Teaching Learning Centre Indian Institute of Technology Hyderabad (Under PMMMNMTT, MHRD)

Online Courses

Introduction

Resultent and forces problem

• Force and moments, Centre of mass, Moment of inertia, Trusses, Frames and Machines, Bending moment and shear force diagrams, Friction and its applications, Virtual work principles.
• Engineering Mechanics and important principles in Engineering mechanics. Fundamental quantities.
• Engineering satics and its objectives.
• Resultant of forces. Parallelogram law, Triangular law, Law of cosines, Sine law, Vectorial addition or subtraction.
• Resultant of more than two forces. Repeated use of triangular law, Polygonal law, Vectorial addition/subtraction.
• Components of force. 2D and 3D. Components of forces using directional consines.

Questions :

1. Which of the following is not a principle law of engineering mechanics.
2. A. Law of parallelogram.

B. Newton's second law of motion.

C. Law of conservation of mass.

D. Newton's third law of motion.

3. Which of the following is not a principle law of engineering mechanics.
4. A. 2.

B. 3.

C. 1.

D. 4.

5. What is the resultant vector of 2i + 3j + 4k and -6i + 9j - 4k.
6. A. -4i + 12j + 4k.

B. 4i + -12j - 4k.

C. -4i + 12j.

D. 4i - 12j.

7. If the resultant vector of 5i + 6j + 10k, -3i + yj + 2k and xi + 2j + zk is 8i + 4j -2k. Find x, y, z.
8. A. x=6, y= -4, z=-14.

B. x=-6, y= 4, z=-14.

C. x=6, y= 4, z=-14.

D. x=6, y= -4, z=14.

9. In a triangle ABC if AB=2, BC=5, CA=7 and find ang.B and ang.C.
10. A. ≈120° and ≈10°.

B. ≈110° and ≈18°.

C. ≈110° and ≈25°.

D. ≈100° and ≈18°.

11. If the magnitude of the vector 2i + yj + 8k is 10.2, find y.
12. A. 6.

B. 7.

C. 5.

D. 4.

13. Find the magnitude of the resultant vector of 6i + 6j + 10k, -3i + 6j + 2k and 3i + 2j + 9k.
14. A. 25.

B. 27.

C. 26.

D. 29.

Moments and force couple

Synopsis:

Examples are discussed on resolving forces into its components. Based on concepts of lecture-1.

Support reaction and equilibrium condition

In this lecture we shall discuss about:

• How to calculate moment of forces acting on a body about origin.
• Moment about any arbitrary point and moment of forces about any axis passing through that arbitrary point.
• What is a force couple and moment due to a force couple acting on body.
• Force couple systems.
• How to shift resultant of forces acting at a point to other point of interest by splitting it into a force , couple pair.
• Law of transmissibility.
• Examples on force-couple systems.

Questions :

1. Calculate the moment of force 2i+3j-5k acting at point A: 3i-5j+7k about origin.
2. A. 4i+29j+19k.

B. 4i+19j+29k.

C. 4i-19j+29k.

D. -4i-19j-29k.

3. Moment of force about point 'O', when force is acting at 'O' is: .
4. A. infinite.

B. indeterminate.

C. zero.

D. cant' be found.

5. Calculate the moment of force i+5k acting at P(1,2,3) about the point (3,2,1) about an axis directed along 3i+4j+5k vector.
6. A. 12j.

B. 0i+0j+0k.

C. 6.782k.

D. 6.782j.

7. Calculate the magnitude of couple due to forces of 9i+5j-4k & -(9i+5j-4k) , first one acting at origin and the other at (3,4,5).
8. A. 5.

B. 78.10.

C. 80.

D. 73.43.

9. Couple due to two concurrent forces is: .
10. A. zero.

B. infinite.

C. magnitude of force acting.

D. Data insufficient.

11. Replace a force of 10j N acting at 7i+8j-9k by a force moment pair at origin .
12. A. F=10j N and Moment = 0 at origin.

B. F=-10j N and Moment = 114Nm at origin.

C. F=114 N and Moment = 0 at origin.

D. F=10j N and Moment = 114 Nm at origin.

13. If F and M are force, moment acting on a body, according to the law of transmissibility, if the point of application of load is shifted by a distance 'd' without altering the line of application, which of the following is true if the state of body remains same before and after shifting the force.
14. A. Magnitude of moment is to be divided by d.

B. Direction of moment should be reversed.

C. Magnitude of moment is to be multiplied by d.

D. Both direction and magnitude of moment should be constant.

Centroid of curve, area, volume

Introduction

• Till now we have seen how to find resultant of forces , moment acting on a body and how to find equivalent force, couple system for given force configuration acting on the system.
• Now let us discuss where will this resultant force and moment will act on the body.
• What is centroid of a given body?
• What is area moment of inertia?
• Law of transmissibility.
• How do we calculate the centroid of a given curve, area, volume? with an example for each.

Questions :

1. Centroid of a body can lie outside of the body.
2. A. True.

B. False.

C. Impossible to decide.

3. Position of centroid of body varies with coordinate system.
4. A. True for all cases.

B. False for all cases.

C. Depends on geometry of body.

5. Calculate the location of centroid of quarter cylinder of length 100cm, radius= 30cm with its base on xy plane and axis along Z.
6. A. (12.738cm,12.738cm,50cm).

B. (12.378,12.378,50cm).

C. (12.738,30cm,50cm).

D. (12.738cm,12.738cm,-50cm).

7. If certain amount of mass to uniformly added to a 3D object. Its centroid: .
8. A. will not change its position.

B. will change its position depending on amount of matter added.

C. will shift to origin.

D. position can't be determined.

Intorduction of Moment of Inertia

Introduction

• Cartesian moment of inertia of an area element.
• Cartesian moment of inertia of curve with a constant width.
• Polar moment of inertia of an area element.
• Moment of inertia of volume elements.

Concepts: Radius of gyration. In both Cartesian system and polar system. Radius of gyration about a polar axis.

Theorems:

• Theorem of Pappus and Guldinus. Curve of revolution, surface of revolution. Product of inertia.
• Parallel axis theorem.

Questions :

1. What is X-area moment of inertia of an area dA present at a distance of k units from x-axis?
2. A. ∫A2dA.

C. ∫k2dA.

C. ∫kdA.

3. If dl is the length of the curve element having the constant and small width of 'h' units then, Iy=?
4. A. ∫hx2dy.

B. h∫x2dy.

C. h∫x2dl.

D. ∫hy2dl.

5. If an area element dA is at a distance of y units from x-axis and x units from y-axis, then what is its polar moment of inertia.
6. A. ∫x2dA + ∫y2dA .

B. ∫xydA + ∫x2dA.

C. ∫y2dA + ∫xydA.

D. ∫√x2+y2dA.

7. Moment of inertia of a circle with radius r about its central axis perpendicular to its plane. .
8. A. (πr2)/8

)

B. (πr2)/4.

C. 4πr4.

D. (πr4)/4.

10. A. kx2 + ky2.

B. 2*√kx2 + ky2.

C. kx2 + ky2 + 2kxky.

D. ∫√kx2+ky2dA.

11. Parallel axis theorem sates that moment of inertia of an element is equal to.
12. A. Moment of inertia of element about its centroid minus product of area and square of distance between centroid and that point.

B. Moment of inertia of element about its centroid plus product of area and square of distance between centroid and that point.

C. Moment of inertia of element about its centroid minus product of area and distance between centroid and that point.

D. D. Moment of inertia of element about its centroid plus product of area and distance between centroid and that point.

Intorduction of Moment of Inertia

Introduction

• We discuss about what is area moment of inertia.
• How to calculate an area element for various geometries.
• Three methods that are used to calculate area moment which are differentiated based on area element chosen:.
1. about X axis i.e, Ix.
2. about Y axis i.e, IY.
• How to calculate combined area moment i.e, Ixy in two different ways
• Ixy for a symmetric body.
• Examples:
1. Calculating Ix IY Ixy a circular arc of radius R.

Questions :

1. What is elemental area of a circular disc:
2. A. r*dθ*dr.

B. r*dθ.

C. dθ*dr.

C. r*dθ.

3. What is elemental area for a cylinder:
4. A. dz*r*dθ*dr.

B. dz*r*dθ.

C. dz*dr*dθ.

D. r*dr*dθ.

5. Ixy for semi circle = a* ( Ixy for full circle) .
6. A. a=2.

B. b=4.

C. a=16.

D. a can be any finite value.

7. Area moment depends on:. .
8. A. Total area.

)

B. Distribution of area.

C. Does not depend on any thing.

D. Cant be predicted.

9. Units of area moment of inertia:
10. A. m2.

B. m.

C. m2.

D. No units.

Concept of mohr circle and it's application

Introduction

• In previous lecture we discussed about finding moment of inertia about an arbitrary axis on a given plane. Now let us look at other methods in doing this. One possible way is using Mohr's circle.
• In this lecture we shall discuss about Mohr's circle and its application.
• Mohr's circle provides geometric means to calculate area moment of inertia IX' , IY' , IXY' about an arbitrary axis given IX ,IY ,IXY.
• Definition of principle axis.
• Deriving the equation for Mohr's circle.
• Understand how to construct a Mohr's circle step wise

Questions :

1. If Ix ,Iy , Ixy are principal area moment of inertia. Calculate area moment of inertia in X'Y' plane if it is obtained by rotating XY plane by angle θ.
2. A. Ixy' =-Ixy and Ix'=Iy , Iy'=Ix.

B. Ixy' =Ixy and Ix'=Iy , Iy'=Ix.

C. Ixy' =-Ixy and Ix'=Ix , Ix'=Ix.

D. Data insufficient.

3. Which of the following conditions define principle axis:.
4. A. Ix=0 and Iy=0.

B. Ix=Iy ≠ 0.

C. Ixy=0.

D. Ixy'=0.

5. Mohr's circle is applied for.
6. A. Moment of inertia of thin wire.

B. Moment of inertia of area.

C. Moment of inertia of volume.

D. None of these.

7. To calculate moment of inertia Ix' , Iy' and Ixy' on a plane X'Y' which makes an angle of θ in counter clock wise direction. We move.
8. A. An angle of 2θ around the Mohr's circle in counter clock wise direction.

B. An angle of θ around the Mohr's circle in clock wise direction.

C. An angle of 2θ around the Mohr's circle in clock wise direction.

D. An angle of θ around the Mohr's circle in counter clock wise direction.

9. Mohr's circle can't lie completely in X < 0 half plane .
10. A. True.

B. False.

Area moment of inertia about arbitrary axis

Mass moment of inertia

Introduction

In previous lectures we talked about area moment of inertia. Now let's talk about mass moment of inertia:

Defining moment of inertia about X (Ix) ,Y (IY) and Z (Iz )axis :

0m(x2+y2)dm = lz

0m(y2+z2)dm = lx

0m(x2+z2)dm = ly

Defining product of inertia on different planes: ∫0m(xy)dm = lxy    ∫0m(yz)dm = lyz    ∫0m(zx)dm = lzx

How to find mass moment of inertia about an arbitrary axis λ given Ix Iy Iz Ixy Iyz Izx values if direction λ is known.

Questions :

1. To calculate mass moment of inertia about an arbitrary axis 'λ' ,which of the following quantities are to be known:
2. A. lx ly l z.

B. lxy lyz l zx.

C. Both A and B.

D. None.

3. Calculate the mass moment of inertia of a hemisphere of radius 'r' and mass 'm' about z-axis whose centre coincides with origin and flat surface is placed on +XY plane. .
4. A. 2/5mr2.

B. 1/5mr2.

C. mr2.

D. Can't be Find.

5. Formula for moment of inertia about an arbitrary axis 'λ' is: (given Ix Iy Iz Ixy Iyz Izx) .
6. A. λx2Ix + λy2Iy + λz2Iz - 2 λxλy   Ixy - 2 λyλz   Iyz - 2 λzλx   Izx.

B. λx2Ix + λy2Iy + λz2Iz + 2 λxλy   Ixy + 2 λyλz   Iyz + 2 λzλx   Izx.

C. λx2Ix + λy2Iy + λz2Iz.

D. 2 λxλy   Ixy + 2 λyλz   Iyz + 2 λzλx   Izx .

7. Calculate the mass moment of inertia of point mass 'm' located at a distance 'r' from z axis about z axis is.
8. A. m r2.

B. Zero.

C. mr.

D. Can't be determined.

Application of theorems of Pappus and Gulidinus

In this lecture we discuss about:

• The curve of revolution.
• Then mass moment of inertia is derived for a thin cylinder using curve of revolution.
• The surface of revolution.
• Then we calculate mass moment of inertia for solid cylinder using surface of revolution.

Questions :

1. Calculate mass moment of inertia for thin cylinder about its axis having a mass of 'm' and radius 'r'.
2. A. m r2.

B. 0.5 m r2 .

C. Zero.

D. 0.5 mr.

3. Choose the volume element generated by revolving a rectangular surface about X-axis with one of its sides along X axis : where 'r' is distance from X axis and dA is rectangular area element on rectangular surface .
4. A. 2 π dA.

B. 2 π r dA.

C. 2 π r dr.

D. dr dA.

5. Mass moment of inertia for solid cylinder of mass 'm' and radius 'r' is:
6. A. m r2.

B. 0.5 m r2.

C. Zero.

D. 0.5 mr.

Trusses, Frames and Machines

In this lecture we discuss about:

• Till now we discussed about the application of forces on point masses and rigid bodies now let us discuss about application of forces on combination of bodies.
• Trusses, frames and machines are combination of many bodies which are connected either to support load or transmit forces.
• We shall discuss about classification of forces acting on a body.
1. Internal forces
2. External forces
• What is truss and what is its purpose?
• Difference between a two force member and a multi force member.
• Various characteristics that define a Truss.
• We discuss about equilibrium conditions that help to calculate forces in truss
• Techniques to find internal forces in members of truss:
1. Method of joints.
2. Method of sections.
• Types of trusses like:
1. Roof trusses.
2. Bridge trusses.

Then we define a simple truss and condition for simple truss, example problems to check if the given truss is simple truss or not.

Questions :

1. Find the number of connecting links in a simple truss if it has 6 joints.
2. A. 6.

B. 15.

C. Data in sufficient.

D. 9.

3. A truss will have.
4. A. Multi force members only.

B. Two force members only.

C. Both two force and multi force members.

D. depends on truss and its application.

5. Equilibrium equations for trusses are written at.

B. Joints.

C. Both.

D. Depends on truss geometry.

7. An internal force can accelerate a body.
8. A. True.

B. False.

C. Can't Say.

10. A. Triangles.

B. Squares.

C. rectangles.

D. Both triangles and rectangles.

Simple trusses, constraints, two force members

In this lecture we discuss about:

Simple trusses and condition for simple trusses

1. Statically determinate system.
2. Statically indeterminate system
3. Under constraint system.

Understand what they mean and conditions for each of them

Deriving conditions for - Statically determinate system, Statically indeterminate system, Under constraint system

1. What is a two force member.
2. Methods of finding a two force member

Questions :

1. Condition for simple truss, if m= number of members in truss & n=number of joints in truss.
2. A. m=2n.

B. m=2n+3.

C. m=2n-3.

D. m=n-3.

3. Condition for static indeterminacy if m=number of links, n=number of joints, r= number of reactions.
4. A. m + n = r.

B. m - r = 2n.

C. m + r = 2n.

D. None.

5. Significance of coefficient '2' on RHS of m + r = 2n is.
6. A. Each joint connects at least two links.

B. Each link in truss is a two force member.

C. There will be at least two support reactions on any truss.

D. No significance.

7. Number of zero force members in a simple truss having 4 joints and 7 links.
8. A. 0.

B. 3.

C. 2.

D. 4.

9. Which of the following conditions need not be satisfied for a zero force members connected by a joint.
10. A. No external forces acting on joint.

B. No support reactions should be provided at joint.

C. For a three member joint, at least two members should be collinear.

D. For a two member joint, the members have to be collinear.

Solving trusses

In this lecture we discuss about:

How to find the forces in the members of truss which are internal forces.

Methods which are used to find forces in truss.

1. Method of joints .
2. It is applied at all joints

Equilibrium equations are: Σ Fx = 0 and Σ Fy = 0 at all joints Useful for truss with small number of members.

3. Method of section
4. It is applied at certain sections to find internal forces only in desired members of truss.

Equilibrium equations are: Σ Fx = 0   Σ Fy = 0 and Σ Mz = 0 on a given section Useful for a truss with large number of members.

Then we discuss examples for both method of joints and method of sections.

Questions :

1. Which of the following is not a equilibrium equation used in method of joints:
2. A. Σ Fx = 0.

B. Σ Fy = 0.

C. Σ Mz = 0.

D. None.

3. Which of the methods is preferable for solving a truss with very large number of members.
4. A. Method of joints.

B. Method of section.

C. Both.

D. None.

5. Number of reaction forces acting on a truss supported by a pin joint and a roller joint in 2D.
6. A. 3.

B. 2.

C. 4.

D. Zero.

7. If forces acting on joints connecting a member act away from them then the force in the member is.
8. A. Compressive.

B. Tensile.

C. Data insufficient.

D. Can be compressive , can be tensile.

Frames and Machines

In this lecture we shall discuss about.

• How do we define frames and check if a given frame is statically determinate or not .
• How do we solve frames (Similar to trusses) , equilibrium equations used in solving.
• Then we will discuss about how we define a machine and equilibrium equations used in solving
• How to classify a given system into one among truss, frame and machine

Questions :

1. A frame should have:
2. A. At least one multi force member.

B. All multi force members.

C. No multi force member.

D. None.

3. A machine is characterised by:
4. A. No moving members.

B. All moving members.

C. At least one moving member.

D. None.

5. Which of the following must have a moving member:
6. A. Truss.

B. Frame.

C. Machine.

D. None.

7. Number of equilibrium equations we have for each member in frame are:.
8. A. 0.

B. 1.

C. 2.

D. 3.

Bending Moment and Shear Foce Diagrams

Types of members: 1. Two force members 2. Multiforce members.

Charactersitics and properties of two force members and multiforce members. Their examples.

Internal forces in each case.

1. Two force member: Axial force.
2. Multi force member: Axial forcce, Shear force and Bending moments.

Bending moment and Shear force. Sign convention in both.

Determination of Shear force and bending moment in a member at any given section by applying equillibium of forces and moments. Finding the relation between Shear force and applied load, Shear force and bending moment.

Example problem: Calculating the bending moment and shear force on a beam with fixed support at one end and rolling support at other end with an external force P acting at the center vertically. Drawing the corresponding Shear force diagram and bending moment diagram.

Questions :

1. Find the two force member in the folowing structure.
2. A. AF.

B. BC.

3. What is the relation between load and the shear force in a beam with a unformly distributed load with supports at both ends:

5. What is the relation between bending moment and the shear force in a beam with a unformly distributed load with supports at both ends.:
6. A. d2M/dx2 = V.

B. MdM/dx = V.

C. M*x = V.

D. D. dM/dx = V.

7. The area of the shear force diagram to the left or to the right of the section is equal to the __________ at that section.
8. When the shear diagram is increasing, the moment diagram is__________.
9. A. Concave downward.

B. Convex upward

C. Concave upward.

D. Convex Downward.

Cables

Cables are another type of load carriers. Loads on the cables can be two types..

Case1. Concentrated loads. Finding shape of the cable under the action of concentrated loads. Finding the tension in the cable. Found by using force equilibrium, moment equilibrium and moment equilibrium at a no-load point on the cable.

Questions :

1. What is the relation between tension at a point (xi,yi) in the cable and the load P1?
2. A. T = P1xi / yicosθ - xisinθ.

B. T = P1yi / yicosθ - xisinθ.

C. T = P1xi / xicosθ - yisinθ.

D. T = P1yi / xicosθ - yisinθ.

Case2a. Uniformly varying load along an axis. Finding shape of the cable under the action of such loads. Finding the tension in the cable.

Taking an infinitesimal small section of the cable at a point and apllying the equllibrium equations to find the shape. Cable attains parabolic shape under such loading.

Tension at each point in cable depends on its position. It is found by moment equillibrium at any of the fixed points.

Relation between position and the tension and load. Finding angle at each position.

Questions :

1. What is the equation of the shape of the cable obtained when hanged from two fixed ends and applied with horizantally uniform load. T0 is the horizantal cmponent of the tension in the cable.
2. A. Y = (w/4T0)*x2.

B. Y = (w/2T0)*x.

C. Y = (w/2T0)*x2.

D. Y = (w/4T0)*x.

3. Angle at any point in the cable is given by:
4. A. Tan-1(2y/x).

B. Tan-1(y/2x).

C. Tan-1(2y2/x).

D. Tan-1(y2/2x).

Shape and tension of cable under varrying load

Case2b. Uniformly varying load along the cable. Finding shape of the cable under the action of such loads. Finding the tension in the cable.

Taking an infinitesimal small section of the cable at a point and applying the equilibrium equations to find the shape. Cable attains hyperbolic shape under such loading.

Questions :

1. Shape of the cable when uniformly loaded along the cable
2. A. y = c cosh(x/c).

B. y = cosh(x/c).

C. y = c cosh(cx).

D. y = cosh(x/c).

Concept of Friction

Concept of friction explained. Relation between applied force and friction acting to resist the relative motion. Static friction and Kinetic friction.

Applications of friction.

1. Wedge
2. Bearings
3. Power Screw
4. Rolling resistance

Two types of bearings: End or collar bearing, Journal bearing.

1. Wedge: Finding the force required to move an object using wedge action. Finding the frictional forces involved in it. Finding the normal reaction between the surfaces of contact.

Questions :

1. Static friction ________ with the increase in force applied to move the body.
2. A. Increases.

B. Decreases.

C. Remains a positive constant.

D. Remains zero.

3. Coefficient of friction for static is ____________ the kinetic friction.
4. A. Less than.

B. Greater than.

C. Same as that of.

D. None of the above.

5. How many forces are involved in a wedge action to move a body.
6. A. 6.

B. 7.

C. 8.

D. 5.

Application of Friction

Power screw. Finding relation between applied force/moment and load to be lifted in terms of screw parameters.

End/Collar bearing. Finding relation between applied moment and load carried by the bearing in terms of bearing parameters.

First the relation for general bearing then deriving the relations from it for the end bearing and collar bearing.

Questions :

1. What is the distance measured parallel to the axis of the screw from a point on one thread to the corresponding point on the adjacent thread called?
2. A. Pitch.

C. Nominal Diameter.

3. Relation between the lead and the pitch for an n screw thread is
4. A. l = p^n.

B. l = n*p.

C. l = n^p.

D. l = p/n.

5. The force require to lift the object with a power screw increases with _________ the helix of the power screw.
6. A. Decrease in.

B. Increase in.

C. Square of .

D. None of the above.

7. In a collar bearing what is the moment required to oppose the frictional force?
8. A. (2wμsR2)/3.

B. (2wμsR2)/5.

C. (5wμsR2)/3 .

Application of Friction : Journal bearing and rolling friction

In this lecture we shall discuss about the friction application - Journal Bearing.

We will find the minimum moment required to start the bearing roll.

then discuss an example for - Rolling friction and an example where both the friction exist

Questions :

1. The kind of friction between rotating shaft and bearing is :
2. A. Static friction.

B. Kinetic friction.

C. No friction.

3. Radius of friction in a journal bearing is: if R - radius of rotating shaft, μk - Friction coefficient
4. A. rf = μK X R .

B. rf = μK / R .

C. rf = μK .

D. None.

5. For a rolling body(rotating clock wise), the normal reaction shifts towards.
6. A. Front of centre line.

B. Back of centre line.

C. Does not shift .

D. Can't say.

Application of Friction : belt and pulley

Belt pulley system. Relation between the tensions on the two sides of the belt in terms of the coefficient of friction.

V shaped pulley. Relation between the tensions on the two sides of the belt in terms of the coefficient of friction and the pulley V-angle.

Driven Pulley and Driving pulley.

Questions :

1. What is the relation between the tensions on the two sides of the belt in terms of the coefficient of friction and the pulley V-angle α.
2. A. T2 = T1*e^(μsβ/sin(α/2)).

B. T2 = T1*e^(μsβ/sin(α)).

C. T2 = T1*e^(μs2β/sin(α/2)).

D. T2 = T1*e^(μs2β/sin(α)).

3. If T1 and T2 are tensions in belt driven by a pulley, then T1 = T2.
4. A. True.

B. False.

C. It depends on the rotational direction.

5. In a driver-driven pulley system. Which of the following are correct?
6. A. In driving pulley, the direction of rotation is along the tighter side.

B. In driving pulley, the direction of rotation is opposite to the tighter side.

C. We can't say .

D. Both of them.

7. If the tensions in the two sides of belt passing through the pulley are T1 , T2 and T1=T2, then:
8. A. Angle of contact =180 deg.

B. Friction coefficient between pulley and belt =0.

C. Pulley is rotating clockwise.

D. None.

hELLO

Principle of Virtual work

Principle of virtual work can be used to find the equilibrium position of a system under the action of external forces

It states that under the action of external forces and moment, under the equilibrium the virtual work done is define

dW = F . δu + M . δθ =0 where:

δu - virtual displacement

δθ - virtual rotation/ angular displacement

then we shall discuss two examples to understand how to apply the principle of virtual work.

Relation of virtual work and potential energy:

Under the effect of conservative forces, we can say that : dW = -dU

We shall discuss about how to find if a system is in stable equilibrium / unstable equilibrium and an example problem to apply the relation of virtual work and potential energy.

Questions :

1. Which of the following is the result of relation of virtual work and potential energy: if P = f(h) where P= potential energy.
2. A. dP/dh = 0.

B. P/h = 0.

C. P = 0.

D. None.

3. Which of the following systems are in neutral equilibrium.
4. A. Ball on the top of inverted hemisphere.

B. Ball kept inside a hemisphere .

C. Ball on the table.

D. None.

5. Condition of stable equilibrium: where U = potential energy and U = f(h).
6. A. d2U/dx2 = 0.

B. d2U/dx2 < 0.

C. d2U/dx2 > 0.

D. None.

7. Which of the following are not conservative forces:
8. A. Friction.

B. Gravity.

C. force between electrons.

D. All of them.