SOLUTION OF PRESEMESTER EXAMINATION: ENGINEERING MECHANICS SOLUTION OF PRESEMESTER EXAMINATION: ENGINEERING MECHANICS©subhankar_karmakar

GROUP-A Q.1 Answer the following questions as per the instructions 2x20=20 Choose the correct answer of the following questions: (i) A truss hinged at one end, supported on rollers at the other, is subjected to horizontals load only. Its reaction at the hinged end will be (a) Horizontal;                   (b) Vertical (c) Both horizontal & vertical (d) None of the above. Ans: (c) both horizontal & vertical (ii) The moment of inertia of a circular section of diameter D about its centroidal axis is given by the expression (a) π(D)4/16           (b) π(D)4/32 (c) π(D)4/64           (d) π(D)4/4 Ans: (c) π(D)4/64 Fill in the blanks in the following questions: (iii)The distance of the centroid of an equilateral triangle with each side(a) is …………. from any of the three sides. Ans a /(2√3) (iv)Poisson’s ratio is defined as the ratio between ……. and ………… . Ans: Lateral Strain, Longitudinal Strain (v)If two forces of equal magnitudes P having an angle 2Ө between them, then their resultant force will be equal to ________ . Ans: 2P CosӨ Choose the correct word/s. (vi) Two equal and opposite force acting at different points of a rigid body is termed as (Bending Moment/ Torque/ Couple). Ans: Couple

Choose correct answer for the following parts: (vii) Statement 1: In stress strain graph of a ductile material, yield point starts at the end of the elastic limit. Statement 2: At yielding point, the deformation becomes plastic by nature. (a) Statement 1 is true, Statement 2 is true. (b) Statement 1 is true, Statement 2 is true and they are unrelated with each other (c) Statement 1 is true, statement 2 is false. (d) Statement 1 is false, Statement 2 is false. Ans: (a) Statement 1 is true, Statement 2 is true. (viii) Statement 1: It is easier to pull a body on a rough surface than to push the body on the same surface. Statement 2: Frictional force always depends upon the magnitude of the normal force. (a) Statement 1 is true, Statement 2 is true. (b) Statement 1 is true, Statement 2 is true and they are unrelated (c) Statement 1 is true, statement 2 is false. (d) Statement 1 is false, Statement 2 is false. Ans: (a) Statement 1 is true, Statement 2 is true. (ix) In a cantilever bending moment is maximum at (a) free end            (b) fixed end (c) at the mid span (d) none of these Ans: (b) fixed end (x) The relationship between linear velocity and angular velocity of a cycle (a) exists under all conditions (b) does not exist under all conditions (c) exists only when it moves on horizontal plane. (d) none of these Ans: (a) exists under all conditions

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SOLUTION OF PRESEMESTER EXAMINATION: ENGINEERING MECHANICS PART-II ©subhankar_karmakar

MOMENTUM : AN IMPORTANT CONCEPT

NEWTON'S LAW OF MECHANICS:

Although we know there are three laws of motion proposed by Issac Newton, but it can be shown that the 2nd law of motion is the fundamental laws of motion, and the other two laws are nothing but special cases of second law.

The second law states that the rate of change of momentum is equal to force, which is another physical quantity and it is a vector.

SO WHAT DOES MOMENTUM MEAN?

MOMENTUM

Tuesday, 16. November, 02:41

Objects in motion are said to have a momentum. This momentum is a vector. It has a size and a direction. The size of the momentum is equal to the mass of the object multiplied by the size of the object's velocity. The direction of the momentum is the same as the direction of the object's velocity.

Momentum is a conserved quantity in physics. This means that if you have several objects in a system, perhaps interacting with each other, but not being influenced by forces from outside of the system, then the total momentum of the system does not change over time. However, the separate momenta of each object within the system may change. One object might change momentum, say losing some momentum, as another object changes momentum in an opposite manner, picking up the momentum that was lost by the first.

IMPORTANCE OF MOMENTUM

Momentum is a corner stone concept in Physics. It is a conserved quantity. That is, within a closed system of interacting objects, the total momentum of that system does not change value. This allows one to calculate and predict the outcomes when objects bounce into one another. Or, by knowing the outcome of a collision, one can reason what was the initial state of the system.

MOMENTUM IS MASS TIMES VELOCITY

When an object is moving, it has a non-zero momentum. If an object is standing still, then its momentum is zero. To calculate the momentum of a moving object multiply the mass of the object times its velocity. The symbol for momentum is a small p.

MOMENTUM IS A VECTOR QUANTITY

Momentum is a vector. That means, of course, that momentum is a quantity that has a magnitude, or size, and a direction. The above problem is a one dimensional problem. That is, the object is moving along a straight line. In situations like this the momentum is usually stated to be positive, i.e., to the right, or negative, i.e., to the left.

MOMENTUM IS NOT VELOCITY

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Sometimes the concept of momentum is confused with the concept of velocity. Do not do this. Momentum is related to velocity. In fact, they both have the same direction. That is, if an object has a velocity that is aimed toward the right, then its momentum will also be directed to the right. However, momentum is made up of both mass and velocity. One must take the mass and multiply it by the velocity to get the momentum.

MOMENTUM IS DIRECTLY PROPORTIONAL TO VELOCITY

If the mass is kept constant, then the momentum of an object is directly proportional to its velocity. In the example at the left, the mass is kept constant at a value of 2.0 kg. The velocity changes from 0 m/s to 10 m/s while the momentum changes from 0 kg-m/s to 20 kg-m/s. This creates a straight line graph when momentum is plotted as a function of velocity. (The symbol for momentum as a function of velocity would be p(v).) The straight line graph demonstrates the direct proportion between momentum and velocity.


That is, if one were to double the velocity of an object, then the momentum of that object would also double. And, if one were to change the velocity of an object by a factor of 1/4, then the momentum of that object would also change by a factor of 1/4.

MOMENTUM IS DIRECTLY PROPORTIONAL TO MASS

If the velocity is kept constant, then the momentum of an object is directly proportional to its mass. In the example at the left, the velocity is kept constant at a value of 3.0 m/s. The mass changes from 0 kg to 10 kg while the momentum changes from 0 kg-m/s to 30 kg-m/s. This creates a straight line graph when momentum is plotted as a function of mass. (The symbol for momentum as a function of mass would be p(m).) The straight line graph demonstrates the direct proportion between momentum and mass.

That is, if one were to triple the mass of an object, then the momentum of that object would also triple. And, if one were to change the mass of an object by a factor of 1/2, then the momentum of that object would also change by a factor of 1/2.

TWO DIMENSIONAL FORCE SYSTEM

Q: WHAT DO YOU UNDERSTAND BY THE TERM "FORCE"? WHAT IS THE EFFECT OF FORCE ON A PARTICLE AND A RIGID BODY? EXPLAIN WITH SUITABLE EXAMPLES.

Answer:

FORMAL DEFINITION:

A FORCE is that which can cause an object with mass to ACCELERATE. Force has both MAGNITUDE and DIRECTION, making it a vector quantity. According to Newton's second law, an object with constant mass will accelerate in proportion to the net force acting upon it and in INVERSE PROPORTION TO ITS MASS (M). An equivalent formulation is that the net force on an object is equal to the RATE OF CHANGE OF MOMENTUM it experiences. Forces acting on three-dimensional objects may also cause them to rotate or deform, or result in a change in pressure. The tendency of a force to cause angular acceleration about an axis is called TORQUE. Deformation and pressure are the result of stress forces within an object.


EXPLANATION OF MECHANICAL FORCE AND IT'S EFFECT ON A PARTICLE:

CHANGE IN POSITION:

To know force well, first we have to understand what do we mean by Change. What does it mean when we say the position of the body has been changed? Whenever we find the state of object becomes different than that of the same object before some time say Δt, then we say that there exists a change in the state of the object. Suppose the change occurs in the position of the body. But to find the initial position of a body, we need a co-ordinate system.

THE CAUSE OF CHANGE:

It has been seen that to induced a change or to make a change in the position of an object we must have to change the energy possess by the body. To transfer energy into the object we shall have to apply FORCE on the body. Therefore Force is the agency that makes a change in position of a body.

THE CONCLUSION: GALILEO'S LAW OF INERTIA OR NEWTON'S FIRST LAW OF MOTION.

So, if there is no force on an object the position of the object won't change with respect to time. It means if a body at rest would remain at rest and a body at uniform motion would remain in a steady motion. This law is known as Galileo's Law of Inertia or Newton's first law of motion.

• 2 DIMENSIONAL FORCE

In physics, force is a vector quantity that is used to describe the interaction between two objects. In a two-dimensional system, forces can act in two different directions, which are typically labeled as the x-axis and the y-axis.

When dealing with two-dimensional force, it is essential to use vector addition to determine the net force acting on an object. The net force is the vector sum of all the forces acting on the object. The direction of the net force is determined by the angle of the resultant force vector.

To calculate the net force in two dimensions, we must first break down each force into its x and y components. The x-component of a force is the amount of force acting in the x-direction, and the y-component is the amount of force acting in the y-direction. Once we have the x and y components for each force, we can add them together to find the net force.

The magnitude of the net force can be calculated using the Pythagorean theorem, which states that the square of the hypotenuse of a right triangle is equal to the sum of the squares of the other two sides. In this case, the hypotenuse is the magnitude of the net force, and the other two sides are the x and y components of the net force.

In summary, when dealing with two-dimensional force, it is essential to use vector addition to determine the net force acting on an object. To calculate the net force, we must first break down each force into its x and y components and then add them together. The magnitude and direction of the net force can be determined using trigonometry.

• ORTHOGONAL RESOLUTION OF A FORCE

Orthogonal resolution of a force is a technique used in physics to break down a force vector into its components along two orthogonal axes, typically the x and y axes. This technique is useful in analyzing the motion of an object under the influence of a force and can help determine the net force acting on an object.

To perform orthogonal resolution of a force, we first need to identify the angle that the force vector makes with respect to one of the axes, usually the x-axis. We can then use trigonometry to determine the components of the force vector along the x and y axes.

If the angle between the force vector and the x-axis is θ, the x-component of the force can be found using the equation F_x = F cos(θ), where F is the magnitude of the force. Similarly, the y-component of the force can be found using the equation F_y = F sin(θ).

Once we have the x and y components of the force, we can use vector addition to determine the net force acting on an object. The net force is the vector sum of all the forces acting on the object and can be found by adding the x and y components of each force separately.

Orthogonal resolution of a force is a powerful technique that is used in many areas of physics, including mechanics, electromagnetism, and fluid dynamics. By breaking down a force vector into its components, we can better understand the forces acting on an object and predict its motion under different conditions.

WHAT IS A FORCE SYSTEM? CAN WE CLASSIFY FORCE SYSTEMS?

ANSWER:
                         A force system may be defined as a system where more than one force act on the body. It means that whenever multiple forces act on a body, we term the forces as a force system. We can further classify force system into different sub-categories depending upon the nature of forces and the point of application of the forces. Almost any system of known forces can be resolved into a single force called a resultant force or simply a Resultant. The resultant is a representative force which has the same effect on the body as the group of forces it replaces. (A couple is an exception to this) It, as one single force, can represent any number of forces and is very useful when resolving multiple groups of forces. It is important to note that for any given system of forces, there is only one resultant.


Different types of force system:

• (i) COPLANAR FORCES:

If two or more forces rest on a plane, then they are called coplanar forces. There are many ways in which forces can be manipulated. It is often easier to work with a large, complicated system of forces by reducing it an ever decreasing number of smaller problems. This is called the "resolution" of forces or force systems. This is one way to simplify what may otherwise seem to be an impossible system of forces acting on a body. Certain systems of forces are easier to resolve than others. Coplanar force systems have all the forces acting in in one plane. They may be concurrent, parallel, non-concurrent or non-parallel. All of these systems can be resolved by using graphic statics or algebra.

• (ii) CONCURRENT FORCES:

A concurrent coplanar force system is a system of two or more forces whose lines of action ALL intersect at a common point. However, all of the individual vectors might not actually be in contact with the common point. These are the most simple force systems to resolve with any one of many graphical or algebraic options. If the line of actions of two or more forces passes through a certain point simultaneously then they are called concurrent forces. concurrent forces may or may not be coplanar.

• (iii) LIKE FORCES:

A parallel coplanar force system consists of two or more forces whose lines of action are ALL parallel. This is commonly the situation when simple beams are analyzed under gravity loads. These can be solved graphically, but are combined most easily using algebraic methods. If the lines of action of two or more forces are parallel to each other, they are called parallel forces and if their directions are same, then they are called LIKE FORCES.

• (iv) UNLIKE FORCES:

If the parallel forces are such that their directions are opposite to each other, then they are termed as "UNLIKE FORCE".

• (v) NON COPLANAR FORCES:

The last illustration is of a "non-concurrent and non-parallel system". This consists of a number of vectors that do not meet at a single point and none of them are parallel. These systems are essentially a jumble of forces and take considerable care to resolve.

on 20th November, 2010: ©subhankar

CENTROID OF COMPLEX GEOMETRIC FIGURES:

So in this articles, we are going to discuss the concepts of centroid for one dimensional as well as two dimensional objects. 

Let's first discuss about 1D and 2D objects, one by one, an 1D object is a line, practically a metallic rod will be considered as a linear, 1D object. Where as any thin plate of negligible thickness can be considered a 2D body. Suppose we have a thin metallic rectangular plate. If it is considered as a 2D rectangular area of b X h.

The concept of centroid has been developed on the basis of resultant of several areas. We know that an area can be represented as the cross product of two vectors, hence it is also an vector. Suppose we have an area A, in a cartesian 2D coordinate system. We just divide the area into n parts, and represent them as a₁, a₂, a₃, ..... a_n.

Let the elemental areas are at a distance x₁, x₂, x₃, ..... x_n, from Y axis and y₁, y₂, y₃, ...y_n from X axis.

The total moments produced about Y axis will be equal to the summation of all the individual moments produced by n elemental areas. 

Now moment is a vector quantity and we know vectors of same kind can be added together, therefore, all the n moment vectors can be added to get a single value of Resultant Moment. 

We also know this resultant moment's position vector. Let the resultant moment passes through a point G. The point through which resultant moment passes through is called Center of the Area or Centroid.

How can we find out the point G, whose coordinates are (X_g,Y_g)?

As moment of an area also obeys VARIGNON'S THEOREM OF MOMENT, then sum of all the moments produced by individual elemental areas will exactly be equal to the moment produced by the total area, i.e. the resultant of all those elemental areas. Now if all the areas are added to have the resultant area which will pass through the centroid G such that it produces a moment of X_gA about Y axis and Y_gA about X axis.

But Varignon's theorem states us that, for a vector system, resultant vector produces the moment about a point, is exactly equal to the sum of all the moments produced by all elemental areas about the same point and in the same plane. Hence, we can write now that,

Sum(a₁x₁+ a₂x₂+ + +a_nx_n) = AX_g
we can use summation sign ∑ to represent these equations,
∑a_ix_i = (∑a_i)X_g
=> X_g = (∑a_ix_i)/((∑a_i)

Sum(a₁y₁+ a₂y₂+ + +a_ny_n) = AY_g
∑a_iy_i = (∑a_i)Y_g
=> Y_g = (∑a_iy_i)/((∑a_i)

Algorithm to find out the Centroid G(X_g, Y_g) of a Complex Geometric Figure.

Step1:
Take a complex 2D figure like an Area or Lamina.

Step2:
Try to identify the basic figures whose algebraic combination produces our problem figure, whose centroid we shall find out.

Step3:
Choose a coordinate system, and make it as our frame of reference. All the distances and coordinate must be define with respect to our frame of reference.

Step4:
Compute the area (a_i), coordinates of their own centroid G_i (x_i, y_i) for each and every elemental areas. While measuring the centroids, all the measurements will be based on according to our chosen Axes.

Step5:
If any particular area has to subtracted to get the complex figure, the area will be negative, where as any area addition will be positive area.

Step6:
If the Centroid of the complex figure be G(X_g,Y_g)then,

=> X_g = (∑a_ix_i)/((∑a_i)

=> Y_g = (∑a_iy_i)/((∑a_i)

Here G₁ is the centroid of the part one where G₂ is the centroid of the circular area that has to be removed where as G₃ is the centroid of the triangular area that has to be removed also.

If we are asked to find moment of inertia of an area, which is nothing but the "second moment of area" then we shall have to find the centroidal moment of inertia first. Then we shall transfer the Moment of Inertia to another axis ie we shall apply parallel axis theorem to transfer moment of inertia from one axis (here centroidal axis) to another parallel axis.