuss the importance of zeroth law. discuss its role in temperature measurement


(9) What is internal energy? Prove that internal energy is a property of the system. 


(10) What is the difference between flow work and pdV or displacement work. Also explain the term enthalpy.


(11) Classify thermodynamic system with example.discuss each of them briefly.


(12) What is SSSF energy equation? explain it and derive the equation.












First Law of Thermodynamics:

Statement: When a closed system executes a complete cycle the sum of heat interactions is equal to the sum of work interactions.
Mathematically, ΣQ=Σ W
The summations being over the entire cycle.

QUESTION BANK : ENGINEERING MECHANICS PART-2

TOPICS: NUMERICALS ON FORCE SYSTEM- UNIT-1

5) A bar of AB 12 m long rests in horizontal position on two smooth planes as shown in the figure. Find the distance X at which 100 kN is to be placed to keep the bar in equilibrium.

 

6) A light string ABCDE whose extremity A is fixed, has weights W₁ & W₂ attached to it at B & C. It passes round a small smooth pulley at D carrying a weight of 300 N at the free end E as shown in figure. If in the equilibrium position, BC is horizontal and AB & CD make 150° and 120° with BC, find (i) Tensions in the strings and (ii) magnitudes of W₁ & W₂  

 

7) Find reactions at all the contact points if weight of P is 200 N & diameter is 100 mm, where as weight of Q is 500 N and diameter is 180 mm.

 

8) Determine the force P required to begin rolling the uniform cylinder of mass (m) over the obstacle of height (h) as shown in the figure.  

 

9) A roller of weight 500 N has a radius of 120 mm and is pulled over a step of height 60 mm by a horizontal force P. Find magnitudes of P to just start the roller over the step.

 

10) Two identical rollers each of weight 100 N are supported by an inclined plane of 30° with horizontal and a vertical wall as shown in the figure. Find all the reactions at each contact point.

 

11) A smooth cylinder of radius 500 mm rests on a horizontal plane and is kept from rolling by a rope OA of 1000 mm length. A bar AB of length 1500 mm and weight 1000 N is hinged at point A and placed against the cylinder of negligible weight. Determine the tension in the rope.

 

12)      A flat belt connects pulley B, which drives a pulley A; attached to an electric motor. μ_s =  0.25 and μ_k = 0.2 between both the pulleys and the belt. If maximum allowable tension in the belt is 600 N, determine the largest torque which can be exerted by belt on pulley B.

      

13)       Two blocks of mass MA & MB are kept at equilibrium as shown in the figure. The friction between the block B & the floor is 0.35 and between the blocks is 0.3, then find the minimum force P to just move the block B.

QUESTION BANK : ENGINEERING MECHANICS

Compiled by Er. Subhankar Karmakar

Unit: 1 (Force System)

VERY SHORT QUESTIONS (2 marks):

1)      What is force & force system?

2)      What is equilibrium? What are the conditions of equilibrium?

3)      Distinguish between coplanar concurrent & coplanar non-concurrent forces.

4)      State & explain the law of transmissibility of forces.

5)      What is moment? What is couple?

6)      State and explain the Varignon’s theorem of moment.

7)      What are the characteristics of couple?

8)      What are parallel forces?

9)      State Lami’s theorem of equilibrium of forces?

10)  Where do we can apply Lami’s theorem?

11)  Explain the terms Resultant & equilibrant?

12)  What is friction? On what factors frictional force depends?

13)  Explain the term Ladder Friction?

14)  What is angle of friction, angle of repose, co-efficient of friction?

15)  Explain the concept of Limiting friction.

16)  Explain the cone of friction. Also discuss it’s physical significance?

17)  State Coulomb’s law of dry friction?

18)  What is a belt drive? Explain the term angle of contact or lap angle.

19)  Explain the term Torque, Couple and Moment?

20)  Distinguish between static and dynamic friction.

BROAD QUESTIONS (5 marks / 10 marks)

1)      Classify Force systems with proper examples.

2)      State & prove Lami’s theorem.

3)      State & prove Varignon’s theorem of moment.

4)      What is the resultant of a force system? How can we determine the resultant?

5)      What are support reactions? Describe different types of loading.

6)      What is distributed loading? Explain UDL & UVL with examples.

7)      What is a fixed supports? Explain the term reaction moment?

8)      State & explain the principle of superposition.

9)      Show that a force at a point can be reduced to a force-couple system at another point on the same plane.

10)  What is the meaning of Equivalent System?

11)  What is Free Body Diagram or FBD?

12)  Prove that for a belt drive, (T₁/T₂) = e^μθ

 

NUMERICALS: (10 marks)

(1) Two rigid rod BA & CA are joined together at point A as shown in the figure. Now a weight of 200 kN is placed at point A. Find the reaction forces in the rod AB & AC.

    [ R_AC = 292.38 kN,  R_AB = 155.57 kN]

 

(2) A weight of 40 kN is suspended by two cables as shown in the figure. Find the tensions T₁ & T₂ in the cables. 

 3) On a square plate ABCD four forces are applied as shown in the figure. Find the resultant of the force system.

         

          4) A uniform bar AB length L and weight W lies in a plane with its ends resting on two smooth surfaces on OA and OB. Find angle θ for equilibrium of bar.

             [θ = 30° ]